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Chapter 9 Molecular Geometry and Covalent Bonding Models

In Chapter 8 "Ionic versus Covalent Bonding", we described the interactions that hold atoms together in chemical substances, focusing on the lattice energy of ionic compounds and the bond energy of covalent compounds. In the process, we introduced Lewis electron structures, which provide a simple method for predicting the number of bonds in common substances. As you learned in Chapter 8 "Ionic versus Covalent Bonding", the dots in Lewis structures represent the valence electrons of the constituent atoms and are paired according to the octet rule. As you will soon discover, however, the bonding in more complex molecules, such as those with multiple bonds or an odd number of electrons, cannot be explained with this simple approach. The purpose of this chapter is to introduce you to conceptual models used by chemists to describe the bonding in more complex compounds.

An experimental image of a covalent bond. This image shows that the bonding electrons on the copper atom in Cu2O occupy dz2 orbitals that point toward the oxygen atoms located at the center and corners of a cube.

In this chapter, we begin with a general method for predicting the structures of simple covalent molecules and polyatomic ions; then we discuss the actual distribution of electrons in covalent bonds. We apply two distinct approaches for describing covalent bonds: (1) a localized model to describe bonding in molecules with two or more atoms attached to a central atom and (2) a delocalized model to explain and predict which diatomic species exist and which do not exist. We conclude by describing more complex molecules and ions with multiple bonds. The tools you acquire in this chapter will enable you to explain why Ca2 is too unstable to exist in nature and why the unpaired electrons on O2 are crucial to the existence of life as we know it. You will also discover why carbon, the basic component of all organic compounds, forms four bonds despite having only two unpaired electrons in its valence electron configuration and how the structure of retinal, the key light-sensing component in our eyes, allows us to detect visible light.

9.2 Localized Bonding and Hybrid Atomic Orbitals

Learning Objective

  1. To describe the bonding in simple compounds using valence bond theory.

Although the VSEPR model is a simple and useful method for qualitatively predicting the structures of a wide range of compounds, it is not infallible. It predicts, for example, that H2S and PH3 should have structures similar to those of H2O and NH3, respectively. In fact, structural studies have shown that the H–S–H and H–P–H angles are more than 12° smaller than the corresponding bond angles in H2O and NH3. More disturbing, the VSEPR model predicts that the simple group 2 halides (MX2), which have four valence electrons, should all have linear X–M–X geometries. Instead, many of these species, including SrF2 and BaF2, are significantly bent. A more sophisticated treatment of bonding is needed for systems such as these. In this section, we present a quantum mechanical description of bonding, in which bonding electrons are viewed as being localized between the nuclei of the bonded atoms. The overlap of bonding orbitals is substantially increased through a process called hybridization, which results in the formation of stronger bonds.

Valence Bond Theory: A Localized Bonding Approach

In , you learned that as two hydrogen atoms approach each other from an infinite distance, the energy of the system reaches a minimum. This region of minimum energy in the energy diagram corresponds to the formation of a covalent bond between the two atoms at an H–H distance of 74 pm (). According to quantum mechanics, bonds form between atoms because their atomic orbitals overlap, with each region of overlap accommodating a maximum of two electrons with opposite spin, in accordance with the Pauli principle. In this case, a bond forms between the two hydrogen atoms when the singly occupied 1s atomic orbital of one hydrogen atom overlaps with the singly occupied 1s atomic orbital of a second hydrogen atom. Electron density between the nuclei is increased because of this orbital overlap and results in a localized electron-pair bond ().

Figure 9.10 Overlap of Two Singly Occupied Hydrogen 1s Atomic Orbitals Produces an H–H Bond in H2

The formation of H2 from two hydrogen atoms, each with a single electron in a 1s orbital, occurs as the electrons are shared to form an electron-pair bond, as indicated schematically by the gray spheres and black arrows. The orange electron density distributions show that the formation of an H2 molecule increases the electron density in the region between the two positively charged nuclei.

Although Lewis and VSEPR structures also contain localized electron-pair bonds, neither description uses an atomic orbital approach to predict the stability of the bond. Doing so forms the basis for a description of chemical bonding known as valence bond theoryA localized bonding model that assumes that the strength of a covalent bond is proportional to the amount of overlap between atomic orbitals and that an atom can use different combinations of atomic orbitals (hybrids) to maximize the overlap between bonded atoms., which is built on two assumptions:

  1. The strength of a covalent bond is proportional to the amount of overlap between atomic orbitals; that is, the greater the overlap, the more stable the bond.
  2. An atom can use different combinations of atomic orbitals to maximize the overlap of orbitals used by bonded atoms.

shows an electron-pair bond formed by the overlap of two ns atomic orbitals, two np atomic orbitals, and an ns and an np orbital where n = 2. Maximum overlap occurs between orbitals with the same spatial orientation and similar energies.

Figure 9.11 Three Different Ways to Form an Electron-Pair Bond

An electron-pair bond can be formed by the overlap of any of the following combinations of two singly occupied atomic orbitals: two ns atomic orbitals (a), an ns and an np atomic orbital (b), and two np atomic orbitals (c) where n = 2. The positive lobe is indicated in yellow, and the negative lobe is in blue.

Let’s examine the bonds in BeH2, for example. According to the VSEPR model, BeH2 is a linear compound with four valence electrons and two Be–H bonds. Its bonding can also be described using an atomic orbital approach. Beryllium has a 1s22s2 electron configuration, and each H atom has a 1s1 electron configuration. Because the Be atom has a filled 2s subshell, however, it has no singly occupied orbitals available to overlap with the singly occupied 1s orbitals on the H atoms. If a singly occupied 1s orbital on hydrogen were to overlap with a filled 2s orbital on beryllium, the resulting bonding orbital would contain three electrons, but the maximum allowed by quantum mechanics is two. How then is beryllium able to bond to two hydrogen atoms? One way would be to add enough energy to excite one of its 2s electrons into an empty 2p orbital and reverse its spin, in a process called promotionThe excitation of an electron from a filled ns2 atomic orbital to an empty np or (n1)d valence orbital.:

In this excited state, the Be atom would have two singly occupied atomic orbitals (the 2s and one of the 2p orbitals), each of which could overlap with a singly occupied 1s orbital of an H atom to form an electron-pair bond. Although this would produce BeH2, the two Be–H bonds would not be equivalent: the 1s orbital of one hydrogen atom would overlap with a Be 2s orbital, and the 1s orbital of the other hydrogen atom would overlap with an orbital of a different energy, a Be 2p orbital. Experimental evidence indicates, however, that the two Be–H bonds have identical energies. To resolve this discrepancy and explain how molecules such as BeH2 form, scientists developed the concept of hybridization.

Hybridization of s and p Orbitals

The localized bonding approach uses a process called hybridizationA process in which two or more atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds., in which atomic orbitals that are similar in energy but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitalsNew atomic orbitals formed from the process of hybridization. because they are produced by combining (hybridizing) two or more atomic orbitals from the same atom.

In BeH2, we can generate two equivalent orbitals by combining the 2s orbital of beryllium and any one of the three degenerate 2p orbitals. By taking the sum and the difference of Be 2s and 2pz atomic orbitals, for example, we produce two new orbitals with major and minor lobes oriented along the z-axes, as shown in .Because the difference A − B can also be written as A + (−B), in and subsequent figures we have reversed the phase(s) of the orbital being subtracted, which is the same as multiplying it by −1 and adding. This gives us , where the value 12 is needed mathematically to indicate that the 2s and 2p orbitals contribute equally to each hybrid orbital.

Equation 9.1

sp=12(2s+2pz) and sp=12(2s2pz)

The position of the atomic nucleus with respect to an sp hybrid orbital. The nucleus is actually located slightly inside the minor lobe, not at the node separating the major and minor lobes.

Figure 9.12 The Formation of sp Hybrid Orbitals

Taking the mathematical sum and difference of an ns and an np atomic orbital where n = 2 gives two equivalent sp hybrid orbitals oriented at 180° to each other.

The nucleus resides just inside the minor lobe of each orbital. In this case, the new orbitals are called sp hybrids because they are formed from one s and one p orbital. The two new orbitals are equivalent in energy, and their energy is between the energy values associated with pure s and p orbitals, as illustrated in this diagram:

Each singly occupied sp hybrid orbitalThe two equivalent hybrid orbitals that result when one ns orbital and one np orbital are combined (hybridized). The two sp hybrid orbitals are oriented at 180° from each other. They are equivalent in energy, and their energy is between the energy values associated with pure s and pure p orbitals. can now form an electron-pair bond with the singly occupied 1s atomic orbital of one of the H atoms. As shown in , each sp orbital on Be has the correct orientation for the major lobes to overlap with the 1s atomic orbital of an H atom. The formation of two energetically equivalent Be–H bonds produces a linear BeH2 molecule. Thus valence bond theory does what neither the Lewis electron structure nor the VSEPR model is able to do; it explains why the bonds in BeH2 are equivalent in energy and why BeH2 has a linear geometry.

Figure 9.13 Explanation of the Bonding in BeH2 Using sp Hybrid Orbitals

Each singly occupied sp hybrid orbital on beryllium can form an electron-pair bond with the singly occupied 1s orbital of a hydrogen atom. Because the two sp hybrid orbitals are oriented at a 180° angle, the BeH2 molecule is linear.

Because both promotion and hybridization require an input of energy, the formation of a set of singly occupied hybrid atomic orbitals is energetically uphill. The overall process of forming a compound with hybrid orbitals will be energetically favorable only if the amount of energy released by the formation of covalent bonds is greater than the amount of energy used to form the hybrid orbitals (). As we will see, some compounds are highly unstable or do not exist because the amount of energy required to form hybrid orbitals is greater than the amount of energy that would be released by the formation of additional bonds.

Figure 9.14 A Hypothetical Stepwise Process for the Formation of BeH2 from a Gaseous Be Atom and Two Gaseous H Atoms

The promotion of an electron from the 2s orbital of beryllium to one of the 2p orbitals is energetically uphill. The overall process of forming a BeH2 molecule from a Be atom and two H atoms will therefore be energetically favorable only if the amount of energy released by the formation of the two Be–H bonds is greater than the amount of energy required for promotion and hybridization.

The concept of hybridization also explains why boron, with a 2s22p1 valence electron configuration, forms three bonds with fluorine to produce BF3, as predicted by the Lewis and VSEPR approaches. With only a single unpaired electron in its ground state, boron should form only a single covalent bond. By the promotion of one of its 2s electrons to an unoccupied 2p orbital, however, followed by the hybridization of the three singly occupied orbitals (the 2s and two 2p orbitals), boron acquires a set of three equivalent hybrid orbitals with one electron each, as shown here:

The hybrid orbitals are degenerate and are oriented at 120° angles to each other (). Because the hybrid atomic orbitals are formed from one s and two p orbitals, boron is said to be sp2hybridized (pronounced “s-p-two” or “s-p-squared”). The singly occupied sp2 hybrid atomic orbitalsThe three equivalent hybrid orbitals that result when one ns orbital and two np orbitals are combined (hybridized). The three sp2 hybrid orbitals are oriented in a plane at 120° from each other. They are equivalent in energy, and their energy is between the energy values associated with pure s and pure p orbitals. can overlap with the singly occupied orbitals on each of the three F atoms to form a trigonal planar structure with three energetically equivalent B–F bonds.

Figure 9.15 Formation of sp2 Hybrid Orbitals

Combining one ns and two np atomic orbitals gives three equivalent sp2 hybrid orbitals in a trigonal planar arrangement; that is, oriented at 120° to one another.

Looking at the 2s22p2 valence electron configuration of carbon, we might expect carbon to use its two unpaired 2p electrons to form compounds with only two covalent bonds. We know, however, that carbon typically forms compounds with four covalent bonds. We can explain this apparent discrepancy by the hybridization of the 2s orbital and the three 2p orbitals on carbon to give a set of four degenerate sp3 (“s-p-three” or “s-p-cubed”) hybrid orbitals, each with a single electron:

The large lobes of the hybridized orbitals are oriented toward the vertices of a tetrahedron, with 109.5° angles between them (). Like all the hybridized orbitals discussed earlier, the sp3 hybrid atomic orbitalsThe four equivalent hybrid orbitals that result when one ns orbital and three np orbitals are combined (hybridized). The four sp3 hybrid orbitals point at the vertices of a tetrahedron, so they are oriented at 109.5° from each other. They are equivalent in energy, and their energy is between the energy values associated with pure s and pure p orbitals. are predicted to be equal in energy.

Figure 9.16 Formation of sp3 Hybrid Orbitals

Combining one ns and three np atomic orbitals results in four sp3 hybrid orbitals oriented at 109.5° to one another in a tetrahedral arrangement.

In addition to explaining why some elements form more bonds than would be expected based on their valence electron configurations, and why the bonds formed are equal in energy, valence bond theory explains why these compounds are so stable: the amount of energy released increases with the number of bonds formed. In the case of carbon, for example, much more energy is released in the formation of four bonds than two, so compounds of carbon with four bonds tend to be more stable than those with only two. Carbon does form compounds with only two covalent bonds (such as CH2 or CF2), but these species are highly reactive, unstable intermediates that form in only certain chemical reactions.

Note the Pattern

Valence bond theory explains the number of bonds formed in a compound and the relative bond strengths.

The bonding in molecules such as NH3 or H2O, which have lone pairs on the central atom, can also be described in terms of hybrid atomic orbitals. In NH3, for example, N, with a 2s22p3 valence electron configuration, can hybridize its 2s and 2p orbitals to produce four sp3 hybrid orbitals. Placing five valence electrons in the four hybrid orbitals, we obtain three that are singly occupied and one with a pair of electrons:

The three singly occupied sp3 lobes can form bonds with three H atoms, while the fourth orbital accommodates the lone pair of electrons. Similarly, H2O has an sp3 hybridized oxygen atom that uses two singly occupied sp3 lobes to bond to two H atoms, and two to accommodate the two lone pairs predicted by the VSEPR model. Such descriptions explain the approximately tetrahedral distribution of electron pairs on the central atom in NH3 and H2O. Unfortunately, however, recent experimental evidence indicates that in CH4 and NH3, the hybridized orbitals are not entirely equivalent in energy, making this bonding model an active area of research.

Example 5

Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen.

  1. H2S
  2. CHCl3

Given: two chemical compounds

Asked for: number of electron pairs and molecular geometry, hybridization, and bonding

Strategy:

A Using the approach from Example 1, determine the number of electron pairs and the molecular geometry of the molecule.

B From the valence electron configuration of the central atom, predict the number and type of hybrid orbitals that can be produced. Fill these hybrid orbitals with the total number of valence electrons around the central atom and describe the hybridization.

Solution:

  1. A H2S has four electron pairs around the sulfur atom with two bonded atoms, so the VSEPR model predicts a molecular geometry that is bent, or V shaped. B Sulfur has a 3s23p4 valence electron configuration with six electrons, but by hybridizing its 3s and 3p orbitals, it can produce four sp3 hybrids. If the six valence electrons are placed in these orbitals, two have electron pairs and two are singly occupied. The two sp3 hybrid orbitals that are singly occupied are used to form S–H bonds, whereas the other two have lone pairs of electrons. Together, the four sp3 hybrid orbitals produce an approximately tetrahedral arrangement of electron pairs, which agrees with the molecular geometry predicted by the VSEPR model.
  2. A The CHCl3 molecule has four valence electrons around the central atom. In the VSEPR model, the carbon atom has four electron pairs, and the molecular geometry is tetrahedral. B Carbon has a 2s22p2 valence electron configuration. By hybridizing its 2s and 2p orbitals, it can form four sp3 hybridized orbitals that are equal in energy. Eight electrons around the central atom (four from C, one from H, and one from each of the three Cl atoms) fill three sp3 hybrid orbitals to form C–Cl bonds, and one forms a C–H bond. Similarly, the Cl atoms, with seven electrons each in their 3s and 3p valence subshells, can be viewed as sp3 hybridized. Each Cl atom uses a singly occupied sp3 hybrid orbital to form a C–Cl bond and three hybrid orbitals to accommodate lone pairs.

Exercise

Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen.

  1. the BF4 ion
  2. hydrazine (H2N–NH2)

Answer:

  1. B is sp3 hybridized; F is also sp3 hybridized so it can accommodate one B–F bond and three lone pairs. The molecular geometry is tetrahedral.
  2. Each N atom is sp3 hybridized and uses one sp3 hybrid orbital to form the N–N bond, two to form N–H bonds, and one to accommodate a lone pair. The molecular geometry about each N is trigonal pyramidal.

Note the Pattern

The number of hybrid orbitals used by the central atom is the same as the number of electron pairs around the central atom.

Hybridization Using d Orbitals

Hybridization is not restricted to the ns and np atomic orbitals. The bonding in compounds with central atoms in the period 3 and below can also be described using hybrid atomic orbitals. In these cases, the central atom can use its valence (n − 1)d orbitals as well as its ns and np orbitals to form hybrid atomic orbitals, which allows it to accommodate five or more bonded atoms (as in PF5 and SF6). Using the ns orbital, all three np orbitals, and one (n − 1)d orbital gives a set of five sp3d hybrid orbitalsThe five hybrid orbitals that result when one ns, three np, and one (n1)d orbitals are combined (hybridized). that point toward the vertices of a trigonal bipyramid (part (a) in ). In this case, the five hybrid orbitals are not all equivalent: three form a triangular array oriented at 120° angles, and the other two are oriented at 90° to the first three and at 180° to each other.

Similarly, the combination of the ns orbital, all three np orbitals, and two nd orbitals gives a set of six equivalent sp3d2 hybrid orbitalsThe six equivalent hybrid orbitals that result when one ns, three np, and two (n1)d orbitals are combined (hybridized). oriented toward the vertices of an octahedron (part (b) in ). In the VSEPR model, PF5 and SF6 are predicted to be trigonal bipyramidal and octahedral, respectively, which agrees with a valence bond description in which sp3d or sp3d2 hybrid orbitals are used for bonding.

Figure 9.17 Hybrid Orbitals Involving d Orbitals

The formation of a set of (a) five sp3d hybrid orbitals and (b) six sp3d2 hybrid orbitals from ns, np, and nd atomic orbitals where n = 4.

Example 6

What is the hybridization of the central atom in each species? Describe the bonding in each species.

  1. XeF4
  2. SO42−
  3. SF4

Given: three chemical species

Asked for: hybridization of the central atom

Strategy:

A Determine the geometry of the molecule using the strategy in Example 1. From the valence electron configuration of the central atom and the number of electron pairs, determine the hybridization.

B Place the total number of electrons around the central atom in the hybrid orbitals and describe the bonding.

Solution:

  1. A Using the VSEPR model, we find that Xe in XeF4 forms four bonds and has two lone pairs, so its structure is square planar and it has six electron pairs. The six electron pairs form an octahedral arrangement, so the Xe must be sp3d2 hybridized. B With 12 electrons around Xe, four of the six sp3d2 hybrid orbitals form Xe–F bonds, and two are occupied by lone pairs of electrons.
  2. A The S in the SO42− ion has four electron pairs and has four bonded atoms, so the structure is tetrahedral. The sulfur must be sp3 hybridized to generate four S–O bonds. B Filling the sp3 hybrid orbitals with eight electrons from four bonds produces four filled sp3 hybrid orbitals.
  3. A The S atom in SF4 contains five electron pairs and four bonded atoms. The molecule has a seesaw structure with one lone pair:

    To accommodate five electron pairs, the sulfur atom must be sp3d hybridized. B Filling these orbitals with 10 electrons gives four sp3d hybrid orbitals forming S–F bonds and one with a lone pair of electrons.

Exercise

What is the hybridization of the central atom in each species? Describe the bonding.

  1. PCl4+
  2. BrF3
  3. SiF62−

Answer:

  1. sp3 with four P–Cl bonds
  2. sp3d with three Br–F bonds and two lone pairs
  3. sp3d2 with six Si–F bonds

Hybridization using d orbitals allows chemists to explain the structures and properties of many molecules and ions. Like most such models, however, it is not universally accepted. Nonetheless, it does explain a fundamental difference between the chemistry of the elements in the period 2 (C, N, and O) and those in period 3 and below (such as Si, P, and S).

Period 2 elements do not form compounds in which the central atom is covalently bonded to five or more atoms, although such compounds are common for the heavier elements. Thus whereas carbon and silicon both form tetrafluorides (CF4 and SiF4), only SiF4 reacts with F to give a stable hexafluoro dianion, SiF62−. Because there are no 2d atomic orbitals, the formation of octahedral CF62− would require hybrid orbitals created from 2s, 2p, and 3d atomic orbitals. The 3d orbitals of carbon are so high in energy that the amount of energy needed to form a set of sp3d2 hybrid orbitals cannot be equaled by the energy released in the formation of two additional C–F bonds. These additional bonds are expected to be weak because the carbon atom (and other atoms in period 2) is so small that it cannot accommodate five or six F atoms at normal C–F bond lengths due to repulsions between electrons on adjacent fluorine atoms. Perhaps not surprisingly, then, species such as CF62− have never been prepared.

Example 7

What is the hybridization of the oxygen atom in OF4? Is OF4 likely to exist?

Given: chemical compound

Asked for: hybridization and stability

Strategy:

A Predict the geometry of OF4 using the VSEPR model.

B From the number of electron pairs around O in OF4, predict the hybridization of O. Compare the number of hybrid orbitals with the number of electron pairs to decide whether the molecule is likely to exist.

Solution:

A The VSEPR model predicts that OF4 will have five electron pairs, resulting in a trigonal bipyramidal geometry with four bonding pairs and one lone pair. B To accommodate five electron pairs, the O atom would have to be sp3d hybridized. The only d orbital available for forming a set of sp3d hybrid orbitals is a 3d orbital, which is much higher in energy than the 2s and 2p valence orbitals of oxygen. As a result, the OF4 molecule is unlikely to exist. In fact, it has not been detected.

Exercise

What is the hybridization of the boron atom in BF63−? Is this ion likely to exist?

Answer:  sp3d2 hybridization; no

Summary

The localized bonding model (called valence bond theory) assumes that covalent bonds are formed when atomic orbitals overlap and that the strength of a covalent bond is proportional to the amount of overlap. It also assumes that atoms use combinations of atomic orbitals (hybrids) to maximize the overlap with adjacent atoms. The formation of hybrid atomic orbitals can be viewed as occurring via promotion of an electron from a filled ns2 subshell to an empty np or (n − 1)d valence orbital, followed by hybridization, the combination of the orbitals to give a new set of (usually) equivalent orbitals that are oriented properly to form bonds. The combination of an ns and an np orbital gives rise to two equivalent sp hybrids oriented at 180°, whereas the combination of an ns and two or three np orbitals produces three equivalent sp2 hybrids or four equivalent sp3 hybrids, respectively. The bonding in molecules with more than an octet of electrons around a central atom can be explained by invoking the participation of one or two (n − 1)d orbitals to give sets of five sp3d or six sp3d2 hybrid orbitals, capable of forming five or six bonds, respectively. The spatial orientation of the hybrid atomic orbitals is consistent with the geometries predicted using the VSEPR model.

Key Takeaway

  • Hybridization increases the overlap of bonding orbitals and explains the molecular geometries of many species whose geometry cannot be explained using a VSEPR approach.

Conceptual Problems

  1. Arrange sp, sp3, and sp2 in order of increasing strength of the bond formed to a hydrogen atom. Explain your reasoning.

  2. What atomic orbitals are combined to form sp3, sp, sp3d2, and sp3d? What is the maximum number of electron-pair bonds that can be formed using each set of hybrid orbitals?

  3. Why is it incorrect to say that an atom with sp2 hybridization will form only three bonds? The carbon atom in the carbonate anion is sp2 hybridized. How many bonds to carbon are present in the carbonate ion? Which orbitals on carbon are used to form each bond?

  4. If hybridization did not occur, how many bonds would N, O, C, and B form in a neutral molecule, and what would be the approximate molecular geometry?

  5. How are hybridization and molecular geometry related? Which has a stronger correlation—molecular geometry and hybridization or Lewis structures and hybridization?

  6. In the valence bond approach to bonding in BeF2, which step(s) require(s) an energy input, and which release(s) energy?

  7. The energies of hybrid orbitals are intermediate between the energies of the atomic orbitals from which they are formed. Why?

  8. How are lone pairs on the central atom treated using hybrid orbitals?

  9. Because nitrogen bonds to only three hydrogen atoms in ammonia, why doesn’t the nitrogen atom use sp2 hybrid orbitals instead of sp3 hybrids?

  10. Using arguments based on orbital hybridization, explain why the CCl62− ion does not exist.

  11. Species such as NF52− and OF42− are unknown. If 3d atomic orbitals were much lower energy, low enough to be involved in hybrid orbital formation, what effect would this have on the stability of such species? Why? What molecular geometry, electron-pair geometry, and hybridization would be expected for each molecule?

Numerical Problems

  1. Draw an energy-level diagram showing promotion and hybridization to describe the bonding in CH3. How does your diagram compare with that for methane? What is the molecular geometry?

  2. Draw an energy-level diagram showing promotion and hybridization to describe the bonding in CH3+. How does your diagram compare with that for methane? What is the molecular geometry?

  3. Draw the molecular structure, including any lone pairs on the central atom, state the hybridization of the central atom, and determine the molecular geometry for each molecule.

    1. BBr3
    2. PCl3
    3. NO3
  4. Draw the molecular structure, including any lone pairs on the central atom, state the hybridization of the central atom, and determine the molecular geometry for each species.

    1. AsBr3
    2. CF3+
    3. H2O
  5. What is the hybridization of the central atom in each of the following?

    1. CF4
    2. CCl22−
    3. IO3
    4. SiH4
  6. What is the hybridization of the central atom in each of the following?

    1. CCl3+
    2. CBr2O
    3. CO32−
    4. IBr2
  7. What is the hybridization of the central atom in PF6? Is this ion likely to exist? Why or why not? What would be the shape of the molecule?

  8. What is the hybridization of the central atom in SF5? Is this ion likely to exist? Why or why not? What would be the shape of the molecule?

Answers

  1.  

    The promotion and hybridization process is exactly the same as shown for CH4 in the chapter. The only difference is that the C atom uses the four singly occupied sp3 hybrid orbitals to form electron-pair bonds with only three H atoms, and an electron is added to the fourth hybrid orbital to give a charge of 1–. The electron-pair geometry is tetrahedral, but the molecular geometry is pyramidal, as in NH3.

  2.  

    1.  

      sp2, trigonal planar

    2.  

      sp3, pyramidal

    3.  

      sp2, trigonal planar

  3. The central atoms in CF4, CCl22–, IO3, and SiH4 are all sp3 hybridized.

  4. The phosphorus atom in the PF6 ion is sp3d2 hybridized, and the ion is octahedral. The PF6 ion is isoelectronic with SF6 and has essentially the same structure. It should therefore be a stable species.

9.3 Delocalized Bonding and Molecular Orbitals

Learning Objective

  1. To use molecular orbital theory to predict bond order.

None of the approaches we have described so far can adequately explain why some compounds are colored and others are not, why some substances with unpaired electrons are stable, and why others are effective semiconductors. (For more information on semiconductors, see , .) These approaches also cannot describe the nature of resonance. Such limitations led to the development of a new approach to bonding in which electrons are not viewed as being localized between the nuclei of bonded atoms but are instead delocalized throughout the entire molecule. Just as with the valence bond theory, the approach we are about to discuss is based on a quantum mechanical model.

In , we described the electrons in isolated atoms as having certain spatial distributions, called orbitals, each with a particular orbital energy. Just as the positions and energies of electrons in atoms can be described in terms of atomic orbitals (AOs), the positions and energies of electrons in molecules can be described in terms of molecular orbitals (MOs)A particular spatial distribution of electrons in a molecule that is associated with a particular orbital energy.—a spatial distribution of electrons in a molecule that is associated with a particular orbital energy. As the name suggests, molecular orbitals are not localized on a single atom but extend over the entire molecule. Consequently, the molecular orbital approach, called molecular orbital theoryA delocalized bonding model in which molecular orbitals are created from the linear combination of atomic orbitals (LCAOs)., is a delocalized approach to bonding.

Note the Pattern

Molecular orbital theory is a delocalized bonding approach that explains the colors of compounds, their stability, and resonance.

Molecular Orbital Theory: A Delocalized Bonding Approach

Although the molecular orbital theory is computationally demanding, the principles on which it is based are similar to those we used to determine electron configurations for atoms. The key difference is that in molecular orbitals, the electrons are allowed to interact with more than one atomic nucleus at a time. Just as with atomic orbitals, we create an energy-level diagram by listing the molecular orbitals in order of increasing energy. We then fill the orbitals with the required number of valence electrons according to the Pauli principle. This means that each molecular orbital can accommodate a maximum of two electrons with opposite spins.

Molecular Orbitals Involving Only ns Atomic Orbitals

We begin our discussion of molecular orbitals with the simplest molecule, H2, formed from two isolated hydrogen atoms, each with a 1s1 electron configuration. As we explained in , electrons can behave like waves. In the molecular orbital approach, the overlapping atomic orbitals are described by mathematical equations called wave functions. (For more information on wave functions, see , .) The 1s atomic orbitals on the two hydrogen atoms interact to form two new molecular orbitals, one produced by taking the sum of the two H 1s wave functions, and the other produced by taking their difference:

Equation 9.2

MO(1)=AO(atom A)+AO(atom B)MO(2)=AO(atom A)AO(atom B)

The molecular orbitals created from are called linear combinations of atomic orbitals (LCAOs)Molecular orbitals created from the sum and the difference of two wave functions (atomic orbitals).. A molecule must have as many molecular orbitals as there are atomic orbitals.

Adding two atomic orbitals corresponds to constructive interference between two waves, thus reinforcing their intensity; the internuclear electron probability density is increased. The molecular orbital corresponding to the sum of the two H 1s orbitals is called a σ1s combination (pronounced “sigma one ess”) (part (a) and part (b) in ). In a sigma (σ) orbitalA bonding molecular orbital in which the electron density along the internuclear axis and between the nuclei has cylindrical symmetry., the electron density along the internuclear axis and between the nuclei has cylindrical symmetry; that is, all cross-sections perpendicular to the internuclear axis are circles. The subscript 1s denotes the atomic orbitals from which the molecular orbital was derived:The ≈ sign is used rather than an = sign because we are ignoring certain constants that are not important to our argument.

Figure 9.18 Molecular Orbitals for the H2 Molecule

(a) This diagram shows the formation of a bonding σ1s molecular orbital for H2 as the sum of the wave functions (Ψ) of two H 1s atomic orbitals. (b) This plot of the square of the wave function (Ψ2) for the bonding σ1s molecular orbital illustrates the increased electron probability density between the two hydrogen nuclei. (Recall from that the probability density is proportional to the square of the wave function.) (c) This diagram shows the formation of an antibonding σ1s* molecular orbital for H2 as the difference of the wave functions (Ψ) of two H 1s atomic orbitals. (d) This plot of the square of the wave function (Ψ2) for the antibonding σ1s* molecular orbital illustrates the node corresponding to zero electron probability density between the two hydrogen nuclei.

Equation 9.3

σ1s ≈ 1s(A) + 1s(B)

Conversely, subtracting one atomic orbital from another corresponds to destructive interference between two waves, which reduces their intensity and causes a decrease in the internuclear electron probability density (part (c) and part (d) in ). The resulting pattern contains a node where the electron density is zero. The molecular orbital corresponding to the difference is called σ1s* (“sigma one ess star”). In a sigma star (σ*) orbitalAn antibonding molecular orbital in which there is a region of zero electron probability (a nodal plane) perpendicular to the internuclear axis., there is a region of zero electron probability, a nodal plane, perpendicular to the internuclear axis:

Equation 9.4

σ1s*1s(A)1s(B)

Note the Pattern

A molecule must have as many molecular orbitals as there are atomic orbitals.

The electron density in the σ1s molecular orbital is greatest between the two positively charged nuclei, and the resulting electron–nucleus electrostatic attractions reduce repulsions between the nuclei. Thus the σ1s orbital represents a bonding molecular orbitalA molecular orbital that forms when atomic orbitals or orbital lobes with the same sign interact to give increased electron probability between the nuclei due to constructive reinforcement of the wave functions.. In contrast, electrons in the σ1s* orbital are generally found in the space outside the internuclear region. Because this allows the positively charged nuclei to repel one another, the σ1s* orbital is an antibonding molecular orbitalA molecular orbital that forms when atomic orbitals or orbital lobes of opposite sign interact to give decreased electron probability between the nuclei due to destructuve reinforcement of the wave functions..

Note the Pattern

Antibonding orbitals contain a node perpendicular to the internuclear axis; bonding orbitals do not.

Energy-Level Diagrams

Because electrons in the σ1s orbital interact simultaneously with both nuclei, they have a lower energy than electrons that interact with only one nucleus. This means that the σ1s molecular orbital has a lower energy than either of the hydrogen 1s atomic orbitals. Conversely, electrons in the σ1s* orbital interact with only one hydrogen nucleus at a time. In addition, they are farther away from the nucleus than they were in the parent hydrogen 1s atomic orbitals. Consequently, the σ1s* molecular orbital has a higher energy than either of the hydrogen 1s atomic orbitals. The σ1s (bonding) molecular orbital is stabilized relative to the 1s atomic orbitals, and the σ1s* (antibonding) molecular orbital is destabilized. The relative energy levels of these orbitals are shown in the energy-level diagramA schematic drawing that compares the energies of the molecular orbitals (bonding, antibonding, and nonbonding) with the energies of the parent atomic orbitals. in .

Note the Pattern

A bonding molecular orbital is always lower in energy (more stable) than the component atomic orbitals, whereas an antibonding molecular orbital is always higher in energy (less stable).

Figure 9.19 Molecular Orbital Energy-Level Diagram for H2

The two available electrons (one from each H atom) in this diagram fill the bonding σ1s molecular orbital. Because the energy of the σ1s molecular orbital is lower than that of the two H 1s atomic orbitals, the H2 molecule is more stable (at a lower energy) than the two isolated H atoms.

To describe the bonding in a homonuclear diatomic moleculeA molecule that consists of two atoms of the same element. such as H2, we use molecular orbitals; that is, for a molecule in which two identical atoms interact, we insert the total number of valence electrons into the energy-level diagram (). We fill the orbitals according to the Pauli principle and Hund’s rule: each orbital can accommodate a maximum of two electrons with opposite spins, and the orbitals are filled in order of increasing energy. Because each H atom contributes one valence electron, the resulting two electrons are exactly enough to fill the σ1s bonding molecular orbital. The two electrons enter an orbital whose energy is lower than that of the parent atomic orbitals, so the H2 molecule is more stable than the two isolated hydrogen atoms. Thus molecular orbital theory correctly predicts that H2 is a stable molecule. Because bonds form when electrons are concentrated in the space between nuclei, this approach is also consistent with our earlier discussion of electron-pair bonds.

Bond Order in Molecular Orbital Theory

In the Lewis electron structures described in , the number of electron pairs holding two atoms together was called the bond order. In the molecular orbital approach, bond orderOne-half the net number of bonding electrons in a molecule. is defined as one-half the net number of bonding electrons:

Equation 9.5

bond order=number of bonding electronsnumber of antibonding electrons2

To calculate the bond order of H2, we see from that the σ1s (bonding) molecular orbital contains two electrons, while the σ1s* (antibonding) molecular orbital is empty. The bond order of H2 is therefore

Equation 9.6

202=1

This result corresponds to the single covalent bond predicted by Lewis dot symbols. Thus molecular orbital theory and the Lewis electron-pair approach agree that a single bond containing two electrons has a bond order of 1. Double and triple bonds contain four or six electrons, respectively, and correspond to bond orders of 2 and 3.

We can use energy-level diagrams such as the one in to describe the bonding in other pairs of atoms and ions where n = 1, such as the H2+ ion, the He2+ ion, and the He2 molecule. Again, we fill the lowest-energy molecular orbitals first while being sure not to violate the Pauli principle or Hund’s rule.

Part (a) in shows the energy-level diagram for the H2+ ion, which contains two protons and only one electron. The single electron occupies the σ1s bonding molecular orbital, giving a (σ1s)1 electron configuration. The number of electrons in an orbital is indicated by a superscript. In this case, the bond order is (10)÷2=12. Because the bond order is greater than zero, the H2+ ion should be more stable than an isolated H atom and a proton. We can therefore use a molecular orbital energy-level diagram and the calculated bond order to predict the relative stability of species such as H2+. With a bond order of only 12, the bond in H2+ should be weaker than in the H2 molecule, and the H–H bond should be longer. As shown in , these predictions agree with the experimental data.

Part (b) in is the molecular orbital energy-level diagram for He2+. This ion has a total of three valence electrons. Because the first two electrons completely fill the σ1s molecular orbital, the Pauli principle states that the third electron must be in the σ1s* antibonding orbital, giving a (σ1s)2(σ1s*)1 electron configuration. This electron configuration gives a bond order of (21)÷2=12. As with H2+, the He2+ ion should be stable, but the He–He bond should be weaker and longer than in H2. In fact, the He2+ ion can be prepared, and its properties are consistent with our predictions ().

Figure 9.20 Molecular Orbital Energy-Level Diagrams for Diatomic Molecules with Only 1s Atomic Orbitals

(a) The H2+ ion, (b) the He2+ ion, and (c) the He2 molecule are shown here.

Table 9.1 Molecular Orbital Electron Configurations, Bond Orders, Bond Lengths, and Bond Energies for some Simple Homonuclear Diatomic Molecules and Ions

Molecule or Ion Electron Configuration Bond Order Bond Length (pm) Bond Energy (kJ/mol)
H2+ 1s)1 12 106 269
H2 1s)2 1 74 436
He2+ (σ1s)2(σ1s*)1 12 108 251
He2 (σ1s)2(σ1s*)2 0 not observed not observed

Finally, we examine the He2 molecule, formed from two He atoms with 1s2 electron configurations. Part (c) in is the molecular orbital energy-level diagram for He2. With a total of four valence electrons, both the σ1s bonding and σ1s* antibonding orbitals must contain two electrons. This gives a (σ1s)2(σ1s*)2 electron configuration, with a predicted bond order of (2 − 2) ÷ 2 = 0, which indicates that the He2 molecule has no net bond and is not a stable species. Experiments show that the He2 molecule is actually less stable than two isolated He atoms due to unfavorable electron–electron and nucleus–nucleus interactions.

In molecular orbital theory, electrons in antibonding orbitals effectively cancel the stabilization resulting from electrons in bonding orbitals. Consequently, any system that has equal numbers of bonding and antibonding electrons will have a bond order of 0, and it is predicted to be unstable and therefore not to exist in nature. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory is able to accommodate systems with an odd number of electrons, such as the H2+ ion.

Note the Pattern

In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory can accommodate systems with an odd number of electrons.

Example 8

Use a molecular orbital energy-level diagram, such as those in , to predict the bond order in the He22+ ion. Is this a stable species?

Given: chemical species

Asked for: molecular orbital energy-level diagram, bond order, and stability

Strategy:

A Combine the two He valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for the system.

B Determine the total number of valence electrons in the He22+ ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.

C Calculate the bond order and predict whether the species is stable.

Solution:

A Two He 1s atomic orbitals combine to give two molecular orbitals: a σ1s bonding orbital at lower energy than the atomic orbitals and a σ1s* antibonding orbital at higher energy. The bonding in any diatomic molecule with two He atoms can be described using the following molecular orbital diagram:

B The He22+ ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can also view He22+ as being formed from two He+ ions, each of which has a single valence electron in the 1s atomic orbital. We can now fill the molecular orbital diagram:

The two electrons occupy the lowest-energy molecular orbital, which is the bonding (σ1s) orbital, giving a (σ1s)2 electron configuration. To avoid violating the Pauli principle, the electron spins must be paired. C So the bond order is

202=1

He22+ is therefore predicted to contain a single He–He bond. Thus it should be a stable species.

Exercise

Use a molecular orbital energy-level diagram to predict the valence-electron configuration and bond order of the H22− ion. Is this a stable species?

Answer: H22− has a valence electron configuration of (σ1s)2(σ1s*)2 with a bond order of 0. It is therefore predicted to be unstable.

So far, our discussion of molecular orbitals has been confined to the interaction of valence orbitals, which tend to lie farthest from the nucleus. When two atoms are close enough for their valence orbitals to overlap significantly, the filled inner electron shells are largely unperturbed; hence they do not need to be considered in a molecular orbital scheme. Also, when the inner orbitals are completely filled, they contain exactly enough electrons to completely fill both the bonding and antibonding molecular orbitals that arise from their interaction. Thus the interaction of filled shells always gives a bond order of 0, so filled shells are not a factor when predicting the stability of a species. This means that we can focus our attention on the molecular orbitals derived from valence atomic orbitals.

A molecular orbital diagram that can be applied to any homonuclear diatomic molecule with two identical alkali metal atoms (Li2 and Cs2, for example) is shown in part (a) in , where M represents the metal atom. Only two energy levels are important for describing the valence electron molecular orbitals of these species: a σns bonding molecular orbital and a σns* antibonding molecular orbital. Because each alkali metal (M) has an ns1 valence electron configuration, the M2 molecule has two valence electrons that fill the σns bonding orbital. As a result, a bond order of 1 is predicted for all homonuclear diatomic species formed from the alkali metals (Li2, Na2, K2, Rb2, and Cs2). The general features of these M2 diagrams are identical to the diagram for the H2 molecule in . Experimentally, all are found to be stable in the gas phase, and some are even stable in solution.

Figure 9.21 Molecular Orbital Energy-Level Diagrams for Alkali Metal and Alkaline Earth Metal Diatomic (M2) Molecules

(a) For alkali metal diatomic molecules, the two valence electrons are enough to fill the σns (bonding) level, giving a bond order of 1. (b) For alkaline earth metal diatomic molecules, the four valence electrons fill both the σns (bonding) and the σns* (nonbonding) levels, leading to a predicted bond order of 0.

Similarly, the molecular orbital diagrams for homonuclear diatomic compounds of the alkaline earth metals (such as Be2), in which each metal atom has an ns2 valence electron configuration, resemble the diagram for the He2 molecule in part (c) in . As shown in part (b) in , this is indeed the case. All the homonuclear alkaline earth diatomic molecules have four valence electrons, which fill both the σns bonding orbital and the σns* antibonding orbital and give a bond order of 0. Thus Be2, Mg2, Ca2, Sr2, and Ba2 are all expected to be unstable, in agreement with experimental data.In the solid state, however, all the alkali metals and the alkaline earth metals exist as extended lattices held together by metallic bonding. (For more information on metallic bonding, see , .) At low temperatures, Be2 is stable.

Example 9

Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Na2 ion.

Given: chemical species

Asked for: molecular orbital energy-level diagram, valence electron configuration, bond order, and stability

Strategy:

A Combine the two sodium valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for this system.

B Determine the total number of valence electrons in the Na2 ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.

C Calculate the bond order and predict whether the species is stable.

Solution:

A Because sodium has a [Ne]3s1 electron configuration, the molecular orbital energy-level diagram is qualitatively identical to the diagram for the interaction of two 1s atomic orbitals. B The Na2 ion has a total of three valence electrons (one from each Na atom and one for the negative charge), resulting in a filled σ3s molecular orbital, a half-filled σ3s* molecular orbital, and a (σ3s)2(σ3s*)1 electron configuration.

C The bond order is (21)÷2=12. With a fractional bond order, we predict that the Na2 ion exists but is highly reactive.

Exercise

Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Ca2+ ion.

Answer: Ca2+ has a (σ4s)2(σ4s*)1 electron configuration and a bond order of 12 and should exist.

Molecular Orbitals Formed from ns and np Atomic Orbitals

Atomic orbitals other than ns orbitals can also interact to form molecular orbitals. Because individual p, d, and f orbitals are not spherically symmetrical, however, we need to define a coordinate system so we know which lobes are interacting in three-dimensional space. Recall from , that for each np subshell, for example, there are npx, npy, and npz orbitals (). All have the same energy and are therefore degenerate, but they have different spatial orientations.

Just as with ns orbitals, we can form molecular orbitals from np orbitals by taking their mathematical sum and difference. When two positive lobes with the appropriate spatial orientation overlap, as illustrated for two npz atomic orbitals in part (a) in , it is the mathematical difference of their wave functions that results in constructive interference, which in turn increases the electron probability density between the two atoms. The difference therefore corresponds to a molecular orbital called a σnpz bonding molecular orbital because, just as with the σ orbitals discussed previously, it is symmetrical about the internuclear axis (in this case, the z-axis):

Equation 9.7

σnpz=npz(A)npz(B)

The other possible combination of the two npz orbitals is the mathematical sum:

Equation 9.8

σnpz=npz(A)+npz(B)

In this combination, shown in part (b) in , the positive lobe of one npz atomic orbital overlaps the negative lobe of the other, leading to destructive interference of the two waves and creating a node between the two atoms. Hence this is an antibonding molecular orbital. Because it, too, is symmetrical about the internuclear axis, this molecular orbital is called a σnpz antibonding molecular orbital. Whenever orbitals combine, the bonding combination is always lower in energy (more stable) than the atomic orbitals from which it was derived, and the antibonding combination is higher in energy (less stable).

Figure 9.22 Formation of Molecular Orbitals from npz Atomic Orbitals on Adjacent Atoms

(a) By convention, in a linear molecule or ion, the z-axis always corresponds to the internuclear axis, with +z to the right. As a result, the signs of the lobes of the npz atomic orbitals on the two atoms alternate − + − +, from left to right. In this case, the σ (bonding) molecular orbital corresponds to the mathematical difference, in which the overlap of lobes with the same sign results in increased probability density between the nuclei. (b) In contrast, the σ* (antibonding) molecular orbital corresponds to the mathematical sum, in which the overlap of lobes with opposite signs results in a nodal plane of zero probability density perpendicular to the internuclear axis.

Note the Pattern

Overlap of atomic orbital lobes with the same sign produces a bonding molecular orbital, regardless of whether it corresponds to the sum or the difference of the atomic orbitals.

The remaining p orbitals on each of the two atoms, npx and npy, do not point directly toward each other. Instead, they are perpendicular to the internuclear axis. If we arbitrarily label the axes as shown in , we see that we have two pairs of np orbitals: the two npx orbitals lying in the plane of the page, and two npy orbitals perpendicular to the plane. Although these two pairs are equivalent in energy, the npx orbital on one atom can interact with only the npx orbital on the other, and the npy orbital on one atom can interact with only the npy on the other. These interactions are side-to-side rather than the head-to-head interactions characteristic of σ orbitals. Each pair of overlapping atomic orbitals again forms two molecular orbitals: one corresponds to the arithmetic sum of the two atomic orbitals and one to the difference. The sum of these side-to-side interactions increases the electron probability in the region above and below a line connecting the nuclei, so it is a bonding molecular orbital that is called a pi (π) orbitalA bonding molecular orbital formed from the side-to-side interactions of two or more parallel np atomic orbitals.. The difference results in the overlap of orbital lobes with opposite signs, which produces a nodal plane perpendicular to the internuclear axis; hence it is an antibonding molecular orbital, called a pi star (π*) orbitalAn antibonding molecular orbital formed from the difference of the side-to-side interactions of two or more parallel np atomic orbitals, creating a nodal plane perpendicular to the internuclear axis..

Equation 9.9

πnpx=npx(A)+npx(B)

Equation 9.10

πnpx=npx(A)npx(B)

The two npy orbitals can also combine using side-to-side interactions to produce a bonding πnpy molecular orbital and an antibonding πnpy molecular orbital. Because the npx and npy atomic orbitals interact in the same way (side-to-side) and have the same energy, the πnpx and πnpy molecular orbitals are a degenerate pair, as are the πnpy and πnpy molecular orbitals.

Figure 9.23 Formation of π Molecular Orbitals from npx and npy Atomic Orbitals on Adjacent Atoms

(a) Because the signs of the lobes of both the npx and the npy atomic orbitals on adjacent atoms are the same, in both cases the mathematical sum corresponds to a π (bonding) molecular orbital. (b) In contrast, in both cases, the mathematical difference corresponds to a π* (antibonding) molecular orbital, with a nodal plane of zero probability density perpendicular to the internuclear axis.

is an energy-level diagram that can be applied to two identical interacting atoms that have three np atomic orbitals each. There are six degenerate p atomic orbitals (three from each atom) that combine to form six molecular orbitals, three bonding and three antibonding. The bonding molecular orbitals are lower in energy than the atomic orbitals because of the increased stability associated with the formation of a bond. Conversely, the antibonding molecular orbitals are higher in energy, as shown. The energy difference between the σ and σ* molecular orbitals is significantly greater than the difference between the two π and π* sets. The reason for this is that the atomic orbital overlap and thus the strength of the interaction are greater for a σ bond than a π bond, which means that the σ molecular orbital is more stable (lower in energy) than the π molecular orbitals.

Figure 9.24 The Relative Energies of the σ and π Molecular Orbitals Derived from npx, npy, and npz Orbitals on Identical Adjacent Atoms

Because the two npz orbitals point directly at each other, their orbital overlap is greater, so the difference in energy between the σ and σ* molecular orbitals is greater than the energy difference between the π and π* orbitals.

Although many combinations of atomic orbitals form molecular orbitals, we will discuss only one other interaction: an ns atomic orbital on one atom with an npz atomic orbital on another. As shown in , the sum of the two atomic wave functions (ns + npz) produces a σ bonding molecular orbital. Their difference (ns − npz) produces a σ* antibonding molecular orbital, which has a nodal plane of zero probability density perpendicular to the internuclear axis.

Figure 9.25 Formation of Molecular Orbitals from an ns Atomic Orbital on One Atom and an npz Atomic Orbital on an Adjacent Atom

(a) The mathematical sum results in a σ (bonding) molecular orbital, with increased probability density between the nuclei. (b) The mathematical difference results in a σ* (antibonding) molecular orbital, with a nodal plane of zero probability density perpendicular to the internuclear axis.

Molecular Orbital Diagrams for Period 2 Homonuclear Diatomic Molecules

We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N2, O2, and F2. When we draw a molecular orbital diagram for a molecule, there are four key points to remember:

  1. The number of molecular orbitals produced is the same as the number of atomic orbitals used to create them (the law of conservation of orbitalsA law that states that the number of molecular orbitals produced is the same as the number of atomic orbitals used to create them.).
  2. As the overlap between two atomic orbitals increases, the difference in energy between the resulting bonding and antibonding molecular orbitals increases.
  3. When two atomic orbitals combine to form a pair of molecular orbitals, the bonding molecular orbital is stabilized about as much as the antibonding molecular orbital is destabilized.
  4. The interaction between atomic orbitals is greatest when they have the same energy.

Note the Pattern

The number of molecular orbitals is always equal to the total number of atomic orbitals we started with.

We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F2. We use the diagram in part (a) in ; the n = 1 orbitals (σ1s and σ1s*) are located well below those of the n = 2 level and are not shown. As illustrated in the diagram, the σ2s and σ2s* molecular orbitals are much lower in energy than the molecular orbitals derived from the 2p atomic orbitals because of the large difference in energy between the 2s and 2p atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2p orbitals on each F is σ2pz, and the next most stable are the two degenerate orbitals, π2px and π2py. For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the σ2pz* orbital is higher in energy than either of the degenerate π2px* and π2py* orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy.

Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F2 molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ2s and σ2s* orbitals, 2 fill the σ2pz orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π* orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F2 to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F2 is predicted to have a stable F–F single bond, in agreement with experimental data.

We now turn to a molecular orbital description of the bonding in O2. It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. To obtain the molecular orbital energy-level diagram for O2, we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in part (b) in . We again fill the orbitals according to Hund’s rule and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ2s and σ2s orbitals, 2 more to fill the σ2pz orbital, and 4 to fill the degenerate π2px and π2py orbitals. According to Hund’s rule, the last 2 electrons must be placed in separate π* orbitals with their spins parallel, giving two unpaired electrons. This leads to a predicted bond order of (8 − 4) ÷ 2 = 2, which corresponds to a double bond, in agreement with experimental data (): the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K.

Figure 9.26 Molecular Orbital Energy-Level Diagrams for Homonuclear Diatomic Molecules

(a) For F2, with 14 valence electrons (7 from each F atom), all of the energy levels except the highest, σ2pz, are filled. This diagram shows 8 electrons in bonding orbitals and 6 in antibonding orbitals, resulting in a bond order of 1. (b) For O2, with 12 valence electrons (6 from each O atom), there are only 2 electrons to place in the (π2px*, π2py*) pair of orbitals. Hund’s rule dictates that one electron occupies each orbital, and their spins are parallel, giving the O2 molecule two unpaired electrons. This diagram shows 8 electrons in bonding orbitals and 4 in antibonding orbitals, resulting in a predicted bond order of 2.

None of the other bonding models can predict the presence of two unpaired electrons in O2. Chemists had long wondered why, unlike most other substances, liquid O2 is attracted into a magnetic field. As shown in , it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for O2 to have unpaired electrons, making it paramagnetic, exactly as predicted by molecular orbital theory. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches we have discussed.

Figure 9.27 Liquid O2 Suspended between the Poles of a Magnet

Because the O2 molecule has two unpaired electrons, it is paramagnetic. Consequently, it is attracted into a magnetic field, which allows it to remain suspended between the poles of a powerful magnet until it evaporates.

The magnetic properties of O2 are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H2O, CO2, and N2 in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H2O, CO2, and N2, have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of O2 with organic compounds to give H2O, CO2, and N2 would require that at least one of the electrons on O2 change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a spin barrier. Consequently, reactions of this type are usually exceedingly slow. If they were not so slow, all organic substances, including this book and you, would disappear in a puff of smoke!

For period 2 diatomic molecules to the left of N2 in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the σ2pz molecular orbital is slightly higher in energy than the degenerate π2px and π2py orbitals. The difference in energy between the 2s and 2p atomic orbitals increases from Li2 to F2 due to increasing nuclear charge and poor screening of the 2s electrons by electrons in the 2p subshell. The bonding interaction between the 2s orbital on one atom and the 2pz orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ2s orbital and increases the energy of the σ2pz orbital. Thus for Li2, Be2, B2, C2, and N2, the σ2pz orbital is higher in energy than the σ3pz orbitals, as shown in . Experimentally, it is found that the energy gap between the ns and np atomic orbitals increases as the nuclear charge increases (). Thus for example, the σ2pz molecular orbital is at a lower energy than the π2px,y pair.

Figure 9.28 Molecular Orbital Energy-Level Diagrams for the Diatomic Molecules of the Period 2 Elements

Unlike earlier diagrams, only the molecular orbital energy levels for the molecules are shown here. For simplicity, the atomic orbital energy levels for the component atoms have been omitted. For Li2 through N2, the σ2pz orbital is higher in energy than the π2px,y orbitals. In contrast, the σ2pz orbital is lower in energy than the π2px,y orbitals for O2 and F2 due to the increase in the energy difference between the 2s and 2p atomic orbitals as the nuclear charge increases across the row.

Completing the diagram for N2 in the same manner as demonstrated previously, we find that the 10 valence electrons result in 8 bonding electrons and 2 antibonding electrons, for a predicted bond order of 3, a triple bond. Experimental data show that the N–N bond is significantly shorter than the F–F bond (109.8 pm in N2 versus 141.2 pm in F2), and the bond energy is much greater for N2 than for F2 (945.3 kJ/mol versus 158.8 kJ/mol, respectively). Thus the N2 bond is much shorter and stronger than the F2 bond, consistent with what we would expect when comparing a triple bond with a single bond.

Example 10

Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S2, a bright blue gas at high temperatures.

Given: chemical species

Asked for: molecular orbital energy-level diagram, bond order, and number of unpaired electrons

Strategy:

A Write the valence electron configuration of sulfur and determine the type of molecular orbitals formed in S2. Predict the relative energies of the molecular orbitals based on how close in energy the valence atomic orbitals are to one another.

B Draw the molecular orbital energy-level diagram for this system and determine the total number of valence electrons in S2.

C Fill the molecular orbitals in order of increasing energy, being sure to obey the Pauli principle and Hund’s rule.

D Calculate the bond order and describe the bonding.

Solution:

A Sulfur has a [Ne]3s23p4 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in and , we need to know how close in energy the 3s and 3p atomic orbitals are because their energy separation will determine whether the π3px,y or the σ3pz molecular orbital is higher in energy. Because the nsnp energy gap increases as the nuclear charge increases (), the σ3pz molecular orbital will be lower in energy than the π3px,y pair.

B The molecular orbital energy-level diagram is as follows:

Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons.

C Ten valence electrons are used to fill the orbitals through π3px and π3py, leaving 2 electrons to occupy the degenerate π3px* and π3py* pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S2 is (σ3s)2(σ3s*)2(σ3pz)2(π3px,y)4(π3px,y)2 with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond.

Exercise

Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O22−).

Answer: (σ2s)2(σ2s*)2(σ2pz)2(π2px,y)4(π2px,y)4; bond order of 1; no unpaired electrons

Molecular Orbitals for Heteronuclear Diatomic Molecules

Diatomic molecules with two different atoms are called heteronuclear diatomic moleculesA molecule that consists of two atoms of different elements.. When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χB > χA), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in . The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond.

Figure 9.29 Molecular Orbital Energy-Level Diagram for a Heteronuclear Diatomic Molecule AB, Where χB > χA

The bonding molecular orbitals are closer in energy to the atomic orbitals of the more electronegative B atom. Consequently, the electrons in the bonding orbitals are not shared equally between the two atoms. On average, they are closer to the B atom, resulting in a polar covalent bond.

Note the Pattern

A molecular orbital energy-level diagram is always skewed toward the more electronegative atom.

An Odd Number of Valence Electrons: NO

Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O2 with N2 at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O2 to produce NO2, which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals.

Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO () shows that the general pattern is similar to that for the O2 molecule (see ). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2p atomic orbitals, the 11th electron must occupy one of the degenerate π* orbitals. The predicted bond order for NO is therefore (83)÷2=212. Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N2 and O2 molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot.

Figure 9.30 Molecular Orbital Energy-Level Diagram for NO

Because NO has 11 valence electrons, it is paramagnetic, with a single electron occupying the (π2px, π2py) pair of orbitals.

Molecular orbital theory can also tell us something about the chemistry of NO. As indicated in the energy-level diagram in , NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, NO is easily oxidized to the NO+ cation, which is isoelectronic with N2 and has a bond order of 3, corresponding to an N≡O triple bond.

Nonbonding Molecular Orbitals

Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from that the 1s orbital of atomic hydrogen is closest in energy to the 3p orbitals of chlorine. Consequently, the filled Cl 3s atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1s and Cl 3p orbitals. Of the three p orbitals, only one, designated as 3pz, can interact with the H 1s orbital. The 3px and 3py atomic orbitals have no net overlap with the 1s orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3s, 3px, and 3py orbitals do not change when HCl forms, they are called nonbonding molecular orbitalsA molecular orbital that forms when atomic orbitals or orbital lobes interact only very weakly, creating essentially no change in the electron probability density between the nuclei.. A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3pz than to the H 1s atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give Hδ+–Clδ as described in .

Figure 9.31 Molecular Orbital Energy-Level Diagram for HCl

The hydrogen 1s atomic orbital interacts most strongly with the 3pz orbital on chlorine, producing a bonding/antibonding pair of molecular orbitals. The other electrons on Cl are best viewed as nonbonding. As a result, only the bonding σ orbital is occupied by electrons, giving a bond order of 1.

Note the Pattern

Electrons in nonbonding molecular orbitals have no effect on bond order.

Example 11

Use a “skewed” molecular orbital energy-level diagram like the one in to describe the bonding in the cyanide ion (CN). What is the bond order?

Given: chemical species

Asked for: “skewed” molecular orbital energy-level diagram, bonding description, and bond order

Strategy:

A Calculate the total number of valence electrons in CN. Then place these electrons in a molecular orbital energy-level diagram like in order of increasing energy. Be sure to obey the Pauli principle and Hund’s rule while doing so.

B Calculate the bond order and describe the bonding in CN.

Solution:

A The CN ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like fills the five lowest-energy orbitals, as shown here:

Because χN > χC, the atomic orbitals of N (on the right) are lower in energy than those of C. B The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN ion has a triple bond, analogous to that in N2.

Exercise

Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl). What is the bond order?

Answer: All molecular orbitals except the highest-energy σ* are filled, giving a bond order of 1.

Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods.

Summary

A molecular orbital (MO) is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital (AO), which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the molecular orbital theory of bonding is a delocalized approach. Molecular orbitals are constructed using linear combinations of atomic orbitals (LCAOs), which are usually the mathematical sums and differences of wave functions that describe overlapping atomic orbitals. Atomic orbitals interact to form three types of molecular orbitals.

  1. Orbitals or orbital lobes with the same sign interact to give increased electron probability along the plane of the internuclear axis because of constructive reinforcement of the wave functions. Consequently, electrons in such molecular orbitals help to hold the positively charged nuclei together. Such orbitals are bonding molecular orbitals, and they are always lower in energy than the parent atomic orbitals.
  2. Orbitals or orbital lobes with opposite signs interact to give decreased electron probability density between the nuclei because of destructive interference of the wave functions. Consequently, electrons in such molecular orbitals are primarily located outside the internuclear region, leading to increased repulsions between the positively charged nuclei. These orbitals are called antibonding molecular orbitals, and they are always higher in energy than the parent atomic orbitals.
  3. Some atomic orbitals interact only very weakly, and the resulting molecular orbitals give essentially no change in the electron probability density between the nuclei. Hence electrons in such orbitals have no effect on the bonding in a molecule or ion. These orbitals are nonbonding molecular orbitals, and they have approximately the same energy as the parent atomic orbitals.

A completely bonding molecular orbital contains no nodes (regions of zero electron probability) perpendicular to the internuclear axis, whereas a completely antibonding molecular orbital contains at least one node perpendicular to the internuclear axis. A sigma (σ) orbital (bonding) or a sigma star (σ*) orbital (antibonding) is symmetrical about the internuclear axis. Hence all cross-sections perpendicular to that axis are circular. Both a pi (π) orbital (bonding) and a pi star (π*) orbital (antibonding) possess a nodal plane that contains the nuclei, with electron density localized on both sides of the plane.

The energies of the molecular orbitals versus those of the parent atomic orbitals can be shown schematically in an energy-level diagram. The electron configuration of a molecule is shown by placing the correct number of electrons in the appropriate energy-level diagram, starting with the lowest-energy orbital and obeying the Pauli principle; that is, placing only two electrons with opposite spin in each orbital. From the completed energy-level diagram, we can calculate the bond order, defined as one-half the net number of bonding electrons. In bond orders, electrons in antibonding molecular orbitals cancel electrons in bonding molecular orbitals, while electrons in nonbonding orbitals have no effect and are not counted. Bond orders of 1, 2, and 3 correspond to single, double, and triple bonds, respectively. Molecules with predicted bond orders of 0 are generally less stable than the isolated atoms and do not normally exist.

Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the parent atomic orbitals increases. With such an approach, the electronic structures of virtually all commonly encountered homonuclear diatomic molecules, molecules with two identical atoms, can be understood. The molecular orbital approach correctly predicts that the O2 molecule has two unpaired electrons and hence is attracted into a magnetic field. In contrast, most substances have only paired electrons. A similar procedure can be applied to molecules with two dissimilar atoms, called heteronuclear diatomic molecules, using a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element. Molecular orbital theory is able to describe the bonding in a molecule with an odd number of electrons such as NO and even to predict something about its chemistry.

Key Takeaway

  • Molecular orbital theory, a delocalized approach to bonding, can often explain a compound’s color, why a compound with unpaired electrons is stable, semiconductor behavior, and resonance, none of which can be explained using a localized approach.

Conceptual Problems

  1. What is the distinction between an atomic orbital and a molecular orbital? How many electrons can a molecular orbital accommodate?

  2. Why is the molecular orbital approach to bonding called a delocalized approach?

  3. How is the energy of an electron affected by interacting with more than one positively charged atomic nucleus at a time? Does the energy of the system increase, decrease, or remain unchanged? Why?

  4. Constructive and destructive interference of waves can be used to understand how bonding and antibonding molecular orbitals are formed from atomic orbitals. Does constructive interference of waves result in increased or decreased electron probability density between the nuclei? Is the result of constructive interference best described as a bonding molecular orbital or an antibonding molecular orbital?

  5. What is a “node” in molecular orbital theory? How is it similar to the nodes found in atomic orbitals?

  6. What is the difference between an s orbital and a σ orbital? How are the two similar?

  7. Why is a σ1s molecular orbital lower in energy than the two s atomic orbitals from which it is derived? Why is a σ1s* molecular orbital higher in energy than the two s atomic orbitals from which it is derived?

  8. What is meant by the term bond order in molecular orbital theory? How is the bond order determined from molecular orbital theory different from the bond order obtained using Lewis electron structures? How is it similar?

  9. What is the effect of placing an electron in an antibonding orbital on the bond order, the stability of the molecule, and the reactivity of a molecule?

  10. How can the molecular orbital approach to bonding be used to predict a molecule’s stability? What advantages does this method have over the Lewis electron-pair approach to bonding?

  11. What is the relationship between bond length and bond order? What effect do antibonding electrons have on bond length? on bond strength?

  12. Draw a diagram that illustrates how atomic p orbitals can form both σ and π molecular orbitals. Which type of molecular orbital typically results in a stronger bond?

  13. What is the minimum number of nodes in σ, π, σ*, and π*? How are the nodes in bonding orbitals different from the nodes in antibonding orbitals?

  14. It is possible to form both σ and π molecular orbitals with the overlap of a d orbital with a p orbital, yet it is possible to form only σ molecular orbitals between s and d orbitals. Illustrate why this is so with a diagram showing the three types of overlap between this set of orbitals. Include a fourth image that shows why s and d orbitals cannot combine to form a π molecular orbital.

  15. Is it possible for an npx orbital on one atom to interact with an npy orbital on another atom to produce molecular orbitals? Why or why not? Can the same be said of npy and npz orbitals on adjacent atoms?

  16. What is meant by degenerate orbitals in molecular orbital theory? Is it possible for σ molecular orbitals to form a degenerate pair? Explain your answer.

  17. Why are bonding molecular orbitals lower in energy than the parent atomic orbitals? Why are antibonding molecular orbitals higher in energy than the parent atomic orbitals?

  18. What is meant by the law of conservation of orbitals?

  19. Atomic orbitals on different atoms have different energies. When atomic orbitals from nonidentical atoms are combined to form molecular orbitals, what is the effect of this difference in energy on the resulting molecular orbitals?

  20. If two atomic orbitals have different energies, how does this affect the orbital overlap and the molecular orbitals formed by combining the atomic orbitals?

  21. Are the Al–Cl bonds in AlCl3 stronger, the same strength, or weaker than the Al–Br bonds in AlBr3? Why?

  22. Are the Ga–Cl bonds in GaCl3 stronger, the same strength, or weaker than the Sb–Cl bonds in SbCl3? Why?

  23. What is meant by a nonbonding molecular orbital, and how is it formed? How does the energy of a nonbonding orbital compare with the energy of bonding or antibonding molecular orbitals derived from the same atomic orbitals?

  24. Many features of molecular orbital theory have analogs in Lewis electron structures. How do Lewis electron structures represent

    1. nonbonding electrons?
    2. electrons in bonding molecular orbitals?
  25. How does electron screening affect the energy difference between the 2s and 2p atomic orbitals of the period 2 elements? How does the energy difference between the 2s and 2p atomic orbitals depend on the effective nuclear charge?

  26. For σ versus π, π versus σ*, and σ* versus π*, which of the resulting molecular orbitals is lower in energy?

  27. The energy of a σ molecular orbital is usually lower than the energy of a π molecular orbital derived from the same set of atomic orbitals. Under specific conditions, however, the order can be reversed. What causes this reversal? In which portion of the periodic table is this kind of orbital energy reversal most likely to be observed?

  28. Is the σ2pz molecular orbital stabilized or destabilized by interaction with the σ2s molecular orbital in N2? in O2? In which molecule is this interaction most important?

  29. Explain how the Lewis electron-pair approach and molecular orbital theory differ in their treatment of bonding in O2.

  30. Why is it crucial to our existence that O2 is paramagnetic?

  31. Will NO or CO react more quickly with O2? Explain your answer.

  32. How is the energy-level diagram of a heteronuclear diatomic molecule, such as CO, different from that of a homonuclear diatomic molecule, such as N2?

  33. How does molecular orbital theory describe the existence of polar bonds? How is this apparent in the molecular orbital diagram of HCl?

Answers

  1. An atomic orbital is a region of space around an atom that has a non-zero probability for an electron with a particular energy. Analogously, a molecular orbital is a region of space in a molecule that has a non-zero probability for an electron with a particular energy. Both an atomic orbital and a molecular orbital can contain two electrons.

  2. No. Because an npx orbital on one atom is perpendicular to an npy orbital on an adjacent atom, the net overlap between the two is zero. This is also true for npy and npz orbitals on adjacent atoms.

Numerical Problems

  1. Use a qualitative molecular orbital energy-level diagram to describe the bonding in S22−. What is the bond order? How many unpaired electrons does it have?

  2. Use a qualitative molecular orbital energy-level diagram to describe the bonding in F22+. What is the bond order? How many unpaired electrons does it have?

  3. If three atomic orbitals combine to form molecular orbitals, how many molecular orbitals are generated? How many molecular orbitals result from the combination of four atomic orbitals? From five?

  4. If two atoms interact to form a bond, and each atom has four atomic orbitals, how many molecular orbitals will form?

  5. Sketch the possible ways of combining two 1s orbitals on adjacent atoms. How many molecular orbitals can be formed by this combination? Be sure to indicate any nodal planes.

  6. Sketch the four possible ways of combining two 2p orbitals on adjacent atoms. How many molecular orbitals can be formed by this combination? Be sure to indicate any nodal planes.

  7. If a diatomic molecule has a bond order of 2 and six bonding electrons, how many antibonding electrons must it have? What would be the corresponding Lewis electron structure (disregarding lone pairs)? What would be the effect of a one-electron reduction on the bond distance?

  8. What is the bond order of a diatomic molecule with six bonding electrons and no antibonding electrons? If an analogous diatomic molecule has six bonding electrons and four antibonding electrons, which has the stronger bond? the shorter bond distance? If the highest occupied molecular orbital in both molecules is bonding, how will a one-electron oxidation affect the bond length?

  9. Qualitatively discuss how the bond distance in a diatomic molecule would be affected by adding an electron to

    1. an antibonding orbital.
    2. a bonding orbital.
  10. Explain why the oxidation of O2 decreases the bond distance, whereas the oxidation of N2increases the N–N distance. Could Lewis electron structures be employed to answer this problem?

  11. Draw a molecular orbital energy-level diagram for Na2+. What is the bond order in this ion? Is this ion likely to be a stable species? If not, would you recommend an oxidation or a reduction to improve stability? Explain your answer. Based on your answers, will Na2+, Na2, or Na2 be the most stable? Why?

  12. Draw a molecular orbital energy-level diagram for Xe2+, showing only the valence orbitals and electrons. What is the bond order in this ion? Is this ion likely to be a stable species? If not, would you recommend an oxidation or a reduction to improve stability? Explain your answer. Based on your answers, will Xe22+, Xe2+, or Xe2 be most stable? Why?

  13. Draw a molecular orbital energy-level diagram for O22− and predict its valence electron configuration, bond order, and stability.

  14. Draw a molecular orbital energy-level diagram for C22– and predict its valence electron configuration, bond order, and stability.

  15. If all the p orbitals in the valence shells of two atoms interact, how many molecular orbitals are formed? Why is it not possible to form three π orbitals (and the corresponding antibonding orbitals) from the set of six p orbitals?

  16. Draw a complete energy-level diagram for B2. Determine the bond order and whether the molecule is paramagnetic or diamagnetic. Explain your rationale for the order of the molecular orbitals.

  17. Sketch a molecular orbital energy-level diagram for each ion. Based on your diagram, what is the bond order of each species?

    1. NO+
    2. NO
  18. The diatomic molecule BN has never been detected. Assume that its molecular orbital diagram would be similar to that shown for CN in but that the σ2pz molecular orbital is higher in energy than the π2pz,y molecular orbitals.

    1. Sketch a molecular orbital diagram for BN.
    2. Based on your diagram, what would be the bond order of this molecule?
    3. Would you expect BN to be stable? Why or why not?
  19. Of the species BN, CO, C2, and N2, which are isoelectronic?

  20. Of the species CN, NO+, B22−, and O2+, which are isoelectronic?

Answers

  1.  

    The bond order is 1, and the ion has no unpaired electrons.

  2. The number of molecular orbitals is always equal to the number of atomic orbitals you start with. Thus, combining three atomic orbitals gives three molecular orbitals, and combining four or five atomic orbitals will give four or five molecular orbitals, respectively.

  3.  

    Combining two atomic s orbitals gives two molecular orbitals, a σ (bonding) orbital with no nodal planes, and a σ* (antibonding) orbital with a nodal plane perpendicular to the internuclear axis.

    1. Adding an electron to an antibonding molecular orbital will decrease the bond order, thereby increasing the bond distance.
    2. Adding an electron to a bonding molecular orbital will increase the bond order, thereby decreasing the bond distance.
  4. Sodium contains only a single valence electron in its 3s atomic orbital. Combining two 3s atomic orbitals gives two molecular orbitals; as shown in the diagram, these are a σ (bonding) orbital and a σ* (antibonding) orbital.

    Although each sodium atom contributes one valence electron, the +1 charge indicates that one electron has been removed. Placing the single electron in the lowest energy molecular orbital gives a σ3s1 electronic configuration and a bond order of 0.5. Consequently, Na2+ should be a stable species. Oxidizing Na2+ by one electron to give Na22+ would remove the electron in the σ3s molecular orbital, giving a bond order of 0. Conversely, reducing Na2+ by one electron to give Na2 would put an additional electron into the σ3s molecular orbital, giving a bond order of 1. Thus, reduction to Na2 would produce a more stable species than oxidation to Na22+. The Na2 ion would have two electrons in the bonding σ3s molecular orbital and one electron in the antibonding σ3s* molecular orbital, giving a bond order of 0.5. Thus, Na2 is the most stable of the three species.

  5.  

    1. The NO+ ion has 10 valence electrons, which fill all the molecular orbitals up to and including the σ2p. With eight electrons in bonding molecular orbitals and two electrons in antibonding orbitals, the bond order in NO+ is (8 − 2)/2 = 3.
    2. The NO ion contains two more electrons, which fill the σ2p* molecular orbital. The bond order in NO is (8 − 4)/2 = 2.
  6. BN and C2 are isoelectronic, with 12 valence electrons, while N2 and CO are isoelectronic, with 14 valence electrons.

9.4 Polyatomic Systems with Multiple Bonds

Learning Objective

  1. To explain resonance structures using molecular orbitals.

So far in our molecular orbital descriptions we have not dealt with polyatomic systems with multiple bonds. To do so, we can use an approach in which we describe σ bonding using localized electron-pair bonds formed by hybrid atomic orbitals, and π bonding using molecular orbitals formed by unhybridized np atomic orbitals.

Multiple Bonds

We begin our discussion by considering the bonding in ethylene (C2H4). Experimentally, we know that the H–C–H and H–C–C angles in ethylene are approximately 120°. This angle suggests that the carbon atoms are sp2 hybridized, which means that a singly occupied sp2 orbital on one carbon overlaps with a singly occupied s orbital on each H and a singly occupied sp2 lobe on the other C. Thus each carbon forms a set of three σ bonds: two C–H (sp2 + s) and one C–C (sp2 + sp2) (part (a) in ). The sp2 hybridization can be represented as follows:

Figure 9.32 Bonding in Ethylene

(a) The σ-bonded framework is formed by the overlap of two sets of singly occupied carbon sp2 hybrid orbitals and four singly occupied hydrogen 1s orbitals to form electron-pair bonds. This uses 10 of the 12 valence electrons to form a total of five σ bonds (four C–H bonds and one C–C bond). (b) One singly occupied unhybridized 2pz orbital remains on each carbon atom to form a carbon–carbon π bond. (Note: by convention, in planar molecules the axis perpendicular to the molecular plane is the z-axis.)

After hybridization, each carbon still has one unhybridized 2pz orbital that is perpendicular to the hybridized lobes and contains a single electron (part (b) in ). The two singly occupied 2pz orbitals can overlap to form a π bonding orbital and a π* antibonding orbital, which produces the energy-level diagram shown in . With the formation of a π bonding orbital, electron density increases in the plane between the carbon nuclei. The π* orbital lies outside the internuclear region and has a nodal plane perpendicular to the internuclear axis. Because each 2pz orbital has a single electron, there are only two electrons, enough to fill only the bonding (π) level, leaving the π* orbital empty. Consequently, the C–C bond in ethylene consists of a σ bond and a π bond, which together give a C=C double bond. Our model is supported by the facts that the measured carbon–carbon bond is shorter than that in ethane (133.9 pm versus 153.5 pm) and the bond is stronger (728 kJ/mol versus 376 kJ/mol in ethane). The two CH2 fragments are coplanar, which maximizes the overlap of the two singly occupied 2pz orbitals.

Figure 9.33 Molecular Orbital Energy-Level Diagram for π Bonding in Ethylene

As in the diatomic molecules discussed previously, the singly occupied 2pz orbitals in ethylene can overlap to form a bonding/antibonding pair of π molecular orbitals. The two electrons remaining are enough to fill only the bonding π orbital. With one σ bond plus one π bond, the carbon–carbon bond order in ethylene is 2.

Triple bonds, as in acetylene (C2H2), can also be explained using a combination of hybrid atomic orbitals and molecular orbitals. The four atoms of acetylene are collinear, which suggests that each carbon is sp hybridized. If one sp lobe on each carbon atom is used to form a C–C σ bond and one is used to form the C–H σ bond, then each carbon will still have two unhybridized 2p orbitals (a 2px,y pair), each with one electron (part (a) in ).

The two 2p orbitals on each carbon can align with the corresponding 2p orbitals on the adjacent carbon to simultaneously form a pair of π bonds (part (b) in ). Because each of the unhybridized 2p orbitals has a single electron, four electrons are available for π bonding, which is enough to occupy only the bonding molecular orbitals. Acetylene must therefore have a carbon–carbon triple bond, which consists of a C–C σ bond and two mutually perpendicular π bonds. Acetylene does in fact have a shorter carbon–carbon bond (120.3 pm) and a higher bond energy (965 kJ/mol) than ethane and ethylene, as we would expect for a triple bond.

Figure 9.34 Bonding in Acetylene

(a) In the formation of the σ-bonded framework, two sets of singly occupied carbon sp hybrid orbitals and two singly occupied hydrogen 1s orbitals overlap. (b) In the formation of two carbon–carbon π bonds in acetylene, two singly occupied unhybridized 2px,y orbitals on each carbon atom overlap. With one σ bond plus two π bonds, the carbon–carbon bond order in acetylene is 3.

Note the Pattern

In complex molecules, hybrid orbitals and valence bond theory can be used to describe σ bonding, and unhybridized π orbitals and molecular orbital theory can be used to describe π bonding.

Example 12

Describe the bonding in HCN using a combination of hybrid atomic orbitals and molecular orbitals. The HCN molecule is linear.

Given: chemical compound and molecular geometry

Asked for: bonding description using hybrid atomic orbitals and molecular orbitals

Strategy:

A From the geometry given, predict the hybridization in HCN. Use the hybrid orbitals to form the σ-bonded framework of the molecule and determine the number of valence electrons that are used for σ bonding.

B Determine the number of remaining valence electrons. Use any remaining unhybridized p orbitals to form π and π* orbitals.

C Fill the orbitals with the remaining electrons in order of increasing energy. Describe the bonding in HCN.

Solution:

A Because HCN is a linear molecule, it is likely that the bonding can be described in terms of sp hybridization at carbon. Because the nitrogen atom can also be described as sp hybridized, we can use one sp hybrid on each atom to form a C–N σ bond. This leaves one sp hybrid on each atom to either bond to hydrogen (C) or hold a lone pair of electrons (N). Of 10 valence electrons (5 from N, 4 from C, and 1 from H), 4 are used for σ bonding:

B We are now left with 2 electrons on N (5 valence electrons minus 1 bonding electron minus 2 electrons in the lone pair) and 2 electrons on C (4 valence electrons minus 2 bonding electrons). We have two unhybridized 2p atomic orbitals left on carbon and two on nitrogen, each occupied by a single electron. These four 2p atomic orbitals can be combined to give four molecular orbitals: two π (bonding) orbitals and two π* (antibonding) orbitals. C With 4 electrons available, only the π orbitals are filled. The overall result is a triple bond (1 σ and 2 π) between C and N.

Exercise

Describe the bonding in formaldehyde (H2C=O), a trigonal planar molecule, using a combination of hybrid atomic orbitals and molecular orbitals.

Answer:

σ-bonding framework: Carbon and oxygen are sp2 hybridized. Two sp2 hybrid orbitals on oxygen have lone pairs, two sp2 hybrid orbitals on carbon form C–H bonds, and one sp2 hybrid orbital on C and O forms a C–O σ bond.

π bonding: Unhybridized, singly occupied 2p atomic orbitals on carbon and oxygen interact to form π (bonding) and π* (antibonding) molecular orbitals. With two electrons, only the π (bonding) orbital is occupied.

Molecular Orbitals and Resonance Structures

In , we used resonance structures to describe the bonding in molecules such as ozone (O3) and the nitrite ion (NO2). We showed that ozone can be represented by either of these Lewis electron structures:

Although the VSEPR model correctly predicts that both species are bent, it gives no information about their bond orders.

Figure 9.35 Bonding in Ozone

(a) In the formation of the σ-bonded framework, three sets of oxygen sp2 hybrid orbitals overlap to give two O–O σ bonds and five lone pairs, two on each terminal O and one on the central O. The σ bonds and lone pairs account for 14 of the 18 valence electrons of O3. (b) One unhybridized 2pz orbital remains on each oxygen atom that is available for π bonding. The unhybridized 2pz orbital on each terminal O atom has a single electron, whereas the unhybridized 2pz orbital on the central O atom has 2 electrons.

Experimental evidence indicates that ozone has a bond angle of 117.5°. Because this angle is close to 120°, it is likely that the central oxygen atom in ozone is trigonal planar and sp2 hybridized. If we assume that the terminal oxygen atoms are also sp2 hybridized, then we obtain the σ-bonded framework shown in . Two of the three sp2 lobes on the central O are used to form O–O sigma bonds, and the third has a lone pair of electrons. Each terminal oxygen atom has two lone pairs of electrons that are also in sp2 lobes. In addition, each oxygen atom has one unhybridized 2p orbital perpendicular to the molecular plane. The σ bonds and lone pairs account for a total of 14 electrons (five lone pairs and two σ bonds, each containing 2 electrons). Each oxygen atom in ozone has 6 valence electrons, so O3 has a total of 18 valence electrons. Subtracting 14 electrons from the total gives us 4 electrons that must occupy the three unhybridized 2p orbitals.

With a molecular orbital approach to describe the π bonding, three 2p atomic orbitals give us three molecular orbitals, as shown in . One of the molecular orbitals is a π bonding molecular orbital, which is shown as a banana-shaped region of electron density above and below the molecular plane. This region has no nodes perpendicular to the O3 plane. The molecular orbital with the highest energy has two nodes that bisect the O–O σ bonds; it is a π* antibonding orbital. The third molecular orbital contains a single node that is perpendicular to the O3 plane and passes through the central O atom; it is a nonbonding molecular orbital. Because electrons in nonbonding orbitals are neither bonding nor antibonding, they are ignored in calculating bond orders.

Figure 9.36 π Bonding in Ozone

The three unhybridized 2pz atomic orbitals interact with one another to form three molecular orbitals: one π bonding orbital at lower energy, one π* antibonding orbital at higher energy, and a nonbonding orbital in between. Placing four electrons in this diagram fills the bonding and nonbonding orbitals. With one filled π bonding orbital holding three atoms together, the net π bond order is 12 per O–O bond. The combined σ/π bond order is thus 112 for each O–O bond.

We can now place the remaining four electrons in the three energy levels shown in , thereby filling the π bonding and the nonbonding levels. The result is a single π bond holding three oxygen atoms together, or 12π bond per O–O. We therefore predict the overall O–O bond order to be 112 (12π bond plus 1 σ bond), just as predicted using resonance structures. The molecular orbital approach, however, shows that the π nonbonding orbital is localized on the terminal O atoms, which suggests that they are more electron rich than the central O atom. The reactivity of ozone is consistent with the predicted charge localization.

Note the Pattern

Resonance structures are a crude way of describing molecular orbitals that extend over more than two atoms.

Example 13

Describe the bonding in the nitrite ion in terms of a combination of hybrid atomic orbitals and molecular orbitals. Lewis dot structures and the VSEPR model predict that the NO2 ion is bent.

Given: chemical species and molecular geometry

Asked for: bonding description using hybrid atomic orbitals and molecular orbitals

Strategy:

A Calculate the number of valence electrons in NO2. From the structure, predict the type of atomic orbital hybridization in the ion.

B Predict the number and type of molecular orbitals that form during bonding. Use valence electrons to fill these orbitals and then calculate the number of electrons that remain.

C If there are unhybridized orbitals, place the remaining electrons in these orbitals in order of increasing energy. Calculate the bond order and describe the bonding.

Solution:

A The lone pair of electrons on nitrogen and a bent structure suggest that the bonding in NO2 is similar to the bonding in ozone. This conclusion is supported by the fact that nitrite also contains 18 valence electrons (5 from N and 6 from each O, plus 1 for the −1 charge). The bent structure implies that the nitrogen is sp2 hybridized.

B If we assume that the oxygen atoms are sp2 hybridized as well, then we can use two sp2 hybrid orbitals on each oxygen and one sp2 hybrid orbital on nitrogen to accommodate the five lone pairs of electrons. Two sp2 hybrid orbitals on nitrogen form σ bonds with the remaining sp2 hybrid orbital on each oxygen. The σ bonds and lone pairs account for 14 electrons. We are left with three unhybridized 2p orbitals, one on each atom, perpendicular to the plane of the molecule, and 4 electrons. Just as with ozone, these three 2p orbitals interact to form bonding, nonbonding, and antibonding π molecular orbitals. The bonding molecular orbital is spread over the nitrogen and both oxygen atoms.

C Placing 4 electrons in the energy-level diagram fills both the bonding and nonbonding molecular orbitals and gives a π bond order of 1/2 per N–O bond. The overall N–O bond order is 112, consistent with a resonance structure.

Exercise

Describe the bonding in the formate ion (HCO2), in terms of a combination of hybrid atomic orbitals and molecular orbitals.

Answer: Like nitrite, formate is a planar polyatomic ion with 18 valence electrons. The σ bonding framework can be described in terms of sp2 hybridized carbon and oxygen, which account for 14 electrons. The three unhybridized 2p orbitals (on C and both O atoms) form three π molecular orbitals, and the remaining 4 electrons occupy both the bonding and nonbonding π molecular orbitals. The overall C–O bond order is therefore 112.

The Chemistry of Vision

Hydrocarbons in which two or more carbon–carbon double bonds are directly linked by carbon–carbon single bonds are generally more stable than expected because of resonance. Because the double bonds are close enough to interact electronically with one another, the π electrons are shared over all the carbon atoms, as illustrated for 1,3-butadiene in . As the number of interacting atomic orbitals increases, the number of molecular orbitals increases, the energy spacing between molecular orbitals decreases, and the systems become more stable (). Thus as a chain of alternating double and single bonds becomes longer, the energy required to excite an electron from the highest-energy occupied (bonding) orbital to the lowest-energy unoccupied (antibonding) orbital decreases. If the chain is long enough, the amount of energy required to excite an electron corresponds to the energy of visible light. For example, vitamin A is yellow because its chain of five alternating double bonds is able to absorb violet light. Many of the colors we associate with dyes result from this same phenomenon; most dyes are organic compounds with alternating double bonds.

Figure 9.37 π Bonding in 1,3-Butadiene

(a) If each carbon atom is assumed to be sp2 hybridized, we can construct a σ-bonded framework that accounts for the C–H and C–C single bonds, leaving four singly occupied 2pz orbitals, one on each carbon atom. (b) As in ozone, these orbitals can interact, in this case to form four molecular orbitals. The molecular orbital at lowest energy is a bonding orbital with 0 nodes, the one at highest energy is antibonding with 3 nodes, and the two in the middle have 1 node and 2 nodes and are somewhere between bonding or antibonding and nonbonding, respectively. The energy of the molecular orbital increases with the number of nodes. With four electrons, only the two bonding orbitals are filled, consistent with the presence of two π bonds.

Figure 9.38 Molecular Orbital Energy-Level Diagrams for a Chain of n Like Orbitals That Interact (n ≤ 5)

As the number of atomic orbitals increases, the difference in energy between the resulting molecular orbital energy levels decreases, which allows light of lower energy to be absorbed. As a result, organic compounds with long chains of carbon atoms and alternating single and double bonds tend to become more deeply colored as the number of double bonds increases.

Note the Pattern

As the number of interacting atomic orbitals increases, the energy separation between the resulting molecular orbitals steadily decreases.

A derivative of vitamin A called retinal is used by the human eye to detect light and has a structure with alternating C=C double bonds. When visible light strikes retinal, the energy separation between the molecular orbitals is sufficiently close that the energy absorbed corresponds to the energy required to change one double bond in the molecule from cis, where like groups are on the same side of the double bond, to trans, where they are on opposite sides, initiating a process that causes a signal to be sent to the brain. If this mechanism is defective, we lose our vision in dim light. Once again, a molecular orbital approach to bonding explains a process that cannot be explained using any of the other approaches we have described.

Summary

To describe the bonding in more complex molecules with multiple bonds, we can use an approach that uses hybrid atomic orbitals to describe the σ bonding and molecular orbitals to describe the π bonding. In this approach, unhybridized np orbitals on atoms bonded to one another are allowed to interact to produce bonding, antibonding, or nonbonding combinations. For π bonds between two atoms (as in ethylene or acetylene), the resulting molecular orbitals are virtually identical to the π molecular orbitals in diatomic molecules such as O2 and N2. Applying the same approach to π bonding between three or four atoms requires combining three or four unhybridized np orbitals on adjacent atoms to generate π bonding, antibonding, and nonbonding molecular orbitals extending over all of the atoms. Filling the resulting energy-level diagram with the appropriate number of electrons explains the bonding in molecules or ions that previously required the use of resonance structures in the Lewis electron-pair approach.

Key Takeaway

  • Polyatomic systems with multiple bonds can be described using hybrid atomic orbitals for σ bonding and molecular orbitals to describe π bonding.

Conceptual Problems

  1. What information is obtained by using the molecular orbital approach to bonding in O3 that is not obtained using the VSEPR model? Can this information be obtained using a Lewis electron-pair approach?

  2. How is resonance explained using the molecular orbital approach?

  3. Indicate what information can be obtained by each method:

    Lewis Electron Structures VSEPR Model Valence Bond Theory Molecular Orbital Theory
    Geometry
    Resonance
    Orbital Hybridization
    Reactivity
    Expanded Valences
    Bond Order

Numerical Problems

  1. Using both a hybrid atomic orbital and molecular orbital approaches, describe the bonding in BCl3 and CS32−.

  2. Use both a hybrid atomic orbital and molecular orbital approaches to describe the bonding in CO2 and N3.

9.5 End-of-Chapter Material

Application Problems

    Problems marked with a ♦ involve multiple concepts.

  1. ♦ Sulfur hexafluoride (SF6) is a very stable gas that is used in a wide range of applications because it is nontoxic, nonflammable, and noncorrosive. Unfortunately, it is also a very powerful “greenhouse gas” that is about 22,000 times more effective at causing global warming than the same mass of CO2.

    1. Draw the Lewis electron structure of SF6 and determine the number of electron groups around the central atom, the molecular geometry, and the hybridization of the central atom.
    2. Suggest a reason for the extremely high stability of SF6.
    3. Despite its rather high molecular mass (146.06 g/mol) and highly polar S–F bonds, SF6 is a gas at room temperature (boiling point = −63°C). Why?
  2. ♦ The elevated concentrations of chlorine monoxide (ClO) that accompany ozone depletions in Earth’s atmosphere can be explained by a sequence of reactions. In the first step, chlorine gas is split into chlorine atoms by sunlight. Each chlorine atom then catalyzes the decomposition of ozone through a chlorine monoxide intermediate.

    1. Write balanced chemical equations showing this sequence of reactions.
    2. Sketch the molecular orbital energy-level diagram of ClO.
    3. Does ClO contain any unpaired electrons?
    4. Based on your molecular orbital diagram, is ClO likely to be a stable species? Explain your answer.
  3. ♦ Saccharin is an artificial sweetener that was discovered in 1879. For several decades, it was used by people who had to limit their intake of sugar for medical reasons. Because it was implicated as a carcinogen in 1977, however, warning labels are now required on foods and beverages containing saccharin. The structure of this sweetener is as follows:

    1. Give the hybridization of all five atoms shown in bold in the structure. Note: all five atoms in the five-membered ring are coplanar.
    2. The carbon–oxygen bond is drawn as a double bond. If the nitrogen and the carbon attached to the C=O group each contribute one electron to the bonding, use both a Lewis electron structure and a hybrid orbital approach to explain the presence of the double bond.
    3. If sulfur and carbon each contribute one electron to nitrogen, how many lone pairs are present on the nitrogen?
    4. What is the geometry of the sulfur atom?
    5. The Lewis electron structure supports a single bond between the carbon and the nitrogen and a double bond between the carbon and the oxygen. In actuality, the C–O bond is longer than expected for a double bond, and the C–N bond is shorter. The nitrogen is also planar. Based on this information, what is the likely hybridization of the nitrogen? Using the concepts of molecular orbital theory, propose an explanation for this observation.
  4. ♦ Pheromones are chemical signals used for communication between members of the same species. For example, the bark beetle uses an aggregation pheromone to signal other bark beetles to congregate at a particular site in a tree. Bark beetle infestations can cause severe damage because the beetles carry a fungal infection that spreads rapidly and can kill the tree. One of the components of this aggregation pheromone has the following structure:

    1. Give the hybridization of all atoms except hydrogen in this pheromone.
    2. How many σ bonds are present in this molecule? How many π bonds are there?
    3. Describe the bonding in this molecule using a combination of the localized and delocalized approaches.
  5. Carbon monoxide is highly poisonous because it binds more strongly than O2 to the iron in red blood cells, which transport oxygen in the blood. Consequently, a victim of CO poisoning suffocates from a lack of oxygen. Draw a molecular orbital energy-level diagram for CO. What is the highest occupied molecular orbital? Are any of the molecular orbitals degenerate? If so, which ones?

Answer

  1.  

    1. There are six electron groups, the molecular geometry is octahedral, and the hybridization of S is sp3d2.

    2. With six fluorine atoms packed around the central sulfur atom, there is no room for another species to approach the sulfur to initiate a reaction. The polar S–F bonds are also expected to be quite strong, so breaking an S–F bond to initiate a reaction is unlikely under most conditions.
    3. SF6 is a gas at room temperature because it has no net dipole moment; the individual S–F bond dipoles cancel one another in this highly symmetrical structure. The absence of a dipole moment results in very weak interactions between SF6 molecules, and as a result SF6 is a gas rather than a liquid or a solid at room temperature.

Chapter 10 Gases

In Chapter 6 "The Structure of Atoms" through Chapter 9 "Molecular Geometry and Covalent Bonding Models" we focused on the microscopic properties of matter—the properties of individual atoms, ions, and molecules—and how the electronic structures of atoms and ions determine the stoichiometry and three-dimensional geometry of the compounds they form. We will now focus on macroscopic properties—the behavior of aggregates with large numbers of atoms, ions, or molecules. An understanding of macroscopic properties is central to an understanding of chemistry. Why, for example, are many substances gases under normal pressures and temperatures (1.0 atm, 25°C), whereas others are liquids or solids? We will examine each form of matter—gases, liquids, and solids—as well as the nature of the forces, such as hydrogen bonding and electrostatic interactions, that hold molecular and ionic compounds together in these three states.

Hot-air balloons being prepared for flight. As the air inside each balloon is heated, the volume of the air increases, filling the balloon. The lower density of air in the balloons allows the balloons to ascend through a substance with higher density—the cooler air.

In Chapter 10 "Gases", we explore the relationships among pressure, temperature, volume, and the amount of gases. You will learn how to use these relationships to describe the physical behavior of a sample of both a pure gaseous substance and mixtures of gases. By the end of this chapter, your understanding of the gas laws and the model used to explain the behavior of gases will allow you to explain how straws and hot-air balloons work, why hand pumps cannot be used in wells beyond a certain depth, why helium-filled balloons deflate so rapidly, and how a gas can be liquefied for use in preserving biological tissue.

10.1 Gaseous Elements and Compounds

Learning Objective

  1. To describe the characteristics of a gas.

The three common phases (or states) of matter are gases, liquids, and solids. Gases have the lowest density of the three, are highly compressible, and completely fill any container in which they are placed. Gases behave this way because their intermolecular forces are relatively weak, so their molecules are constantly moving independently of the other molecules present. Solids, in contrast, are relatively dense, rigid, and incompressible because their intermolecular forces are so strong that the molecules are essentially locked in place. Liquids are relatively dense and incompressible, like solids, but they flow readily to adapt to the shape of their containers, like gases. We can therefore conclude that the sum of the intermolecular forces in liquids are between those of gases and solids. compares the three states of matter and illustrates the differences at the molecular level.

Figure 10.1 A Diatomic Substance (O2) in the Solid, Liquid, and Gaseous States

(a) Solid O2 has a fixed volume and shape, and the molecules are packed tightly together. (b) Liquid O2 conforms to the shape of its container but has a fixed volume; it contains relatively densely packed molecules. (c) Gaseous O2 fills its container completely—regardless of the container’s size or shape—and consists of widely separated molecules.

The state of a given substance depends strongly on conditions. For example, H2O is commonly found in all three states: solid ice, liquid water, and water vapor (its gaseous form). Under most conditions, we encounter water as the liquid that is essential for life; we drink it, cook with it, and bathe in it. When the temperature is cold enough to transform the liquid to ice, we can ski or skate on it, pack it into a snowball or snow cone, and even build dwellings with it. Water vaporThe distinction between a gas and a vapor is subtle: the term vapor refers to the gaseous form of a substance that is a liquid or a solid under normal conditions (25°C, 1.0 atm). Nitrogen (N2) and oxygen (O2) are thus referred to as gases, but gaseous water in the atmosphere is called water vapor. is a component of the air we breathe, and it is produced whenever we heat water for cooking food or making coffee or tea. Water vapor at temperatures greater than 100°C is called steam. Steam is used to drive large machinery, including turbines that generate electricity. The properties of the three states of water are summarized in .

Table 10.1 Properties of Water at 1.0 atm

Temperature State Density (g/cm3)
≤0°C solid (ice) 0.9167 (at 0.0°C)
0°C–100°C liquid (water) 0.9997 (at 4.0°C)
≥100°C vapor (steam) 0.005476 (at 127°C)

The geometric structure and the physical and chemical properties of atoms, ions, and molecules usually do not depend on their physical state; the individual water molecules in ice, liquid water, and steam, for example, are all identical. In contrast, the macroscopic properties of a substance depend strongly on its physical state, which is determined by intermolecular forces and conditions such as temperature and pressure.

shows the locations in the periodic table of those elements that are commonly found in the gaseous, liquid, and solid states. Except for hydrogen, the elements that occur naturally as gases are on the right side of the periodic table. Of these, all the noble gases (group 18) are monatomic gases, whereas the other gaseous elements are diatomic molecules (H2, N2, O2, F2, and Cl2). Oxygen can also form a second allotrope, the highly reactive triatomic molecule ozone (O3), which is also a gas. In contrast, bromine (as Br2) and mercury (Hg) are liquids under normal conditions (25°C and 1.0 atm, commonly referred to as “room temperature and pressure”). Gallium (Ga), which melts at only 29.76°C, can be converted to a liquid simply by holding a container of it in your hand or keeping it in a non-air-conditioned room on a hot summer day. The rest of the elements are all solids under normal conditions.

Figure 10.2 Elements That Occur Naturally as Gases, Liquids, and Solids at 25°C and 1 atm

The noble gases and mercury occur as monatomic species, whereas all other gases and bromine are diatomic molecules.

Many of the elements and compounds we have encountered so far are typically found as gases; some of the more common ones are listed in . Gaseous substances include many binary hydrides, such as the hydrogen halides (HX); hydrides of the chalcogens; hydrides of the group 15 elements N, P, and As; hydrides of the group 14 elements C, Si, and Ge; and diborane (B2H6). In addition, many of the simple covalent oxides of the nonmetals are gases, such as CO, CO2, NO, NO2, SO2, SO3, and ClO2. Many low-molecular-mass organic compounds are gases as well, including all the hydrocarbons with four or fewer carbon atoms and simple molecules such as dimethyl ether [(CH3)2O], methyl chloride (CH3Cl), formaldehyde (CH2O), and acetaldehyde (CH3CHO). Finally, most of the commonly used refrigerants, such as the chlorofluorocarbons (CFCs) and the hydrochlorofluorocarbons (HCFCs), which were discussed in , are gases.

Table 10.2 Some Common Substances That Are Gases at 25°C and 1.0 atm

Elements Compounds
He (helium) HF (hydrogen fluoride) C2H4 (ethylene)
Ne (neon) HCl (hydrogen chloride) C2H2 (acetylene)
Ar (argon) HBr (hydrogen bromide) C3H8 (propane)
Kr (krypton) HI (hydrogen iodide) C4H10 (butane)
Xe (xenon) HCN (hydrogen cyanide)* CO (carbon monoxide)
Rn (radon) H2S (hydrogen sulfide) CO2 (carbon dioxide)
H2 (hydrogen) NH3 (ammonia) NO (nitric oxide)
N2 (nitrogen) PH3 (phosphine) N2O (nitrous oxide)
O2 (oxygen) CH4 (methane) NO2 (nitrogen dioxide)
O3 (ozone) C2H6 (ethane) SO2 (sulfur dioxide)
F2 (fluorine)
Cl2 (chlorine)
*HCN boils at 26°C at 1 atm, so it is included in this table.

All of the gaseous substances mentioned previously (other than the monatomic noble gases) contain covalent or polar covalent bonds and are nonpolar or polar molecules. In contrast, the strong electrostatic attractions in ionic compounds, such as NaBr (boiling point = 1390°C) or LiF (boiling point = 1673°C), prevent them from existing as gases at room temperature and pressure. In addition, the lightest members of any given family of compounds are most likely gases, and the boiling points of polar compounds are generally greater than those of nonpolar compounds of similar molecular mass. Therefore, in a given series of compounds, the lightest and least polar members are the ones most likely to be gases. With relatively few exceptions, however, compounds with more than about five atoms from period 2 or below are too heavy to exist as gases under normal conditions.

Note the Pattern

Gaseous substances often contain covalent or polar covalent bonds, exist as nonpolar or slightly polar molecules, have relatively low molecular masses, and contain five or fewer atoms from periods 1 or 2.

While gases have a wide array of uses, a particularly grim use of a gaseous substance is believed to have been employed by the Persians on the Roman city of Dura in eastern Syria in the third century AD. The Persians dug a tunnel underneath the city wall to enter and conquer the city. Archeological evidence suggests that when the Romans responded with counter-tunnels to stop the siege, the Persians ignited bitumen and sulfur crystals to produce a dense, poisonous gas. It is likely that bellows or chimneys distributed the toxic fumes. The remains of about 20 Roman soldiers were discovered at the base of the city wall at the entrance to a tunnel that was less than 2 m high and 11 m long. Because it is highly unlikely that the Persians could have slaughtered so many Romans at the entrance to such a confined space, archeologists speculate that the ancient Persians used chemical warfare to successfully conquer the city.

Example 1

Which compounds would you predict to be gases at room temperature and pressure?

  1. cyclohexene
  2. lithium carbonate
  3. cyclobutane
  4. vanadium(III) oxide
  5. benzoic acid (C6H5CO2H)

Given: compounds

Asked for: physical state

Strategy:

A Decide whether each compound is ionic or covalent. An ionic compound is most likely a solid at room temperature and pressure, whereas a covalent compound may be a solid, a liquid, or a gas.

B Among the covalent compounds, those that are relatively nonpolar and have low molecular masses are most likely gases at room temperature and pressure.

Solution:

A Lithium carbonate is Li2CO3, containing Li+ and CO32− ions, and vanadium(III) oxide is V2O3, containing V3+ and O2− ions. Both are primarily ionic compounds that are expected to be solids. The remaining three compounds are all covalent.

B Benzoic acid has more than four carbon atoms and is polar, so it is not likely to be a gas. Both cyclohexene and cyclobutane are essentially nonpolar molecules, but cyclobutane (C4H8) has a significantly lower molecular mass than cyclohexene (C6H10), which again has more than four carbon atoms. We therefore predict that cyclobutane is most likely a gas at room temperature and pressure, while cyclohexene is a liquid. In fact, with a boiling point of only 12°C, compared to 83°C for cyclohexene, cyclobutane is indeed a gas at room temperature and pressure.

Exercise

Which compounds would you predict to be gases at room temperature and pressure?

  1. n-butanol
  2. ammonium fluoride (NH4F)
  3. ClF
  4. ethylene oxide

  5. HClO4

Answer: c; d

Summary

Bulk matter can exist in three states: gas, liquid, and solid. Gases have the lowest density of the three, are highly compressible, and fill their containers completely. Elements that exist as gases at room temperature and pressure are clustered on the right side of the periodic table; they occur as either monatomic gases (the noble gases) or diatomic molecules (some halogens, N2, O2). Many inorganic and organic compounds with four or fewer nonhydrogen atoms are also gases at room temperature and pressure. All gaseous substances are characterized by weak interactions between the constituent molecules or atoms.

Key Takeaway

  • The molecules in gaseous substances often contain covalent or polar covalent bonds, are nonpolar or slightly polar molecules, and have relatively low molecular masses.

Conceptual Problems

  1. Explain the differences between the microscopic and the macroscopic properties of matter. Is the boiling point of a compound a microscopic or macroscopic property? molecular mass? Why?

  2. Determine whether the melting point, the dipole moment, and electrical conductivity are macroscopic or microscopic properties of matter and explain your reasoning.

  3. How do the microscopic properties of matter influence the macroscopic properties? Can you relate molecular mass to boiling point? Why or why not? Can polarity be related to boiling point?

  4. For a substance that has gas, liquid, and solid phases, arrange these phases in order of increasing

    1. density.
    2. strength of intermolecular interactions.
    3. compressibility.
    4. molecular motion.
    5. order in the arrangement of the molecules or atoms.
  5. Explain what is wrong with this statement: “The state of matter largely determines the molecular properties of a substance.”

  6. Describe the most important factors that determine the state of a given compound. What external conditions influence whether a substance exists in any one of the three states of matter?

  7. Which elements of the periodic table exist as gases at room temperature and pressure? Of these, which are diatomic molecules and which are monatomic? Which elements are liquids at room temperature and pressure? Which portion of the periodic table contains elements whose binary hydrides are most likely gases at room temperature?

  8. Is the following observation correct? “Almost all nonmetal binary hydrides are gases at room temperature, but metal hydrides are all solids.” Explain your reasoning.

  9. Is the following observation correct? “All the hydrides of the chalcogens are gases at room temperature and pressure except the binary hydride of oxygen, which is a liquid.” Explain your reasoning. Would you expect 1-chloropropane to be a gas? iodopropane? Why?

  10. Explain why ionic compounds are not gases under normal conditions.

Answers

  1. The molecular properties of a substance control its state of matter under a given set of conditions, not the other way around. The presence of strong intermolecular forces favors a condensed state of matter (liquid or solid), while very weak intermolecular interaction favor the gaseous state. In addition, the shape of the molecules dictates whether a condensed phase is a liquid or a solid.

  2. Elements that exist as gases are mainly found in the upper right corner and on the right side of the periodic table. The following elements exist as gases: H, He, N, O, F, Ne, Cl, Ar, Kr, Xe, and Rn. Thus, half of the halogens, all of the noble gases, and the lightest chalcogens and picnogens are gases. Of these, all except the noble gases exist as diatomic molecules. Only two elements exist as liquids at a normal room temperature of 20°C–25°C: mercury and bromine. The upper right portion of the periodic table also includes most of the elements whose binary hydrides are gases. In addition, the binary hydrides of the elements of Groups 14–16 are gases.

10.2 Gas Pressure

Learning Objective

  1. To describe and measure the pressure of a gas.

At the macroscopic level, a complete physical description of a sample of a gas requires four quantities: temperature (expressed in kelvins), volume (expressed in liters), amount (expressed in moles), and pressure (in atmospheres). As we explain in this section and , these variables are not independent. If we know the values of any three of these quantities, we can calculate the fourth and thereby obtain a full physical description of the gas. Temperature, volume, and amount have been discussed in previous chapters. We now discuss pressure and its units of measurement.

Units of Pressure

Any object, whether it is your computer, a person, or a sample of gas, exerts a force on any surface with which it comes in contact. The air in a balloon, for example, exerts a force against the interior surface of the balloon, and a liquid injected into a mold exerts a force against the interior surface of the mold, just as a chair exerts a force against the floor because of its mass and the effects of gravity. If the air in a balloon is heated, the increased kinetic energy of the gas eventually causes the balloon to burst because of the increased pressure(P)The amount of force (F) exerted on a given area (A) of surface: P=F/A. of the gas, the force (F) per unit area (A) of surface:

Equation 10.1

P=FA

Pressure is dependent on both the force exerted and the size of the area to which the force is applied. We know from that applying the same force to a smaller area produces a higher pressure. When we use a hose to wash a car, for example, we can increase the pressure of the water by reducing the size of the opening of the hose with a thumb.

The units of pressure are derived from the units used to measure force and area. In the English system, the units of force are pounds and the units of area are square inches, so we often see pressure expressed in pounds per square inch (lb/in2, or psi), particularly among engineers. For scientific measurements, however, the SI units for force are preferred. The SI unit for pressure, derived from the SI units for force (newtons) and area (square meters), is the newton per square meter (N/m2), which is called the pascal (Pa)The SI unit for pressure. The pascal is newtons per square meter: N/m2., after the French mathematician Blaise Pascal (1623–1662):

Equation 10.2

1 Pa = 1 N/m2

To convert from pounds per square inch to pascals, multiply psi by 6894.757 [1 Pa = 1 psi (6894.757)].

Blaise Pascal (1623–1662)

In addition to his talents in mathematics (he invented modern probability theory), Pascal did research in physics and was an author and a religious philosopher as well. His accomplishments include invention of the first syringe and the first digital calculator and development of the principle of hydraulic pressure transmission now used in brake systems and hydraulic lifts.

Example 2

Assuming a paperback book has a mass of 2.00 kg, a length of 27.0 cm, a width of 21.0 cm, and a thickness of 4.5 cm, what pressure does it exert on a surface if it is

  1. lying flat?
  2. standing on edge in a bookcase?

Given: mass and dimensions of object

Asked for: pressure

Strategy:

A Calculate the force exerted by the book and then compute the area that is in contact with a surface.

B Substitute these two values into to find the pressure exerted on the surface in each orientation.

Solution:

The force exerted by the book does not depend on its orientation. Recall from that the force exerted by an object is F = ma, where m is its mass and a is its acceleration. In Earth’s gravitational field, the acceleration is due to gravity (9.8067 m/s2 at Earth’s surface). In SI units, the force exerted by the book is therefore

F = ma = (2.00 kg)(9.8067 m/s2) = 19.6 (kg·m)/s2 = 19.6 N
  1. A We calculated the force as 19.6 N. When the book is lying flat, the area is (0.270 m)(0.210 m) = 0.0567 m2. B The pressure exerted by the text lying flat is thus

    P=19.6  N0.0567 m2=3.46×102 Pa
  2. A If the book is standing on its end, the force remains the same, but the area decreases:

    (21.0 cm)(4.5 cm) = (0.210 m)(0.045 m) = 9.5 × 10−3 m2

    B The pressure exerted by the book in this position is thus

    P=19.6 N9.5×103 m2=2.1×103 Pa

    Thus the pressure exerted by the book varies by a factor of about six depending on its orientation, although the force exerted by the book does not vary.

Exercise

What pressure does a 60.0 kg student exert on the floor

  1. when standing flat-footed in the laboratory in a pair of tennis shoes (the surface area of the soles is approximately 180 cm2)?
  2. as she steps heel-first onto a dance floor wearing high-heeled shoes (the area of the heel = 1.0 cm2)?

Answers:

  1. 3.27 × 104 Pa (4.74 lb/in.2)
  2. 5.9 × 106 Pa (8.5 × 102 lb/in.2)

Atmospheric Pressure

Just as we exert pressure on a surface because of gravity, so does our atmosphere. We live at the bottom of an ocean of gases that becomes progressively less dense with increasing altitude. Approximately 99% of the mass of the atmosphere lies within 30 km of Earth’s surface, and half of it is within the first 5.5 km (). Every point on Earth’s surface experiences a net pressure called atmospheric pressure. The pressure exerted by the atmosphere is considerable: a 1.0 m2 column, measured from sea level to the top of the atmosphere, has a mass of about 10,000 kg, which gives a pressure of about 100 kPa:

Equation 10.3

pressure=(1.0×104 kg)(9.807 m/s2)1.0 m2=0.98×105 Pa=98 kPa

Figure 10.3 Atmospheric Pressure

Each square meter of Earth’s surface supports a column of air that is more than 200 km high and weighs about 10,000 kg at Earth’s surface, resulting in a pressure at the surface of 1.01 × 105 N/m2. This corresponds to a pressure of 101 kPa = 760 mmHg = 1 atm.

In English units, this is about 14 lb/in.2, but we are so accustomed to living under this pressure that we never notice it. Instead, what we notice are changes in the pressure, such as when our ears pop in fast elevators in skyscrapers or in airplanes during rapid changes in altitude. We make use of atmospheric pressure in many ways. We can use a drinking straw because sucking on it removes air and thereby reduces the pressure inside the straw. The atmospheric pressure pushing down on the liquid in the glass then forces the liquid up the straw.

Atmospheric pressure can be measured using a barometerA device used to measure atmospheric pressure., a device invented in 1643 by one of Galileo’s students, Evangelista Torricelli (1608–1647). A barometer may be constructed from a long glass tube that is closed at one end. It is filled with mercury and placed upside down in a dish of mercury without allowing any air to enter the tube. Some of the mercury will run out of the tube, but a relatively tall column remains inside (). Why doesn’t all the mercury run out? Gravity is certainly exerting a downward force on the mercury in the tube, but it is opposed by the pressure of the atmosphere pushing down on the surface of the mercury in the dish, which has the net effect of pushing the mercury up into the tube. Because there is no air above the mercury inside the tube in a properly filled barometer (it contains a vacuum), there is no pressure pushing down on the column. Thus the mercury runs out of the tube until the pressure exerted by the mercury column itself exactly balances the pressure of the atmosphere. Under normal weather conditions at sea level, the two forces are balanced when the top of the mercury column is approximately 760 mm above the level of the mercury in the dish, as shown in . This value varies with meteorological conditions and altitude. In Denver, Colorado, for example, at an elevation of about 1 mile, or 1609 m (5280 ft), the height of the mercury column is 630 mm rather than 760 mm.

Figure 10.4 A Mercury Barometer

The pressure exerted by the atmosphere on the surface of the pool of mercury supports a column of mercury in the tube that is about 760 mm tall. Because the boiling point of mercury is quite high (356.73°C), there is very little mercury vapor in the space above the mercury column.

Mercury barometers have been used to measure atmospheric pressure for so long that they have their own unit for pressure: the millimeter of mercury (mmHg)A unit of pressure, often called the torr., often called the torrA unit of pressure. One torr is the same as 1 mmHg., after Torricelli. Standard atmospheric pressureThe atmospheric pressure required to support a column of mercury exactly 760 mm tall, which is also referred to as 1 atmosphere (atm). is the atmospheric pressure required to support a column of mercury exactly 760 mm tall; this pressure is also referred to as 1 atmosphere (atm)Also referred to as standard atmospheric pressure, it is the atmospheric pressure required to support a column of mercury exactly 760 mm tall.. These units are also related to the pascal:

Equation 10.4

1 atm = 760 mmHg = 760 torr = 1.01325 × 105 Pa = 101.325 kPa

Thus a pressure of 1 atm equals 760 mmHg exactly and is approximately equal to 100 kPa.

Example 3

One of the authors visited Rocky Mountain National Park several years ago. After departing from an airport at sea level in the eastern United States, he arrived in Denver (altitude 5280 ft), rented a car, and drove to the top of the highway outside Estes Park (elevation 14,000 ft). He noticed that even slight exertion was very difficult at this altitude, where the atmospheric pressure is only 454 mmHg. Convert this pressure to

  1. atmospheres.
  2. kilopascals.

Given: pressure in millimeters of mercury

Asked for: pressure in atmospheres and kilopascals

Strategy:

Use the conversion factors in to convert from millimeters of mercury to atmospheres and kilopascals.

Solution:

From , we have 1 atm = 760 mmHg = 101.325 kPa. The pressure at 14,000 ft in atm is thus

P=(454 mmHg)(1 atm760 mmHg)=0.597  atm

The pressure in kPa is given by

P=(0.597 atm)(101.325 kPa1 atm)=60.5 kPa

Exercise

Mt. Everest, at 29,028 ft above sea level, is the world’s tallest mountain. The normal atmospheric pressure at this altitude is about 0.308 atm. Convert this pressure to

  1. millimeters of mercury.
  2. kilopascals.

Answer: a. 234 mmHg; b. 31.2 kPa

Manometers

Barometers measure atmospheric pressure, but manometersA device used to measure the pressures of samples of gases contained in an apparatus. measure the pressures of samples of gases contained in an apparatus. The key feature of a manometer is a U-shaped tube containing mercury (or occasionally another nonvolatile liquid). A closed-end manometer is shown schematically in part (a) in . When the bulb contains no gas (i.e., when its interior is a near vacuum), the heights of the two columns of mercury are the same because the space above the mercury on the left is a near vacuum (it contains only traces of mercury vapor). If a gas is released into the bulb on the right, it will exert a pressure on the mercury in the right column, and the two columns of mercury will no longer be the same height. The difference between the heights of the two columns is equal to the pressure of the gas.

Figure 10.5 The Two Types of Manometer

(a) In a closed-end manometer, the space above the mercury column on the left (the reference arm) is essentially a vacuum (P ≈ 0), and the difference in the heights of the two columns gives the pressure of the gas contained in the bulb directly. (b) In an open-end manometer, the left (reference) arm is open to the atmosphere (P ≈ 1 atm), and the difference in the heights of the two columns gives the difference between atmospheric pressure and the pressure of the gas in the bulb.

If the tube is open to the atmosphere instead of closed, as in the open-end manometer shown in part (b) in , then the two columns of mercury have the same height only if the gas in the bulb has a pressure equal to the atmospheric pressure. If the gas in the bulb has a higher pressure, the mercury in the open tube will be forced up by the gas pushing down on the mercury in the other arm of the U-shaped tube. The pressure of the gas in the bulb is therefore the sum of the atmospheric pressure (measured with a barometer) and the difference in the heights of the two columns. If the gas in the bulb has a pressure less than that of the atmosphere, then the height of the mercury will be greater in the arm attached to the bulb. In this case, the pressure of the gas in the bulb is the atmospheric pressure minus the difference in the heights of the two columns.

Example 4

Suppose you want to construct a closed-end manometer to measure gas pressures in the range 0.000–0.200 atm. Because of the toxicity of mercury, you decide to use water rather than mercury. How tall a column of water do you need? (At 25°C, the density of water is 0.9970 g/cm3; the density of mercury is 13.53 g/cm3.)

Given: pressure range and densities of water and mercury

Asked for: column height

Strategy:

A Calculate the height of a column of mercury corresponding to 0.200 atm in millimeters of mercury. This is the height needed for a mercury-filled column.

B From the given densities, use a proportion to compute the height needed for a water-filled column.

Solution:

A In millimeters of mercury, a gas pressure of 0.200 atm is

P=(0.200 atm)(760 mmHg1 atm)=152 mmHg

Using a mercury manometer, you would need a mercury column at least 152 mm high.

B Because water is less dense than mercury, you need a taller column of water to achieve the same pressure as a given column of mercury. The height needed for a water-filled column corresponding to a pressure of 0.200 atm is proportional to the ratio of the density of mercury (dHg) to the density of water (dH2O):

(heightH2O)(dH2O)=(heightHg)(dHg)heightH2O=(heightHg)(dHgdH2O)=(152 mm)(13.53 g/cm30.9970 g/cm3)=2.06×103 mm H2O=2.06 m H2O

This answer makes sense: it takes a taller column of a less dense liquid to achieve the same pressure.

Exercise

Suppose you want to design a barometer to measure atmospheric pressure in an environment that is always hotter than 30°C. To avoid using mercury, you decide to use gallium, which melts at 29.76°C; the density of liquid gallium at 25°C is 6.114 g/cm3. How tall a column of gallium do you need if P = 1.00 atm?

Answer: 1.68 m

The answer to Example 4 also tells us the maximum depth of a farmer’s well if a simple suction pump will be used to get the water out. If a column of water 2.06 m high corresponds to 0.200 atm, then 1.00 atm corresponds to a column height of

h2.06 m=1.00 atm0.200 atmh=10.3 m

A suction pump is just a more sophisticated version of a straw: it creates a vacuum above a liquid and relies on atmospheric pressure to force the liquid up a tube. If 1 atm pressure corresponds to a 10.3 m (33.8 ft) column of water, then it is physically impossible for atmospheric pressure to raise the water in a well higher than this. Until electric pumps were invented to push water mechanically from greater depths, this factor greatly limited where people could live because obtaining water from wells deeper than about 33 ft was difficult.

Summary

Four quantities must be known for a complete physical description of a sample of a gas: temperature, volume, amount, and pressure. Pressure is force per unit area of surface; the SI unit for pressure is the pascal (Pa), defined as 1 newton per square meter (N/m2). The pressure exerted by an object is proportional to the force it exerts and inversely proportional to the area on which the force is exerted. The pressure exerted by Earth’s atmosphere, called atmospheric pressure, is about 101 kPa or 14.7 lb/in.2 at sea level. Atmospheric pressure can be measured with a barometer, a closed, inverted tube filled with mercury. The height of the mercury column is proportional to atmospheric pressure, which is often reported in units of millimeters of mercury (mmHg), also called torr. Standard atmospheric pressure, the pressure required to support a column of mercury 760 mm tall, is yet another unit of pressure: 1 atmosphere (atm). A manometer is an apparatus used to measure the pressure of a sample of a gas.

Key Takeaway

  • Pressure is defined as the force exerted per unit area; it can be measured using a barometer or manometer.

Key Equation

Definition of pressure

: P=FA

Conceptual Problems

  1. What four quantities must be known to completely describe a sample of a gas? What units are commonly used for each quantity?

  2. If the applied force is constant, how does the pressure exerted by an object change as the area on which the force is exerted decreases? In the real world, how does this relationship apply to the ease of driving a small nail versus a large nail?

  3. As the force on a fixed area increases, does the pressure increase or decrease? With this in mind, would you expect a heavy person to need smaller or larger snowshoes than a lighter person? Explain.

  4. What do we mean by atmospheric pressure? Is the atmospheric pressure at the summit of Mt. Rainier greater than or less than the pressure in Miami, Florida? Why?

  5. Which has the highest atmospheric pressure—a cave in the Himalayas, a mine in South Africa, or a beach house in Florida? Which has the lowest?

  6. Mars has an average atmospheric pressure of 0.007 atm. Would it be easier or harder to drink liquid from a straw on Mars than on Earth? Explain your answer.

  7. Is the pressure exerted by a 1.0 kg mass on a 2.0 m2 area greater than or less than the pressure exerted by a 1.0 kg mass on a 1.0 m2 area? What is the difference, if any, between the pressure of the atmosphere exerted on a 1.0 m2 piston and a 2.0 m2 piston?

  8. If you used water in a barometer instead of mercury, what would be the major difference in the instrument?

Answer

  1. Because pressure is defined as the force per unit area (P = F/A), increasing the force on a given area increases the pressure. A heavy person requires larger snowshoes than a lighter person. Spreading the force exerted on the heavier person by gravity (that is, their weight) over a larger area decreases the pressure exerted per unit of area, such as a square inch, and makes them less likely to sink into the snow.

Numerical Problems

  1. Calculate the pressure in atmospheres and kilopascals exerted by a fish tank that is 2.0 ft long, 1.0 ft wide, and 2.5 ft high and contains 25.0 gal of water in a room that is at 20°C; the tank itself weighs 15 lb (dH2O = 1.00 g/cm3 at 20°C). If the tank were 1 ft long, 1 ft wide, and 5 ft high, would it exert the same pressure? Explain your answer.

  2. Calculate the pressure in pascals and in atmospheres exerted by a carton of milk that weighs 1.5 kg and has a base of 7.0 cm × 7.0 cm. If the carton were lying on its side (height = 25 cm), would it exert more or less pressure? Explain your reasoning.

  3. If atmospheric pressure at sea level is 1.0 × 105 Pa, what is the mass of air in kilograms above a 1.0 cm2 area of your skin as you lie on the beach? If atmospheric pressure is 8.2 × 104 Pa on a mountaintop, what is the mass of air in kilograms above a 4.0 cm2 patch of skin?

  4. Complete the following table:

    atm kPa mmHg torr
    1.40
    723
    43.2
  5. The SI unit of pressure is the pascal, which is equal to 1 N/m2. Show how the product of the mass of an object and the acceleration due to gravity result in a force that, when exerted on a given area, leads to a pressure in the correct SI units. What mass in kilograms applied to a 1.0 cm2 area is required to produce a pressure of

    1. 1.0 atm?
    2. 1.0 torr?
    3. 1 mmHg?
    4. 1 kPa?
  6. If you constructed a manometer to measure gas pressures over the range 0.60–1.40 atm using the liquids given in the following table, how tall a column would you need for each liquid? The density of mercury is 13.5 g/cm3. Based on your results, explain why mercury is still used in barometers, despite its toxicity.

    Liquid Density (20°C) Column Height (m)
    isopropanol 0.785
    coconut oil 0.924
    glycerine 1.259

Answer

  1. 5.4 kPa or 5.3 × 10−2 atm; 11 kPa, 1.1 × 10−3 atm; the same force acting on a smaller area results in a greater pressure.

10.3 Relationships among Pressure, Temperature, Volume, and Amount

Learning Objective

  1. To understand the relationships among pressure, temperature, volume, and the amount of a gas.

Early scientists explored the relationships among the pressure of a gas (P) and its temperature (T), volume (V), and amount (n) by holding two of the four variables constant (amount and temperature, for example), varying a third (such as pressure), and measuring the effect of the change on the fourth (in this case, volume). The history of their discoveries provides several excellent examples of the scientific method as presented in .

The Relationship between Pressure and Volume

As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal.

Robert Boyle (1627–1691)

Boyle, the youngest (and 14th!) child of the Earl of Cork, was an important early figure in chemistry whose views were often at odds with accepted wisdom. Boyle’s studies of gases are reported to have utilized a very tall J-tube that he set up in the entryway of his house, which was several stories tall. He is known for the gas law that bears his name and for his book, The Sceptical Chymist, which was published in 1661 and influenced chemists for many years after his death. In addition, one of Boyle’s early essays on morals is said to have inspired Jonathan Swift to write Gulliver’s Travels.

The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in . In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (). A simple plot of V versus P gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows:

Equation 10.5

PV = constant

Figure 10.6 Boyle’s Experiment Using a J-Shaped Tube to Determine the Relationship between Gas Pressure and Volume

(a) Initially the gas is at a pressure of 1 atm = 760 mmHg (the mercury is at the same height in both the arm containing the sample and the arm open to the atmosphere); its volume is V. (b) If enough mercury is added to the right side to give a difference in height of 760 mmHg between the two arms, the pressure of the gas is 760 mmHg (atmospheric pressure) + 760 mmHg = 1520 mmHg and the volume is V/2. (c) If an additional 760 mmHg is added to the column on the right, the total pressure on the gas increases to 2280 mmHg, and the volume of the gas decreases to V/3.

Figure 10.7 Plots of Boyle’s Data

(a) Here are actual data from a typical experiment conducted by Boyle. Boyle used non-SI units to measure the volume (in.3 rather than cm3) and the pressure (in. Hg rather than mmHg). (b) This plot of pressure versus volume is a hyperbola. Because PV is a constant, decreasing the pressure by a factor of two results in a twofold increase in volume and vice versa. (c) A plot of volume versus 1/pressure for the same data shows the inverse linear relationship between the two quantities, as expressed by the equation V = constant/P.

Dividing both sides by P gives an equation illustrating the inverse relationship between P and V:

Equation 10.6

V=constantP=constant(1P) or V1P

where the ∝ symbol is read “is proportional to.” A plot of V versus 1/P is thus a straight line whose slope is equal to the constant in and . Dividing both sides of by V instead of P gives a similar relationship between P and 1/V. The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s lawA law that states that at constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure., after its discoverer, and can be stated as follows: At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure.

The Relationship between Temperature and Volume

Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air.

The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of V versus T was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C ().A sample of gas cannot really have a volume of zero because any sample of matter must have some volume. Furthermore, at 1 atm pressure all gases liquefy at temperatures well above −273.15°C. Note from part (a) in that the slope of the plot of V versus T varies for the same gas at different pressures but that the intercept remains constant at −273.15°C. Similarly, as shown in part (b) in , plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis.

Jacques Alexandre César Charles (1746–1823) and Joseph-Louis Gay-Lussac (1778–1850)

In 1783, Charles filled a balloon (“aerostatic globe”) with hydrogen (generated by the reaction of iron with more than 200 kg of acid over several days) and flew successfully for almost an hour. When the balloon descended in a nearby village, however, the terrified townspeople destroyed it. In 1804, Gay-Lussac managed to ascend to 23,000 ft (more than 7000 m) to collect samples of the atmosphere to analyze its composition as a function of altitude. In the process, he had trouble breathing and nearly froze to death, but he set an altitude record that endured for decades. (To put Gay-Lussac’s achievement in perspective, recall that modern jetliners cruise at only 35,000 ft!)

The significance of the invariant T intercept in plots of V versus T was recognized in 1848 by the British physicist William Thomson (1824–1907), later named Lord Kelvin. He postulated that −273.15°C was the lowest possible temperature that could theoretically be achieved, for which he coined the term absolute zero (0 K)The lowest possible temperature that can be theoretically achieved; it corresponds to −273.15°C..

We can state Charles’s and Gay-Lussac’s findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship, illustrated in part (b) in , is often referred to as Charles’s lawA law that states that at constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). and is stated mathematically as

Equation 10.7

V=(constant)[T(in K)] or VT(in K, at constant P)

Charles’s law is valid for virtually all gases at temperatures well above their boiling points. Note that the temperature must be expressed in kelvins, not in degrees Celsius.

Figure 10.8 The Relationship between Volume and Temperature

(a) In these plots of volume versus temperature for equal-sized samples of H2 at three different pressures, the solid lines show the experimentally measured data down to −100°C, and the broken lines show the extrapolation of the data to V = 0. The temperature scale is given in both degrees Celsius and kelvins. Although the slopes of the lines decrease with increasing pressure, all of the lines extrapolate to the same temperature at V = 0 (−273.15°C = 0 K). (b) In these plots of volume versus temperature for different amounts of selected gases at 1 atm pressure, all the plots extrapolate to a value of V = 0 at −273.15°C, regardless of the identity or the amount of the gas.

The Relationship between Amount and Volume

We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (). (This is the historic “Avogadro’s hypothesis” introduced in .) A logical corollary, sometimes called Avogadro’s lawA law that states that at constant temperature and pressure, the volume of a sample of gas is directly proportional to the number of moles of gas in the sample., describes the relationship between the volume and the amount of a gas: At constant temperature and pressure, the volume of a sample of gas is directly proportional to the number of moles of gas in the sample. Stated mathematically,

Equation 10.8

V=(constant)(n) or Vn (constant T and P)

This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures.

Note the Pattern

For a sample of gas,

V increases as P decreases (and vice versa)

V increases as T increases (and vice versa)

V increases as n increases (and vice versa)

Figure 10.9 Avogadro’s Hypothesis

Equal volumes of four different gases at the same temperature and pressure contain the same number of gaseous particles. Because the molar mass of each gas is different, the mass of each gas sample is different even though all contain 1 mol of gas.

The relationships among the volume of a gas and its pressure, temperature, and amount are summarized in . Volume increases with increasing temperature or amount but decreases with increasing pressure.

Figure 10.10 The Empirically Determined Relationships among Pressure, Volume, Temperature, and Amount of a Gas

The thermometer and pressure gauge indicate the temperature and the pressure qualitatively, the level in the flask indicates the volume, and the number of particles in each flask indicates relative amounts.

Summary

Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure (Charles’s law), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present (Avogadro’s law). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is absolute zero (0 K), the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature.

Key Takeaway

  • The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas.

Conceptual Problems

  1. Sketch a graph of the volume of a gas versus the pressure on the gas. What would the graph of V versus P look like if volume was directly proportional to pressure?

  2. What properties of a gas are described by Boyle’s law, Charles’s law, and Avogadro’s law? In each law, what quantities are held constant? Why does the constant in Boyle’s law depend on the amount of gas used and the temperature at which the experiments are carried out?

  3. Use Charles’s law to explain why cooler air sinks.

  4. Use Boyle’s law to explain why it is dangerous to heat even a small quantity of water in a sealed container.

Answer

Numerical Problems

  1. A 1.00 mol sample of gas at 25°C and 1.0 atm has an initial volume of 22.4 L. Calculate the results of each change, assuming all the other conditions remain constant.

    1. The pressure is changed to 85.7 mmHg. How many milliliters does the gas occupy?
    2. The volume is reduced to 275 mL. What is the pressure in millimeters of mercury?
    3. The pressure is increased to 25.3 atm. What is the temperature in degrees Celsius?
    4. The sample is heated to 30°C. What is the volume in liters?
    5. The sample is compressed to 1255 mL, and the pressure is increased to 2555 torr. What is the temperature of the gas in kelvins?
  2. A 1.00 mol sample of gas is at 300 K and 4.11 atm. What is the volume of the gas under these conditions? The sample is compressed to 6.0 atm at constant temperature, giving a volume of 3.99 L. Is this result consistent with Boyle’s law?

Answer

    1. 1.99 × 105 mL
    2. 6.19 × 104 mmHg
    3. 7270°C
    4. 22.8 L
    5. 51.4 K

10.4 The Ideal Gas Law

Learning Objective

  1. To use the ideal gas law to describe the behavior of a gas.

In , you learned how the volume of a gas changes when its pressure, temperature, or amount is changed, as long as the other two variables are held constant. In this section, we describe how these relationships can be combined to give a general expression that describes the behavior of a gas.

Deriving the Ideal Gas Law

Any set of relationships between a single quantity (such as V) and several other variables (P, T, and n) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions derived in are as follows:

 

Boyle’s law

V1P(at constant nT)

 

Charles’s law

VT(at constant nP)

 

Avogadro’s law

Vn(at constant TP)

Combining these three expressions gives

Equation 10.9

VnTP

which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as

Equation 10.10

V=(constant)(nTP)

By convention, the proportionality constant in is called the gas constantA proportionality constant that is used in the ideal gas law., which is represented by the letter R. Inserting R into gives

Equation 10.11

V=RnTP=nRTP

Clearing the fractions by multiplying both sides of by P gives

Equation 10.12

PV = nRT

This equation is known as the ideal gas lawA law relating pressure, temperature, volume, and the amount of an ideal gas..

An ideal gasA hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces. is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. As you will learn in , the ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed.

Note the Pattern

Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures.

Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then

Equation 10.13

R = 0.082057 (L·atm)/(K·mol)

Because the product PV has the units of energy, as described in , and Essential Skills 4 (, ), R can also have units of J/(K·mol) or cal/(K·mol):

Equation 10.14

R = 8.3145 J/(K·mol) = 1.9872 cal/(K·mol)

Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and 1 atm pressure, referred to as standard temperature and pressure (STP)The conditions 0°C (273.15 K) and 1 atm pressure for a gas.. We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in :

Equation 10.15

V=nRTP=(1.000 mol)[0.082057 (L·atm)/(K·mol)](273.15 K)1.000 atm=22.41 L

Thus the volume of 1 mol of an ideal gas at 0°C and 1 atm pressure is 22.41 L, approximately equivalent to the volume of three basketballs. The quantity 22.41 L is called the standard molar volumeThe volume of 1 mol of an ideal gas at STP (0°C and 1 atm pressure), which is 22.41 L. of an ideal gas. The molar volumes of several real gases at STP are given in , which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at STP. The relationships described in as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed.

Table 10.3 Molar Volumes of Selected Gases at Standard Temperature (0°C) and Pressure (1 atm)

Gas Molar Volume (L)
He 22.434
Ar 22.397
H2 22.433
N2 22.402
O2 22.397
CO2 22.260
NH3 22.079

If n, R, and T are all constant in , the equation reduces to

Equation 10.16

V=(constant)(1P) or V1P

which is exactly the same as Boyle’s law in .

Similarly, Charles’s law states that the volume of a fixed quantity of gas is directly proportional to its temperature at constant pressure. If n and P in are fixed, then

Equation 10.17

V=nRTP=(constant)(T) or VT

which is exactly the same as .

Applying the Ideal Gas Law

The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables (P, V, T, and n). It also allows us to predict the final state of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters (P, V, T, and n) are specified for an initial state. Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample.

Example 5

The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft3), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon?

Given: volume, temperature, and pressure

Asked for: amount of gas

Strategy:

A Solve the ideal gas law for the unknown quantity, in this case n.

B Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed.

Solution:

A We are given values for P, T, and V and asked to calculate n. If we solve the ideal gas law () for n, we obtain

n=PVRT

B P and T are given in units that are not compatible with the units of the gas constant [R = 0.082057 (L·atm)/(K·mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres:

P=(745 mmHg)(1 atm760 mmHg)=0.980 atmV=31,150 L(given)T=30+273=303 Κ

Substituting these values into the expression we derived for n, we obtain

n=PVRT=(0.980 atm)(31,150 L)[0.082057(L·atm)/(K·mol)](303 K)=1.23×103 mol H2

Exercise

Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO2. What is the pressure of the gas at 25°C?

Answer: 1.5 atm

In Example 5, we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of changes in any of the specified conditions on any of the other parameters, as shown in Example 6.

Example 6

Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example 5?

Given: temperature, pressure, amount, and volume in August; temperature in January

Asked for: volume in January

Strategy:

A Use the results from Example 5 for August as the initial conditions and then calculate the change in volume due to the change in temperature from 86°F to 14°F. Begin by constructing a table showing the initial and final conditions.

B Rearrange the ideal gas law to isolate those quantities that differ between the initial and final states on one side of the equation, in this case V and T.

C Equate the ratios of those terms that change for the two sets of conditions. Making sure to use the appropriate units, insert the quantities and solve for the unknown parameter.

Solution:

A To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions:

August (initial) January (final)
T 30°C = 303 K −10°C = 263 K
P 0.980 atm 0.980 atm
n 1.23 × 103 mol H2 1.23 × 103 mol H2
V 31,150 L ?

Thus we are asked to calculate the effect of a change in temperature on the volume of a fixed amount of gas at constant pressure.

B Recall that we can rearrange the ideal gas law to give

V=(nRP)(T)

Both n and P are the same in both cases, which means that nR/P is a constant. Dividing both sides by T gives

VT=nRP=constant

This is the relationship first noted by Charles.

C We see from this expression that under conditions where the amount (n) of gas and the pressure (P) do not change, the ratio V/T also does not change. If we have two sets of conditions for the same amount of gas at the same pressure, we can therefore write

V1T1=V2T2

where the subscripts 1 and 2 refer to the initial and final conditions, respectively. Solving for V2 and inserting the given quantities in the appropriate units, we obtain

V2=V1T2T1=(31,350 L)(263 K)303 K=2.70×104 L

It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change.

Exercise

At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon?

Answer: 0.52 L

Example 6 illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle (PV = constant) and the relationship between volume and amount observed by Avogadro (V/n = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example 6 can be applied in any such case, as we demonstrate in Example 7 (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion).

Example 7

Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise 5 (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)?

Given: initial volume, amount, temperature, and pressure; final temperature

Asked for: final pressure

Strategy:

Follow the strategy outlined in Example 6.

Solution:

Prepare a table to determine which parameters change and which are held constant:

Initial Final
V 0.406 L 0.406 L
n 0.025 mol 0.025 mol
T 25°C = 298 K 750°C = 1023 K
P 1.5 atm ?

Once again, two parameters are constant while one is varied, and we are asked to calculate the fourth. As before, we begin with the ideal gas law and rearrange it as necessary to get all the constant quantities on one side. In this case, because V and n are constant, we rearrange to obtain

P=(nRV)(T)=(constant)(T)

Dividing both sides by T, we obtain an equation analogous to the one in Example 6, P/T = nR/V = constant. Thus the ratio of P to T does not change if the amount and volume of a gas are held constant. We can thus write the relationship between any two sets of values of P and T for the same sample of gas at the same volume as

P1T1=P2T2

In this example, P1 = 1.5 atm, T1 = 298 K, and T2 = 1023 K, and we are asked to find P2. Solving for P2 and substituting the appropriate values, we obtain

P2=P1T2T1=(1.5 atm)(1023 K)298 K=5.1 atm

This pressure is more than enough to rupture a thin sheet metal container and cause an explosion!

Exercise

Suppose that a fire extinguisher, filled with CO2 to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher?

Answer: 23.4 atm

In Example 10.6 and Example 10.7, two of the four parameters (P, V, T, and n) were fixed while one was allowed to vary, and we were interested in the effect on the value of the fourth. In fact, we often encounter cases where two of the variables P, V, and T are allowed to vary for a given sample of gas (hence n is constant), and we are interested in the change in the value of the third under the new conditions. If we rearrange the ideal gas law so that P, V, and T, the quantities that change, are on one side and the constant terms (R and n for a given sample of gas) are on the other, we obtain

Equation 10.18

PVT=nR=constant

Thus the quantity PV/T is constant if the total amount of gas is constant. We can therefore write the relationship between any two sets of parameters for a sample of gas as follows:

Equation 10.19

P1V1T1=P2V2T2

This equation can be solved for any of the quantities P2, V2, or T2 if the initial conditions are known, as shown in Example 8.

Example 8

We saw in Example 5 that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 103 mol of H2 gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude?

Given: initial pressure, temperature, amount, and volume; final pressure and temperature

Asked for: final volume

Strategy:

Follow the strategy outlined in Example 6.

Solution:

Begin by setting up a table of the two sets of conditions:

Initial Final
P 745 mmHg = 0.980 atm 312 mmHg = 0.411 atm
T 30°C = 303 K −30°C = 243 K
n 1.23 × 103 mol H2 1.23 × 103 mol H2
V 31,150 L ?

Thus all the quantities except V2 are known. Solving for V2 and substituting the appropriate values give

V2=V1(P1T2P2T1)=(31,150 L)[(0.980 atm)(243 K)(0.411 atm)(303 K)]=5.96×104L

Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to increase the volume of the gas, while decreasing the temperature tends to decrease the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both T and 1/P, the variable that changes the most will have the greatest effect on V. In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation.

We could also have solved this problem by solving the ideal gas law for V and then substituting the relevant parameters for an altitude of 23,000 ft:

V=nRTP=(1.23×103 mol)[0.082057 (L·atm)/(K·mol)](243 K)0.411 atm=5.97×104L

Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. There is often more than one “right” way to solve chemical problems.

Exercise

A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.)

Answer: 4.07 × 103

Using the Ideal Gas Law to Calculate Gas Densities and Molar Masses

The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain

Equation 10.20

nV=PRT

The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass (m, in grams) divided by its molar mass (M, in grams per mole):

Equation 10.21

n=mM

Substituting this expression for n into gives

Equation 10.22

mMV=PRT

Because m/V is the density d of a substance, we can replace m/V by d and rearrange to give

Equation 10.23

d=PMRT

The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL).

Example 9

Calculate the density of butane at 25°C and a pressure of 750 mmHg.

Given: compound, temperature, and pressure

Asked for: density

Strategy:

A Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant.

B Substitute these values into to obtain the density.

Solution:

A The molar mass of butane (C4H10) is

(4)(12.011) + (10)(1.0079) = 58.123 g/mol

Using 0.082057 (L·atm)/(K·mol) for R means that we need to convert the temperature from degrees Celsius to kelvins (T = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres:

(750 mmHg)(1 atm760 mmHg)=0.987 atm

B Substituting these values into gives

d=PMRT=(0.987 atm)(58.123 g/mol)[0.082057 (L·atm)/(K·mol)](298 K)=2.35 g/L

Exercise

Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics.

Answer: radon, 9.23 g/L; N2, 1.17 g/L

A common use of is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example 10.

Example 10

The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound.

Given: pressure, temperature, mass, and volume

Asked for: molar mass and chemical formula

Strategy:

A Solve for the molar mass of the gas and then calculate the density of the gas from the information given.

B Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass.

C Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas.

Solution:

A Solving for the molar mass gives

M=dRTP

Density is the mass of the gas divided by its volume:

d=mV=0.289  g0.157  L=1.84  g/L

B We must convert the other quantities to the appropriate units before inserting them into the equation:

T=18+273=291  KP=(727 mmHg)(1 atm760 mmHg)=0.957 atm

The molar mass of the unknown gas is thus

M=dRTP=(1.84 g/L)[0.082057(L·atm)/(K·mol)](291 K)0.957 atm=45.9 g/mol

C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations:

NO=14 + 16=30 g/molN2O=(2)(14)+16=44 g/molNO2=14+(2)(16)=46 g/mol

The most likely choice is NO2 which is in agreement with the data. The red-brown color of smog also results from the presence of NO2 gas.

Exercise

You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it.

Answer: 44 g/mol; CO2

Summary

The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant and has the value 0.08206 (L·atm)/(K·mol), 8.3145 J/(K·mol), or 1.9872 cal/(K·mol), depending on the units used. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the standard molar volume. All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity (P, V, T, or n) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of P, V, T, and n) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured.

Key Takeaway

  • The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known.

Key Equations

Ideal gas law

: PV = nRT

Relationship between initial and final conditions

: P1V1T1=P2V2T2   n is constant

Density of a gas

: d=PMRT

Conceptual Problems

  1. For an ideal gas, is volume directly proportional or inversely proportional to temperature? What is the volume of an ideal gas at absolute zero?

  2. What is meant by STP? If a gas is at STP, what further information is required to completely describe the state of the gas?

  3. For a given amount of a gas, the volume, temperature, and pressure under any one set of conditions are related to the volume, the temperature, and the pressure under any other set of conditions by the equation P1V1T1=P2V2T2. Derive this equation from the ideal gas law. At constant temperature, this equation reduces to one of the laws discussed in ; which one? At constant pressure, this equation reduces to one of the laws discussed in ; which one?

  4. Predict the effect of each change on one variable if the other variables are held constant.

    1. If the number of moles of gas increases, what is the effect on the temperature of the gas?
    2. If the temperature of a gas decreases, what is the effect on the pressure of the gas?
    3. If the volume of a gas increases, what is the effect on the temperature of the gas?
    4. If the pressure of a gas increases, what is the effect on the number of moles of the gas?
  5. What would the ideal gas law be if the following were true?

    1. volume were proportional to pressure
    2. temperature were proportional to amount
    3. pressure were inversely proportional to temperature
    4. volume were inversely proportional to temperature
    5. both pressure and volume were inversely proportional to temperature
  6. Given the following initial and final values, what additional information is needed to solve the problem using the ideal gas law?

    Given Solve for
    V1, T1, T2, n1 n 2
    P1, P2, T2, n2 n 1
    T1, T2 V 2
    P1, n1 P 2
  7. Given the following information and using the ideal gas law, what equation would you use to solve the problem?

    Given Solve for
    P1, P2, T1 T 2
    V1, n1, n2 V 2
    T1, T2, V1, V2, n2 n 1
  8. Using the ideal gas law as a starting point, derive the relationship between the density of a gas and its molar mass. Which would you expect to be denser—nitrogen or oxygen? Why does radon gas accumulate in basements and mine shafts?

  9. Use the ideal gas law to derive an equation that relates the remaining variables for a sample of an ideal gas if the following are held constant.

    1. amount and volume
    2. pressure and amount
    3. temperature and volume
    4. temperature and amount
    5. pressure and temperature
  10. Tennis balls that are made for Denver, Colorado, feel soft and do not bounce well at lower altitudes. Use the ideal gas law to explain this observation. Will a tennis ball designed to be used at sea level be harder or softer and bounce better or worse at higher altitudes?

Answer

    1. P/T = constant
    2. V/T = constant (Charles’ law)
    3. P/n = constant
    4. PV = constant (Boyle’s law)
    5. V/n = constant (Avogadro’s law)

Numerical Problems

  1. Calculate the number of moles in each sample at STP.

    1. 1580 mL of NO2
    2. 847 cm3 of HCl
    3. 4.792 L of H2
    4. a 15.0 cm × 6.7 cm × 7.5 cm container of ethane
  2. Calculate the number of moles in each sample at STP.

    1. 2200 cm3 of CO2
    2. 1200 cm3 of N2
    3. 3800 mL of SO2
    4. 13.75 L of NH3
  3. Calculate the mass of each sample at STP.

    1. 36 mL of HI
    2. 550 L of H2S
    3. 1380 cm3 of CH4
  4. Calculate the mass of each sample at STP.

    1. 3.2 L of N2O
    2. 65 cm3 of Cl2
    3. 3600 mL of HBr
  5. Calculate the volume in liters of each sample at STP.

    1. 1.68 g of Kr
    2. 2.97 kg of propane (C3H8)
    3. 0.643 mg of (CH3)2O
  6. Calculate the volume in liters of each sample at STP.

    1. 3.2 g of Xe
    2. 465 mg of CS2
    3. 5.34 kg of acetylene (C2H2)
  7. Calculate the volume of each gas at STP.

    1. 1.7 L at 28°C and 96.4 kPa
    2. 38.0 mL at 17°C and 103.4 torr
    3. 650 mL at −15°C and 723 mmHg
  8. Calculate the volume of each gas at STP.

    1. 2.30 L at 23°C and 740 mmHg
    2. 320 mL at 13°C and 97.2 kPa
    3. 100.5 mL at 35°C and 1.4 atm
  9. A 8.60 L tank of nitrogen gas at a pressure of 455 mmHg is connected to an empty tank with a volume of 5.35 L. What is the final pressure in the system after the valve connecting the two tanks is opened? Assume that the temperature is constant.

  10. At constant temperature, what pressure in atmospheres is needed to compress 14.2 L of gas initially at 25.2 atm to a volume of 12.4 L? What pressure is needed to compress 27.8 L of gas to 20.6 L under similar conditions?

  11. One method for preparing hydrogen gas is to pass HCl gas over hot aluminum; the other product of the reaction is AlCl3. If you wanted to use this reaction to fill a balloon with a volume of 28,500 L at sea level and a temperature of 78°F, what mass of aluminum would you need? What volume of HCl at STP would you need?

  12. An 3.50 g sample of acetylene is burned in excess oxygen according to the following reaction:

    2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)

    At STP, what volume of CO2(g) is produced?

  13. Calculate the density of ethylene (C2H4) under each set of conditions.

    1. 7.8 g at 0.89 atm and 26°C
    2. 6.3 mol at 102.6 kPa and 38°C
    3. 9.8 g at 3.1 atm and −45°C
  14. Determine the density of O2 under each set of conditions.

    1. 42 g at 1.1 atm and 25°C
    2. 0.87 mol at 820 mmHg and 45°C
    3. 16.7 g at 2.4 atm and 67°C
  15. At 140°C, the pressure of a diatomic gas in a 3.0 L flask is 635 kPa. The mass of the gas is 88.7 g. What is the most likely identity of the gas?

  16. What volume must a balloon have to hold 6.20 kg of H2 for an ascent from sea level to an elevation of 20,320 ft, where the temperature is −37°C and the pressure is 369 mmHg?

  17. What must be the volume of a balloon that can hold 313.0 g of helium gas and ascend from sea level to an elevation of 1.5 km, where the temperature is 10.0°C and the pressure is 635.4 mmHg?

  18. A typical automobile tire is inflated to a pressure of 28.0 lb/in.2 Assume that the tire is inflated when the air temperature is 20°C; the car is then driven at high speeds, which increases the temperature of the tire to 43°C. What is the pressure in the tire? If the volume of the tire had increased by 8% at the higher temperature, what would the pressure be?

  19. The average respiratory rate for adult humans is 20 breaths per minute. If each breath has a volume of 310 mL of air at 20°C and 0.997 atm, how many moles of air does a person inhale each day? If the density of air is 1.19 kg/m3, what is the average molecular mass of air?

  20. Kerosene has a self-ignition temperature of 255°C. It is a common accelerant used by arsonists, but its presence is easily detected in fire debris by a variety of methods. If a 1.0 L glass bottle containing a mixture of air and kerosene vapor at an initial pressure of 1 atm and an initial temperature of 23°C is pressurized, at what pressure would the kerosene vapor ignite?

Answers

    1. 7.05 × 10−2 mol
    2. 3.78 × 10−2 mol
    3. 0.2138 mol
    4. 3.4 × 10−2 mol
    1. 0.21 g HI;
    2. 840 g H2S;
    3. 0.988 g CH4
    1. 0.449 L Kr
    2. 1510 L C3H8
    3. 3.13 × 10−4 L (CH3)2O
    1. 1.5 L
    2. 4.87 mL
    3. 650 mL
  1. 281 mmHg

  2. 20.9 kg Al, 5.20 × 104 L HCl

    1. 1.0 g/L
    2. 1.1 g/L
    3. 4.6 g/L
  3. 2174 L

10.5 Mixtures of Gases

Learning Objective

  1. To determine the contribution of each component gas to the total pressure of a mixture of gases.

In our use of the ideal gas law thus far, we have focused entirely on the properties of pure gases with only a single chemical species. But what happens when two or more gases are mixed? In this section, we describe how to determine the contribution of each gas present to the total pressure of the mixture.

Partial Pressures

The ideal gas law assumes that all gases behave identically and that their behavior is independent of attractive and repulsive forces. If volume and temperature are held constant, the ideal gas equation can be rearranged to show that the pressure of a sample of gas is directly proportional to the number of moles of gas present:

Equation 10.24

P=n(RTV)=n(constant)

Nothing in the equation depends on the nature of the gas—only the amount.

With this assumption, let’s suppose we have a mixture of two ideal gases that are present in equal amounts. What is the total pressure of the mixture? Because the pressure depends on only the total number of particles of gas present, the total pressure of the mixture will simply be twice the pressure of either component. More generally, the total pressure exerted by a mixture of gases at a given temperature and volume is the sum of the pressures exerted by each gas alone. Furthermore, if we know the volume, the temperature, and the number of moles of each gas in a mixture, then we can calculate the pressure exerted by each gas individually, which is its partial pressureThe pressure a gas in a mixture would exert if it were the only one present (at the same temperature and volume)., the pressure the gas would exert if it were the only one present (at the same temperature and volume).

To summarize, the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. This law was first discovered by John Dalton, the father of the atomic theory of matter. It is now known as Dalton’s law of partial pressuresA law that states that the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases.. We can write it mathematically as

Equation 10.25

Pt=P1+P2+P3++Pi

where Pt is the total pressure and the other terms are the partial pressures of the individual gases ().

Figure 10.11 Dalton’s Law

The total pressure of a mixture of gases is the sum of the partial pressures of the individual gases.

For a mixture of two ideal gases, A and B, we can write an expression for the total pressure:

Equation 10.26

Pt=PA+PB=nA(RTV)+nB(RTV)=(nA+nB)(RTV)

More generally, for a mixture of i components, the total pressure is given by

Equation 10.27

Pt=(n1+n2+n3++ni)(RTV)

restates in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of dozens or even hundreds of gaseous species. For to be valid, the identity of the particles present cannot have an effect. Thus an ideal gas must be one whose properties are not affected by either the size of the particles or their intermolecular interactions because both will vary from one gas to another. The calculation of total and partial pressures for mixtures of gases is illustrated in Example 11.

Example 11

For reasons that we will examine in , deep-sea divers must use special gas mixtures in their tanks, rather than compressed air, to avoid serious problems, most notably a condition called “the bends.” At depths of about 350 ft, divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of O2 and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature?

Given: masses of components, total volume, and temperature

Asked for: partial pressures and total pressure

Strategy:

A Calculate the number of moles of He and O2 present.

B Use the ideal gas law to calculate the partial pressure of each gas. Then add together the partial pressures to obtain the total pressure of the gaseous mixture.

Solution:

A The number of moles of He is

(326.4 g)(1.000 mol4.0026 g) = 81.55 mol

The number of moles of O2 is

(51.2 g)(1.000 mol32.00 g) = 1.60 mol

B We can now use the ideal gas law to calculate the partial pressure of each:

PO2=nO2(RTV)=(1.60 mol){[0.082057 (L·atm)/(K·mol)](293.15 K)10.0 L}=3.85 atmPHe=nHe(RTV)=(81.55 mol){[0.082057 (L·atm)]/(K·mol)(293.15 K)10.0 L}=196 atm

The total pressure is the sum of the two partial pressures:

Pt=PO2+PHe=3.85 atm + 196 atm=200 atm

Exercise

A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder.

Answer: PCH4=137  atm; PC2H6=13.4  atm; Pt=151  atm

Mole Fractions of Gas Mixtures

The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction (X)The ratio of the number of moles of any component of a mixture to the total number of moles of all species present in the mixture. of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture (nt):

Equation 10.28

mole fraction of A=XA=moles Atotal moles=nAnt

The mole fraction is a dimensionless quantity between 0 and 1. If XA = 1.0, then the sample is pure A, not a mixture. If XA = 0, then no A is present in the mixture. The sum of the mole fractions of all the components present must equal 1.

To see how mole fractions can help us understand the properties of gas mixtures, let’s evaluate the ratio of the pressure of a gas A to the total pressure of a gas mixture that contains A. We can use the ideal gas law to describe the pressures of both gas A and the mixture: PA = nART/V and Pt = ntRT/V. The ratio of the two is thus

Equation 10.29

PAPt=nART/VntRT/V=nAnt=XA

Rearranging this equation gives

Equation 10.30

PA = XAPt

That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas. This conclusion is a direct result of the ideal gas law, which assumes that all gas particles behave ideally. Consequently, the pressure of a gas in a mixture depends on only the percentage of particles in the mixture that are of that type, not their specific physical or chemical properties. Recall from () that by volume, Earth’s atmosphere is about 78% N2, 21% O2, and 0.9% Ar, with trace amounts of gases such as CO2, H2O, and others. This means that 78% of the particles present in the atmosphere are N2; hence the mole fraction of N2 is 78%/100% = 0.78. Similarly, the mole fractions of O2 and Ar are 0.21 and 0.009, respectively. Using , we therefore know that the partial pressure of N2 is 0.78 atm (assuming an atmospheric pressure of exactly 760 mmHg) and, similarly, the partial pressures of O2 and Ar are 0.21 and 0.009 atm, respectively.

Example 12

We have just calculated the partial pressures of the major gases in the air we inhale. Experiments that measure the composition of the air we exhale yield different results, however. The following table gives the measured pressures of the major gases in both inhaled and exhaled air. Calculate the mole fractions of the gases in exhaled air.

Inhaled Air (mmHg) Exhaled Air (mmHg)
PN2 597 568
PO2 158 116
PCO2 0.3 28
PH2O 5 48
P Ar 8 8
P t 767 767

Given: pressures of gases in inhaled and exhaled air

Asked for: mole fractions of gases in exhaled air

Strategy:

Calculate the mole fraction of each gas using .

Solution:

The mole fraction of any gas A is given by

XA=PAPt

where PA is the partial pressure of A and Pt is the total pressure. In this case,

Pt=(767 mmHg)(1 atm760 mmHg)=1.01 atm

The following table gives the values of PA and XA for exhaled air.

P A X A
PN2 (568 mmHg)(1 atm760 mmHg)=0.747 atm 0.747 atm1.01 atm=0.740
PO2 (116 mmHg)(1 atm760 mmHg)=0.153 atm 0.153 atm1.01 atm=0.151
PCO2 (28 mmHg)(1 atm760 mmHg)=0.037  atm 0.031 atm1.01 atm=0.037
PH2O (48 mmHg)(1 atm760 mmHg)=0.063  atm 0.063 atm1.01 atm=0.062
P Ar (8 mmHg)(1 atm760 mmHg)=0.011  atm 0.011 atm1.01 atm=0.011

Exercise

We saw in Example 10 that Venus is an inhospitable place, with a surface temperature of 560°C and a surface pressure of 90 atm. The atmosphere consists of about 96% CO2 and 3% N2, with trace amounts of other gases, including water, sulfur dioxide, and sulfuric acid. Calculate the partial pressures of CO2 and N2.

Answer: PCO2=86 atm; PN2=3 atm

Summary

The pressure exerted by each gas in a gas mixture (its partial pressure) is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components (Dalton’s law of partial pressures). The amount of gas present in a mixture may be described by its partial pressure or its mole fraction. The mole fraction of any component of a mixture is the ratio of the number of moles of that substance to the total number of moles of all substances present. In a mixture of gases, the partial pressure of each gas is the product of the total pressure and the mole fraction of that gas.

Key Takeaway

  • The partial pressure of each gas in a mixture is proportional to its mole fraction.

Key Equations

Mole fraction

: XA=moles Atotal moles=nAnt

Relationship between partial pressure and mole fraction

: PA = XAPt

Conceptual Problems

  1. Dalton’s law of partial pressures makes one key assumption about the nature of the intermolecular interactions in a mixture of gases. What is it?

  2. What is the relationship between the partial pressure of a gas and its mole fraction in a mixture?

Numerical Problems

  1. What is the partial pressure of each gas if the following amounts of substances are placed in a 25.0 L container at 25°C? What is the total pressure of each mixture?

    1. 1.570 mol of CH4 and 0.870 mol of CO2
    2. 2.63 g of CO and 1.24 g of NO2
    3. 1.78 kg of CH3Cl and 0.92 kg of SO2
  2. What is the partial pressure of each gas in the following 3.0 L mixtures at 37°C as well as the total pressure?

    1. 0.128 mol of SO2 and 0.098 mol of methane (CH4)
    2. 3.40 g of acetylene (C2H2) and 1.54 g of He
    3. 0.267 g of NO, 4.3 g of Ar, and 0.872 g of SO2
  3. In a mixture of helium, oxygen, and methane in a 2.00 L container, the partial pressures of He and O2 are 13.6 kPa and 29.2 kPa, respectively, and the total pressure inside the container is 95.4 kPa. What is the partial pressure of methane? If the methane is ignited to initiate its combustion with oxygen and the system is then cooled to the original temperature of 30°C, what is the final pressure inside the container (in kilopascals)?

  4. A 2.00 L flask originally contains 1.00 g of ethane (C2H6) and 32.0 g of oxygen at 21°C. During ignition, the ethane reacts completely with oxygen to produce CO2 and water vapor, and the temperature of the flask increases to 200°C. Determine the total pressure and the partial pressure of each gas before and after the reaction.

  5. If a 20.0 L cylinder at 19°C is charged with 5.0 g each of sulfur dioxide and oxygen, what is the partial pressure of each gas? The sulfur dioxide is ignited in the oxygen to produce sulfur trioxide gas, and the mixture is allowed to cool to 19°C at constant pressure. What is the final volume of the cylinder? What is the partial pressure of each gas in the piston?

  6. The highest point on the continent of Europe is Mt. Elbrus in Russia, with an elevation of 18,476 ft. The highest point on the continent of South America is Mt. Aconcagua in Argentina, with an elevation of 22,841 ft.

    1. The following table shows the variation of atmospheric pressure with elevation. Use the data in the table to construct a plot of pressure versus elevation.

      Elevation (km) Pressure in Summer (mmHg) Pressure in Winter (mmHg)
      0.0 760.0 760.0
      1.0 674.8 670.6
      1.5 635.4 629.6
      2.0 598.0 590.8
      3.0 528.9 519.7
      5.0 410.6 398.7
      7.0 314.9 301.6
      9.0 237.8 224.1
    2. Use your graph to estimate the pressures in millimeters of mercury during the summer and the winter at the top of both mountains in both atmospheres and kilopascals.
    3. Given that air is 20.95% O2 by volume, what is the partial pressure of oxygen in atmospheres during the summer at each location?

Answers

    1. PCH4 = 1.54 atm, PCO2 = 0.851 atm, PT = 2.39 atm
    2. PCO = 0.0918 atm, PNO2 = 0.0264 atm, PT = 0.1182 atm
    3. PCH3Cl = 34.5 atm, PSO2 = 14 atm, PT = 49 atm
  1. 52.6 kPa, 66.2 kPa

10.6 Gas Volumes and Stoichiometry

Learning Objective

  1. To relate the amount of gas consumed or released in a chemical reaction to the stoichiometry of the reaction.

With the ideal gas law, we can use the relationship between the amounts of gases (in moles) and their volumes (in liters) to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. This is important for several reasons. Many reactions that are carried out in the laboratory involve the formation or reaction of a gas, so chemists must be able to quantitatively treat gaseous products and reactants as readily as they quantitatively treat solids or solutions. Furthermore, many, if not most, industrially important reactions are carried out in the gas phase for practical reasons. Gases mix readily, are easily heated or cooled, and can be transferred from one place to another in a manufacturing facility via simple pumps and plumbing. As a chemical engineer said to one of the authors, “Gases always go where you want them to, liquids sometimes do, but solids almost never do.”

Example 13

Sulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give SO2, followed by the reaction of SO2 with O2 in the presence of a catalyst to give SO3, which reacts with water to give H2SO4. The overall chemical equation is as follows:

+ 32O2+H2OH2SO4

What volume of O2 (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton of H2SO4?

Given: reaction, temperature, pressure, and mass of one product

Asked for: volume of gaseous reactant

Strategy:

A Calculate the number of moles of H2SO4 in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of O2 required.

B Use the ideal gas law to determine the volume of O2 required under the given conditions. Be sure that all quantities are expressed in the appropriate units.

Solution:

We can see from the stoichiometry of the reaction that 32 mol of O2 is required to produce 1 mol of H2SO4. This is a standard stoichiometry problem of the type presented in , except this problem asks for the volume of one of the reactants (O2) rather than its mass. We proceed exactly as in , using the strategy

mass of H2SO4 → moles H2SO4 → moles O2 → liters O2

A We begin by calculating the number of moles of H2SO4 in 1.00 tn:

moles  H2SO4=(1.00 tn  H2SO4)(2000 lb1 tn)(453.6 g1 lb)(1 mol H2SO498.08 g)=9250 mol H2SO4

We next calculate the number of moles of O2 required:

moles  O2=(9250 mol  H2SO4)(1.5 mol  O21 mol  H2SO4)=1.39×104 mol  O2

B After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O2:

T=273+22=295 KP=(745 mmHg)(1 atm760 mmHg)=0.980  atmVolume of  O2=nO2(RTP)=(1.39×104 mol O2)[0.08206 (L·atm)/(K·mol)(295 K)0.980 atm]=3.43×105 L  O2

The answer means that more than 300,000 L of oxygen gas are needed to produce 1 tn of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry.

Exercise

In Example 5, we saw that Charles used a balloon containing approximately 31,150 L of H2 for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation:

Fe(s) + 2 HCl(aq) → H2(g) + FeCl2(aq)

How much iron (in kilograms) was needed to produce this volume of H2 if the temperature was 30°C and the atmospheric pressure was 745 mmHg?

Answer: 68.6 kg of Fe (approximately 150 lb)

Many of the advances made in chemistry during the 18th and 19th centuries were the result of careful experiments done to determine the identity and quantity of gases produced in chemical reactions. For example, in 1774, Joseph Priestley was able to isolate oxygen gas by the thermal decomposition of mercuric oxide (HgO). In the 1780s, Antoine Lavoisier conducted experiments that showed that combustion reactions, which require oxygen, produce what we now know to be carbon dioxide. Both sets of experiments required the scientists to collect and manipulate gases produced in chemical reactions, and both used a simple technique that is still used in chemical laboratories today: collecting a gas by the displacement of water. As shown in , the gas produced in a reaction can be channeled through a tube into inverted bottles filled with water. Because the gas is less dense than liquid water, it bubbles to the top of the bottle, displacing the water. Eventually, all the water is forced out and the bottle contains only gas. If a calibrated bottle is used (i.e., one with markings to indicate the volume of the gas) and the bottle is raised or lowered until the level of the water is the same both inside and outside, then the pressure within the bottle will exactly equal the atmospheric pressure measured separately with a barometer.

Figure 10.12 An Apparatus for Collecting Gases by the Displacement of Water

When KClO3(s) is heated, O2 is produced according to the equation KClO3(s)KCl(s)+32O2(g). The oxygen gas travels through the tube, bubbles up through the water, and is collected in a bottle as shown.

The only gases that cannot be collected using this technique are those that readily dissolve in water (e.g., NH3, H2S, and CO2) and those that react rapidly with water (such as F2 and NO2). Remember, however, when calculating the amount of gas formed in the reaction, the gas collected inside the bottle is not pure. Instead, it is a mixture of the product gas and water vapor. As we will discuss in , all liquids (including water) have a measurable amount of vapor in equilibrium with the liquid because molecules of the liquid are continuously escaping from the liquid’s surface, while other molecules from the vapor phase collide with the surface and return to the liquid. The vapor thus exerts a pressure above the liquid, which is called the liquid’s vapor pressure. In the case shown in , the bottle is therefore actually filled with a mixture of O2 and water vapor, and the total pressure is, by Dalton’s law of partial pressures, the sum of the pressures of the two components:

Pt=PO2+PH2O

If we want to know the pressure of the gas generated in the reaction to calculate the amount of gas formed, we must first subtract the pressure due to water vapor from the total pressure. This is done by referring to tabulated values of the vapor pressure of water as a function of temperature (). As shown in , the vapor pressure of water increases rapidly with increasing temperature, and at the normal boiling point (100°C), the vapor pressure is exactly 1 atm. The methodology is illustrated in Example 14.

Table 10.4 Vapor Pressure of Water at Various Temperatures

T (°C) P (in mmHg) T P T P T P
0 4.58 21 18.66 35 42.2 92 567.2
5 6.54 22 19.84 40 55.4 94 611.0
10 9.21 23 21.08 45 71.9 96 657.7
12 10.52 24 22.39 50 92.6 98 707.3
14 11.99 25 23.77 55 118.1 100 760.0
16 13.64 26 25.22 60 149.5 102 815.8
17 14.54 27 26.75 65 187.7 104 875.1
18 15.48 28 28.37 70 233.8 106 937.8
19 16.48 29 30.06 80 355.3 108 1004.2
20 17.54 30 31.84 90 525.9 110 1074.4

Figure 10.13 A Plot of the Vapor Pressure of Water versus Temperature

The vapor pressure is very low (but not zero) at 0°C and reaches 1 atm = 760 mmHg at the normal boiling point, 100°C.

Example 14

Sodium azide (NaN3) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation:

2NaN3(s)Δ2Na(s)+3N2(g)

This reaction is used to inflate the air bags that cushion passengers during automobile collisions. The reaction is initiated in air bags by an electrical impulse and results in the rapid evolution of gas. If the N2 gas that results from the decomposition of a 5.00 g sample of NaN3 could be collected by displacing water from an inverted flask, as in , what volume of gas would be produced at 22°C and 762 mmHg?

Given: reaction, mass of compound, temperature, and pressure

Asked for: volume of nitrogen gas produced

Strategy:

A Calculate the number of moles of N2 gas produced. From the data in , determine the partial pressure of N2 gas in the flask.

B Use the ideal gas law to find the volume of N2 gas produced.

Solution:

A Because we know the mass of the reactant and the stoichiometry of the reaction, our first step is to calculate the number of moles of N2 gas produced:

moles  N2=(5.00 g  NaN3)(1 mol  NaN365.01 g  NaN3)(3  mol  N22 mol  NaN3)=0.115 mol  N2

The pressure given (762 mmHg) is the total pressure in the flask, which is the sum of the pressures due to the N2 gas and the water vapor present. tells us that the vapor pressure of water is 19.84 mmHg at 22°C (295 K), so the partial pressure of the N2 gas in the flask is only 762 − 19.84 = 742 mmHg = 0.976 atm.

B Solving the ideal gas law for V and substituting the other quantities (in the appropriate units), we get

V=nRTP=(0.115 mol  N2)[0.08206 (L·atm)/(K·mol)](295 K)0.976 atm=2.85 L

Exercise

A 1.00 g sample of zinc metal is added to a solution of dilute hydrochloric acid. It dissolves to produce H2 gas according to the equation Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq). The resulting H2 gas is collected in a water-filled bottle at 30°C and an atmospheric pressure of 760 mmHg. What volume does it occupy?

Answer: 0.397 L

Summary

The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can then be calculated from the volume of water displaced and the atmospheric pressure. A gas collected in such a way is not pure, however, but contains a significant amount of water vapor. The measured pressure must therefore be corrected for the vapor pressure of water, which depends strongly on the temperature.

Key Takeaway

  • The ideal gas equation and the stoichiometry of a reaction can be used to calculate the volume of gas produced or consumed in a reaction.

Conceptual Problems

  1. Why are so many industrially important reactions carried out in the gas phase?

  2. The volume of gas produced during a chemical reaction can be measured by collecting the gas in an inverted container filled with water. The gas forces water out of the container, and the volume of liquid displaced is a measure of the volume of gas. What additional information must be considered to determine the number of moles of gas produced? The volume of some gases cannot be measured using this method. What property of a gas precludes the use of this method?

  3. Equal masses of two solid compounds (A and B) are placed in separate sealed flasks filled with air at 1 atm and heated to 50°C for 10 hours. After cooling to room temperature, the pressure in the flask containing A was 1.5 atm. In contrast, the pressure in the flask containing B was 0.87 atm. Suggest an explanation for these observations. Would the masses of samples A and B still be equal after the experiment? Why or why not?

Numerical Problems

  1. Balance each chemical equation and then determine the volume of the indicated reactant at STP that is required for complete reaction. Assuming complete reaction, what is the volume of the products?

    1. SO2(g) + O2(g) → SO3(g) given 2.4 mol of O2
    2. H2(g) + Cl2(g) → HCl(g) given 0.78 g of H2
    3. C2H6(g) + O2(g) → CO2(g) + H2O(g) given 1.91 mol of O2
  2. During the smelting of iron, carbon reacts with oxygen to produce carbon monoxide, which then reacts with iron(III) oxide to produce iron metal and carbon dioxide. If 1.82 L of CO2 at STP is produced,

    1. what mass of CO is consumed?
    2. what volume of CO at STP is consumed?
    3. how much O2 (in liters) at STP is used?
    4. what mass of carbon is consumed?
    5. how much iron metal (in grams) is produced?
  3. Complete decomposition of a sample of potassium chlorate produced 1.34 g of potassium chloride and oxygen gas.

    1. What is the mass of KClO3 in the original sample?
    2. What mass of oxygen is produced?
    3. What is the volume of oxygen produced at STP?
  4. The combustion of a 100.0 mg sample of an herbicide in excess oxygen produced 83.16 mL of CO2 and 72.9 mL of H2O vapor at STP. A separate analysis showed that the sample contained 16.44 mg of chlorine. If the sample is known to contain only C, H, Cl, and N, determine the percent composition and the empirical formula of the herbicide.

  5. The combustion of a 300.0 mg sample of an antidepressant in excess oxygen produced 326 mL of CO2 and 164 mL of H2O vapor at STP. A separate analysis showed that the sample contained 23.28% oxygen. If the sample is known to contain only C, H, O, and N, determine the percent composition and the empirical formula of the antidepressant.

Answers

    1. 2.20 g KClO3
    2. 0.863 g O2
    3. 604 mL O2
  1. Percent composition: 58.3% C, 4.93% H, 23.28% O, and 13.5% N; empirical formula: C10H10O3N2

10.8 The Behavior of Real Gases

Learning Objective

  1. To recognize the differences between the behavior of an ideal gas and a real gas.

The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases.

Pressure, Volume, and Temperature Relationships in Real Gases

For an ideal gas, a plot of PV/nRT versus P gives a horizontal line with an intercept of 1 on the PV/nRT axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (part (a) in ). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (part (b) in ). Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in for N2. Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid.

Figure 10.21 Real Gases Do Not Obey the Ideal Gas Law, Especially at High Pressures

(a) In these plots of PV/nRT versus P at 273 K for several common gases, there are large negative deviations observed for C2H4 and CO2 because they liquefy at relatively low pressures. (b) These plots illustrate the relatively good agreement between experimental data for real gases and the ideal gas law at low pressures.

Figure 10.22 The Effect of Temperature on the Behavior of Real Gases

A plot of PV/nRT versus P for nitrogen gas at three temperatures shows that the approximation to ideal gas behavior becomes better as the temperature increases.

Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of PV/nRT is greater than the value predicted by the ideal gas law.

Figure 10.23 The Effect of Nonzero Volume of Gas Particles on the Behavior of Gases at Low and High Pressures

(a) At low pressures, the volume occupied by the molecules themselves is small compared with the volume of the container. (b) At high pressures, the molecules occupy a large portion of the volume of the container, resulting in significantly decreased space in which the molecules can move.

Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is less than expected (). Thus as shown in , at low temperatures, the ratio of PV/nRT is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the PV/nRT versus P plot for many gases.

Note the Pattern

Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted.

At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid).

The van der Waals Equation

The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called van der Waals equationA modification of the ideal gas law designed to describe the behavior of real gases by explicitly including the effects of molecular volume and intermolecular forces.,

Equation 10.40

(P+an2V2)(Vnb)=nRT

a and b are empirical constants that are different for each gas. The values of a and b are listed in for several common gases. The pressure term—P + (an2/V2)—corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, n2/V2 represents the concentration of the gas (n/V) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in . The volume term—V − nb—corrects for the volume occupied by the gaseous molecules.

Table 10.5 van der Waals Constants for Some Common Gases

Gas a (L2·atm)/mol2) b (L/mol)
He 0.03410 0.0238
Ne 0.205 0.0167
Ar 1.337 0.032
H2 0.2420 0.0265
N2 1.352 0.0387
O2 1.364 0.0319
Cl2 6.260 0.0542
NH3 4.170 0.0371
CH4 2.273 0.0430
CO2 3.610 0.0429

Figure 10.24 The Effect of Intermolecular Attractive Forces on the Pressure a Gas Exerts on the Container Walls

(a) At low pressures, there are relatively few attractive intermolecular interactions to lessen the impact of the molecule striking the wall of the container, and the pressure is close to that predicted by the ideal gas law. (b) At high pressures, with the average intermolecular distance relatively small, the effect of intermolecular interactions is to lessen the impact of a given molecule striking the container wall, resulting in a lower pressure than predicted by the ideal gas law.

The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on V and P, respectively. Because nonzero molecular volumes produce a measured volume that is larger than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is less than that expected based on the ideal gas law, so the an2/V2 term must be added to the measured pressure to correct for these effects.

Example 19

You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.0 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)?

Given: volume of cylinder, mass of compound, pressure, and temperature

Asked for: safety

Strategy:

A Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law.

B Obtain a and b values for Cl2 from . Use the van der Waals equation to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture.

Solution:

A We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol):

(500 g)(1 mol70.906 g)=7.05  mol Cl

Using the ideal gas law and the temperature in kelvins (298 K), we calculate the pressure:

P=nRTV=(7.05 mol)[0.08206 (L·atm)/(K·mol)](298 K)4.0 L=43 atm

If chlorine behaves like an ideal gas, you have a real problem!

B Now let’s use the van der Waals equation with the a and b values for Cl2 from . Solving for P gives

P=nRTVnban2V2=(7.05 mol)(0.08206 L·atm/K·mol)(298 K)4.0 L(7.05 mol)(0.0542 L/mol)(6.2602 L2·atm/mol2)(7.05 mol)2(4.0 L)2=47.7  atm19.4 atm=28 atm (to two significant figures)

This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation.

Exercise

A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the

  1. ideal gas law.
  2. van der Waals equation.

Answer: a. 77 atm; b. 67 atm

Liquefaction of Gases

LiquefactionThe condensation of gases into a liquid form. of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure).

Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals a coefficients are relatively easy to liquefy because large a coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small a coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O2, N2, Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points. In , we will consider in more detail the nature of the intermolecular forces that allow gases to liquefy.

Note the Pattern

A large value of a indicates the presence of relatively strong intermolecular attractive interactions.

The ultracold liquids formed from the liquefaction of gases are called cryogenic liquidsAn ultracold liquid formed from the liquefaction of gases., from the Greek kryo, meaning “cold,” and genes, meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called cryosurgery, which selectively destroys tissues with a minimal loss of blood by the use of extreme cold.

Figure 10.25 A Liquid Natural Gas Transport Ship

Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. LNG consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. LPG is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles.

Summary

No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of PV/nRT versus P at a given temperature; for an ideal gas, PV/nRT versus P = 1 under all conditions. At high pressures, most real gases exhibit larger PV/nRT values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit PV/nRT values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, liquefaction occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold cryogenic liquids.

Key Takeaway

  • Molecular volumes and intermolecular attractions cause the properties of real gases to deviate from those predicted by the ideal gas law.

Key Equation

van der Waals equation

: (P+an2V2)(Vnb)=nRT

Conceptual Problems

  1. What factors cause deviations from ideal gas behavior? Use a sketch to explain your answer based on interactions at the molecular level.

  2. Explain the effect of nonzero atomic volume on the ideal gas law at high pressure. Draw a typical graph of volume versus 1/P for an ideal gas and a real gas.

  3. For an ideal gas, the product of pressure and volume should be constant, regardless of the pressure. Experimental data for methane, however, show that the value of PV decreases significantly over the pressure range 0 to 120 atm at 0°C. The decrease in PV over the same pressure range is much smaller at 100°C. Explain why PV decreases with increasing temperature. Why is the decrease less significant at higher temperatures.

  4. What is the effect of intermolecular forces on the liquefaction of a gas? At constant pressure and volume, does it become easier or harder to liquefy a gas as its temperature increases? Explain your reasoning. What is the effect of increasing the pressure on the liquefaction temperature?

  5. Describe qualitatively what a and b, the two empirical constants in the van der Waals equation, represent.

  6. In the van der Waals equation, why is the term that corrects for volume negative and the term that corrects for pressure positive? Why is n/V squared?

  7. Liquefaction of a gas depends strongly on two factors. What are they? As temperature is decreased, which gas will liquefy first—ammonia, methane, or carbon monoxide? Why?

  8. What is a cryogenic liquid? Describe three uses of cryogenic liquids.

  9. Air consists primarily of O2, N2, Ar, Ne, Kr, and Xe. Use the concepts discussed in this chapter to propose two methods by which air can be separated into its components. Which component of air will be isolated first?

  10. How can gas liquefaction facilitate the storage and transport of fossil fuels? What are potential drawbacks to these methods?

Numerical Problems

  1. The van der Waals constants for xenon are a = 4.19 (L2·atm)/mol2 and b = 0.0510 L/mol. If a 0.250 mol sample of xenon in a container with a volume of 3.65 L is cooled to −90°C, what is the pressure of the sample assuming ideal gas behavior? What would be the actual pressure under these conditions?

  2. The van der Waals constants for water vapor are a = 5.46 (L2·atm)/mol2 and b = 0.0305 L/mol. If a 20.0 g sample of water in a container with a volume of 5.0 L is heated to 120°C, what is the pressure of the sample assuming ideal gas behavior? What would be the actual pressure under these conditions?

10.9 Essential Skills 5

Topics

  • Preparing a Graph
  • Interpreting a Graph

Previous Essential Skills sections presented the fundamental mathematical operations you need to know to solve problems by manipulating chemical equations. This section describes how to prepare and interpret graphs, two additional skills that chemistry students must have to understand concepts and solve problems.

Preparing a Graph

A graph is a pictorial representation of a mathematical relationship. It is an extremely effective tool for understanding and communicating the relationship between two or more variables. Each axis is labeled with the name of the variable to which it corresponds, along with the unit in which the variable is measured, and each axis is divided by tic marks or grid lines into segments that represent those units (or multiples). The scale of the divisions should be chosen so that the plotted points are distributed across the entire graph. Whenever possible, data points should be combined with a bar that intersects the data point and indicates the range of error of the measurement, although for simplicity the bars are frequently omitted in undergraduate textbooks. Lines or curves that represent theoretical or computational results are drawn using a “best-fit” approach; that is, data points are not connected as a series of straight-line segments; rather, a smooth line or curve is drawn that provides the best fit to the plotted data.

The independent variable is usually assigned to the horizontal, or x, axis (also called the abscissa), and the dependent variable to the vertical, or y, axis (called the ordinate). Let’s examine, for example, an experiment in which we are interested in plotting the change in the concentration of compound A with time. Because time does not depend on the concentration of A but the concentration of A does depend on the amount of time that has passed during the reaction, time is the independent variable and concentration is the dependent variable. In this case, the time is assigned to the horizontal axis and the concentration of A to the vertical axis.

We may plot more than one dependent variable on a graph, but the lines or curves corresponding to each set of data must be clearly identified with labels, different types of lines (a dashed line, for example), or different symbols for the respective data points (e.g., a triangle versus a circle). When words are used to label a line or curve, either a key identifying the different sets of data or a label placed next to each line or curve is used.

Interpreting a Graph

Two types of graphs are frequently encountered in beginning chemistry courses: linear and log-linear. Here we describe each type.

Linear Graphs

In a linear graph, the plot of the relationship between the variables is a straight line and thus can be expressed by the equation for a straight line:

y = mx + b

where m is the slope of the line and b is the point where the line intersects the vertical axis (where x = 0), called the y-intercept. The slope is calculated using the following formula:

m=y2y1x2x1=ΔyΔx=riserun

For accuracy, two widely separated points should be selected for use in the formula to minimize the effects of any reporting or measurement errors that may have occurred in any given region of the graph. For example, when concentrations are measured, limitations in the sensitivity of an instrument as well as human error may result in measurements being less accurate for samples with low concentrations than for those that are more concentrated. The graph of the change in the concentration of A with time is an example of a linear graph. The key features of a linear plot are shown on the generic example.

It is important to remember that when a graphical procedure is used to calculate a slope, the scale on each axis must be of the same order (they must have the same exponent). For example, although acceleration is a change in velocity over time (Δvt), the slope of a linear plot of velocity versus time only gives the correct value for acceleration (m/s2) if the average acceleration over the interval and the instantaneous acceleration are identical; that is, the acceleration must be constant over the same interval.

Log-Linear Graphs

A log-linear plot is a representation of the following general mathematical relationship:

y = Acmx

Here, y is equal to some value Ac when x = 0. As described in Essential Skills 3 in , , taking the logarithm of both sides produces

log y = log A + mx log c = (m log c)x + log A

When expressed in this form, the equation is that of a straight line (y = mx + b), where the plot of y is on a logarithmic axis and (m log c)x is on a linear axis. This type of graph is known as a log-linear plot. Log-linear plots are particularly useful for graphing changes in pH versus changes in the concentration of another substance. One example of a log-linear plot, where y = [HA] and x = pH, is shown here:

From the linear equation, a log-linear plot has a y-intercept of log A, so the value of A may be obtained directly from the plot if the x axis begins at 0 (in this case it does not, as is often the case in pH plots). Using our example, however, we can calculate [HA] at the y-intercept first by calculating the slope of the line using any two points and the equation log[HA]2log[HA]1pH2pH1:

[HA2]=0.600[HA1]=0.006log[HA2]=0.222 log[HA1]=2.222pH2=2.5 pH1=3.5 m log c=0.222(2.222)2.53.5=2.0001.0=2.0

Using any point along the line (e.g., [HA] = 0.100, pH = 2.9), we can then calculate the y-intercept (pH = 0):

log[HA]=2.0 pH +b   1.00=[(2.0)(2.9)]+b         4.8=b

Thus at a pH of 0.0, log[HA] = 4.8 and [HA] = 6.31 × 104 M. The exercises provide practice in drawing and interpreting graphs.

Skill Builder ES1

  1. The absorbance of light by various aqueous solutions of phosphate was measured and tabulated as follows:

    Absorbance (400 nm) PO43− (mol/L)
    0.16 3.2 × 10−5
    0.38 8.4 × 10−5
    0.62 13.8 × 10−5
    0.88 19.4 × 10−5

    Graph the data with the dependent variable on the y axis and the independent variable on the x axis and then calculate the slope. If a sample has an absorbance of 0.45, what is the phosphate concentration in the sample?

  2. The following table lists the conductivity of three aqueous solutions with varying concentrations. Create a plot from these data and then predict the conductivity of each sample at a concentration of 15.0 × 103 ppm.

    0 ppm 5.00 × 103 ppm 10.00 × 103 ppm
    K2CO3 0.0 7.0 14.0
    Seawater 0.0 8.0 15.5
    Na2SO4 0.0 6.0 11.8

Solution:

  1. Absorbance is the dependent variable, and concentration is the independent variable. We calculate the slope using two widely separated data points:

    m=ΔyΔx=0.620.1613.8×1053.2×105=4.3×103

    According to our graph, the y-intercept, b, is 0.00. Thus when y = 0.45,

    0.45 = (4.3 × 103)(x) + 0.00 x = 10 × 10−5

    This is in good agreement with a graphical determination of the phosphate concentration at an absorbance of 0.45, which gives a value of 10.2 × 10−5 mol/L.

  2. Conductivity is the dependent variable, and concentration is the independent variable. From our graph, at 15.0 × 103 ppm, the conductivity of K2CO3 is predicted to be 21; that of seawater, 24; and that of Na2SO4, 18.

10.10 End-of-Chapter Material

Application Problems

    Be sure you are familiar with the material in Essential Skills 5 () before proceeding to the Application Problems. Problems marked with a ♦ involve multiple concepts.

  1. ♦ Oxalic acid (C2H2O4) is a metabolic product of many molds. Although oxalic acid is toxic to humans if ingested, many plants and vegetables contain significant amounts of oxalic acid or oxalate salts. In solution, oxalic acid can be oxidized by air via the following chemical equation:

    H2C2O4(aq) + O2(g) → H2O2(l) + 2 CO2(g)

    If a plant metabolized enough oxalic acid to produce 3.2 L of CO2 on a day when the temperature was 29°C and the pressure was 752 mmHg, how many grams of oxalic acid were converted to CO2? Given that air is 21% oxygen, what volume of air was needed for the oxidation?

  2. ♦ The decomposition of iron oxide is used to produce gas during the manufacture of porous, expanded materials. These materials have very low densities due to the swelling that occurs during the initial rapid heating. Consequently, they are used as additives to provide insulation in concrete, road building, and other construction materials. Iron oxide decomposes at 1150°C according to the following chemical equation:

    6 Fe2O3(s) → 4 Fe3O4(s) + O2(g)
    1. Explain how this reaction could cause materials containing Fe2O3 to swell on heating.
    2. If you begin with 15.4 g of Fe2O3, what volume of O2 gas at STP is produced?
    3. What is the volume of the same amount of O2 gas at 1200°C, assuming a constant pressure of 1.0 atm?
    4. If 6.3 L of gas were produced at 1200°C and 1 atm, how many kilograms of Fe2O3 were initially used in the reaction? How many kilograms of Fe3O4 are produced?
  3. ♦ A 70 kg man expends 480 kcal of energy per hour shoveling snow. The oxidation of organic nutrients such as glucose during metabolism liberates approximately 3.36 kcal of energy per gram of oxygen consumed. If air is 21% oxygen, what volume of air at STP is needed to produce enough energy for the man to clear snow from a walkway that requires 35 minutes of shoveling?

  4. ♦ Calcium carbonate is an important filler in the processing industry. Its many uses include a reinforcing agent for rubber and improving the whiteness and hiding power of paints. When calcium nitrate is used as a starting material in the manufacture of fertilizers, calcium carbonate is produced according to the following chemical reaction:

    Ca(NO3)2 + 2 NH3 + CO2 + H2O → CaCO3 + 2 NH4NO3
    1. What volume of CO2 at STP is needed to react completely with 28.0 g of calcium nitrate?
    2. What volume of ammonia at STP is needed to react with the amount of calcium nitrate?
    3. If this reaction were carried out in Denver, Colorado (pressure = 630 mmHg), what volumes of CO2 and NH3 at room temperature (20°C) would be needed?
  5. ♦ Calcium nitrate used in the process described in Problem 4 is produced by the reaction of fluoroapatite—Ca5[(PO4)3(F)]—with nitric acid:

    Ca5[(PO4)3(F)] + 10 HNO3 → 5 Ca(NO3)2 + HF + 3 H3PO4 Ca(NO3)2 + 2 NH3 + CO2 + H2O → CaCO3 + 2 NH4NO3
    1. Your lab is in Denver, Colorado, and you have a cylinder with 8.40 L of CO2 at room temperature (20°C) and 4.75 atm pressure. What would be the volume of the CO2 gas at 630 mmHg? How many grams of fluoroapatite could be converted to CaCO3 using the two reactions and this amount of CO2?
    2. If 3.50 L of a 0.753 M HF solution was produced during the conversion of fluoroapatite to calcium nitrate, how many grams of calcium carbonate could you produce, assuming 100% efficiency?
    3. How many liters of ammonia gas at this higher altitude and 20°C are required to convert the calcium nitrate to calcium carbonate and to neutralize all of the HF solution?
    4. If the reaction of fluoroapatite with nitric acid takes place at 330 atm and 80°C, what volumes of NH3 and CO2 are needed to convert 20.5 kg of fluoroapatite to calcium carbonate?
  6. Mars has an average temperature of −47°C; a surface pressure of 500 Pa; and an atmosphere that is 95% carbon dioxide, 3% nitrogen, and 2% argon by mass, with traces of other gases. What is the partial pressure (in atmospheres) of each gas in this atmosphere? A 5.0 L sample is returned to Earth and stored in a laboratory at 19°C and 1 atm. What is the volume of this sample?

  7. ♦ Chlorofluorocarbons (CFCs) are inert substances that were long used as refrigerants. Because CFCs are inert, when they are released into the atmosphere they are not rapidly destroyed in the lower atmosphere. Instead, they are carried into the stratosphere, where they cause ozone depletion. A method for destroying CFC stockpiles passes the CFC through packed sodium oxalate (Na2C2O4) powder at 270°C. The reaction for Freon-12 (CF2Cl2) is as follows:

    CF2Cl2(g) + 2 Na2C2O4(s) → 2 NaF(s) + 2 NaCl + C(s) + 4 CO2(g)
    1. If this reaction produced 11.4 L of CO2 gas at 21°C and 752 mmHg, what mass of sodium oxalate was consumed in the reaction?
    2. If you had to design a reactor to carry out the reaction at a maximum safe pressure of 10.0 atm while destroying 1.0 kg of Freon-12, what volume reactor would you need?
    3. A 2.50 L reaction vessel is charged with 20.0 atm of Freon-12 and excess sodium oxalate at 20°C. The temperature is increased to 270°C, and the pressure is monitored as the reaction progresses. What is the initial pressure at 270°C? What is the final pressure when the reaction has gone to completion?
  8. ♦ The exhaust from a typical six-cylinder car contains the following average compositions of CO and CO2 under different conditions (data reported as percent by volume; rpm = rotations per minute):

    Species Idling (1000 rpm) Accelerating (4000 rpm) Decelerating (800 rpm)
    CO 1.0 1.2 0.60
    CO2 0.80 0.40 0.40
    1. What are the mole fractions of CO and CO2 under each set of conditions?
    2. If the engine takes in 4.70 L of air at STP mixed with fine droplets of gasoline (C8H18) with each rotation, how many grams of gasoline are burned per minute?
  9. Automobile airbags inflate by the decomposition of sodium azide (NaN3), which produces sodium metal and nitrogen gas according to the following chemical equation:

    2 NaN3 → 2 Na(s) + 3 N2(g)

    How many grams of sodium azide are needed to inflate a 15.0 L airbag at 20°C and 760 mmHg? The density of NaN3 is 1.847 g/cm3. What is the volume of the gas produced compared to the solid reactant? Suggest a plausible reason to explain why skin burns can result from the inflation of an airbag during an automobile accident.

  10. Under basic conditions, the reaction of hydrogen peroxide (H2O2) and potassium permanganate (KMnO4) produces oxygen and manganese dioxide. During a laboratory exercise, you carefully weighed out your sample of KMnO4. Unfortunately, however, you lost your data just before mixing the KMnO4 with an H2O2 solution of unknown concentration. Devise a method to determine the mass of your sample of KMnO4 using excess H2O2.

  11. Carbonated beverages are pressurized with CO2. In an attempt to produce another bubbly soda beverage, an intrepid chemist attempted to use three other gases: He, N2, and Xe. Rank the four beverages in order of how fast the drink would go “flat” and explain your reasoning. Which beverage would have the shortest shelf life (i.e., how long will an unopened bottle still be good)? Explain your answer.

  12. ♦ Urea is synthesized industrially by the reaction of ammonia and carbon dioxide to produce ammonium carbamate, followed by dehydration of ammonium carbamate to give urea and water. This process is shown in the following set of chemical equations:

    2 NH3(g) + CO2(g) → NH2CO2NH4(s) NH2CO2NH4(s) → NH2CONH2(s) + H2O(g)
    1. A 50.0 L reaction vessel is charged with 2.5 atm of ammonia and 2.5 atm of CO2 at 50°C, and the vessel is then heated to 150°C. What is the pressure in the vessel when the reaction has gone to completion? If the vessel is then cooled to 20°C, what is the pressure?
    2. An aqueous solution of urea and water is drained from the reaction vessel. What is the molarity of the urea solution? Industrially, this reaction is actually carried out at pressures ranging from 130 to 260 atm and temperatures of approximately 180°C. Give a plausible reason for using these extreme conditions.
  13. Explain what happens to the temperature, the volume, or the pressure of a gas during each operation and give the direction of heat flow, if any.

    1. The gas is allowed to expand from V1 to V2; a heat transfer occurs to maintain a constant gas temperature.
    2. The gas is allowed to expand from V2 to V3 with no concomitant heat transfer.
    3. The gas is compressed from V3 to V4; a heat transfer occurs to maintain a constant gas temperature.
    4. The gas is compressed from V4 to V1; no heat transfer occurs.

    These four processes constitute the cycle used in refrigeration, in which a gas such as Freon is alternately compressed and allowed to expand in the piston of a compressor. Which step eventually causes the food in a refrigerator to cool? Where does the thermal energy go that was removed in the cooling process?

Answers

  1. 5.9 g oxalic acid, 7.8 L

  2. 278 L

Chapter 11 Liquids

In Chapter 10 "Gases", you learned that attractive intermolecular forces cause most gases to condense to liquids at high pressure, low temperature, or both. Substances that normally are liquids are held together by exactly the same forces that are responsible for the liquefaction of gases. One such substance is water, the solvent in which all biochemical reactions take place. Because of its thermal properties, water also modulates Earth’s temperature, maintaining a temperature range suitable for life. Other liquids are used to manufacture objects that we use every day—for example, a solid material is converted to a liquid, the liquid is injected into a mold, and it is then solidified into complex shapes under conditions that are carefully controlled. To understand such processes, our study of the macroscopic properties of matter must include an understanding of the properties of liquids and the interconversion of the three states of matter: gases, liquids, and solids.

Water beading up on the surface of a freshly waxed car. The waxed, nonpolar surface does not interact strongly with the polar water molecules. The absence of attractive interactions causes the water to form round beads.

In this chapter, we look more closely at the intermolecular forces that are responsible for the properties of liquids, describe some of the unique properties of liquids compared with the other states of matter, and then consider changes in state between liquids and gases or solids. By the end of the chapter, you will understand what is happening at the molecular level when you dry yourself with a towel, why you feel cold when you come out of the water, why ice is slippery, and how it is possible to decaffeinate coffee without removing important flavor components. You will also learn how liquid crystal display (LCD) devices in electronic devices function, and how adhesive strips used to measure body temperature change color to indicate a fever.

11.1 The Kinetic Molecular Description of Liquids

Learning Objective

  1. To be familiar with the kinetic molecular description of liquids.

The kinetic molecular theory of gases described in gives a reasonably accurate description of the behavior of gases. A similar model can be applied to liquids, but it must take into account the nonzero volumes of particles and the presence of strong intermolecular attractive forces.

In a gas, the distance between molecules, whether monatomic or polyatomic, is very large compared with the size of the molecules; thus gases have a low density and are highly compressible. In contrast, the molecules in liquids are very close together, with essentially no empty space between them. As in gases, however, the molecules in liquids are in constant motion, and their kinetic energy (and hence their speed) depends on their temperature. We begin our discussion by examining some of the characteristic properties of liquids to see how each is consistent with a modified kinetic molecular description.

Density

The molecules of a liquid are packed relatively close together. Consequently, liquids are much denser than gases. The density of a liquid is typically about the same as the density of the solid state of the substance. Densities of liquids are therefore more commonly measured in units of grams per cubic centimeter (g/cm3) or grams per milliliter (g/mL) than in grams per liter (g/L), the unit commonly used for gases.

Molecular Order

Liquids exhibit short-range order because strong intermolecular attractive forces cause the molecules to pack together rather tightly. Because of their higher kinetic energy compared to the molecules in a solid, however, the molecules in a liquid move rapidly with respect to one another. Thus unlike the ions in the ionic solids discussed in , , the molecules in liquids are not arranged in a repeating three-dimensional array. Unlike the molecules in gases, however, the arrangement of the molecules in a liquid is not completely random.

Compressibility

Liquids have so little empty space between their component molecules that they cannot be readily compressed. Compression would force the atoms on adjacent molecules to occupy the same region of space.

Thermal Expansion

The intermolecular forces in liquids are strong enough to keep them from expanding significantly when heated (typically only a few percent over a 100°C temperature range). Thus the volumes of liquids are somewhat fixed. Notice from that the density of water, for example, changes by only about 3% over a 90-degree temperature range.

Table 11.1 The Density of Water at Various Temperatures

T (°C) Density (g/cm3)
0 0.99984
30 0.99565
60 0.98320
90 0.96535

Diffusion

Molecules in liquids diffuse because they are in constant motion (). A molecule in a liquid cannot move far before colliding with another molecule, however, so the mean free path in liquids is very short, and the rate of diffusion is much slower than in gases.

Figure 11.1 Molecular Diffusion in a Liquid

A drop of an aqueous solution containing a marker dye is added to a larger volume of water. As it diffuses, the color of the dye becomes fainter at the edges.

Fluidity

Liquids can flow, adjusting to the shape of their containers, because their molecules are free to move. This freedom of motion and their close spacing allow the molecules in a liquid to move rapidly into the openings left by other molecules, in turn generating more openings, and so forth ().

Figure 11.2 Why Liquids Flow

Molecules in a liquid are in constant motion. Consequently, when the flask is tilted, molecules move to the left and down due to the force of gravity, and the openings are occupied by other molecules. The result is a net flow of liquid out of the container.

Summary

The properties of liquids can be explained using a modified version of the kinetic molecular theory of gases described in . This model explains the higher density, greater order, and lower compressibility of liquids versus gases; the thermal expansion of liquids; why they diffuse; and why they adopt the shape (but not the volume) of their containers.

Key Takeaway

  • The kinetic molecular description of liquids must take into account both the nonzero volumes of particles and the presence of strong intermolecular attractive forces.

Conceptual Problems

  1. A liquid, unlike a gas, is virtually incompressible. Explain what this means using macroscopic and microscopic descriptions. What general physical properties do liquids share with solids? What properties do liquids share with gases?

  2. Using a kinetic molecular approach, discuss the differences and similarities between liquids and gases with regard to

    1. thermal expansion.
    2. fluidity.
    3. diffusion.
  3. How must the ideal gas law be altered to apply the kinetic molecular theory of gases to liquids? Explain.

  4. Why are the root mean square speeds of molecules in liquids less than the root mean square speeds of molecules in gases?

11.2 Intermolecular Forces

Learning Objective

  1. To describe the intermolecular forces in liquids.

The properties of liquids are intermediate between those of gases and solids but are more similar to solids. In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.

Note the Pattern

The properties of liquids are intermediate between those of gases and solids but are more similar to solids.

Intermolecular forces determine bulk properties such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid.

Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures. (For more information on the behavior of real gases and deviations from the ideal gas law, see , .)

In this section, we explicitly consider three kinds of intermolecular interactions:There are two additional types of electrostatic interaction that you are already familiar with: the ion–ion interactions that are responsible for ionic bonding and the ion–dipole interactions that occur when ionic substances dissolve in a polar substance such as water. (For more information on ionic bonding, see . For more information on the dissolution of ionic substances, see and .) dipole–dipole interactions, London dispersion forces, and hydrogen bonds. The first two are often described collectively as van der Waals forcesThe intermolecular forces known as dipole–dipole interactions and London dispersion forces..

Dipole–Dipole Interactions

Recall from that polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in part (a) in . These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (part (c) in ). Hence dipole–dipole interactionsA kind of intermolecular interaction (force) that results between molecules with net dipole moments., such as those in part (b) in , are attractive intermolecular interactions, whereas those in part (d) in are repulsive intermolecular interactions. Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in . On average, however, the attractive interactions dominate.

Figure 11.3 Attractive and Repulsive Dipole–Dipole Interactions

(a and b) Molecular orientations in which the positive end of one dipole (δ+) is near the negative end of another (δ) (and vice versa) produce attractive interactions. (c and d) Molecular orientations that juxtapose the positive or negative ends of the dipoles on adjacent molecules produce repulsive interactions.

Figure 11.4 Both Attractive and Repulsive Dipole–Dipole Interactions Occur in a Liquid Sample with Many Molecules

Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions we considered in . Recall that the attractive energy between two ions is proportional to 1/r, where r is the distance between the ions. Doubling the distance (r → 2r) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/r6, so doubling the distance between the dipoles decreases the strength of the interaction by 26, or 64-fold. Thus a substance such as HCl, which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure, whereas NaCl, which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in . Using what we learned in about predicting relative bond polarities from the electronegativities of the bonded atoms, we can make educated guesses about the relative boiling points of similar molecules.

Table 11.2 Relationships between the Dipole Moment and the Boiling Point for Organic Compounds of Similar Molar Mass

Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K)
C3H6 (cyclopropane) 42 0 240
CH3OCH3 (dimethyl ether) 46 1.30 248
CH3CN (acetonitrile) 41 3.9 355

Note the Pattern

The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6.

Example 1

Arrange ethyl methyl ether (CH3OCH2CH3), 2-methylpropane [isobutane, (CH3)2CHCH3], and acetone (CH3COCH3) in order of increasing boiling points. Their structures are as follows:

Given: compounds

Asked for: order of increasing boiling points

Strategy:

Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points.

Solution:

The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds. The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point. Ethyl methyl ether has a structure similar to H2O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point. Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point. Thus we predict the following order of boiling points: 2-methylpropane < ethyl methyl ether < acetone. This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.

Exercise

Arrange carbon tetrafluoride (CF4), ethyl methyl sulfide (CH3SC2H5), dimethyl sulfoxide [(CH3)2S=O], and 2-methylbutane [isopentane, (CH3)2CHCH2CH3] in order of decreasing boiling points.

Answer: dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C)

London Dispersion Forces

Thus far we have considered only interactions between polar molecules, but other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both ().

What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole momentsThe short-lived dipole moment in atoms and nonpolar molecules caused by the constant motion of their electrons, which results in an asymmetrical distribution of charge at any given instant., which produce attractive forces called London dispersion forcesA kind of intermolecular interaction (force) that results from temporary fluctuations in the electron distribution within atoms and nonpolar molecules. between otherwise nonpolar substances.

Table 11.3 Normal Melting and Boiling Points of Some Elements and Nonpolar Compounds

Substance Molar Mass (g/mol) Melting Point (°C) Boiling Point (°C)
Ar 40 −189.4 −185.9
Xe 131 −111.8 −108.1
N2 28 −210 −195.8
O2 32 −218.8 −183.0
F2 38 −219.7 −188.1
I2 254 113.7 184.4
CH4 16 −182.5 −161.5

Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in , the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipoleA short-lived dipole moment that is created in atoms and nonpolar molecules adjacent to atoms or molecules with an instantaneous dipole moment., in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/r6. Doubling the distance therefore decreases the attractive energy by 26, or 64-fold.

Figure 11.5 Instantaneous Dipole Moments

The formation of an instantaneous dipole moment on one He atom (a) or an H2 molecule (b) results in the formation of an induced dipole on an adjacent atom or molecule.

Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H2 molecules in part (b) in , tends to become more pronounced as atomic and molecular masses increase (). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. (For more information on shielding, see , .) As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizabilityThe ease of deformation of the electron distribution in an atom or molecule.. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones.

Note the Pattern

For similar substances, London dispersion forces get stronger with increasing molecular size.

The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles, as we shall see when we discuss solutions in . Thus London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in ). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the empirical formula C5H12. Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other n-pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n-pentane (36.1°C).

Figure 11.6 Mass and Surface Area Affect the Strength of London Dispersion Forces

(a) In this series of four simple alkanes, larger molecules have stronger London forces between them than smaller molecules and consequently higher boiling points. (b) Linear n-pentane molecules have a larger surface area and stronger intermolecular forces than spherical neopentane molecules. As a result, neopentane is a gas at room temperature, whereas n-pentane is a volatile liquid.

All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate.

Example 2

Arrange n-butane, propane, 2-methylpropane [isobutene, (CH3)2CHCH3], and n-pentane in order of increasing boiling points.

Given: compounds

Asked for: order of increasing boiling points

Strategy:

Determine the intermolecular forces in the compounds and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.

Solution:

The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n-pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n-butane (−0.5°C) < n-pentane (36.1°C).

Exercise

Arrange GeH4, SiCl4, SiH4, CH4, and GeCl4 in order of decreasing boiling points.

Answer: GeCl4 (87°C) > SiCl4 (57.6°C) > GeH4 (−88.5°C) > SiH4 (−111.8°C) > CH4 (−161°C)

Hydrogen Bonds

Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in . Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H2Te and H2Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C.

Figure 11.7 The Effects of Hydrogen Bonding on Boiling Points

These plots of the boiling points of the covalent hydrides of the elements of groups 14–17 show that the boiling points of the lightest members of each series for which hydrogen bonding is possible (HF, NH3, and H2O) are anomalously high for compounds with such low molecular masses.

Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bondsAn unusually strong dipole-dipole interaction (intermolecular force) that results when hydrogen is bonded to very electronegative elements, such as O, N, and F., as shown for ice in . A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor) and the atom that has the lone pair of electrons (the hydrogen bond acceptor). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two OH hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water rather than sinks.

Figure 11.8 The Hydrogen-Bonded Structure of Ice

Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cagelike structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.

Note the Pattern

Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.

Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” (we will discuss how antifreeze works in ) and why unprotected pipes in houses break if they are allowed to freeze.

Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH3. Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time.

Example 3

Considering CH3OH, C2H6, Xe, and (CH3)3N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.

Given: compounds

Asked for: formation of hydrogen bonds and structure

Strategy:

A Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors.

B Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.

Solution:

A Of the species listed, xenon (Xe), ethane (C2H6), and trimethylamine [(CH3)3N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.

B The one compound that can act as a hydrogen bond donor, methanol (CH3OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows:

Exercise

Considering CH3CO2H, (CH3)3N, NH3, and CH3F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.

Answer: CH3CO2H and NH3;

Example 4

Arrange C60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N2O in order of increasing boiling points.

Given: compounds

Asked for: order of increasing boiling points

Strategy:

Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.

Solution:

Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N2O have very similar molar masses (40 and 44 g/mol, respectively), but N2O is polar while Ar is not. Consequently, N2O should have a higher boiling point. A C60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N2O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C60 should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses: He (−269°C) < Ar (−185.7°C) < N2O (−88.5°C) < C60 (>280°C) < NaCl (1465°C).

Exercise

Arrange 2,4-dimethylheptane, Ne, CS2, Cl2, and KBr in order of decreasing boiling points.

Answer: KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS2 (46.6°C) > Cl2 (−34.6°C) > Ne (−246°C)

Summary

Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid or the liquid and gas phases are due to changes in intermolecular interactions but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces), and hydrogen bonds. Dipole–dipole interactions arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/r6, where r is the distance between dipoles. London dispersion forces are due to the formation of instantaneous dipole moments in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an induced dipole in adjacent molecules. Like dipole–dipole interactions, their energy falls off as 1/r6. Larger atoms tend to be more polarizable than smaller ones because their outer electrons are less tightly bound and are therefore more easily perturbed. Hydrogen bonds are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the hydrogen bond donor) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor). Because of strong OH hydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cagelike structure that is less dense than liquid water.

Key Takeaway

  • Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds.

Conceptual Problems

  1. What is the main difference between intramolecular interactions and intermolecular interactions? Which is typically stronger? How are changes of state affected by these different kinds of interactions?

  2. Describe the three major kinds of intermolecular interactions discussed in this chapter and their major features. The hydrogen bond is actually an example of one of the other two types of interaction. Identify the kind of interaction that includes hydrogen bonds and explain why hydrogen bonds fall into this category.

  3. Which are stronger—dipole–dipole interactions or London dispersion forces? Which are likely to be more important in a molecule with heavy atoms? Explain your answers.

  4. Explain why hydrogen bonds are unusually strong compared to other dipole–dipole interactions. How does the strength of hydrogen bonds compare with the strength of covalent bonds?

  5. Liquid water is essential for life as we know it, but based on its molecular mass, water should be a gas under standard conditions. Why is water a liquid rather than a gas under standard conditions?

  6. Describe the effect of polarity, molecular mass, and hydrogen bonding on the melting point and boiling point of a substance.

  7. Why are intermolecular interactions more important for liquids and solids than for gases? Under what conditions must these interactions be considered for gases?

  8. Using acetic acid as an example, illustrate both attractive and repulsive intermolecular interactions. How does the boiling point of a substance depend on the magnitude of the repulsive intermolecular interactions?

  9. In group 17, elemental fluorine and chlorine are gases, whereas bromine is a liquid and iodine is a solid. Why?

  10. The boiling points of the anhydrous hydrogen halides are as follows: HF, 19°C; HCl, −85°C; HBr, −67°C; and HI, −34°C. Explain any trends in the data, as well as any deviations from that trend.

  11. Identify the most important intermolecular interaction in each of the following.

    1. SO2
    2. HF
    3. CO2
    4. CCl4
    5. CH2Cl2
  12. Identify the most important intermolecular interaction in each of the following.

    1. LiF
    2. I2
    3. ICl
    4. NH3
    5. NH2Cl
  13. Would you expect London dispersion forces to be more important for Xe or Ne? Why? (The atomic radius of Ne is 38 pm, whereas that of Xe is 108 pm.)

  14. Arrange Kr, Cl2, H2, N2, Ne, and O2 in order of increasing polarizability. Explain your reasoning.

  15. Both water and methanol have anomalously high boiling points due to hydrogen bonding, but the boiling point of water is greater than that of methanol despite its lower molecular mass. Why? Draw the structures of these two compounds, including any lone pairs, and indicate potential hydrogen bonds.

  16. The structures of ethanol, ethylene glycol, and glycerin are as follows:

    Arrange these compounds in order of increasing boiling point. Explain your rationale.

  17. Do you expect the boiling point of H2S to be higher or lower than that of H2O? Justify your answer.

  18. Ammonia (NH3), methylamine (CH3NH2), and ethylamine (CH3CH2NH2) are gases at room temperature, while propylamine (CH3CH2CH2NH2) is a liquid at room temperature. Explain these observations.

  19. Why is it not advisable to freeze a sealed glass bottle that is completely filled with water? Use both macroscopic and microscopic models to explain your answer. Is a similar consideration required for a bottle containing pure ethanol? Why or why not?

  20. Which compound in the following pairs will have the higher boiling point? Explain your reasoning.

    1. NH3 or PH3
    2. ethylene glycol (HOCH2CH2OH) or ethanol
    3. 2,2-dimethylpropanol [CH3C(CH3)2CH2OH] or n-butanol (CH3CH2CH2CH2OH)
  21. Some recipes call for vigorous boiling, while others call for gentle simmering. What is the difference in the temperature of the cooking liquid between boiling and simmering? What is the difference in energy input?

  22. Use the melting of a metal such as lead to explain the process of melting in terms of what is happening at the molecular level. As a piece of lead melts, the temperature of the metal remains constant, even though energy is being added continuously. Why?

  23. How does the O–H distance in a hydrogen bond in liquid water compare with the O–H distance in the covalent O–H bond in the H2O molecule? What effect does this have on the structure and density of ice?

    1. Explain why the hydrogen bonds in liquid HF are stronger than the corresponding intermolecular HI interactions in liquid HI.
    2. In which substance are the individual hydrogen bonds stronger: HF or H2O? Explain your reasoning.
    3. For which substance will hydrogen bonding have the greater effect on the boiling point: HF or H2O? Explain your reasoning.

Answers

  1. Water is a liquid under standard conditions because of its unique ability to form four strong hydrogen bonds per molecule.

  2. As the atomic mass of the halogens increases, so does the number of electrons and the average distance of those electrons from the nucleus. Larger atoms with more electrons are more easily polarized than smaller atoms, and the increase in polarizability with atomic number increases the strength of London dispersion forces. These intermolecular interactions are strong enough to favor the condensed states for bromine and iodine under normal conditions of temperature and pressure.

    1. The V-shaped SO2 molecule has a large dipole moment due to the polar S=O bonds, so dipole–dipole interactions will be most important.
    2. The H–F bond is highly polar, and the fluorine atom has three lone pairs of electrons to act as hydrogen bond acceptors; hydrogen bonding will be most important.
    3. Although the C=O bonds are polar, this linear molecule has no net dipole moment; hence, London dispersion forces are most important.
    4. This is a symmetrical molecule that has no net dipole moment, and the Cl atoms are relatively polarizable; thus, London dispersion forces will dominate.
    5. This molecule has a small dipole moment, as well as polarizable Cl atoms. In such a case, dipole–dipole interactions and London dispersion forces are often comparable in magnitude.
  3. Water has two polar O–H bonds with H atoms that can act as hydrogen bond donors, plus two lone pairs of electrons that can act as hydrogen bond acceptors, giving a net of four hydrogen bonds per H2O molecule. Although methanol also has two lone pairs of electrons on oxygen that can act as hydrogen bond acceptors, it only has one O–H bond with an H atom that can act as a hydrogen bond donor. Consequently, methanol can only form two hydrogen bonds per molecule on average, versus four for water. Hydrogen bonding therefore has a much greater effect on the boiling point of water.

  4. Vigorous boiling causes more water molecule to escape into the vapor phase, but does not affect the temperature of the liquid. Vigorous boiling requires a higher energy input than does gentle simmering.

11.3 Unique Properties of Liquids

Learning Objective

  1. To describe the unique properties of liquids.

Although you have been introduced to some of the interactions that hold molecules together in a liquid, we have not yet discussed the consequences of those interactions for the bulk properties of liquids. We now turn our attention to three unique properties of liquids that intimately depend on the nature of intermolecular interactions: surface tension, capillary action, and viscosity.

Surface Tension

We stated in that liquids tend to adopt the shapes of their containers. Why, then, do small amounts of water on a freshly waxed car form raised droplets instead of a thin, continuous film? The answer lies in a property called surface tension, which depends on intermolecular forces.

presents a microscopic view of a liquid droplet. A typical molecule in the interior of the droplet is surrounded by other molecules that exert attractive forces from all directions. Consequently, there is no net force on the molecule that would cause it to move in a particular direction. In contrast, a molecule on the surface experiences a net attraction toward the drop because there are no molecules on the outside to balance the forces exerted by adjacent molecules in the interior. Because a sphere has the smallest possible surface area for a given volume, intermolecular attractive interactions between water molecules cause the droplet to adopt a spherical shape. This maximizes the number of attractive interactions and minimizes the number of water molecules at the surface. Hence raindrops are almost spherical, and drops of water on a waxed (nonpolar) surface, which does not interact strongly with water, form round beads (see the chapter opener photo). A dirty car is covered with a mixture of substances, some of which are polar. Attractive interactions between the polar substances and water cause the water to spread out into a thin film instead of forming beads.

Figure 11.9 A Representation of Surface Tension in a Liquid

Molecules at the surface of water experience a net attraction to other molecules in the liquid, which holds the surface of the bulk sample together. In contrast, those in the interior experience uniform attractive forces..

The same phenomenon holds molecules together at the surface of a bulk sample of water, almost as if they formed a skin. When filling a glass with water, the glass can be overfilled so that the level of the liquid actually extends above the rim. Similarly, a sewing needle or a paper clip can be placed on the surface of a glass of water where it “floats,” even though steel is much denser than water (part (a) in ). Many insects take advantage of this property to walk on the surface of puddles or ponds without sinking (part (b) in ).

Figure 11.10 The Effects of the High Surface Tension of Liquid Water

(a) A paper clip can “float” on water because of surface tension. (b) Surface tension also allows insects such as this water strider to “walk on water.”

Such phenomena are manifestations of surface tensionThe energy required to increase the surface area of a liquid by a certain amount. Surface tension is measured in units of energy per area (e.g., J/m2)., which is defined as the energy required to increase the surface area of a liquid by a specific amount. Surface tension is therefore measured as energy per unit area, such as joules per square meter (J/m2) or dyne per centimeter (dyn/cm), where 1 dyn = 1 × 10−5 N. The values of the surface tension of some representative liquids are listed in . Note the correlation between the surface tension of a liquid and the strength of the intermolecular forces: the stronger the intermolecular forces, the higher the surface tension. For example, water, with its strong intermolecular hydrogen bonding, has one of the highest surface tension values of any liquid, whereas low-boiling-point organic molecules, which have relatively weak intermolecular forces, have much lower surface tensions. Mercury is an apparent anomaly, but its very high surface tension is due to the presence of strong metallic bonding, which we will discuss in more detail in .

Table 11.4 Surface Tension, Viscosity, Vapor Pressure (at 25°C Unless Otherwise Indicated), and Normal Boiling Points of Common Liquids

Substance Surface Tension (× 10−3 J/m2) Viscosity (mPa·s) Vapor Pressure (mmHg) Normal Boiling Point (°C)
Organic Compounds
diethyl ether 17 0.22 531 34.6
n-hexane 18 0.30 149 68.7
acetone 23 0.31 227 56.5
ethanol 22 1.07 59 78.3
ethylene glycol 48 16.1 ~0.08 198.9
Liquid Elements
bromine 41 0.94 218 58.8
mercury 486 1.53 0.0020 357
Water
0°C 75.6 1.79 4.6
20°C 72.8 1.00 17.5
60°C 66.2 0.47 149
100°C 58.9 0.28 760

Adding soaps and detergents that disrupt the intermolecular attractions between adjacent water molecules can reduce the surface tension of water. Because they affect the surface properties of a liquid, soaps and detergents are called surface-active agents, or surfactantsSubstances (surface-active agents), such as soaps and detergents, that disrupt the attractive intermolecular interactions between molecules of a polar liquid, thereby reducing the surface tension of the liquid.. In the 1960s, US Navy researchers developed a method of fighting fires aboard aircraft carriers using “foams,” which are aqueous solutions of fluorinated surfactants. The surfactants reduce the surface tension of water below that of fuel, so the fluorinated solution is able to spread across the burning surface and extinguish the fire. Such foams are now used universally to fight large-scale fires of organic liquids.

Capillary Action

Intermolecular forces also cause a phenomenon called capillary actionThe tendency of a polar liquid to rise against gravity into a small-diameter glass tube., which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a capillary), as shown in . When a glass capillary is put into a dish of water, water is drawn up into the tube. The height to which the water rises depends on the diameter of the tube and the temperature of the water but not on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises.

Figure 11.11 The Phenomenon of Capillary Action

When a glass capillary is placed in liquid water, water rises up into the capillary. The smaller the diameter of the capillary, the higher the water rises. The height of the water does not depend on the angle at which the capillary is tilted.

Capillary action is the net result of two opposing sets of forces: cohesive forcesThe intermolecular forces that hold a liquid together., which are the intermolecular forces that hold a liquid together, and adhesive forcesThe attractive intermolecular forces between a liquid and the substance comprising the surface of a capillary., which are the attractive forces between a liquid and the substance that composes the capillary. Water has both strong adhesion to glass, which contains polar SiOH groups, and strong intermolecular cohesion. When a glass capillary is put into water, the surface tension due to cohesive forces constricts the surface area of water within the tube, while adhesion between the water and the glass creates an upward force that maximizes the amount of glass surface in contact with the water. If the adhesive forces are stronger than the cohesive forces, as is the case for water, then the liquid in the capillary rises to the level where the downward force of gravity exactly balances this upward force. If, however, the cohesive forces are stronger than the adhesive forces, as is the case for mercury and glass, the liquid pulls itself down into the capillary below the surface of the bulk liquid to minimize contact with the glass (part (a) in ). The upper surface of a liquid in a tube is called the meniscusThe upper surface of the liquid in a tube., and the shape of the meniscus depends on the relative strengths of the cohesive and adhesive forces. In liquids such as water, the meniscus is concave; in liquids such as mercury, however, which have very strong cohesive forces and weak adhesion to glass, the meniscus is convex (part (b) in ).

Note the Pattern

Polar substances are drawn up a glass capillary and generally have a concave meniscus.

Figure 11.12 The Effects of Capillary Action

(a) This drawing illustrates the shape of the meniscus and the relative height of a mercury column when a glass capillary is put into liquid mercury. The meniscus is convex and the surface of the liquid inside the tube is lower than the level of the liquid outside the tube. (b) Because water adheres strongly to the polar surface of glass, it has a concave meniscus, whereas mercury, which does not adhere to the glass, has a convex meniscus.

Fluids and nutrients are transported up the stems of plants or the trunks of trees by capillary action. Plants contain tiny rigid tubes composed of cellulose, to which water has strong adhesion. Because of the strong adhesive forces, nutrients can be transported from the roots to the tops of trees that are more than 50 m tall. Cotton towels are also made of cellulose; they absorb water because the tiny tubes act like capillaries and “wick” the water away from your skin. The moisture is absorbed by the entire fabric, not just the layer in contact with your body.

Viscosity

Viscosity (η)The resistance of a liquid to flow. is the resistance of a liquid to flow. Some liquids, such as gasoline, ethanol, and water, flow very readily and hence have a low viscosity. Others, such as motor oil, molasses, and maple syrup, flow very slowly and have a high viscosity. The two most common methods for evaluating the viscosity of a liquid are (1) to measure the time it takes for a quantity of liquid to flow through a narrow vertical tube and (2) to measure the time it takes steel balls to fall through a given volume of the liquid. The higher the viscosity, the slower the liquid flows through the tube and the steel balls fall. Viscosity is expressed in units of the poise (mPa·s); the higher the number, the higher the viscosity. The viscosities of some representative liquids are listed in and show a correlation between viscosity and intermolecular forces. Because a liquid can flow only if the molecules can move past one another with minimal resistance, strong intermolecular attractive forces make it more difficult for molecules to move with respect to one another. The addition of a second hydroxyl group to ethanol, for example, which produces ethylene glycol (HOCH2CH2OH), increases the viscosity 15-fold. This effect is due to the increased number of hydrogen bonds that can form between hydroxyl groups in adjacent molecules, resulting in dramatically stronger intermolecular attractive forces.

There is also a correlation between viscosity and molecular shape. Liquids consisting of long, flexible molecules tend to have higher viscosities than those composed of more spherical or shorter-chain molecules. The longer the molecules, the easier it is for them to become “tangled” with one another, making it more difficult for them to move past one another. London dispersion forces also increase with chain length. Due to a combination of these two effects, long-chain hydrocarbons (such as motor oils) are highly viscous.

Note the Pattern

Viscosity increases as intermolecular interactions or molecular size increases.

Motor oils and other lubricants demonstrate the practical importance of controlling viscosity. The oil in an automobile engine must effectively lubricate under a wide range of conditions, from subzero starting temperatures to the 200°C that oil can reach in an engine in the heat of the Mojave Desert in August. Viscosity decreases rapidly with increasing temperatures because the kinetic energy of the molecules increases, and higher kinetic energy enables the molecules to overcome the attractive forces that prevent the liquid from flowing. As a result, an oil that is thin enough to be a good lubricant in a cold engine will become too “thin” (have too low a viscosity) to be effective at high temperatures. The viscosity of motor oils is described by an SAE (Society of Automotive Engineers) rating ranging from SAE 5 to SAE 50 for engine oils: the lower the number, the lower the viscosity. So-called single-grade oils can cause major problems. If they are viscous enough to work at high operating temperatures (SAE 50, for example), then at low temperatures, they can be so viscous that a car is difficult to start or an engine is not properly lubricated. Consequently, most modern oils are multigrade, with designations such as SAE 20W/50 (a grade used in high-performance sports cars), in which case the oil has the viscosity of an SAE 20 oil at subzero temperatures (hence the W for winter) and the viscosity of an SAE 50 oil at high temperatures. These properties are achieved by a careful blend of additives that modulate the intermolecular interactions in the oil, thereby controlling the temperature dependence of the viscosity. Many of the commercially available oil additives “for improved engine performance” are highly viscous materials that increase the viscosity and effective SAE rating of the oil, but overusing these additives can cause the same problems experienced with highly viscous single-grade oils.

Example 5

Based on the nature and strength of the intermolecular cohesive forces and the probable nature of the liquid–glass adhesive forces, predict what will happen when a glass capillary is put into a beaker of SAE 20 motor oil. Will the oil be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? (Hint: the surface of glass is lined with Si–OH groups.)

Given: substance and composition of the glass surface

Asked for: behavior of oil and the shape of meniscus

Strategy:

A Identify the cohesive forces in the motor oil.

B Determine whether the forces interact with the surface of glass. From the strength of this interaction, predict the behavior of the oil and the shape of the meniscus.

Solution:

A Motor oil is a nonpolar liquid consisting largely of hydrocarbon chains. The cohesive forces responsible for its high boiling point are almost solely London dispersion forces between the hydrocarbon chains. B Such a liquid cannot form strong interactions with the polar Si–OH groups of glass, so the surface of the oil inside the capillary will be lower than the level of the liquid in the beaker. The oil will have a convex meniscus similar to that of mercury.

Exercise

Predict what will happen when a glass capillary is put into a beaker of ethylene glycol. Will the ethylene glycol be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)?

Answer: Capillary action will pull the ethylene glycol up into the capillary. The meniscus will be concave.

Summary

Surface tension is the energy required to increase the surface area of a liquid by a given amount. The stronger the intermolecular interactions, the greater the surface tension. Surfactants are molecules, such as soaps and detergents, that reduce the surface tension of polar liquids like water. Capillary action is the phenomenon in which liquids rise up into a narrow tube called a capillary. It results when cohesive forces, the intermolecular forces in the liquid, are weaker than adhesive forces, the attraction between a liquid and the surface of the capillary. The shape of the meniscus, the upper surface of a liquid in a tube, also reflects the balance between adhesive and cohesive forces. The viscosity of a liquid is its resistance to flow. Liquids that have strong intermolecular forces tend to have high viscosities.

Key Takeaway

  • Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions.

Conceptual Problems

  1. Why is a water droplet round?

  2. How is the environment of molecules on the surface of a liquid droplet different from that of molecules in the interior of the droplet? How is this difference related to the concept of surface tension?

  3. Explain the role of intermolecular and intramolecular forces in surface tension.

  4. A mosquito is able to walk across water without sinking, but if a few drops of detergent are added to the water, the insect will sink. Why?

  5. Explain how soaps or surfactants decrease the surface tension of a liquid. How does the meniscus of an aqueous solution in a capillary change if a surfactant is added? Illustrate your answer with a diagram.

  6. Of CH2Cl2, hexane, and ethanol, which has the lowest viscosity? Which has the highest surface tension? Explain your reasoning in each case.

  7. At 25°C, cyclohexanol has a surface tension of 32.92 mN/m2, whereas the surface tension of cyclohexanone, which is very similar chemically, is only 25.45 mN/m2. Why is the surface tension of cyclohexanone so much less than that of cyclohexanol?

  8. What is the relationship between

    1. surface tension and temperature?
    2. viscosity and temperature?

    Explain your answers in terms of a microscopic picture.

  9. What two opposing forces are responsible for capillary action? How do these forces determine the shape of the meniscus?

  10. Which of the following liquids will have a concave meniscus in a glass capillary? Explain your reasoning.

    1. pentane
    2. diethylene glycol (HOCH2CH2OCH2CH2OH)
    3. carbon tetrachloride
  11. How does viscosity depend on molecular shape? What molecular features make liquids highly viscous?

Answers

  1. Adding a soap or a surfactant to water disrupts the attractive intermolecular interactions between water molecules, thereby decreasing the surface tension. Because water is a polar molecule, one would expect that a soap or a surfactant would also disrupt the attractive interactions responsible for adhesion of water to the surface of a glass capillary. As shown in the sketch, this would decrease the height of the water column inside the capillary, as well as making the meniscus less concave.

  2. As the structures indicate, cyclohexanol is a polar substance that can engage in hydrogen bonding, much like methanol or ethanol; consequently, it is expected to have a higher surface tension due to stronger intermolecular interactions.

  3. Cohesive forces are the intermolecular forces that hold the molecules of the liquid together, while adhesive forces are the attractive forces between the molecules of the liquid and the walls of the capillary. If the adhesive forces are stronger than the cohesive forces, the liquid is pulled up into the capillary and the meniscus is concave. Conversely, if the cohesive forces are stronger than the adhesive forces, the level of the liquid inside the capillary will be lower than the level outside the capillary, and the meniscus will be convex.

  4. Viscous substances often consist of molecules that are much longer than they are wide and whose structures are often rather flexible. As a result, the molecules tend to become tangled with one another (much like overcooked spaghetti), which decreases the rate at which they can move through the liquid.

Numerical Problems

  1. The viscosities of five liquids at 25°C are given in the following table. Explain the observed trends in viscosity.

    Compound Molecular Formula Viscosity (mPa·s)
    benzene C6H6 0.604
    aniline C6H5NH2 3.847
    1,2-dichloroethane C2H4Cl2 0.779
    heptane C7H16 0.357
    1-heptanol C7H15OH 5.810
  2. The following table gives values for the viscosity, boiling point, and surface tension of four substances. Examine these data carefully to see whether the data for each compound are internally consistent and point out any obvious errors or inconsistencies. Explain your reasoning.

    Compound Viscosity (mPa·s at 20°C) Boiling Point (°C) Surface Tension (dyn/cm at 25°C)
    A 0.41 61 27.16
    B 0.55 65 22.55
    C 0.92 105 36.76
    D 0.59 110 28.53
  3. Surface tension data (in dyn/cm) for propanoic acid (C3H6O2), and 2-propanol (C3H8O), as a function of temperature, are given in the following table. Plot the data for each compound and explain the differences between the two graphs. Based on these data, which molecule is more polar?

    Compound 25°C 50°C 75°C
    propanoic acid 26.20 23.72 21.23
    2-propanol 20.93 18.96 16.98

Answer

  1.  

    The plots of surface tension versus temperature for propionic acid and isopropanol have essentially the same slope, but at all temperatures the surface tension of propionic acid is about 30% greater than for isopropanol. Because surface tension is a measure of the cohesive forces in a liquid, these data suggest that the cohesive forces for propionic acid are significantly greater than for isopropanol. Both substances consist of polar molecules with similar molecular masses, and the most important intermolecular interactions are likely to be dipole–dipole interactions. Consequently, these data suggest that propionic acid is more polar than isopropanol.

11.4 Vapor Pressure

Learning Objective

  1. To know how and why the vapor pressure of a liquid varies with temperature.

Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressureThe pressure created over a liquid by the molecules of a liquid substance that have enough kinetic energy to escape to the vapor phase. of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid.

Evaporation and Condensation

Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy (KE) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (), just as we did for a gas (). As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy (E0) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than E0. The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than E0 has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization)The physical process by which atoms or molecules in the liquid phase enter the gas or vapor phase., where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure.

Figure 11.13 The Distribution of the Kinetic Energies of the Molecules of a Liquid at Two Temperatures

Just as with gases, increasing the temperature shifts the peak to a higher energy and broadens the curve. Only molecules with a kinetic energy greater than E0 can escape from the liquid to enter the vapor phase, and the proportion of molecules with KE > E0 is greater at the higher temperature.

To understand the causes of vapor pressure, consider the apparatus shown in . When a liquid is introduced into an evacuated chamber (part (a) in ), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with KE > E0 will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase.

Figure 11.14 Vapor Pressure

(a) When a liquid is introduced into an evacuated chamber, molecules with sufficient kinetic energy escape from the surface and enter the vapor phase, causing the pressure in the chamber to increase. (b) When sufficient molecules are in the vapor phase for a given temperature, the rate of condensation equals the rate of evaporation (a steady state is reached), and the pressure in the container becomes constant.

As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensationThe physical process by which atoms or molecules in the vapor phase enter the liquid phase. (part (b) in ). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a steady state will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in .

Figure 11.15 The Relative Rates of Evaporation and Condensation as a Function of Time after a Liquid Is Introduced into a Sealed Chamber

The rate of evaporation depends only on the surface area of the liquid and is essentially constant. The rate of condensation depends on the number of molecules in the vapor phase and increases steadily until it equals the rate of evaporation.

Equilibrium Vapor Pressure

Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no net change in a system, constitute a dynamic equilibriumA state in which two opposing processes occur at the same rate, thus producing no net change in the system.. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressureThe pressure exerted by a vapor in dynamic equilibrium with its liquid. of the liquid.

If a liquid is in an open container, however, most of the molecules that escape into the vapor phase will not collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquidsA liquid with a relatively high vapor pressure. have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquidsA liquid with a relatively low vapor pressure. have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water () are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile.

The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point (). It does not depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in . Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is much steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release.

Figure 11.16 The Vapor Pressures of Several Liquids as a Function of Temperature

The point at which the vapor pressure curve crosses the P = 1 atm line (dashed) is the normal boiling point of the liquid.

Note the Pattern

Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions.

The exponential rise in vapor pressure with increasing temperature in allows us to use natural logarithms to express the nonlinear relationship as a linear one.For a review of natural logarithms, refer to Essential Skills 6 in .

Equation 11.1

ln P=ΔHvapR(1T)+CEquation for a straight line:y=mx+b

where ln P is the natural logarithm of the vapor pressure, ΔHvap is the enthalpy of vaporization, R is the universal gas constant [8.314 J/(mol·K)], T is the temperature in kelvins, and C is the y-intercept, which is a constant for any given line. A plot of ln P versus the inverse of the absolute temperature (1/T) is a straight line with a slope of −ΔHvap/R. , called the Clausius–Clapeyron equationA linear relationship that expresses the nonlinear relationship between the vapor pressure of a liquid and temperature: ln P=ΔHvap/RT+C, where P is pressure, ΔHvap is the heat of vaporization, R is the universal gas constant, T is the absolute temperature, and C is a constant. The Clausius–Clapeyron equation can be used to calculate the heat of vaporization of a liquid from its measured vapor pressure at two or more temperatures., can be used to calculate the ΔHvap of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine ΔHvap is to measure the vapor pressure of a liquid at two temperatures and insert the values of P and T for these points into , which is derived from the Clausius–Clapeyron equation:

Equation 11.2

ln(P2P1)=ΔHvapR(1T21T1)

Conversely, if we know ΔHvap and the vapor pressure P1 at any temperature T1, we can use to calculate the vapor pressure P2 at any other temperature T2, as shown in Example 6.

Example 6

The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table:

T (°C) 80.0 100 120 140
P (torr) 0.0888 0.2729 0.7457 1.845

From these data, calculate the enthalpy of vaporization (ΔHvap) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.)

Given: vapor pressures at four temperatures

Asked for: ΔHvap of mercury and vapor pressure at 160°C

Strategy:

A Use to obtain ΔHvap directly from two pairs of values in the table, making sure to convert all values to the appropriate units.

B Substitute the calculated value of ΔHvap into to obtain the unknown pressure (P2).

Solution:

A The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using and find the value of ΔHvap from the slope of the line, an alternative approach is to use to obtain ΔHvap directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvins because the equation requires absolute temperatures. Substituting the values measured at 80.0°C (T1) and 120.0°C (T2) into gives

 ln(0.7457 torr0.0888 torr)=ΔHvap8.314  J/(mol·K)[1(120+273)K1(80.0+273) K]     ln(8.398)=ΔHvap8.314  J·mol1·K1(2.88×104K1)             2.13=ΔHvap(0.346×104) J1·molΔHvap=61,400  J/mol=61.4 kJ/mol

B We can now use this value of ΔHvap to calculate the vapor pressure of the liquid (P2) at 160.0°C (T2):

ln(P20.0888  torr)=61,400  J·mol18.314  J·mol1·K1[1(160+273) K1(80.0+273) K]=3.86

Using the relationship elnx = x, we have

ln(P20.0888  torr)=3.86       P20.0888  torr=e3.86=47.5                       P2=4.21  torr

At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect.

Exercise

The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr?

Answer: 1896°C

Boiling Points

As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling pointThe temperature at which a substance boils at a pressure of 1 atm. of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the actual boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C.

Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked.

Note the Pattern

As pressure increases, the boiling point of a liquid increases and vice versa.

Table 11.5 The Boiling Points of Water at Various Locations on Earth

Place Altitude above Sea Level (ft) Atmospheric Pressure (mmHg) Boiling Point of Water (°C)
Mt. Everest, Nepal/Tibet 29,028 240 70
Bogota, Colombia 11,490 495 88
Denver, Colorado 5280 633 95
Washington, DC 25 759 100
Dead Sea, Israel/Jordan −1312 799 101.4

Example 7

Use to estimate the following.

  1. the boiling point of water in a pressure cooker operating at 1000 mmHg
  2. the pressure required for mercury to boil at 250°C

Given: data in , pressure, and boiling point

Asked for: corresponding boiling point and pressure

Strategy:

A To estimate the boiling point of water at 1000 mmHg, refer to and find the point where the vapor pressure curve of water intersects the line corresponding to a pressure of 1000 mmHg.

B To estimate the pressure required for mercury to boil at 250°C, find the point where the vapor pressure curve of mercury intersects the line corresponding to a temperature of 250°C.

Solution:

  1. A The vapor pressure curve of water intersects the P = 1000 mmHg line at about 110°C; this is therefore the boiling point of water at 1000 mmHg.
  2. B The vertical line corresponding to 250°C intersects the vapor pressure curve of mercury at P ≈ 75 mmHg. Hence this is the pressure required for mercury to boil at 250°C.

Exercise

Use the data in to estimate the following.

  1. the normal boiling point of ethylene glycol
  2. the pressure required for diethyl ether to boil at 20°C.

Answer:

  1. 200°C
  2. 450 mmHg

Summary

Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation. Eventually, a steady state is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of dynamic equilibrium. Under these conditions, a liquid exhibits a characteristic equilibrium vapor pressure that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the Clausius–Clapeyron equation. This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. Volatile liquids are liquids with high vapor pressures, which tend to evaporate readily from an open container; nonvolatile liquids have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its normal boiling point.

Key Takeaways

  • The equilibrium vapor pressure of a liquid depends on the temperature and the intermolecular forces present.
  • The relationship between pressure, enthalpy of vaporization, and temperature is given by the Clausius-Clapeyron equation.

Key Equations

Clausius–Clapeyron equation

: ln P=ΔHvapR(1T)+C

Using vapor pressure at two temperatures to calculate Δ H vap

: ln (P2P1)=ΔHvapR(1T21T1)

Conceptual Problems

  1. What is the relationship between the boiling point, vapor pressure, and temperature of a substance and atmospheric pressure?

  2. What is the difference between a volatile liquid and a nonvolatile liquid? Suppose that two liquid substances have the same molecular mass, but one is volatile and the other is nonvolatile. What differences in the molecular structures of the two substances could account for the differences in volatility?

  3. An “old wives’ tale” states that applying ethanol to the wrists of a child with a very high fever will help to reduce the fever because blood vessels in the wrists are close to the skin. Is there a scientific basis for this recommendation? Would water be as effective as ethanol?

  4. Why is the air over a strip of grass significantly cooler than the air over a sandy beach only a few feet away?

  5. If gasoline is allowed to sit in an open container, it often feels much colder than the surrounding air. Explain this observation. Describe the flow of heat into or out of the system, as well as any transfer of mass that occurs. Would the temperature of a sealed can of gasoline be higher, lower, or the same as that of the open can? Explain your answer.

  6. What is the relationship between the vapor pressure of a liquid and

    1. its temperature?
    2. the surface area of the liquid?
    3. the pressure of other gases on the liquid?
    4. its viscosity?
  7. At 25°C, benzene has a vapor pressure of 12.5 kPa, whereas the vapor pressure of acetic acid is 2.1 kPa. Which is more volatile? Based on the intermolecular interactions in the two liquids, explain why acetic acid has the lower vapor pressure.

Numerical Problems

  1. Acetylene (C2H2), which is used for industrial welding, is transported in pressurized cylinders. Its vapor pressure at various temperatures is given in the following table. Plot the data and use your graph to estimate the vapor pressure of acetylene at 293 K. Then use your graph to determine the value of ΔHvap for acetylene. How much energy is required to vaporize 2.00 g of acetylene at 250 K?

    T (K) 145 155 175 200 225 250 300
    P (mmHg) 1.3 7.8 32.2 190 579 1370 5093
  2. The following table gives the vapor pressure of water at various temperatures. Plot the data and use your graph to estimate the vapor pressure of water at 25°C and at 75°C. What is the vapor pressure of water at 110°C? Use these data to determine the value of ΔHvap for water.

    T (°C) 0 10 30 50 60 80 100
    P (mmHg) 4.6 9.2 31.8 92.6 150 355 760
  3. The ΔHvap of carbon tetrachloride is 29.8 kJ/mol, and its normal boiling point is 76.8°C. What is its boiling point at 0.100 atm?

  4. The normal boiling point of sodium is 883°C. If ΔHvap is 97.4 kJ/mol, what is the vapor pressure (in millimeters of mercury) of liquid sodium at 300°C?

  5. An unknown liquid has a vapor pressure of 0.860 atm at 63.7°C and a vapor pressure of 0.330 atm at 35.1°C. Use the data in in to identify the liquid.

  6. An unknown liquid has a boiling point of 75.8°C at 0.910 atm and a boiling point of 57.2°C at 0.430 atm. Use the data in in to identify the liquid.

  7. If the vapor pressure of a liquid is 0.850 atm at 20°C and 0.897 atm at 25°C, what is the normal boiling point of the liquid?

  8. If the vapor pressure of a liquid is 0.799 atm at 99.0°C and 0.842 atm at 111°C, what is the normal boiling point of the liquid?

  9. The vapor pressure of liquid SO2 is 33.4 torr at −63.4°C and 100.0 torr at −47.7 K.

    1. What is the ΔHvap of SO2?
    2. What is its vapor pressure at −24.5 K?
    3. At what temperature is the vapor pressure equal to 220 torr?
  10. The vapor pressure of CO2 at various temperatures is given in the following table:

    T (°C) −120 −110 −100 −90
    P (torr) 9.81 34.63 104.81 279.5
    1. What is ΔHvap over this temperature range?
    2. What is the vapor pressure of CO2 at −70°C?
    3. At what temperature does CO2 have a vapor pressure of 310 torr?

Answers

  1.  

    vapor pressure at 273 K is 3050 mmHg; ΔHvap = 18.7 kJ/mol, 1.44 kJ

  2. 12.5°C

  3. ΔHvap = 28.9 kJ/mol, n-hexane

  4. ΔHvap = 7.81 kJ/mol, 36°C

11.5 Changes of State

Learning Objective

  1. To calculate the energy changes that accompany phase changes.

We take advantage of changes between the gas, liquid, and solid states to cool a drink with ice cubes (solid to liquid), cool our bodies by perspiration (liquid to gas), and cool food inside a refrigerator (gas to liquid and vice versa). We use dry ice, which is solid CO2, as a refrigerant (solid to gas), and we make artificial snow for skiing and snowboarding by transforming a liquid to a solid. In this section, we examine what happens when any of the three forms of matter is converted to either of the other two. These changes of state are often called phase changesA change of state that occurs when any of the three forms of matter (solids, liquids, and gases) is converted to either of the other two.. The six most common phase changes are shown in .

Figure 11.17 The Three Phases of Matter and the Processes That Interconvert Them When the Temperature Is Changed

Enthalpy changes that accompany phase transitions are indicated by purple and green arrows.

Energy Changes That Accompany Phase Changes

Phase changes are always accompanied by a change in the energy of a system. For example, converting a liquid, in which the molecules are close together, to a gas, in which the molecules are, on average, far apart, requires an input of energy (heat) to give the molecules enough kinetic energy to allow them to overcome the intermolecular attractive forces. The stronger the attractive forces, the more energy is needed to overcome them. Solids, which are highly ordered, have the strongest intermolecular interactions, whereas gases, which are very disordered, have the weakest. Thus any transition from a more ordered to a less ordered state (solid to liquid, liquid to gas, or solid to gas) requires an input of energy; it is endothermic. Conversely, any transition from a less ordered to a more ordered state (liquid to solid, gas to liquid, or gas to solid) releases energy; it is exothermic. The energy change associated with each common phase change is shown in .

In , we defined the enthalpy changes associated with various chemical and physical processes. The melting points and molar enthalpies of fusion (ΔHfus), the energy required to convert from a solid to a liquid, a process known as fusion (or melting)The conversion of a solid to a liquid., as well as the normal boiling points and enthalpies of vaporization (ΔHvap) of selected compounds are listed in . The substances with the highest melting points usually have the highest enthalpies of fusion; they tend to be ionic compounds that are held together by very strong electrostatic interactions. Substances with high boiling points are those with strong intermolecular interactions that must be overcome to convert a liquid to a gas, resulting in high enthalpies of vaporization. The enthalpy of vaporization of a given substance is much greater than its enthalpy of fusion because it takes more energy to completely separate molecules (conversion from a liquid to a gas) than to enable them only to move past one another freely (conversion from a solid to a liquid).

Table 11.6 Melting and Boiling Points and Enthalpies of Fusion and Vaporization for Selected Substances

Substance Melting Point (°C) ΔHfus (kJ/mol) Boiling Point (°C) ΔHvap (kJ/mol)
N2 −210.0 0.71 −195.8 5.6
HCl −114.2 2.00 −85.1 16.2
Br2 −7.2 10.6 58.8 30.0
CCl4 −22.6 2.56 76.8 29.8
CH3CH2OH (ethanol) −114.1 4.93 78.3 38.6
CH3(CH2)4CH3 (n-hexane) −95.4 13.1 68.7 28.9
H2O 0 6.01 100 40.7
Na 97.8 2.6 883 97.4
NaF 996 33.4 1704 176.1

Note the Pattern

ΔH is positive for any transition from a more ordered to a less ordered state and negative for a transition from a less ordered to a more ordered state.

The direct conversion of a solid to a gas, without an intervening liquid phase, is called sublimationThe conversion of a solid directly to a gas (without an intervening liquid phase).. The amount of energy required to sublime 1 mol of a pure solid is the enthalpy of sublimation (ΔHsub)The enthalpy change that accompanies the conversion of a solid directly to a gas.. Common substances that sublime at standard temperature and pressure (STP; 0°C, 1 atm) include CO2 (dry ice); iodine (); naphthalene, a substance used to protect woolen clothing against moths; and 1,4-dichlorobenzene. As shown in , the enthalpy of sublimation of a substance is the sum of its enthalpies of fusion and vaporization provided all values are at the same T; this is an application of Hess’s law. (For more information about Hess’s law, see , ).

Equation 11.3

ΔHsub = ΔHfus + ΔHvap

Figure 11.18 The Sublimation of Solid Iodine

When solid iodine is heated at ordinary atmospheric pressure, it sublimes. When the I2 vapor comes in contact with a cold surface, it deposits I2 crystals.

Fusion, vaporization, and sublimation are endothermic processes; they occur only with the absorption of heat. Anyone who has ever stepped out of a swimming pool on a cool, breezy day has felt the heat loss that accompanies the evaporation of water from the skin. Our bodies use this same phenomenon to maintain a constant temperature: we perspire continuously, even when at rest, losing about 600 mL of water daily by evaporation from the skin. We also lose about 400 mL of water as water vapor in the air we exhale, which also contributes to cooling. Refrigerators and air-conditioners operate on a similar principle: heat is absorbed from the object or area to be cooled and used to vaporize a low-boiling-point liquid, such as ammonia or the chlorofluorocarbons (CFCs) and the hydrofluorocarbons (HCFCs) discussed in in connection with the ozone layer. The vapor is then transported to a different location and compressed, thus releasing and dissipating the heat. Likewise, ice cubes efficiently cool a drink not because of their low temperature but because heat is required to convert ice at 0°C to liquid water at 0°C, as demonstrated later in Example 8.

Temperature Curves

The processes on the right side of —freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve.

Heating Curves

shows a heating curveA plot of the temperature of a substance versus the heat added or versus the heating time at a constant rate of heating., a plot of temperature versus heating time, for a 75 g sample of water. The sample is initially ice at 1 atm and −23°C; as heat is added, the temperature of the ice increases linearly with time. The slope of the line depends on both the mass of the ice and the specific heat (Cs)The number of joules required to raise the temperature of 1 g of a substance by 1°C. of ice, which is the number of joules required to raise the temperature of 1 g of ice by 1°C. As the temperature of the ice increases, the water molecules in the ice crystal absorb more and more energy and vibrate more vigorously. At the melting point, they have enough kinetic energy to overcome attractive forces and move with respect to one another. As more heat is added, the temperature of the system does not increase further but remains constant at 0°C until all the ice has melted. Once all the ice has been converted to liquid water, the temperature of the water again begins to increase. Now, however, the temperature increases more slowly than before because the specific heat capacity of water is greater than that of ice. When the temperature of the water reaches 100°C, the water begins to boil. Here, too, the temperature remains constant at 100°C until all the water has been converted to steam. At this point, the temperature again begins to rise, but at a faster rate than seen in the other phases because the heat capacity of steam is less than that of ice or water.

Figure 11.19 A Heating Curve for Water

This plot of temperature shows what happens to a 75 g sample of ice initially at 1 atm and −23°C as heat is added at a constant rate: A–B: heating solid ice; B–C: melting ice; C–D: heating liquid water; D–E: vaporizing water; E–F: heating steam.

Thus the temperature of a system does not change during a phase change. In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process, and as long as even a small amount of liquid water is present, the temperature of the system remains at 100°C during the boiling process. The rate at which heat is added does not affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils.

Note the Pattern

The temperature of a sample does not change during a phase change.

If heat is added at a constant rate, as in , then the length of the horizontal lines, which represents the time during which the temperature does not change, is directly proportional to the magnitude of the enthalpies associated with the phase changes. In , the horizontal line at 100°C is much longer than the line at 0°C because the enthalpy of vaporization of water is several times greater than the enthalpy of fusion.

A superheated liquidAn unstable liquid at a temperature and pressure at which it should be a gas. is a sample of a liquid at the temperature and pressure at which it should be a gas. Superheated liquids are not stable; the liquid will eventually boil, sometimes violently. The phenomenon of superheating causes “bumping” when a liquid is heated in the laboratory. When a test tube containing water is heated over a Bunsen burner, for example, one portion of the liquid can easily become too hot. When the superheated liquid converts to a gas, it can push or “bump” the rest of the liquid out of the test tube. Placing a stirring rod or a small piece of ceramic (a “boiling chip”) in the test tube allows bubbles of vapor to form on the surface of the object so the liquid boils instead of becoming superheated. Superheating is the reason a liquid heated in a smooth cup in a microwave oven may not boil until the cup is moved, when the motion of the cup allows bubbles to form.

Cooling Curves

The cooling curveA plot of the temperature of a substance versus the heat removed or versus the cooling time at a constant rate of cooling., a plot of temperature versus cooling time, in plots temperature versus time as a 75 g sample of steam, initially at 1 atm and 200°C, is cooled. Although we might expect the cooling curve to be the mirror image of the heating curve in , the cooling curve is not an identical mirror image. As heat is removed from the steam, the temperature falls until it reaches 100°C. At this temperature, the steam begins to condense to liquid water. No further temperature change occurs until all the steam is converted to the liquid; then the temperature again decreases as the water is cooled. We might expect to reach another plateau at 0°C, where the water is converted to ice; in reality, however, this does not always occur. Instead, the temperature often drops below the freezing point for some time, as shown by the little dip in the cooling curve below 0°C. This region corresponds to an unstable form of the liquid, a supercooled liquidA metastable liquid phase that exists below the normal melting point of a substance.. If the liquid is allowed to stand, if cooling is continued, or if a small crystal of the solid phase is added (a seed crystalA solid sample of a substance that can be added to a supercooled liquid or a supersaturated solution to help induce crystallization.), the supercooled liquid will convert to a solid, sometimes quite suddenly. As the water freezes, the temperature increases slightly due to the heat evolved during the freezing process and then holds constant at the melting point as the rest of the water freezes. Subsequently, the temperature of the ice decreases again as more heat is removed from the system.

Figure 11.20 A Cooling Curve for Water

This plot of temperature shows what happens to a 75 g sample of steam initially at 1 atm and 200°C as heat is removed at a constant rate: A–B: cooling steam; B–C: condensing steam; C–D: cooling liquid water to give a supercooled liquid; D–E: warming the liquid as it begins to freeze; E–F: freezing liquid water; F–G: cooling ice.

Supercooling effects have a huge impact on Earth’s climate. For example, supercooling of water droplets in clouds can prevent the clouds from releasing precipitation over regions that are persistently arid as a result. Clouds consist of tiny droplets of water, which in principle should be dense enough to fall as rain. In fact, however, the droplets must aggregate to reach a certain size before they can fall to the ground. Usually a small particle (a nucleus) is required for the droplets to aggregate; the nucleus can be a dust particle, an ice crystal, or a particle of silver iodide dispersed in a cloud during seeding (a method of inducing rain). Unfortunately, the small droplets of water generally remain as a supercooled liquid down to about −10°C, rather than freezing into ice crystals that are more suitable nuclei for raindrop formation. One approach to producing rainfall from an existing cloud is to cool the water droplets so that they crystallize to provide nuclei around which raindrops can grow. This is best done by dispersing small granules of solid CO2 (dry ice) into the cloud from an airplane. Solid CO2 sublimes directly to the gas at pressures of 1 atm or lower, and the enthalpy of sublimation is substantial (25.3 kJ/mol). As the CO2 sublimes, it absorbs heat from the cloud, often with the desired results.

Example 8

If a 50.0 g ice cube at 0.0°C is added to 500 mL of tea at 20.0°C, what is the temperature of the tea when the ice cube has just melted? Assume that no heat is transferred to or from the surroundings. The density of water (and iced tea) is 1.00 g/mL over the range 0°C–20°C, the specific heats of liquid water and ice are 4.184 J/(g·°C) and 2.062 J/(g·°C), respectively, and the enthalpy of fusion of ice is 6.01 kJ/mol.

Given: mass, volume, initial temperature, density, specific heats, and ΔHfus

Asked for: final temperature

Strategy:

Substitute the values given into the general equation relating heat gained to heat lost () to obtain the final temperature of the mixture.

Solution:

Recall from that when two substances or objects at different temperatures are brought into contact, heat will flow from the warmer one to the cooler. The amount of heat that flows is given by

q = mCsΔT

where q is heat, m is mass, Cs is the specific heat, and ΔT is the temperature change. Eventually, the temperatures of the two substances will become equal at a value somewhere between their initial temperatures. Calculating the temperature of iced tea after adding an ice cube is slightly more complicated. The general equation relating heat gained and heat lost is still valid, but in this case we also have to take into account the amount of heat required to melt the ice cube from ice at 0.0°C to liquid water at 0.0°C:

qlost=qgained(miced tea)[Cs(H2O)](ΔTiced tea)={(mice)[Cs(H2O)](ΔTice)+(molice)ΔHfus(ice)}(500 g)[4.184 J/(g·°C)](Tf20.0°C)=[(50.0 g)[4.184 J/(g·°C)](Tf0.0°C)]     + (50.0 g18.0 g/mol)(6.01×103 J/mol)(2090 J/°C)(Tf)4.18×104 J=[(209 J/°C)(Tf)+1.67×104 J]2.53×104 J=(2310 J/°C)Tf11.0°C=Tf

Exercise

Suppose you are overtaken by a blizzard while ski touring and you take refuge in a tent. You are thirsty, but you forgot to bring liquid water. You have a choice of eating a few handfuls of snow (say 400 g) at −5.0°C immediately to quench your thirst or setting up your propane stove, melting the snow, and heating the water to body temperature before drinking it. You recall that the survival guide you leafed through at the hotel said something about not eating snow, but you can’t remember why—after all, it’s just frozen water. To understand the guide’s recommendation, calculate the amount of heat that your body will have to supply to bring 400 g of snow at −5.0°C to your body’s internal temperature of 37°C. Use the data in Example 8

Answer: 200 kJ (4.1 kJ to bring the ice from −5.0°C to 0.0°C, 133.6 kJ to melt the ice at 0.0°C, and 61.9 kJ to bring the water from 0.0°C to 37°C), which is energy that would not have been expended had you first melted the snow.

Summary

Changes of state are examples of phase changes, or phase transitions. All phase changes are accompanied by changes in the energy of a system. Changes from a more-ordered state to a less-ordered state (such as a liquid to a gas) are endothermic. Changes from a less-ordered state to a more-ordered state (such as a liquid to a solid) are always exothermic. The conversion of a solid to a liquid is called fusion (or melting). The energy required to melt 1 mol of a substance is its enthalpy of fusion (ΔHfus). The energy change required to vaporize 1 mol of a substance is the enthalpy of vaporization (ΔHvap). The direct conversion of a solid to a gas is sublimation. The amount of energy needed to sublime 1 mol of a substance is its enthalpy of sublimation (ΔHsub) and is the sum of the enthalpies of fusion and vaporization. Plots of the temperature of a substance versus heat added or versus heating time at a constant rate of heating are called heating curves. Heating curves relate temperature changes to phase transitions. A superheated liquid, a liquid at a temperature and pressure at which it should be a gas, is not stable. A cooling curve is not exactly the reverse of the heating curve because many liquids do not freeze at the expected temperature. Instead, they form a supercooled liquid, a metastable liquid phase that exists below the normal melting point. Supercooled liquids usually crystallize on standing, or adding a seed crystal of the same or another substance can induce crystallization.

Key Takeaway

  • Fusion, vaporization, and sublimation are endothermic processes, whereas freezing, condensation, and deposition are exothermic processes.

Conceptual Problems

  1. In extremely cold climates, snow can disappear with no evidence of its melting. How can this happen? What change(s) in state are taking place? Would you expect this phenomenon to be more common at high or low altitudes? Explain your answer.

  2. Why do car manufacturers recommend that an automobile should not be left standing in subzero temperatures if its radiator contains only water? Car manufacturers also warn car owners that they should check the fluid level in a radiator only when the engine is cool. What is the basis for this warning? What is likely to happen if it is ignored?

  3. Use Hess’s law and a thermochemical cycle to show that, for any solid, the enthalpy of sublimation is equal to the sum of the enthalpy of fusion of the solid and the enthalpy of vaporization of the resulting liquid.

  4. Three distinct processes occur when an ice cube at −10°C is used to cool a glass of water at 20°C. What are they? Which causes the greatest temperature change in the water?

  5. When frost forms on a piece of glass, crystals of ice are deposited from water vapor in the air. How is this process related to sublimation? Describe the energy changes that take place as the water vapor is converted to frost.

  6. What phase changes are involved in each process? Which processes are exothermic, and which are endothermic?

    1. ice melting
    2. distillation
    3. condensation forming on a window
    4. the use of dry ice to create a cloud for a theatrical production
  7. What phase changes are involved in each process? Which processes are exothermic, and which are endothermic?

    1. evaporation of methanol
    2. crystallization
    3. liquefaction of natural gas
    4. the use of naphthalene crystals to repel moths
  8. Why do substances with high enthalpies of fusion tend to have high melting points?

  9. Why is the enthalpy of vaporization of a compound invariably much larger than its enthalpy of fusion?

  10. What is the opposite of fusion, sublimation, and condensation? Describe the phase change in each pair of opposing processes and state whether each phase change is exothermic or endothermic.

  11. Draw a typical heating curve (temperature versus amount of heat added at a constant rate) for conversion of a solid to a liquid and then to a gas. What causes some regions of the plot to have a positive slope? What is happening in the regions of the plot where the curve is horizontal, meaning that the temperature does not change even though heat is being added?

  12. If you know the mass of a sample of a substance, how could you use a heating curve to calculate the specific heat of the substance, as well as the change in enthalpy associated with a phase change?

  13. Draw the heating curve for a liquid that has been superheated. How does this differ from a normal heating curve for a liquid? Draw the cooling curve for a liquid that has been supercooled. How does this differ from a normal cooling curve for a liquid?

Answers

  1. When snow disappears without melting, it must be subliming directly from the solid state to the vapor state. The rate at which this will occur depends solely on the partial pressure of water, not on the total pressure due to other gases. Consequently, altitude (and changes in atmospheric pressure) will not affect the rate of sublimation directly.

  2. The general equations and enthalpy changes for the changes of state involved in converting a solid to a gas are:

    solidliquidΔHfusliquidgasΔHvapsolidgasΔHsub=ΔHfus+ΔHvap

    The relationship between these enthalpy changes is shown schematically in the thermochemical cycle below:

  3. The formation of frost on a surface is an example of deposition, which is the reverse of sublimation. The change in enthalpy for deposition is equal in magnitude, but opposite in sign, to ΔHsub, which is a positive number: ΔHsub = ΔHfus + ΔHvap.

    1. liquid + heat → vapor: endothermic
    2. liquid → solid + heat: exothermic
    3. gas → liquid + heat: exothermic
    4. solid + heat → vapor: endothermic
  4. The enthalpy of vaporization is larger than the enthalpy of fusion because vaporization requires the addition of enough energy to disrupt all intermolecular interactions and create a gas in which the molecules move essentially independently. In contrast, fusion requires much less energy, because the intermolecular interactions in a liquid and a solid are similar in magnitude in all condensed phases. Fusion requires only enough energy to overcome the intermolecular interactions that lock molecules in place in a lattice, thereby allowing them to move more freely.

  5.  

    The portions of the curve with a positive slope correspond to heating a single phase, while the horizontal portions of the curve correspond to phase changes. During a phase change, the temperature of the system does not change, because the added heat is melting the solid at its melting point or evaporating the liquid at its boiling point.

  6.  

    A superheated liquid exists temporarily as liquid with a temperature above the normal boiling point of the liquid. When a supercooled liquid boils, the temperature drops as the liquid is converted to vapor.

    Conversely, a supercooled liquid exists temporarily as a liquid with a temperature lower than the normal melting point of the solid. As shown below, when a supercooled liquid crystallizes, the temperature increases as the liquid is converted to a solid.

Numerical Problems

  1. The density of oxygen at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of oxygen.

    T (K) 60 70 80 90 100 120 140
    d (mol/L) 40.1 38.6 37.2 35.6 0.123 0.102 0.087
  2. The density of propane at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of propane.

    T (K) 100 125 150 175 200 225 250 275
    d (mol/L) 16.3 15.7 15.0 14.4 13.8 13.2 0.049 0.044
  3. Draw the cooling curve for a sample of the vapor of a compound that has a melting point of 34°C and a boiling point of 77°C as it is cooled from 100°C to 0°C.

  4. Propionic acid has a melting point of −20.8°C and a boiling point of 141°C. Draw a heating curve showing the temperature versus time as heat is added at a constant rate to show the behavior of a sample of propionic acid as it is heated from −50°C to its boiling point. What happens above 141°C?

  5. A 0.542 g sample of I2 requires 96.1 J of energy to be converted to vapor. What is the enthalpy of sublimation of I2?

  6. A 2.0 L sample of gas at 210°C and 0.762 atm condenses to give 1.20 mL of liquid, and 476 J of heat is released during the process. What is the enthalpy of vaporization of the compound?

  7. One fuel used for jet engines and rockets is aluminum borohydride [Al(BH4)3], a liquid that readily reacts with water to produce hydrogen. The liquid has a boiling point of 44.5°C. How much energy is needed to vaporize 1.0 kg of aluminum borohydride at 20°C, given a ΔHvap of 30 kJ/mol and a molar heat capacity (Cp) of 194.6 J/(mol·K)?

  8. How much energy is released when freezing 100.0 g of dimethyl disulfide (C2H6S2) initially at 20°C? Use the following information: melting point = −84.7°C, ΔHfus = 9.19 kJ/mol, Cp = 118.1 J/(mol·K).

    The following four problems use the following information (the subscript p indicates measurements taken at constant pressure): ΔHfus(H2O) = 6.01 kJ/mol, ΔHvap(H2O) = 40.66 kJ/mol, Cp(s)(crystalline H2O) = 38.02 J/(mol·K), Cp(l)(liquid H2O) = 75.35 J/(mol·K), and Cp(g)(H2O gas) = 33.60 J/(mol·K).

  1. How much heat is released in the conversion of 1.00 L of steam at 21.9 atm and 200°C to ice at −6.0°C and 1 atm?

  2. How much heat must be applied to convert a 1.00 g piece of ice at −10°C to steam at 120°C?

  3. How many grams of boiling water must be added to a glass with 25.0 g of ice at −3°C to obtain a liquid with a temperature of 45°C?

  4. How many grams of ice at −5.0°C must be added to 150.0 g of water at 22°C to give a final temperature of 15°C?

Answers

  1.  

    The transition from a liquid to a gaseous phase is accompanied by a drastic decrease in density. According to the data in the table and the plot, the boiling point of liquid oxygen is between 90 and 100 K (actually 90.2 K).

  2.  

  3. 45.0 kJ/mol

  4. 488 kJ

  1. 32.6 kJ

  2. 57 g

11.6 Critical Temperature and Pressure

Learning Objective

  1. To know what is meant by the critical temperature and pressure of a liquid.

In , , we saw that a combination of high pressure and low temperature allows gases to be liquefied. As we increase the temperature of a gas, liquefaction becomes more and more difficult because higher and higher pressures are required to overcome the increased kinetic energy of the molecules. In fact, for every substance, there is some temperature above which the gas can no longer be liquefied, regardless of pressure. This temperature is the critical temperature (Tc)The highest temperature at which a substance can exist as a liquid, regardless of the applied pressure., the highest temperature at which a substance can exist as a liquid. Above the critical temperature, the molecules have too much kinetic energy for the intermolecular attractive forces to hold them together in a separate liquid phase. Instead, the substance forms a single phase that completely occupies the volume of the container. Substances with strong intermolecular forces tend to form a liquid phase over a very large temperature range and therefore have high critical temperatures. Conversely, substances with weak intermolecular interactions have relatively low critical temperatures. Each substance also has a critical pressure (Pc)The minimum pressure needed to liquefy a substance at its critical temperature., the minimum pressure needed to liquefy it at the critical temperature. The combination of critical temperature and critical pressure is called the critical pointThe combination of the critical temperature and the critical pressure of a substance. of a substance. The critical temperatures and pressures of several common substances are listed in .

Note the Pattern

High-boiling-point, nonvolatile liquids have high critical temperatures and vice versa.

Table 11.7 Critical Temperatures and Pressures of Some Simple Substances

Substance Tc (°C) Pc (atm)
NH3 132.4 113.5
CO2 31.0 73.8
CH3CH2OH (ethanol) 240.9 61.4
He −267.96 2.27
Hg 1477 1587
CH4 −82.6 46.0
N2 −146.9 33.9
H2O 374.0 217.7

Supercritical Fluids

To understand what happens at the critical point, consider the effects of temperature and pressure on the densities of liquids and gases, respectively. As the temperature of a liquid increases, its density decreases. As the pressure of a gas increases, its density increases. At the critical point, the liquid and gas phases have exactly the same density, and only a single phase exists. This single phase is called a supercritical fluidThe single, dense fluid phase that exists above the critical temperature of a substance., which exhibits many of the properties of a gas but has a density more typical of a liquid. For example, the density of water at its critical point (T = 374°C, P = 217.7 atm) is 0.32 g/mL, about one-third that of liquid water at room temperature but much greater than that of water vapor under most conditions. The transition between a liquid/gas mixture and a supercritical phase is demonstrated for a sample of benzene in . At the critical temperature, the meniscus separating the liquid and gas phases disappears.

Figure 11.21 Supercritical Benzene

Below the critical temperature of benzene (Tc = 289°C), the meniscus between the liquid and gas phases is apparent. At the critical temperature, the meniscus disappears because the density of the vapor is equal to the density of the liquid. Above Tc, a dense homogeneous fluid fills the tube.

In the last few years, supercritical fluids have evolved from laboratory curiosities to substances with important commercial applications. For example, carbon dioxide has a low critical temperature (31°C), a comparatively low critical pressure (73 atm), and low toxicity, making it easy to contain and relatively safe to manipulate. Because many substances are quite soluble in supercritical CO2, commercial processes that use it as a solvent are now well established in the oil industry, the food industry, and others. Supercritical CO2 is pumped into oil wells that are no longer producing much oil to dissolve the residual oil in the underground reservoirs. The less-viscous solution is then pumped to the surface, where the oil can be recovered by evaporation (and recycling) of the CO2. In the food, flavor, and fragrance industry, supercritical CO2 is used to extract components from natural substances for use in perfumes, remove objectionable organic acids from hops prior to making beer, and selectively extract caffeine from whole coffee beans without removing important flavor components. The latter process was patented in 1974, and now virtually all decaffeinated coffee is produced this way. The earlier method used volatile organic solvents such as methylene chloride (dichloromethane [CH2Cl2], boiling point = 40°C), which is difficult to remove completely from the beans and is known to cause cancer in laboratory animals at high doses.

Example 9

Arrange methanol, n-butane, n-pentane, and N2O in order of increasing critical temperatures.

Given: compounds

Asked for: order of increasing critical temperatures

Strategy:

A Identify the intermolecular forces in each molecule and then assess the strengths of those forces.

B Arrange the compounds in order of increasing critical temperatures.

Solution:

A The critical temperature depends on the strength of the intermolecular interactions that hold a substance together as a liquid. In N2O, a slightly polar substance, weak dipole–dipole interactions and London dispersion forces are important. Butane (C4H10) and pentane (C5H12) are larger, nonpolar molecules that exhibit only London dispersion forces. Methanol, in contrast, should have substantial intermolecular hydrogen bonding interactions. Because hydrogen bonds are stronger than the other intermolecular forces, methanol will have the highest Tc. London forces are more important for pentane than for butane because of its larger size, so n-pentane will have a higher Tc than n-butane. The only remaining question is whether N2O is polar enough to have stronger intermolecular interactions than pentane or butane. Because the electronegativities of O and N are quite similar, the answer is probably no, so N2O should have the lowest Tc. B We therefore predict the order of increasing critical temperatures as N2O < n-butane < n-pentane < methanol. The actual values are N2O (36.9°C) < n-butane (152.0°C) < n-pentane (196.9°C) < methanol (239.9°C). This is the same order as their normal boiling points—N2O (−88.7°C) < n-butane (−0.2°C) < n-pentane (36.0°C) < methanol (65°C)—because both critical temperature and boiling point depend on the relative strengths of the intermolecular interactions.

Exercise

Arrange ethanol, methanethiol (CH3SH), ethane, and n-hexanol in order of increasing critical temperatures.

Answer: ethane (32.3°C) < methanethiol (196.9°C) < ethanol (240.9°C) < n-hexanol (336.9°C)

Molten Salts and Ionic Liquids

Heating a salt to its melting point produces a molten saltA salt that has been heated to its melting point.. If we heated a sample of solid NaCl to its melting point of 801°C, for example, it would melt to give a stable liquid that conducts electricity. The characteristics of molten salts other than electrical conductivity are their high heat capacity, ability to attain very high temperatures (over 700°C) as a liquid, and utility as solvents because of their relatively low toxicity.

Molten salts have many uses in industry and the laboratory. For example, in solar power towers in the desert of California, mirrors collect and focus sunlight to melt a mixture of sodium nitrite and sodium nitrate. The heat stored in the molten salt is used to produce steam that drives a steam turbine and a generator, thereby producing electricity from the sun for southern California.

Due to their low toxicity and high thermal efficiency, molten salts have also been used in nuclear reactors to enable operation at temperatures greater than 750°C. One prototype reactor tested in the 1950s used a fuel and a coolant consisting of molten fluoride salts, including NaF, ZrF4, and UF4. Molten salts are also useful in catalytic processes such as coal gasification, in which carbon and water react at high temperatures to form CO and H2.

Note the Pattern

Molten salts are good electrical conductors, have a high heat capacity, can maintain a high temperature as a liquid, and are relatively nontoxic.

Although molten salts have proven highly useful, more recently chemists have been studying the characteristics of ionic liquidsIonic substances that are liquids at room temperature and pressure and that consist of small, symmetrical anions combined with larger, symmetrical organic cations that prevent the formation of a highly organized structure., ionic substances that are liquid at room temperature and pressure. These substances consist of small, symmetrical anions, such as PF6 and BF4, combined with larger, asymmetrical organic cations that prevent the formation of a highly organized structure, resulting in a low melting point. By varying the cation and the anion, chemists can tailor the liquid to specific needs, such as using a solvent in a given reaction or extracting specific molecules from a solution. For example, an ionic liquid consisting of a bulky cation and anions that bind metal contaminants such as mercury and cadmium ions can remove those toxic metals from the environment. A similar approach has been applied to removing uranium and americium from water contaminated by nuclear waste.

Note the Pattern

Ionic liquids consist of small, symmetrical anions combined with larger asymmetrical cations, which produce a highly polar substance that is a liquid at room temperature and pressure.

The initial interest in ionic liquids centered on their use as a low-temperature alternative to molten salts in batteries for missiles, nuclear warheads, and space probes. Further research revealed that ionic liquids had other useful properties—for example, some could dissolve the black rubber of discarded tires, allowing it to be recovered for recycling. Others could be used to produce commercially important organic compounds with high molecular mass, such as Styrofoam and Plexiglas, at rates 10 times faster than traditional methods.

Summary

A substance cannot form a liquid above its critical temperature, regardless of the applied pressure. Above the critical temperature, the molecules have enough kinetic energy to overcome the intermolecular attractive forces. The minimum pressure needed to liquefy a substance at its critical temperature is its critical pressure. The combination of the critical temperature and critical pressure of a substance is its critical point. Above the critical temperature and pressure, a substance exists as a dense fluid called a supercritical fluid, which resembles a gas in that it completely fills its container but has a density comparable to that of a liquid. A molten salt is a salt heated to its melting point, giving a stable liquid that conducts electricity. Ionic liquids are ionic substances that are liquids at room temperature. Their disorganized structure results in a low melting point.

Key Takeaway

  • The critical temperature and critical pressure of a substance define its critical point, beyond which the substance forms a supercritical fluid.

Conceptual Problems

  1. Describe the changes that take place when a liquid is heated above its critical temperature. How does this affect the physical properties?

  2. What is meant by the term critical pressure? What is the effect of increasing the pressure on a gas to above its critical pressure? Would it make any difference if the temperature of the gas was greater than its critical temperature?

  3. Do you expect the physical properties of a supercritical fluid to be more like those of the gas or the liquid phase? Explain. Can an ideal gas form a supercritical fluid? Why or why not?

  4. What are the limitations in using supercritical fluids to extract organic materials? What are the advantages?

  5. Describe the differences between a molten salt and an ionic liquid. Under what circumstances would an ionic liquid be preferred over a molten salt?

11.7 Phase Diagrams

Learning Objective

  1. To understand the general features of a phase diagram.

The state exhibited by a given sample of matter depends on the identity, temperature, and pressure of the sample. A phase diagramA graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system. is a graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system.

A typical phase diagram consists of discrete regions that represent the different phases exhibited by a substance (). Each region corresponds to the range of combinations of temperature and pressure over which that phase is stable. The combination of high pressure and low temperature (upper left of ) corresponds to the solid phase, whereas the gas phase is favored at high temperature and low pressure (lower right). The combination of high temperature and high pressure (upper right) corresponds to a supercritical fluid.

Figure 11.22 A Typical Phase Diagram for a Substance That Exhibits Three Phases—Solid, Liquid, and Gas—and a Supercritical Region

Note the Pattern

The solid phase is favored at low temperature and high pressure; the gas phase is favored at high temperature and low pressure.

General Features of a Phase Diagram

The lines in a phase diagram correspond to the combinations of temperature and pressure at which two phases can coexist in equilibrium. In , the line that connects points A and D separates the solid and liquid phases and shows how the melting point of a solid varies with pressure. The solid and liquid phases are in equilibrium all along this line; crossing the line horizontally corresponds to melting or freezing. The line that connects points A and B is the vapor pressure curve of the liquid, which we discussed in . It ends at the critical point, beyond which the substance exists as a supercritical fluid. The line that connects points A and C is the vapor pressure curve of the solid phase. Along this line, the solid is in equilibrium with the vapor phase through sublimation and deposition. Finally, point A, where the solid/liquid, liquid/gas, and solid/gas lines intersect, is the triple pointThe point in a phase diagram where the solid/liquid, liquid/gas, and solid/gas lines intersect; it represents the only combination of temperature and pressure at which all three phases are in equilibrium and can therefore exist simultaneously.; it is the only combination of temperature and pressure at which all three phases (solid, liquid, and gas) are in equilibrium and can therefore exist simultaneously. Because no more than three phases can ever coexist, a phase diagram can never have more than three lines intersecting at a single point.

Remember that a phase diagram, such as the one in , is for a single pure substance in a closed system, not for a liquid in an open beaker in contact with air at 1 atm pressure. In practice, however, the conclusions reached about the behavior of a substance in a closed system can usually be extrapolated to an open system without a great deal of error.

The Phase Diagram of Water

shows the phase diagram of water and illustrates that the triple point of water occurs at 0.01°C and 0.00604 atm (4.59 mmHg). Far more reproducible than the melting point of ice, which depends on the amount of dissolved air and the atmospheric pressure, the triple point (273.16 K) is used to define the absolute (Kelvin) temperature scale. The triple point also represents the lowest pressure at which a liquid phase can exist in equilibrium with the solid or vapor. At pressures less than 0.00604 atm, therefore, ice does not melt to a liquid as the temperature increases; the solid sublimes directly to water vapor. Sublimation of water at low temperature and pressure can be used to “freeze-dry” foods and beverages. The food or beverage is first cooled to subzero temperatures and placed in a container in which the pressure is maintained below 0.00604 atm. Then, as the temperature is increased, the water sublimes, leaving the dehydrated food (such as that used by backpackers or astronauts) or the powdered beverage (as with freeze-dried coffee).

The phase diagram for water illustrated in part (b) in shows the boundary between ice and water on an expanded scale. The melting curve of ice slopes up and slightly to the left rather than up and to the right as in ; that is, the melting point of ice decreases with increasing pressure; at 100 MPa (987 atm), ice melts at −9°C. Water behaves this way because it is one of the few known substances for which the crystalline solid is less dense than the liquid (others include antimony and bismuth). Increasing the pressure of ice that is in equilibrium with water at 0°C and 1 atm tends to push some of the molecules closer together, thus decreasing the volume of the sample. The decrease in volume (and corresponding increase in density) is smaller for a solid or a liquid than for a gas, but it is sufficient to melt some of the ice.

Figure 11.23 Two Versions of the Phase Diagram of Water

(a) In this graph with linear temperature and pressure axes, the boundary between ice and liquid water is almost vertical. (b) This graph with an expanded scale illustrates the decrease in melting point with increasing pressure. (The letters refer to points discussed in Example 10.)

In part (b) in , point A is located at P = 1 atm and T = −1.0°C, within the solid (ice) region of the phase diagram. As the pressure increases to 150 atm while the temperature remains the same, the line from point A crosses the ice/water boundary to point B, which lies in the liquid water region. Consequently, applying a pressure of 150 atm will melt ice at −1.0°C. We have already indicated that the pressure dependence of the melting point of water is of vital importance. If the solid/liquid boundary in the phase diagram of water were to slant up and to the right rather than to the left, ice would be denser than water, ice cubes would sink, water pipes would not burst when they freeze, and antifreeze would be unnecessary in automobile engines.

Until recently, many textbooks described ice skating as being possible because the pressure generated by the skater’s blade is high enough to melt the ice under the blade, thereby creating a lubricating layer of liquid water that enables the blade to slide across the ice. Although this explanation is intuitively satisfying, it is incorrect, as we can show by a simple calculation. Recall from that pressure (P) is the force (F) applied per unit area (A):

Equation 11.4

P=FA

To calculate the pressure an ice skater exerts on the ice, we need to calculate only the force exerted and the area of the skate blade. If we assume a 75.0 kg (165 lb) skater, then the force exerted by the skater on the ice due to gravity is

Equation 11.5

F = mg

where m is the mass and g is the acceleration due to Earth’s gravity (9.81 m/s2). Thus the force is

Equation 11.6

F = (75.0 kg)(9.81 m/s2) = 736 (kg·m)/s2 = 736 N

If we assume that the skate blades are 2.0 mm wide and 25 cm long, then the area of the bottom of each blade is

Equation 11.7

A = (2.0 × 10−3 m)(25 × 10−2 m) = 5.0 × 10−4 m2

If the skater is gliding on one foot, the pressure exerted on the ice is

Equation 11.8

P=736 N5.0×104 m2=1.5×106 N/m2=1.5×106 Pa=15 atm

The pressure is much lower than the pressure needed to decrease the melting point of ice by even 1°C, and experience indicates that it is possible to skate even when the temperature is well below freezing. Thus pressure-induced melting of the ice cannot explain the low friction that enables skaters (and hockey pucks) to glide. Recent research indicates that the surface of ice, where the ordered array of water molecules meets the air, consists of one or more layers of almost liquid water. These layers, together with melting induced by friction as a skater pushes forward, appear to account for both the ease with which a skater glides and the fact that skating becomes more difficult below about −7°C, when the number of lubricating surface water layers decreases.

The Phase Diagram of Carbon Dioxide

In contrast to the phase diagram of water, the phase diagram of CO2 () has a more typical melting curve, sloping up and to the right. The triple point is −56.6°C and 5.11 atm, which means that liquid CO2 cannot exist at pressures lower than 5.11 atm. At 1 atm, therefore, solid CO2 sublimes directly to the vapor while maintaining a temperature of −78.5°C, the normal sublimation temperature. Solid CO2 is generally known as dry ice because it is a cold solid with no liquid phase observed when it is warmed. Also notice the critical point at 30.98°C and 72.79 atm. In addition to the uses discussed in , supercritical carbon dioxide is emerging as a natural refrigerant, making it a low carbon (and thus a more environmentally friendly) solution for domestic heat pumps.

Figure 11.24 The Phase Diagram of Carbon Dioxide

Note the critical point, the triple point, and the normal sublimation temperature in this diagram.

Example 10

Referring to the phase diagram of water in ,

  1. predict the physical form of a sample of water at 400°C and 150 atm.
  2. describe the changes that occur as the sample in part (a) is slowly allowed to cool to −50°C at a constant pressure of 150 atm.

Given: phase diagram, temperature, and pressure

Asked for: physical form and physical changes

Strategy:

A Identify the region of the phase diagram corresponding to the initial conditions and identify the phase that exists in this region.

B Draw a line corresponding to the given pressure. Move along that line in the appropriate direction (in this case cooling) and describe the phase changes.

Solution:

  1. A Locate the starting point on the phase diagram in part (a) in . The initial conditions correspond to point A, which lies in the region of the phase diagram representing water vapor. Thus water at T = 400°C and P = 150 atm is a gas.
  2. B Cooling the sample at constant pressure corresponds to moving left along the horizontal line in part (a) in . At about 340°C (point B), we cross the vapor pressure curve, at which point water vapor will begin to condense and the sample will consist of a mixture of vapor and liquid. When all of the vapor has condensed, the temperature drops further, and we enter the region corresponding to liquid water (indicated by point C). Further cooling brings us to the melting curve, the line that separates the liquid and solid phases at a little below 0°C (point D), at which point the sample will consist of a mixture of liquid and solid water (ice). When all of the water has frozen, cooling the sample to −50°C takes us along the horizontal line to point E, which lies within the region corresponding to solid water. At P = 150 atm and T = −50°C, therefore, the sample is solid ice.

Exercise

Referring to the phase diagram of water in , predict the physical form of a sample of water at −0.0050°C as the pressure is gradually increased from 1.0 mmHg to 218 atm.

Answer: The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid.

Summary

The states of matter exhibited by a substance under different temperatures and pressures can be summarized graphically in a phase diagram, which is a plot of pressure versus temperature. Phase diagrams contain discrete regions corresponding to the solid, liquid, and gas phases. The solid and liquid regions are separated by the melting curve of the substance, and the liquid and gas regions are separated by its vapor pressure curve, which ends at the critical point. Within a given region, only a single phase is stable, but along the lines that separate the regions, two phases are in equilibrium at a given temperature and pressure. The lines separating the three phases intersect at a single point, the triple point, which is the only combination of temperature and pressure at which all three phases can coexist in equilibrium. Water has an unusual phase diagram: its melting point decreases with increasing pressure because ice is less dense than liquid water. The phase diagram of carbon dioxide shows that liquid carbon dioxide cannot exist at atmospheric pressure. Consequently, solid carbon dioxide sublimes directly to a gas.

Key Takeaway

  • A phase diagram is a graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system. It shows the triple point, the critical point, and four regions: solid, liquid, gas, and a supercritical region.

Conceptual Problems

  1. A phase diagram is a graphic representation of the stable phase of a substance at any combination of temperature and pressure. What do the lines separating different regions in a phase diagram indicate? What information does the slope of a line in a phase diagram convey about the physical properties of the phases it separates? Can a phase diagram have more than one point where three lines intersect?

  2. If the slope of the line corresponding to the solid/liquid boundary in the phase diagram of water were positive rather than negative, what would be the effect on aquatic life during periods of subzero temperatures? Explain your answer.

Answer

  1. The lines in a phase diagram represent boundaries between different phases; at any combination of temperature and pressure that lies on a line, two phases are in equilibrium. It is physically impossible for more than three phases to coexist at any combination of temperature and pressure, but in principle there can be more than one triple point in a phase diagram. The slope of the line separating two phases depends upon their relative densities. For example, if the solid–liquid line slopes up and to the right, the liquid is less dense than the solid, while if it slopes up and to the left, the liquid is denser than the solid.

Numerical Problems

  1. Naphthalene (C10H8) is the key ingredient in mothballs. It has normal melting and boiling points of 81°C and 218°C, respectively. The triple point of naphthalene is 80°C at 1000 Pa. Use these data to construct a phase diagram for naphthalene and label all the regions of your diagram.

  2. Argon is an inert gas used in welding. It has normal boiling and freezing points of 87.3 K and 83.8 K, respectively. The triple point of argon is 83.8 K at 0.68 atm. Use these data to construct a phase diagram for argon and label all the regions of your diagram.

11.8 Liquid Crystals

Learning Objective

  1. To describe the properties of liquid crystals.

When cooled, most liquids undergo a simple phase transitionAnother name for a phase change. to an ordered crystalline solid, a relatively rigid substance that has a fixed shape and volume. (For more information on the characteristics of matter, see , .) In the phase diagrams for these liquids, there are no regions between the liquid and solid phases. Thousands of substances are known, however, that exhibit one or more phases intermediate between the liquid state, in which the molecules are free to tumble and move past one another, and the solid state, in which the molecules or ions are rigidly locked into place. In these intermediate phases, the molecules have an ordered arrangement and yet can still flow like a liquid. Hence they are called liquid crystalsA substance that exhibits phases that have properties intermediate between those of a crystalline solid and a normal liquid and possess long-range molecular order but still flow., and their unusual properties have found a wide range of commercial applications. They are used, for example, in the liquid crystal displays (LCDs) in digital watches, calculators, and computer and video displays.

The first documented example of a liquid crystal was reported by the Austrian Frederick Reinitzer in 1888. Reinitzer was studying the properties of a cholesterol derivative, cholesteryl benzoate, and noticed that it behaved strangely as it melted. The white solid first formed a cloudy white liquid phase at 145°C, which reproducibly transformed into a clear liquid at 179°C (). The transitions were completely reversible: cooling molten cholesteryl benzoate below 179°C caused the clear liquid to revert to a milky one, which then crystallized at the melting point of 145°C.

Figure 11.25 Cholesteryl Benzoate

(a) When the temperature is greater than 179°C, the substance is an isotropic liquid through which images can be seen. (b) When the temperature is between 145°C and 179°C, the substance is in the cholesteric liquid crystalline phase and is an opaque, milky liquid.

In a normal liquid, the molecules possess enough thermal energy to overcome the intermolecular attractive forces and tumble freely. This arrangement of the molecules is described as isotropicThe arrangement of molecules that is equally disordered in all directions., which means that it is equally disordered in all directions. Liquid crystals, in contrast, are anisotropicAn arrangement of molecules in which their properties depend on the direction they are measured.: their properties depend on the direction in which they are viewed. Hence liquid crystals are not as disordered as a liquid because the molecules have some degree of alignment.

Most substances that exhibit the properties of liquid crystals consist of long, rigid rod- or disk-shaped molecules that are easily polarizable and can orient themselves in one of three different ways, as shown in . In the nematic phaseOne of three different ways that most liquid crystals can orient themselves. Only the long axes of the molecules are aligned, so they are free to rotate or to slide past one another., the molecules are not layered but are pointed in the same direction. As a result, the molecules are free to rotate or slide past one another. In the smectic phaseOne of three different ways that most liquid crystals can orient themselves. The long axes of the molecules are aligned (similar to the nematic phase), but the molecules are arranged in planes, too., the molecules maintain the general order of the nematic phase but are also aligned in layers. Several variants of the smectic phase are known, depending on the angle formed between the molecular axes and the planes of molecules. The simplest such structure is the so-called smectic A phase, in which the molecules can rotate about their long axes within a given plane, but they cannot readily slide past one another. In the cholesteric phaseOne of three different ways that most liquid crystals can orient themselves. The molecules are arranged in planes (similar to the smectic phase), but each layer is rotated by a certain amount with respect to those above and below it, giving it a helical structure., the molecules are directionally oriented and stacked in a helical pattern, with each layer rotated at a slight angle to the ones above and below it. As the degree of molecular ordering increases from the nematic phase to the cholesteric phase, the liquid becomes more opaque, although direct comparisons are somewhat difficult because most compounds form only one of these liquid crystal phases when the solid is melted or the liquid is cooled.

Figure 11.26 The Arrangement of Molecules in the Nematic, Smectic, and Cholesteric Liquid Crystal Phases

In the nematic phase, only the long axes of the molecules are parallel, and the ends are staggered at random intervals. In the smectic phase, the long axes of the molecules are parallel, and the molecules are also arranged in planes. Finally, in the cholesteric phase, the molecules are arranged in layers; each layer is rotated with respect to the ones above and below it to give a spiral structure. The molecular order increases from the nematic phase to the smectic phase to the cholesteric phase, and the phases become increasingly opaque.

Molecules that form liquid crystals tend to be rigid molecules with polar groups that exhibit relatively strong dipole–dipole or dipole–induced dipole interactions, hydrogen bonds, or some combination of both. Some examples of substances that form liquid crystals are listed in along with their characteristic phase transition temperature ranges. In most cases, the intermolecular interactions are due to the presence of polar or polarizable groups. Aromatic rings and multiple bonds between carbon and nitrogen or oxygen are especially common. Moreover, many liquid crystals are composed of molecules with two similar halves connected by a unit having a multiple bond.

Figure 11.27 Structures of Typical Molecules That Form Liquid Crystals*

Because of their anisotropic structures, liquid crystals exhibit unusual optical and electrical properties. The intermolecular forces are rather weak and can be perturbed by an applied electric field. Because the molecules are polar, they interact with an electric field, which causes them to change their orientation slightly. Nematic liquid crystals, for example, tend to be relatively translucent, but many of them become opaque when an electric field is applied and the molecular orientation changes. This behavior is ideal for producing dark images on a light or an opalescent background, and it is used in the LCDs in digital watches; handheld calculators; flat-screen monitors; and car, ship, and aircraft instrumentation. Although each application differs in the details of its construction and operation, the basic principles are similar, as illustrated in .

Note the Pattern

Liquid crystals tend to form from long, rigid molecules with polar groups.

Figure 11.28 Schematic Drawing of an LCD Device, Showing the Various Layers

Applying a voltage to selected segments of the device will produce any of the numbers. The device is a sandwich that contains several very thin layers, consisting of (from top to bottom) a sheet of polarizer to produce polarized light, a transparent electrode, a thin layer of a liquid crystalline substance, a second transparent electrode, a second polarizer, and a screen. Applying an electrical voltage to the liquid crystal changes its orientation slightly, which rotates the plane of the polarized light and makes the area appear dark.

Changes in molecular orientation that are dependent on temperature result in an alteration of the wavelength of reflected light. Changes in reflected light produce a change in color, which can be customized by using either a single type of liquid crystalline material or mixtures. It is therefore possible to build a liquid crystal thermometer that indicates temperature by color () and to use liquid crystals in heat-sensitive films to detect flaws in electronic board connections where overheating can occur.

Figure 11.29 An Inexpensive Fever Thermometer That Uses Liquid Crystals

Each section contains a liquid crystal sample with a different liquid crystalline range. The section whose liquid crystalline range corresponds to the temperature of the body becomes translucent (here shown in green), indicating the temperature.

We also see the effect of liquid crystals in nature. Iridescent green beetles, known as jewel beetles, change color because of the light-reflecting properties of the cells that make up their external skeletons, not because of light absorption from their pigment. The cells form helices with a structure like those found in cholesteric liquid crystals. When the pitch of the helix is close to the wavelength of visible light, the cells reflect light with wavelengths that lead to brilliant metallic colors. Because a color change occurs depending on a person’s angle of view, researchers in New Zealand are studying the beetles to develop a thin material that can be used as a currency security measure. The automobile industry is also interested in exploring such materials for use in paints that would change color at different viewing angles.

With only molecular structure as a guide, one cannot precisely predict which of the various liquid crystalline phases a given compound will actually form. One can, however, identify molecules containing the kinds of structural features that tend to result in liquid crystalline behavior, as demonstrated in Example 11.

Example 11

Which molecule is most likely to form a liquid crystalline phase as the isotropic liquid is cooled?

  1. isooctane (2,2,4-trimethylpentane)
  2. ammonium thiocyanate [NH4(SCN)]
  3. p-azoxyanisole

  4. sodium decanoate {Na[CH3(CH2)8CO2]}

Given: compounds

Asked for: liquid crystalline behavior

Strategy:

Determine which compounds have a rigid structure and contain polar groups. Those that do are likely to exhibit liquid crystal behavior.

Solution:

  1. Isooctane is not long and rigid and contains no polar groups, so it is unlikely to exhibit liquid crystal behavior.
  2. Ammonium thiocyanate is ionic, and ionic compounds tend to have high melting points, so it should not form a liquid crystalline phase. In fact, ionic compounds that form liquid crystals are very rare indeed.
  3. p-Azoxyanisole combines two planar phenyl rings linked through a multiply bonded unit, and it contains polar groups. The combination of a long, rigid shape and polar groups makes it a reasonable candidate for a liquid crystal.
  4. Sodium decanoate is the sodium salt of a straight-chain carboxylic acid. The n-alkyl chain is long, but it is flexible rather than rigid, so the compound is probably not a liquid crystal.

Exercise

Which compound is least likely to form a liquid crystal phase?

Answer: (b) Biphenyl; although it is rather long and rigid, it lacks any polar substituents.

Summary

Many substances exhibit phases that have properties intermediate between those of a crystalline solid and a normal liquid. These substances, which possess long-range molecular order but still flow like liquids, are called liquid crystals. Liquid crystals are typically long, rigid molecules that can interact strongly with one another; they do not have isotropic structures, which are completely disordered, but rather have anisotropic structures, which exhibit different properties when viewed from different directions. In the nematic phase, only the long axes of the molecules are aligned, whereas in the smectic phase, the long axes of the molecules are parallel and the molecules are arranged in planes. In the cholesteric phase, the molecules are arranged in planes, but each layer is rotated by a certain amount with respect to those above and below it, giving a helical structure.

Key Takeaway

  • Liquid crystals tend to consist of rigid molecules with polar groups, and their anisotropic structures exhibit unusual optical and electrical properties.

Conceptual Problems

  1. Describe the common structural features of molecules that form liquid crystals. What kind of intermolecular interactions are most likely to result in a long-chain molecule that exhibits liquid crystalline behavior? Does an electrical field affect these interactions?

  2. What is the difference between an isotropic liquid and an anisotropic liquid? Which is more anisotropic—a cholesteric liquid crystal or a nematic liquid crystal?

11.9 Essential Skills 6

Topics

  • Natural Logarithms
  • Calculations Using Natural Logarithms

Essential Skills 3 in , , introduced the common, or base-10, logarithms and showed how to use the properties of exponents to perform logarithmic calculations. In this section, we describe natural logarithms, their relationship to common logarithms, and how to do calculations with them using the same properties of exponents.

Natural Logarithms

Many natural phenomena exhibit an exponential rate of increase or decrease. Population growth is an example of an exponential rate of increase, whereas a runner’s performance may show an exponential decline if initial improvements are substantially greater than those that occur at later stages of training. Exponential changes are represented logarithmically by ex, where e is an irrational number whose value is approximately 2.7183. The natural logarithm, abbreviated as ln, is the power x to which e must be raised to obtain a particular number. The natural logarithm of e is 1 (ln e = 1).

Some important relationships between base-10 logarithms and natural logarithms are as follows:

101 = 10 = e2.303 ln ex = x ln 10 = ln(e2.303) = 2.303 log 10 = ln e = 1

According to these relationships, ln 10 = 2.303 and log 10 = 1. Because multiplying by 1 does not change an equality,

ln 10 = 2.303 log 10

Substituting any value y for 10 gives

ln y = 2.303 log y

Other important relationships are as follows:

log Ax = x log A ln ex = x ln e = x = elnx

Entering a value x, such as 3.86, into your calculator and pressing the “ln” key gives the value of ln x, which is 1.35 for x = 3.86. Conversely, entering the value 1.35 and pressing “ex” key gives an answer of 3.86.On some calculators, pressing [INV] and then [ln x] is equivalent to pressing [ex]. Hence

eln3.86 = e1.35 = 3.86 ln(e3.86) = 3.86

Skill Builder ES1

Calculate the natural logarithm of each number and express each as a power of the base e.

  1. 0.523
  2. 1.63

Solution:

  1. ln(0.523) = −0.648; e−0.648 = 0.523
  2. ln(1.63) = 0.489; e0.489 = 1.63

Skill Builder ES2

What number is each value the natural logarithm of?

  1. 2.87
  2. 0.030
  3. −1.39

Solution:

  1. ln x = 2.87; x = e2.87 = 17.6 = 18 to two significant figures
  2. ln x = 0.030; x = e0.030 = 1.03 = 1.0 to two significant figures
  3. ln x = −1.39; x = e−1.39 = 0.249 = 0.25

Calculations with Natural Logarithms

Like common logarithms, natural logarithms use the properties of exponents. We can expand in Essential Skills 3 to include natural logarithms:

Operation Exponential Form Logarithm
Multiplication (10a)(10b) = 10a+b log(ab) = log a + log b
(ex)(ey) = ex+y ln(exey) = ln(ex) + ln(ey) = x + y
Division 10a10b=10a  bexey=ex  y log(ab)=log alog bln (xy)=ln xln yln(exey)=ln(ex)ln(ey)=xy
Inverse log(1a)=logaln(1x)=ln x

The number of significant figures in a number is the same as the number of digits after the decimal point in its logarithm. For example, the natural logarithm of 18.45 is 2.9151, which means that e2.9151 is equal to 18.45.

Skill Builder ES3

Calculate the natural logarithm of each number.

  1. 22 × 18.6
  2. 0.512.67
  3. 0.079 × 1.485
  4. 20.50.026

Solution:

  1. ln(22 × 18.6) = ln(22) + ln(18.6) = 3.09 + 2.923 = 6.01. Alternatively, 22 × 18.6 = 410; ln(410) = 6.02.
  2. ln(0.512.67)=ln(0.51)ln(2.67)=0.670.982=1.65. Alternatively, 0.512.67=0.19; ln(0.19) = −1.66.
  3. ln(0.079 × 1.485) = ln(0.079) + ln(1.485) = −2.54 + 0.395 = −2.15. Alternatively, 0.079 × 1.485 = 0.12; ln(0.12) = −2.12.
  4. ln(20.50.026)=ln(20.5)ln(0.026)=3.0204(3.65)=6.67. Alternatively, 20.50.026=790; ln(790) = 6.67.

The answers obtained using the two methods may differ slightly due to rounding errors.

Skill Builder ES4

Calculate the natural logarithm of each number.

  1. 34 × 16.5
  2. 2.100.052
  3. 0.402 × 3.930
  4. 0.16410.7

Solution:

  1. 6.33
  2. 3.70
  3. 0.457
  4. −4.178

11.10 End-of-Chapter Material

Application Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 6 () before proceeding to the Application Problems. Problems marked with a ♦ involve multiple concepts.

  1. During cold periods, workers in the citrus industry often spray water on orange trees to prevent them from being damaged, even though ice forms on the fruit.

    1. Explain the scientific basis for this practice.
    2. To illustrate why the production of ice prevents damage to the fruit during cold weather, calculate the heat released by formation of ice from 1000 L of water at 10°C.
  2. ♦ Relative humidity is the ratio of the actual partial pressure of water in the air to the vapor pressure of water at that temperature (i.e., if the air was completely saturated with water vapor), multiplied by 100 to give a percentage. On a summer day in the Chesapeake, when the temperature was recorded as 35°C, the partial pressure of water was reported to be 33.9 mmHg.

    1. The following table gives the vapor pressure of water at various temperatures. Calculate the relative humidity.

      T (°C) 0 10 30 50 60 80 100
      P (mmHg) 4.6 9.2 31.8 92.6 150 355 760
    2. Why does it seem “drier” in the winter, even though the relative humidity may be the same as in the summer?
  3. ♦ Liquids are frequently classified according to their physical properties, such as surface tension, vapor pressure, and boiling point. Such classifications are useful when substitutes are needed for a liquid that might not be available.

    1. Draw the structure of methanol, benzene, pentane, toluene, cyclohexane, 1-butanol, trichloroethylene, acetic acid, acetone, and chloroform.
    2. Identify the most important kind of intermolecular interaction in each.
    3. Sort the compounds into three groups with similar characteristics.
    4. If you needed a substitute for trimethylpentane, from which group would you make your selection?
  4. ♦ In the process of freeze drying, which is used as a preservation method and to aid in the shipping or storage of fruit and biological samples, a sample is cooled and then placed in a compartment in which a very low pressure is maintained, ≈0.01 atm.

    1. Explain how this process removes water and “dries” the sample.
    2. Identify the phase change that occurs during this process.
    3. Using the Clausius–Clapeyron equation, show why it is possible to remove water and still maintain a low temperature at this pressure.
  5. ♦ Many industrial processes for preparing compounds use “continuous-flow reactors,” which are chemical reaction vessels in which the reactants are mixed and allowed to react as they flow along a tube. The products are removed at a certain distance from the starting point, when the reaction is nearly complete. The key operating parameters in a continuous-flow reactor are temperature, reactor volume, and reactant flow rate. As an industrial chemist, you think you have successfully modified a particular process to produce a higher product yield by substituting one reactant for another. The viscosity of the new reactant is, however, greater than that of the initial reactant.

    1. Which of the operating parameters will be most greatly affected by this change?
    2. What other parameter could be changed to compensate for the substitution?
    3. Predict the possible effects on your reactor and your process if you do not compensate for the substitution.

Chapter 12 Solids

In this chapter, we turn our attention to the structures and properties of solids. The solid state is distinguished from the gas and liquid states by a rigid structure in which the component atoms, ions, or molecules are usually locked into place. In many solids, the components are arranged in extended three-dimensional patterns, producing a wide range of properties that can often be tailored to specific functions. Thus diamond, an allotrope of elemental carbon, is one of the hardest materials known, yet graphite, another allotrope of carbon, is a soft, slippery material used in pencil lead and as a lubricant. Metallic sodium is soft enough to be cut with a dull knife, but crystalline sodium chloride turns into a fine powder when struck with a hammer.

Regular, repeating units in a pattern. This drawing by M. C. Escher shows two possible choices for a repeating unit. Repeating units are typical of crystalline solids.

Solids, also called materials, are so important in today’s technology that the subdisciplines of solid-state chemistry and materials science are among the most active and exciting areas of modern chemical research. After presenting a basic survey of the structures of solids, we will examine how the properties of solids are determined by their composition and structure. We will also explore the principles underlying the electrical properties of metals, insulators, semiconductors (which are at the heart of the modern electronics industry), and superconductors. By the end of the chapter, you will know why some metals “remember” their shape after being bent and why ceramics are used in jet engines. You will also understand why carbon- or boron-fiber materials are used in high-performance golf clubs and tennis rackets, why nylon is used to make parachutes, and how solid electrolytes improve the performance of high-capacity batteries.

12.1 Crystalline and Amorphous Solids

Learning Objective

  1. To know the characteristic properties of crystalline and amorphous solids.

With few exceptions, the particles that compose a solid material, whether ionic, molecular, covalent, or metallic, are held in place by strong attractive forces between them. When we discuss solids, therefore, we consider the positions of the atoms, molecules, or ions, which are essentially fixed in space, rather than their motions (which are more important in liquids and gases). The constituents of a solid can be arranged in two general ways: they can form a regular repeating three-dimensional structure called a crystal latticeA regular repeating three-dimensional structure., thus producing a crystalline solidA solid with a regular repeating three-dimensional structure., or they can aggregate with no particular order, in which case they form an amorphous solidA solid with no particular structural order. (from the Greek ámorphos, meaning “shapeless”).

Crystalline solids, or crystals, have distinctive internal structures that in turn lead to distinctive flat surfaces, or faces. The faces intersect at angles that are characteristic of the substance. When exposed to x-rays, each structure also produces a distinctive pattern that can be used to identify the material (see ). The characteristic angles do not depend on the size of the crystal; they reflect the regular repeating arrangement of the component atoms, molecules, or ions in space. When an ionic crystal is cleaved (), for example, repulsive interactions cause it to break along fixed planes to produce new faces that intersect at the same angles as those in the original crystal. In a covalent solid such as a cut diamond, the angles at which the faces meet are also not arbitrary but are determined by the arrangement of the carbon atoms in the crystal.

Crystalline faces. The faces of crystals can intersect at right angles, as in galena (PbS) and pyrite (FeS2), or at other angles, as in quartz.

Cleavage surfaces of an amorphous solid. Obsidian, a volcanic glass with the same chemical composition as granite (typically KAlSi3O8), tends to have curved, irregular surfaces when cleaved.

Figure 12.1 Cleaving a Crystal of an Ionic Compound along a Plane of Ions

Deformation of the ionic crystal causes one plane of atoms to slide along another. The resulting repulsive interactions between ions with like charges cause the layers to separate.

Crystals tend to have relatively sharp, well-defined melting points because all the component atoms, molecules, or ions are the same distance from the same number and type of neighbors; that is, the regularity of the crystalline lattice creates local environments that are the same. Thus the intermolecular forces holding the solid together are uniform, and the same amount of thermal energy is needed to break every interaction simultaneously.

Amorphous solids have two characteristic properties. When cleaved or broken, they produce fragments with irregular, often curved surfaces; and they have poorly defined patterns when exposed to x-rays because their components are not arranged in a regular array. An amorphous, translucent solid is called a glassAn amorphous, translucent solid. A glass is a solid that has been cooled too quickly to form ordered crystals.. Almost any substance can solidify in amorphous form if the liquid phase is cooled rapidly enough. Some solids, however, are intrinsically amorphous, because either their components cannot fit together well enough to form a stable crystalline lattice or they contain impurities that disrupt the lattice. For example, although the chemical composition and the basic structural units of a quartz crystal and quartz glass are the same—both are SiO2 and both consist of linked SiO4 tetrahedra—the arrangements of the atoms in space are not. Crystalline quartz contains a highly ordered arrangement of silicon and oxygen atoms, but in quartz glass the atoms are arranged almost randomly. When molten SiO2 is cooled rapidly (4 K/min), it forms quartz glass, whereas the large, perfect quartz crystals sold in mineral shops have had cooling times of thousands of years. In contrast, aluminum crystallizes much more rapidly. Amorphous aluminum forms only when the liquid is cooled at the extraordinary rate of 4 × 1013 K/s, which prevents the atoms from arranging themselves into a regular array.

The lattice of crystalline quartz (SiO2). The atoms form a regular arrangement in a structure that consists of linked tetrahedra.

In an amorphous solid, the local environment, including both the distances to neighboring units and the numbers of neighbors, varies throughout the material. Different amounts of thermal energy are needed to overcome these different interactions. Consequently, amorphous solids tend to soften slowly over a wide temperature range rather than having a well-defined melting point like a crystalline solid. If an amorphous solid is maintained at a temperature just below its melting point for long periods of time, the component molecules, atoms, or ions can gradually rearrange into a more highly ordered crystalline form.

Note the Pattern

Crystals have sharp, well-defined melting points; amorphous solids do not.

Summary

Solids are characterized by an extended three-dimensional arrangement of atoms, ions, or molecules in which the components are generally locked into their positions. The components can be arranged in a regular repeating three-dimensional array (a crystal lattice), which results in a crystalline solid, or more or less randomly to produce an amorphous solid. Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points. In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures.

Key Takeaway

  • Crystalline solids have regular ordered arrays of components held together by uniform intermolecular forces, whereas the components of amorphous solids are not arranged in regular arrays.

Conceptual Problems

  1. Compare the solid and liquid states in terms of

    1. rigidity of structure.
    2. long-range order.
    3. short-range order.
  2. How do amorphous solids differ from crystalline solids in each characteristic? Which of the two types of solid is most similar to a liquid?

    1. rigidity of structure
    2. long-range order
    3. short-range order
  3. Why is the arrangement of the constituent atoms or molecules more important in determining the properties of a solid than a liquid or a gas?

  4. Why are the structures of solids usually described in terms of the positions of the constituent atoms rather than their motion?

  5. What physical characteristics distinguish a crystalline solid from an amorphous solid? Describe at least two ways to determine experimentally whether a material is crystalline or amorphous.

  6. Explain why each characteristic would or would not favor the formation of an amorphous solid.

    1. slow cooling of pure molten material
    2. impurities in the liquid from which the solid is formed
    3. weak intermolecular attractive forces
  7. A student obtained a solid product in a laboratory synthesis. To verify the identity of the solid, she measured its melting point and found that the material melted over a 12°C range. After it had cooled, she measured the melting point of the same sample again and found that this time the solid had a sharp melting point at the temperature that is characteristic of the desired product. Why were the two melting points different? What was responsible for the change in the melting point?

Answers

  1. The arrangement of the atoms or molecules is more important in determining the properties of a solid because of the greater persistent long-range order of solids. Gases and liquids cannot readily be described by the spatial arrangement of their components because rapid molecular motion and rearrangement defines many of the properties of liquids and gases.

  2. The initial solid contained the desired compound in an amorphous state, as indicated by the wide temperature range over which melting occurred. Slow cooling of the liquid caused it to crystallize, as evidenced by the sharp second melting point observed at the expected temperature.

12.2 The Arrangement of Atoms in Crystalline Solids

Learning Objectives

  1. To recognize the unit cell of a crystalline solid.
  2. To calculate the density of a solid given its unit cell.

Because a crystalline solid consists of repeating patterns of its components in three dimensions (a crystal lattice), we can represent the entire crystal by drawing the structure of the smallest identical units that, when stacked together, form the crystal. This basic repeating unit is called a unit cellThe smallest repeating unit of a crystal lattice.. For example, the unit cell of a sheet of identical postage stamps is a single stamp, and the unit cell of a stack of bricks is a single brick. In this section, we describe the arrangements of atoms in various unit cells.

Unit cells are easiest to visualize in two dimensions. In many cases, more than one unit cell can be used to represent a given structure, as shown for the Escher drawing in the chapter opener and for a two-dimensional crystal lattice in . Usually the smallest unit cell that completely describes the order is chosen. The only requirement for a valid unit cell is that repeating it in space must produce the regular lattice. Thus the unit cell in part (d) in is not a valid choice because repeating it in space does not produce the desired lattice (there are triangular holes). The concept of unit cells is extended to a three-dimensional lattice in the schematic drawing in .

Figure 12.2 Unit Cells in Two Dimensions

(a–c) Three two-dimensional lattices illustrate the possible choices of the unit cell. The unit cells differ in their relative locations or orientations within the lattice, but they are all valid choices because repeating them in any direction fills the overall pattern of dots. (d) The triangle is not a valid unit cell because repeating it in space fills only half of the space in the pattern.

Figure 12.3 Unit Cells in Three Dimensions

These images show (a) a three-dimensional unit cell and (b) the resulting regular three-dimensional lattice.

The Unit Cell

There are seven fundamentally different kinds of unit cells, which differ in the relative lengths of the edges and the angles between them (). Each unit cell has six sides, and each side is a parallelogram. We focus primarily on the cubic unit cells, in which all sides have the same length and all angles are 90°, but the concepts that we introduce also apply to substances whose unit cells are not cubic.

Figure 12.4 The General Features of the Seven Basic Unit Cells

The lengths of the edges of the unit cells are indicated by a, b, and c, and the angles are defined as follows: α, the angle between b and c; β, the angle between a and c; and γ, the angle between a and b.

If the cubic unit cell consists of eight component atoms, molecules, or ions located at the corners of the cube, then it is called simple cubicA cubic unit cell that consists of eight component atoms, molecules, or ions located at the corners of a cube. (part (a) in ). If the unit cell also contains an identical component in the center of the cube, then it is body-centered cubic (bcc)A cubic unit cell with eight component atoms, molecules, or ions located at the corners of a cube plus an identical component in the center of the cube. (part (b) in ). If there are components in the center of each face in addition to those at the corners of the cube, then the unit cell is face-centered cubic (fcc)A cubic unit cell with eight component atoms, molecules, or ions located at the corners of a cube plus an identical component in the center of each face of the cube. (part (c) in ).

Figure 12.5 The Three Kinds of Cubic Unit Cell

For the three kinds of cubic unit cells, simple cubic (a), body-centered cubic (b), and face-centered cubic (c), there are three representations for each: a ball-and-stick model, a space-filling cutaway model that shows the portion of each atom that lies within the unit cell, and an aggregate of several unit cells.

As indicated in , a solid consists of a large number of unit cells arrayed in three dimensions. Any intensive property of the bulk material, such as its density, must therefore also be related to its unit cell. Because density is the mass of substance per unit volume, we can calculate the density of the bulk material from the density of a single unit cell. To do this, we need to know the size of the unit cell (to obtain its volume), the molar mass of its components, and the number of components per unit cell. When we count atoms or ions in a unit cell, however, those lying on a face, an edge, or a corner contribute to more than one unit cell, as shown in . For example, an atom that lies on a face of a unit cell is shared by two adjacent unit cells and is therefore counted as 12 atom per unit cell. Similarly, an atom that lies on the edge of a unit cell is shared by four adjacent unit cells, so it contributes 14 atom to each. An atom at a corner of a unit cell is shared by all eight adjacent unit cells and therefore contributes 18 atom to each.The statement that atoms lying on an edge or a corner of a unit cell count as 14 or 18 atom per unit cell, respectively, is true for all unit cells except the hexagonal one, in which three unit cells share each vertical edge and six share each corner (), leading to values of 13 and 16 atom per unit cell, respectively, for atoms in these positions. In contrast, atoms that lie entirely within a unit cell, such as the atom in the center of a body-centered cubic unit cell, belong to only that one unit cell.

Note the Pattern

For all unit cells except hexagonal, atoms on the faces contribute 12 atom to each unit cell, atoms on the edges contribute 14 atom to each unit cell, and atoms on the corners contribute 18 atom to each unit cell.

Example 1

Metallic gold has a face-centered cubic unit cell (part (c) in ). How many Au atoms are in each unit cell?

Given: unit cell

Asked for: number of atoms per unit cell

Strategy:

Using , identify the positions of the Au atoms in a face-centered cubic unit cell and then determine how much each Au atom contributes to the unit cell. Add the contributions of all the Au atoms to obtain the total number of Au atoms in a unit cell.

Solution:

As shown in , a face-centered cubic unit cell has eight atoms at the corners of the cube and six atoms on the faces. Because atoms on a face are shared by two unit cells, each counts as 12 atom per unit cell, giving 6×12=3 Au atoms per unit cell. Atoms on a corner are shared by eight unit cells and hence contribute only 18 atom per unit cell, giving 8×18=1 Au atom per unit cell. The total number of Au atoms in each unit cell is thus 3 + 1 = 4.

Exercise

Metallic iron has a body-centered cubic unit cell (part (b) in ). How many Fe atoms are in each unit cell?

Answer: two

Now that we know how to count atoms in unit cells, we can use unit cells to calculate the densities of simple compounds. Note, however, that we are assuming a solid consists of a perfect regular array of unit cells, whereas real substances contain impurities and defects that affect many of their bulk properties, including density. Consequently, the results of our calculations will be close but not necessarily identical to the experimentally obtained values.

Example 2

Calculate the density of metallic iron, which has a body-centered cubic unit cell (part (b) in ) with an edge length of 286.6 pm.

Given: unit cell and edge length

Asked for: density

Strategy:

A Determine the number of iron atoms per unit cell.

B Calculate the mass of iron atoms in the unit cell from the molar mass and Avogadro’s number. Then divide the mass by the volume of the cell.

Solution:

A We know from Example 1 that each unit cell of metallic iron contains two Fe atoms.

B The molar mass of iron is 55.85 g/mol. Because density is mass per unit volume, we need to calculate the mass of the iron atoms in the unit cell from the molar mass and Avogadro’s number and then divide the mass by the volume of the cell (making sure to use suitable units to get density in g/cm3):

mass of Fe=(2 atoms Fe)(1 mol6.022×1023 atoms)(55.85 gmol)=1.855×10−22 gvolume=[(286.6 pm)(1012 mpm)(102 cmm)]3=2.354×10−23 cm3density=1.855×10−22 g2.354×10−23 cm3=7.880 g/cm3

This result compares well with the tabulated experimental value of 7.874 g/cm3.

Exercise

Calculate the density of gold, which has a face-centered cubic unit cell (part (c) in ) with an edge length of 407.8 pm.

Answer: 19.29 g/cm3

Packing of Spheres

Our discussion of the three-dimensional structures of solids has considered only substances in which all the components are identical. As we shall see, such substances can be viewed as consisting of identical spheres packed together in space; the way the components are packed together produces the different unit cells. Most of the substances with structures of this type are metals.

Simple Cubic Structure

The arrangement of the atoms in a solid that has a simple cubic unit cell was shown in part (a) in . Each atom in the lattice has only six nearest neighbors in an octahedral arrangement. Consequently, the simple cubic lattice is an inefficient way to pack atoms together in space: only 52% of the total space is filled by the atoms. The only element that crystallizes in a simple cubic unit cell is polonium. Simple cubic unit cells are, however, common among binary ionic compounds, where each cation is surrounded by six anions and vice versa.

The arrangement of atoms in a simple cubic unit cell. Each atom in the lattice has six nearest neighbors in an octahedral arrangement.

Body-Centered Cubic Structure

The body-centered cubic unit cell is a more efficient way to pack spheres together and is much more common among pure elements. Each atom has eight nearest neighbors in the unit cell, and 68% of the volume is occupied by the atoms. As shown in part (b) in , the body-centered cubic structure consists of a single layer of spheres in contact with each other and aligned so that their centers are at the corners of a square; a second layer of spheres occupies the square-shaped “holes” above the spheres in the first layer. The third layer of spheres occupies the square holes formed by the second layer, so that each lies directly above a sphere in the first layer, and so forth. All the alkali metals, barium, radium, and several of the transition metals have body-centered cubic structures.

Hexagonal Close-Packed and Cubic Close-Packed Structures

The most efficient way to pack spheres is the close-packed arrangement, which has two variants. A single layer of close-packed spheres is shown in part (a) in . Each sphere is surrounded by six others in the same plane to produce a hexagonal arrangement. Above any set of seven spheres are six depressions arranged in a hexagon. In principle, all six sites are the same, and any one of them could be occupied by an atom in the next layer. Actually, however, these six sites can be divided into two sets, labeled B and C in part (a) in . Sites B and C differ because as soon as we place a sphere at a B position, we can no longer place a sphere in any of the three C positions adjacent to A and vice versa.

Figure 12.6 Close-Packed Layers of Spheres

(a) In this single layer of close-packed spheres, each sphere is surrounded by six others in a hexagonal arrangement. (b) Placing an atom at a B position prohibits placing an atom at any of the adjacent C positions and results in all the atoms in the second layer occupying the B positions. (c) Placing the atoms in the third layer over the atoms at A positions in the first layer gives the hexagonal close-packed structure. Placing the third-layer atoms over the C positions gives the cubic close-packed structure.

If we place the second layer of spheres at the B positions in part (a) in , we obtain the two-layered structure shown in part (b) in . There are now two alternatives for placing the first atom of the third layer: we can place it directly over one of the atoms in the first layer (an A position) or at one of the C positions, corresponding to the positions that we did not use for the atoms in the first or second layers (part (c) in ). If we choose the first arrangement and repeat the pattern in succeeding layers, the positions of the atoms alternate from layer to layer in the pattern ABABAB…, resulting in a hexagonal close-packed (hcp) structureOne of two variants of the close-packed arrangement—the most efficient way to pack spheres in a lattice—in which the atomic positions alternate from layer to layer in an ABABAB… pattern. (part (a) in ). If we choose the second arrangement and repeat the pattern indefinitely, the positions of the atoms alternate as ABCABC…, giving a cubic close-packed (ccp) structureOne of two variants of the close-packed arrangement—the most efficient way to pack spheres in a lattice—in which the atomic positions alter from layer to layer in an ABCABC… pattern. (part (b) in ). Because the ccp structure contains hexagonally packed layers, it does not look particularly cubic. As shown in part (b) in , however, simply rotating the structure reveals its cubic nature, which is identical to a fcc structure. The hcp and ccp structures differ only in the way their layers are stacked. Both structures have an overall packing efficiency of 74%, and in both each atom has 12 nearest neighbors (6 in the same plane plus 3 in each of the planes immediately above and below).

Figure 12.7 Close-Packed Structures: hcp and ccp

The illustrations in (a) show an exploded view, a side view, and a top view of the hcp structure. The simple hexagonal unit cell is outlined in the side and top views. Note the similarity to the hexagonal unit cell shown in . The ccp structure in (b) is shown in an exploded view, a side view, and a rotated view. The rotated view emphasizes the fcc nature of the unit cell (outlined). The line that connects the atoms in the first and fourth layers of the ccp structure is the body diagonal of the cube.

compares the packing efficiency and the number of nearest neighbors for the different cubic and close-packed structures; the number of nearest neighbors is called the coordination numberThe number of nearest neighbors in a solid structure.. Most metals have hcp, ccp, or bcc structures, although several metals exhibit both hcp and ccp structures, depending on temperature and pressure.

Table 12.1 Properties of the Common Structures of Metals

Structure Percentage of Space Occupied by Atoms Coordination Number
simple cubic 52 6
body-centered cubic 68 8
hexagonal close packed 74 12
cubic close packed (identical to face-centered cubic) 74 12

Summary

The smallest repeating unit of a crystal lattice is the unit cell. The simple cubic unit cell contains only eight atoms, molecules, or ions at the corners of a cube. A body-centered cubic (bcc) unit cell contains one additional component in the center of the cube. A face-centered cubic (fcc) unit cell contains a component in the center of each face in addition to those at the corners of the cube. Simple cubic and bcc arrangements fill only 52% and 68% of the available space with atoms, respectively. The hexagonal close-packed (hcp) structure has an ABABAB… repeating arrangement, and the cubic close-packed (ccp) structure has an ABCABC… repeating pattern; the latter is identical to an fcc lattice. The hcp and ccp arrangements fill 74% of the available space and have a coordination number of 12 for each atom in the lattice, the number of nearest neighbors. The simple cubic and bcc lattices have coordination numbers of 6 and 8, respectively.

Key Takeaway

  • A crystalline solid can be represented by its unit cell, which is the smallest identical unit that when stacked together produces the characteristic three-dimensional structure.

Conceptual Problems

  1. Why is it valid to represent the structure of a crystalline solid by the structure of its unit cell? What are the most important constraints in selecting a unit cell?

  2. All unit cell structures have six sides. Can crystals of a solid have more than six sides? Explain your answer.

  3. Explain how the intensive properties of a material are reflected in the unit cell. Are all the properties of a bulk material the same as those of its unit cell? Explain your answer.

  4. The experimentally measured density of a bulk material is slightly higher than expected based on the structure of the pure material. Propose two explanations for this observation.

  5. The experimentally determined density of a material is lower than expected based on the arrangement of the atoms in the unit cell, the formula mass, and the size of the atoms. What conclusion(s) can you draw about the material?

  6. Only one element (polonium) crystallizes with a simple cubic unit cell. Why is polonium the only example of an element with this structure?

  7. What is meant by the term coordination number in the structure of a solid? How does the coordination number depend on the structure of the metal?

  8. Arrange the three types of cubic unit cells in order of increasing packing efficiency. What is the difference in packing efficiency between the hcp structure and the ccp structure?

  9. The structures of many metals depend on pressure and temperature. Which structure—bcc or hcp—would be more likely in a given metal at very high pressures? Explain your reasoning.

  10. A metal has two crystalline phases. The transition temperature, the temperature at which one phase is converted to the other, is 95°C at 1 atm and 135°C at 1000 atm. Sketch a phase diagram for this substance. The metal is known to have either a ccp structure or a simple cubic structure. Label the regions in your diagram appropriately and justify your selection for the structure of each phase.

Numerical Problems

  1. Metallic rhodium has an fcc unit cell. How many atoms of rhodium does each unit cell contain?

  2. Chromium has a structure with two atoms per unit cell. Is the structure of this metal simple cubic, bcc, fcc, or hcp?

  3. The density of nickel is 8.908 g/cm3. If the metallic radius of nickel is 125 pm, what is the structure of metallic nickel?

  4. The density of tungsten is 19.3 g/cm3. If the metallic radius of tungsten is 139 pm, what is the structure of metallic tungsten?

  5. An element has a density of 10.25 g/cm3 and a metallic radius of 136.3 pm. The metal crystallizes in a bcc lattice. Identify the element.

  6. A 21.64 g sample of a nonreactive metal is placed in a flask containing 12.00 mL of water; the final volume is 13.81 mL. If the length of the edge of the unit cell is 387 pm and the metallic radius is 137 pm, determine the packing arrangement and identify the element.

  7. A sample of an alkali metal that has a bcc unit cell is found to have a mass of 1.000 g and a volume of 1.0298 cm3. When the metal reacts with excess water, the reaction produces 539.29 mL of hydrogen gas at 0.980 atm and 23°C. Identify the metal, determine the unit cell dimensions, and give the approximate size of the atom in picometers.

  8. A sample of an alkaline earth metal that has a bcc unit cell is found to have a mass 5.000 g and a volume of 1.392 cm3. Complete reaction with chlorine gas requires 848.3 mL of chlorine gas at 1.050 atm and 25°C. Identify the metal, determine the unit cell dimensions, and give the approximate size of the atom in picometers.

  9. Lithium crystallizes in a bcc structure with an edge length of 3.509 Å. Calculate its density. What is the approximate metallic radius of lithium in picometers?

  10. Vanadium is used in the manufacture of rust-resistant vanadium steel. It forms bcc crystals with a density of 6.11 g/cm3 at 18.7°C. What is the length of the edge of the unit cell? What is the approximate metallic radius of the vanadium in picometers?

  11. A simple cubic cell contains one metal atom with a metallic radius of 100 pm.

    1. Determine the volume of the atom(s) contained in one unit cell [the volume of a sphere = (43)πr3].
    2. What is the length of one edge of the unit cell? (Hint: there is no empty space between atoms.)
    3. Calculate the volume of the unit cell.
    4. Determine the packing efficiency for this structure.
    5. Use the steps in Problem 11 to calculate the packing efficiency for a bcc unit cell with a metallic radius of 1.00 Å.

Answers

  1. four

  2. fcc

  3. molybdenum

  4. sodium, unit cell edge = 428 pm, r = 185 pm

  5. d = 0.5335 g/cm3, r =151.9 pm

12.3 Structures of Simple Binary Compounds

Learning Objectives

  1. To use the cation:anion radius ratio to predict the structures of simple binary compounds.
  2. To understand how x-rays are diffracted by crystalline solids.

The structures of most binary compounds can be described using the packing schemes we have just discussed for metals. To do so, we generally focus on the arrangement in space of the largest species present. In ionic solids, this generally means the anions, which are usually arranged in a simple cubic, bcc, fcc, or hcp lattice. (For more information about anions, see , .) Often, however, the anion lattices are not truly “close packed”; because the cations are large enough to prop them apart somewhat, the anions are not actually in contact with one another. In ionic compounds, the cations usually occupy the “holes” between the anions, thus balancing the negative charge. The ratio of cations to anions within a unit cell is required to achieve electrical neutrality and corresponds to the bulk stoichiometry of the compound.

Common Structures of Binary Compounds

As shown in part (a) in , a simple cubic lattice of anions contains only one kind of hole, located in the center of the unit cell. Because this hole is equidistant from all eight atoms at the corners of the unit cell, it is called a cubic holeThe hole located at the center of the simple cubic lattice. The hole is equidistant from all eight atoms or ions at the corners of the unit cell. An atom or ion in a cubic hole has a coordination number of 8.. An atom or ion in a cubic hole therefore has a coordination number of 8. Many ionic compounds with relatively large cations and a 1:1 cation:anion ratio have this structure, which is called the cesium chloride structureThe unit cell for many ionic compounds with relatively large cations and a 1:1 cation:anion ratio. () because CsCl is a common example.Solid-state chemists tend to describe the structures of new compounds in terms of the structure of a well-known reference compound. Hence you will often read statements such as “Compound X possesses the cesium chloride (or sodium chloride, etc.) structure” to describe the structure of compound X. Notice in that the z = 0 and the z = 1.0 planes are always the same. This is because the z = 1.0 plane of one unit cell becomes the z = 0 plane of the succeeding one. The unit cell in CsCl contains a single Cs+ ion as well as 8×18Cl=1Cl ion, for an overall stoichiometry of CsCl. The cesium chloride structure is most common for ionic substances with relatively large cations, in which the ratio of the radius of the cation to the radius of the anion is in the range shown in .

Figure 12.8 Holes in Cubic Lattices

The three illustrations show (a) the cubic hole that is in the center of a simple cubic lattice of anions, (b) the locations of the octahedral holes in a face-centered cubic lattice of anions, and (c) the locations of the tetrahedral holes in a face-centered cubic lattice of anions.

Figure 12.9 The Cesium Chloride Structure

The Cs+ ion occupies the cubic hole in the center of a cube of Cl ions. The drawings at the right are horizontal cross-sections through the unit cell at the bottom (z = 0) and halfway between the bottom and top (z = 0.5). A top cross-section (z = 1) is identical to z = 0. Such cross-sections often help us visualize the arrangement of atoms or ions in the unit cell more easily.

Table 12.2 Relationship between the Cation:Anion Radius Ratio and the Site Occupied by the Cations

Approximate Range of Cation:Anion Radius Ratio Hole Occupied by Cation Cation Coordination Number
0.225–0.414 tetrahedral 4
0.414–0.732 octahedral 6
0.732–1.000 cubic 8

Note the Pattern

Very large cations occupy cubic holes, cations of intermediate size occupy octahedral holes, and small cations occupy tetrahedral holes in the anion lattice.

In contrast, a face-centered cubic (fcc) array of atoms or anions contains two types of holes: octahedral holesOne of two kinds of holes in a face-centered cubic array of atoms or ions (the other is a tetrahedral hole). One octahedral hole is located in the center of the face-centered cubic unit cell, and there is a shared one in the middle of each edge. An atom or ion in an octahedral hole has a coordination number of 6., one in the center of the unit cell plus a shared one in the middle of each edge (part (b) in ), and tetrahedral holesOne of two kinds of holes in a face-centered cubic array of atoms or ions (the other is an octahedral hole). Tetrahedral holes are located between an atom at a corner and the three atoms at the centers of the adjacent faces of the face-centered cubic unit cell. An atom or ion in a tetrahedral hole has a coordination number of 4., located between an atom at a corner and the three atoms at the centers of the adjacent faces (part (c) in ). As shown in , the ratio of the radius of the cation to the radius of the anion is the most important determinant of whether cations occupy the cubic holes in a cubic anion lattice or the octahedral or tetrahedral holes in an fcc lattice of anions. Very large cations occupy cubic holes in a cubic anion lattice, cations of intermediate size tend to occupy the octahedral holes in an fcc anion lattice, and relatively small cations tend to occupy the tetrahedral holes in an fcc anion lattice. In general, larger cations have higher coordination numbers than small cations.

The most common structure based on a fcc lattice is the sodium chloride structureThe solid structure that results when the octahedral holes of an fcc lattice of anions are filled with cations. (), which contains an fcc array of Cl ions with Na+ ions in all the octahedral holes. We can understand the sodium chloride structure by recognizing that filling all the octahedral holes in an fcc lattice of Cl ions with Na+ ions gives a total of 4 Cl ions (one on each face gives 6×12=3 plus one on each corner gives 8×18=1, for a total of 4) and 4 Na+ ions (one on each edge gives 12×14=3 plus one in the middle, for a total of 4). The result is an electrically neutral unit cell and a stoichiometry of NaCl. As shown in , the Na+ ions in the sodium chloride structure also form an fcc lattice. The sodium chloride structure is favored for substances with two atoms or ions in a 1:1 ratio and in which the ratio of the radius of the cation to the radius of the anion is between 0.414 and 0.732. It is observed in many compounds, including MgO and TiC.

Figure 12.10 The Sodium Chloride Structure

In NaCl, the Na+ ions occupy the octahedral holes in an fcc lattice of Cl ions, resulting in an fcc array of Na+ ions as well.

The structure shown in is called the zinc blende structureThe solid structure that results when half of the tetrahedral holes in an fcc lattice of anions are filled with cations with a 1:1 cation:anion ratio and a coordination number of 4., from the common name of the mineral ZnS. It results when the cation in a substance with a 1:1 cation:anion ratio is much smaller than the anion (if the cation:anion radius ratio is less than about 0.414). For example, ZnS contains an fcc lattice of S2− ions, and the cation:anion radius ratio is only about 0.40, so we predict that the cation would occupy either a tetrahedral hole or an octahedral hole. In fact, the relatively small Zn2+ cations occupy the tetrahedral holes in the lattice. If all 8 tetrahedral holes in the unit cell were occupied by Zn2+ ions, however, the unit cell would contain 4 S2− and 8 Zn2+ ions, giving a formula of Zn2S and a net charge of +4 per unit cell. Consequently, the Zn2+ ions occupy every other tetrahedral hole, as shown in , giving a total of 4 Zn2+ and 4 S2− ions per unit cell and a formula of ZnS. The zinc blende structure results in a coordination number of 4 for each Zn2+ ion and a tetrahedral arrangement of the four S2− ions around each Zn2+ ion.

Figure 12.11 The Zinc Blende Structure

Zn2+ ions occupy every other tetrahedral hole in the fcc array of S2− ions. Each Zn2+ ion is surrounded by four S2− ions in a tetrahedral arrangement.

Example 3

  1. If all the tetrahedral holes in an fcc lattice of anions are occupied by cations, what is the stoichiometry of the resulting compound?
  2. Use the ionic radii given in to identify a plausible oxygen-containing compound with this stoichiometry and structure.

Given: lattice, occupancy of tetrahedral holes, and ionic radii

Asked for: stoichiometry and identity

Strategy:

A Use to determine the number and location of the tetrahedral holes in an fcc unit cell of anions and place a cation in each.

B Determine the total number of cations and anions in the unit cell; their ratio is the stoichiometry of the compound.

C From the stoichiometry, suggest reasonable charges for the cation and the anion. Use the data in to identify a cation–anion combination that has a cation:anion radius ratio within a reasonable range.

Solution:

  1. shows that the tetrahedral holes in an fcc unit cell of anions are located entirely within the unit cell, for a total of eight (one near each corner). B Because the tetrahedral holes are located entirely within the unit cell, there are eight cations per unit cell. We calculated previously that an fcc unit cell of anions contains a total of four anions per unit cell. The stoichiometry of the compound is therefore M8Y4 or, reduced to the smallest whole numbers, M2Y.
  2. C The M2Y stoichiometry is consistent with a lattice composed of M+ ions and Y2− ions. If the anion is O2− (ionic radius 140 pm), we need a monocation with a radius no larger than about 140 × 0.414 = 58 pm to fit into the tetrahedral holes. According to , none of the monocations has such a small radius; therefore, the most likely possibility is Li+ at 76 pm. Thus we expect Li2O to have a structure that is an fcc array of O2− anions with Li+ cations in all the tetrahedral holes.

Exercise

If only half the octahedral holes in an fcc lattice of anions are filled by cations, what is the stoichiometry of the resulting compound?

Answer: MX2; an example of such a compound is cadmium chloride (CdCl2), in which the empty cation sites form planes running through the crystal.

We examine only one other structure of the many that are known, the perovskite structureA structure that consists of a bcc array of two metal ions, with one set (M) located at the corners of the cube, and the other set (M′) in the centers of the cube.. Perovskite is the generic name for oxides with two different kinds of metal and have the general formula MM′O3, such as CaTiO3. The structure is a body-centered cubic (bcc) array of two metal ions, with one M (Ca in this case) located at the corners of the cube, and the other M′ (in this case Ti) in the centers of the cube. The oxides are in the centers of the square faces (part (a) in ). The stoichiometry predicted from the unit cell shown in part (a) in agrees with the general formula; each unit cell contains 8×18=1 Ca, 1 Ti, and 6×12=3 O atoms. The Ti and Ca atoms have coordination numbers of 6 and 12, respectively. We will return to the perovskite structure when we discuss high-temperature superconductors in .

Figure 12.12 The Perovskite Structure of CaTiO3

Two equivalent views are shown: (a) a view with the Ti atom at the center and (b) an alternative view with the Ca atom at the center.

X-Ray Diffraction

As you learned in , the wavelengths of x-rays are approximately the same magnitude as the distances between atoms in molecules or ions. Consequently, x-rays are a useful tool for obtaining information about the structures of crystalline substances. In a technique called x-ray diffractionAn technique used to obtain information about the structures of crystalline substances by using x-rays., a beam of x-rays is aimed at a sample of a crystalline material, and the x-rays are diffracted by layers of atoms in the crystalline lattice (part (a) in ). When the beam strikes photographic film, it produces an x-ray diffraction pattern, which consists of dark spots on a light background (part (b) in ). In 1912, the German physicist Max von Laue (1879–1960; Nobel Prize in Physics, 1914) predicted that x-rays should be diffracted by crystals, and his prediction was rapidly confirmed. Within a year, two British physicists, William Henry Bragg (1862–1942) and his son, William Lawrence Bragg (1890–1972), had worked out the mathematics that allows x-ray diffraction to be used to measure interatomic distances in crystals. The Braggs shared the Nobel Prize in Physics in 1915, when the son was only 25 years old. Virtually everything we know today about the detailed structures of solids and molecules in solids is due to the x-ray diffraction technique.

Figure 12.13 X-Ray Diffraction

These illustrations show (a) a schematic drawing of x-ray diffraction and (b) the x-ray diffraction pattern of a zinc blende crystalline solid captured on photographic film.

Recall from that two waves that are in phase interfere constructively, thus reinforcing each other and generating a wave with a greater amplitude. In contrast, two waves that are out of phase interfere destructively, effectively canceling each other. When x-rays interact with the components of a crystalline lattice, they are scattered by the electron clouds associated with each atom. As shown in , , and , the atoms in crystalline solids are typically arranged in planes. illustrates how two adjacent planes of atoms can scatter x-rays in a way that results in constructive interference. If two x-rays that are initially in phase are diffracted by two planes of atoms separated by a distance d, the lower beam travels the extra distance indicated by the lines BC and CD. The angle of incidence, designated as θ, is the angle between the x-ray beam and the planes in the crystal. Because BC = CD = d sin θ, the extra distance that the lower beam in must travel compared with the upper beam is 2d sin θ. For these two x-rays to arrive at a detector in phase, the extra distance traveled must be an integral multiple n of the wavelength λ:

Equation 12.1

2d sin θ = nλ

is the Bragg equationThe equation that describes the relationship between two x-ray beams diffracted from different planes of atoms: 2d sin θ=nλ.. The structures of crystalline substances with both small molecules and ions or very large biological molecules, with molecular masses in excess of 100,000 amu, can now be determined accurately and routinely using x-ray diffraction and the Bragg equation. Example 4 illustrates how to use the Bragg equation to calculate the distance between planes of atoms in crystals.

Figure 12.14 The Reflection of X-Rays from Two Adjacent Planes of Atoms Can Result in Constructive Interference of the X-Rays

(a) The x-ray diffracted by the lower layer of atoms must travel a distance that is longer by 2d sin θ than the distance traveled by the x-ray diffracted by the upper layer of atoms. Only if this distance (BC plus CD) equals an integral number of wavelengths of the x-rays (i.e., only if λ = 2d sin θ) will the x-rays arrive at the detector in phase. (b) In a solid, many different sets of planes of atoms can diffract x-rays. Each has a different interplanar distance and therefore diffracts the x-rays at a different angle θ, which produces a characteristic pattern of spots.

Example 4

X-rays from a copper x-ray tube (λ = 1.54062 Å or 154.062 pm)In x-ray diffraction, the angstrom (Å) is generally used as the unit of wavelength. are diffracted at an angle of 10.89° from a sample of crystalline gold. Assuming that n = 1, what is the distance between the planes that gives rise to this reflection? Give your answer in angstroms and picometers to four significant figures.

Given: wavelength, diffraction angle, and number of wavelengths

Asked for: distance between planes

Strategy:

Substitute the given values into the Bragg equation and solve to obtain the distance between planes.

Solution:

We are given n, θ, and λ and asked to solve for d, so this is a straightforward application of the Bragg equation. For an answer in angstroms, we do not even have to convert units. Solving the Bragg equation for d gives

d=nλ2 sin θ

and substituting values gives

d=(1)(1.54062  Å)2 sin 10.89°=4.077  Å=407.7  pm

This value corresponds to the edge length of the fcc unit cell of elemental gold.

Exercise

X-rays from a molybdenum x-ray tube (λ = 0.709300 Å) are diffracted at an angle of 7.11° from a sample of metallic iron. Assuming that n = 1, what is the distance between the planes that gives rise to this reflection? Give your answer in angstroms and picometers to three significant figures.

Answer: 2.87 Å or 287 pm (corresponding to the edge length of the bcc unit cell of elemental iron)

Summary

The structures of most binary compounds are dictated by the packing arrangement of the largest species present (the anions), with the smaller species (the cations) occupying appropriately sized holes in the anion lattice. A simple cubic lattice of anions contains a single cubic hole in the center of the unit cell. Placing a cation in the cubic hole results in the cesium chloride structure, with a 1:1 cation:anion ratio and a coordination number of 8 for both the cation and the anion. An fcc array of atoms or ions contains both octahedral holes and tetrahedral holes. If the octahedral holes in an fcc lattice of anions are filled with cations, the result is a sodium chloride structure. It also has a 1:1 cation:anion ratio, and each ion has a coordination number of 6. Occupation of half the tetrahedral holes by cations results in the zinc blende structure, with a 1:1 cation:anion ratio and a coordination number of 4 for the cations. More complex structures are possible if there are more than two kinds of atoms in a solid. One example is the perovskite structure, in which the two metal ions form an alternating bcc array with the anions in the centers of the square faces. Because the wavelength of x-ray radiation is comparable to the interatomic distances in most solids, x-ray diffraction can be used to provide information about the structures of crystalline solids. X-rays diffracted from different planes of atoms in a solid reinforce one another if they are in phase, which occurs only if the extra distance they travel corresponds to an integral number of wavelengths. This relationship is described by the Bragg equation: 2d sin θ = nλ.

Key Takeaway

  • The ratio of cations to anions within a unit cell produces electrical neutrality and corresponds to the bulk stoichiometry of a compound, the structure of which can be determined using x-ray diffraction.

Key Equation

Bragg equation

: 2d sin θ = nλ

Conceptual Problems

  1. Using circles or spheres, sketch a unit cell containing an octahedral hole. Which of the basic structural types possess octahedral holes? If an ion were placed in an octahedral hole, what would its coordination number be?

  2. Using circles or spheres, sketch a unit cell containing a tetrahedral hole. Which of the basic structural types possess tetrahedral holes? If an ion were placed in a tetrahedral hole, what would its coordination number be?

  3. How many octahedral holes are there in each unit cell of the sodium chloride structure? Potassium fluoride contains an fcc lattice of F ions that is identical to the arrangement of Cl ions in the sodium chloride structure. Do you expect K+ ions to occupy the tetrahedral or octahedral holes in the fcc lattice of F ions?

  4. The unit cell of cesium chloride consists of a cubic array of chloride ions with a cesium ion in the center. Why then is cesium chloride described as having a simple cubic structure rather than a bcc structure? The unit cell of iron also consists of a cubic array of iron atoms with an iron atom in the center of the cube. Is this a bcc or a simple cubic unit cell? Explain your answer.

  5. Why are x-rays used to determine the structure of crystalline materials? Could gamma rays also be used to determine crystalline structures? Why or why not?

  6. X-rays are higher in energy than most other forms of electromagnetic radiation, including visible light. Why can’t you use visible light to determine the structure of a crystalline material?

  7. When x-rays interact with the atoms in a crystal lattice, what relationship between the distances between planes of atoms in the crystal structure and the wavelength of the x-rays results in the scattered x-rays being exactly in phase with one another? What difference in structure between amorphous materials and crystalline materials makes it difficult to determine the structures of amorphous materials by x-ray diffraction?

  8. It is possible to use different x-ray sources to generate x-rays with different wavelengths. Use the Bragg equation to predict how the diffraction angle would change if a molybdenum x-ray source (x-ray wavelength = 70.93 pm) were used instead of a copper source (x-ray wavelength = 154.1 pm).

  9. Based on the Bragg equation, if crystal A has larger spacing in its diffraction pattern than crystal B, what conclusion can you draw about the spacing between layers of atoms in A compared with B?

Numerical Problems

  1. Thallium bromide crystallizes in the cesium chloride structure. This bcc structure contains a Tl+ ion in the center of the cube with Br ions at the corners. Sketch an alternative unit cell for this compound.

  2. Potassium fluoride has a lattice identical to that of sodium chloride. The potassium ions occupy octahedral holes in an fcc lattice of fluoride ions. Propose an alternative unit cell that can also represent the structure of KF.

  3. Calcium fluoride is used to fluoridate drinking water to promote dental health. Crystalline CaF2 (d = 3.1805 g/cm3) has a structure in which calcium ions are located at each corner and the middle of each edge of the unit cell, which contains eight fluoride ions per unit cell. The length of the edge of this unit cell is 5.463 Å. Use this information to determine Avogadro’s number.

  4. Zinc and oxygen form a compound that is used as both a semiconductor and a paint pigment. This compound has the following structure:

    What is the empirical formula of this compound?

  5. Here are two representations of the perovskite structure:

    Are they identical? What is the empirical formula corresponding to each representation?

  6. The salt MX2 has a cubic close-packed (ccp) structure in which all the tetrahedral holes are filled by anions. What is the coordination number of M? of X?

  7. A compound has a structure based on simple cubic packing of the anions, and the cations occupy half of the cubic holes. What is the empirical formula of this compound? What is the coordination number of the cation?

  8. Barium and fluoride form a compound that crystallizes in the fluorite structure, in which the fluoride ions occupy all the tetrahedral holes in a ccp array of barium ions. This particular compound is used in embalming fluid. What is its empirical formula?

  9. Cadmium chloride is used in paints as a yellow pigment. Is the following structure consistent with an empirical formula of CdCl2? If not, what is the empirical formula of the structure shown?

  10. Use the information in the following table to decide whether the cation will occupy a tetrahedral hole, an octahedral hole, or a cubic hole in each case.

    Cation Radius (pm) Anion Radius (pm)
    78.0 132
    165 133
    81 174
  11. Calculate the angle of diffraction when x-rays from a copper tube (λ = 154 pm) are diffracted by planes of atoms parallel to the faces of the cubic unit cell for Mg (260 pm), Zn (247 pm), and Ni (216 pm). The length on one edge of the unit cell is given in parentheses; assume first-order diffraction (n = 1).

  12. If x-rays from a copper target (λ = 154 pm) are scattered at an angle of 17.23° by a sample of Mg, what is the distance (in picometers) between the planes responsible for this diffraction? How does this distance compare with that in a sample of Ni for which θ = 20.88°?

Answers

  1. d = 3.1805 g/cm3; Avogadro’s number = 6.023 × 1023 mol−1

  2. Both have same stoichiometry, CaTiO3

  3. Stoichiometry is MX2; coordination number of cations is 8

  4. No, the structure shown has an empirical formula of Cd3Cl8.

  5. Mg: 17.2°, Zn: 18.2°, Ni: 20.9°

12.4 Defects in Crystals

Learning Objective

  1. To understand the origin and nature of defects in crystals.

The crystal lattices we have described represent an idealized, simplified system that can be used to understand many of the important principles governing the behavior of solids. In contrast, real crystals contain large numbers of defectsErrors in an idealized crystal lattice. (typically more than 104 per milligram), ranging from variable amounts of impurities to missing or misplaced atoms or ions. These defects occur for three main reasons:

  1. It is impossible to obtain any substance in 100% pure form. Some impurities are always present.
  2. Even if a substance were 100% pure, forming a perfect crystal would require cooling the liquid phase infinitely slowly to allow all atoms, ions, or molecules to find their proper positions. Cooling at more realistic rates usually results in one or more components being trapped in the “wrong” place in a lattice or in areas where two lattices that grew separately intersect.
  3. Applying an external stress to a crystal, such as a hammer blow, can cause microscopic regions of the lattice to move with respect to the rest, thus resulting in imperfect alignment.

In this section, we discuss how defects determine some of the properties of solids. We begin with solids that consist of neutral atoms, specifically metals, and then turn to ionic compounds.

Defects in Metals

Metals can have various types of defects. A point defectA defect in a crystal that affects a single point in the lattice. is any defect that involves only a single particle (a lattice point) or sometimes a very small set of points. A line defectA defect in a crystal that affects a row of points in the lattice. is restricted to a row of lattice points, and a plane defectA defect in a crystal that affects a plane of points in the lattice. involves an entire plane of lattice points in a crystal. A vacancyA point defect that consists of a single atom missing from a site in a crystal. occurs where an atom is missing from the normal crystalline array; it constitutes a tiny void in the middle of a solid (). We focus primarily on point and plane defects in our discussion because they are encountered most frequently.

Figure 12.15 Common Defects in Crystals

In this two-dimensional representation of a crystal lattice containing substitutional and interstitial impurities, vacancies, and line defects, a dashed line and arrows indicate the position of the line defect.

Impurities

Impurities can be classified as interstitial or substitutional. An interstitial impurityA point defect that results when an impurity atom occupies an octahedral hole or a tetrahedral hole in the lattice between atoms. is usually a smaller atom (typically about 45% smaller than the host) that can fit into the octahedral or tetrahedral holes in the metal lattice (). Steels consist of iron with carbon atoms added as interstitial impurities (). The inclusion of one or more transition metals or semimetals can improve the corrosion resistance of steel.

Table 12.3 Compositions, Properties, and Uses of Some Types of Steel

Name of Steel Typical Composition* Properties Applications
low-carbon <0.15% C soft and ductile wire
mild carbon 0.15%–0.25% C malleable and ductile cables, chains, and nails
high-carbon 0.60%–1.5% C hard and brittle knives, cutting tools, drill bits, and springs
stainless 15%–20% Cr, 1%–5% Mn, 5%–10% Ni, 1%–3% Si, 1% C, 0.05% P corrosion resistant cutlery, instruments, and marine fittings
invar 36% Ni low coefficient of thermal expansion measuring tapes and meter sticks
manganese 10%–20% Mn hard and wear resistant armor plate, safes, and rails
high-speed 14%–20% W retains hardness at high temperatures high-speed cutting tools
silicon 1%–5% Si hard, strong, and highly magnetic magnets in electric motors and transformers
*In addition to enough iron to bring the total percentage up to 100%, most steels contain small amounts of carbon (0.5%–1.5%) and manganese (<2%).

In contrast, a substitutional impurityA point defect that results when an impurity atom occupies a normal lattice site. is a different atom of about the same size that simply replaces one of the atoms that compose the host lattice (). Substitutional impurities are usually chemically similar to the substance that constitutes the bulk of the sample, and they generally have atomic radii that are within about 15% of the radius of the host. For example, strontium and calcium are chemically similar and have similar radii, and as a result, strontium is a common impurity in crystalline calcium, with the Sr atoms randomly occupying sites normally occupied by Ca.

Note the Pattern

Interstitial impurities are smaller atoms than the host atom, whereas substitutional impurities are usually chemically similar and are similar in size to the host atom.

Dislocations, Deformations, and Work Hardening

Inserting an extra plane of atoms into a crystal lattice produces an edge dislocationA crystal defect that results from the insertion of an extra plane of atoms into part of the crystal lattice.. A familiar example of an edge dislocation occurs when an ear of corn contains an extra row of kernels between the other rows (). An edge dislocation in a crystal causes the planes of atoms in the lattice to deform where the extra plane of atoms begins (). The edge dislocation frequently determines whether the entire solid will deform and fail under stress.

Figure 12.16 Edge Dislocations

Shown are two examples of edge dislocations: (a) an edge dislocation in an ear of corn and (b) a three-dimensional representation of an edge dislocation in a solid, illustrating how an edge dislocation can be viewed as a simple line defect arising from the insertion of an extra set of atoms into the lattice. In both cases, the origin of the edge dislocation is indicated by the symbol .

DeformationA distortion that occurs when a dislocation moves through a crystal. occurs when a dislocation moves through a crystal. To illustrate the process, suppose you have a heavy rug that is lying a few inches off-center on a nonskid pad. To move the rug to its proper place, you could pick up one end and pull it. Because of the large area of contact between the rug and the pad, however, they will probably move as a unit. Alternatively, you could pick up the rug and try to set it back down exactly where you want it, but that requires a great deal of effort (and probably at least one extra person). An easier solution is to create a small wrinkle at one end of the rug (an edge dislocation) and gradually push the wrinkle across, resulting in a net movement of the rug as a whole (part (a) in ). Moving the wrinkle requires only a small amount of energy because only a small part of the rug is actually moving at any one time. Similarly, in a solid, the contacts between layers are broken in only one place at a time, which facilitates the deformation process.

Figure 12.17 The Role of Dislocation in the Motion of One Planar Object across Another

(a) Pushing a wrinkle across the rug results in a net movement of the rug with relatively little expenditure of energy because at any given time only a very small amount of the rug is not in contact with the floor. (b) A second intersecting wrinkle prevents movement of the first by “pinning” it.

If the rug we have just described has a second wrinkle at a different angle, however, it is very difficult to move the first one where the two wrinkles intersect (part (b) in ); this process is called pinningA process that increases the mechanical strength of a material by introducing multiple defects into a material so that the presence of one defect prevents the motion of another.. Similarly, intersecting dislocations in a solid prevent them from moving, thereby increasing the mechanical strength of the material. In fact, one of the major goals of materials science is to find ways to pin dislocations to strengthen or harden a material.

Pinning can also be achieved by introducing selected impurities in appropriate amounts. Substitutional impurities that are a mismatch in size to the host prevent dislocations from migrating smoothly along a plane. Generally, the higher the concentration of impurities, the more effectively they block migration, and the stronger the material. For example, bronze, which contains about 20% tin and 80% copper by mass, produces a much harder and sharper weapon than does either pure tin or pure copper. Similarly, pure gold is too soft to make durable jewelry, so most gold jewelry contains 75% (18 carat) or 58% (14 carat) gold by mass, with the remainder consisting of copper, silver, or both.

If an interstitial impurity forms polar covalent bonds to the host atoms, the layers are prevented from sliding past one another, even when only a small amount of the impurity is present. For example, because iron forms polar covalent bonds to carbon, the strongest steels need to contain only about 1% carbon by mass to substantially increase their strength ().

Most materials are polycrystalline, which means they consist of many microscopic individual crystals called grains that are randomly oriented with respect to one another. The place where two grains intersect is called a grain boundaryThe place where two grains in a solid intersect.. The movement of a deformation through a solid tends to stop at a grain boundary. Consequently, controlling the grain size in solids is critical for obtaining desirable mechanical properties; fine-grained materials are usually much stronger than coarse-grained ones.

Grain boundaries. As a polycrystalline material solidifies, grains with irregular shapes form. The interfaces between grains constitute grain boundaries. (Squares represent unit cells within grains.)

Work hardeningThe practice of introducing a dense network of dislocations throughout a solid, making it very tough and hard. is the introduction of a dense network of dislocations throughout a solid, which makes it very tough and hard. If all the defects in a single 1 cm3 sample of a work-hardened material were laid end to end, their total length could be 106 km! The legendary blades of the Japanese and Moorish swordsmiths owed much of their strength to repeated work hardening of the steel. As the density of defects increases, however, the metal becomes more brittle (less malleable). For example, bending a paper clip back and forth several times increases its brittleness from work hardening and causes the wire to break.

Memory Metal

The compound NiTi, popularly known as “memory metal” or nitinol (nickel–titanium Naval Ordinance Laboratory, after the site where it was first prepared), illustrates the importance of deformations. If a straight piece of NiTi wire is wound into a spiral, it will remain in the spiral shape indefinitely, unless it is warmed to 50°C–60°C, at which point it will spontaneously straighten out again. The chemistry behind the temperature-induced change in shape is moderately complex, but for our purposes it is sufficient to know that NiTi can exist in two different solid phases.

The high-temperature phase has the cubic cesium chloride structure, in which a Ti atom is embedded in the center of a cube of Ni atoms (or vice versa). The low-temperature phase has a related but kinked structure, in which one of the angles of the unit cell is no longer 90°. Bending an object made of the low-temperature (kinked) phase creates defects that change the pattern of kinks within the structure. If the object is heated to a temperature greater than about 50°C, the material undergoes a transition to the cubic high-temperature phase, causing the object to return to its original shape. The shape of the object above 50°C is controlled by a complex set of defects and dislocations that can be relaxed or changed only by the thermal motion of the atoms.

Memory metal. Flexon is a fatigue-resistant alloy of Ti and Ni that is used as a frame for glasses because of its durability and corrosion resistance.

Memory metal has many other practical applications, such as its use in temperature-sensitive springs that open and close valves in the automatic transmissions of cars. Because NiTi can also undergo pressure- or tension-induced phase transitions, it is used to make wires for straightening teeth in orthodontic braces and in surgical staples that change shape at body temperature to hold broken bones together.

Another flexible, fatigue-resistant alloy composed of titanium and nickel is Flexon. Originally discovered by metallurgists who were creating titanium-based alloys for use in missile heat shields, Flexon is now used as a durable, corrosion-resistant frame for glasses, among other uses.

Example 5

Because steels with at least 4% chromium are much more corrosion resistant than iron, they are collectively sold as “stainless steel.” Referring to the composition of stainless steel in and, if needed, the atomic radii in , predict which type of impurity is represented by each element in stainless steel, excluding iron, that are present in at least 0.05% by mass.

Given: composition of stainless steel and atomic radii

Asked for: type of impurity

Strategy:

Using the data in and the atomic radii in , determine whether the impurities listed are similar in size to an iron atom. Then determine whether each impurity is chemically similar to Fe. If similar in both size and chemistry, the impurity is likely to be a substitutional impurity. If not, it is likely to be an interstitial impurity.

Solution:

According to , stainless steel typically contains about 1% carbon, 1%–5% manganese, 0.05% phosphorus, 1%–3% silicon, 5%–10% nickel, and 15%–20% chromium. The three transition elements (Mn, Ni, and Cr) lie near Fe in the periodic table, so they should be similar to Fe in chemical properties and atomic size (atomic radius = 125 pm). Hence they almost certainly will substitute for iron in the Fe lattice. Carbon is a second-period element that is nonmetallic and much smaller (atomic radius = 77 pm) than iron. Carbon will therefore tend to occupy interstitial sites in the iron lattice. Phosphorus and silicon are chemically quite different from iron (phosphorus is a nonmetal, and silicon is a semimetal), even though they are similar in size (atomic radii of 106 and 111 pm, respectively). Thus they are unlikely to be substitutional impurities in the iron lattice or fit into interstitial sites, but they could aggregate into layers that would constitute plane defects.

Exercise

Consider nitrogen, vanadium, zirconium, and uranium impurities in a sample of titanium metal. Which is most likely to form an interstitial impurity? a substitutional impurity?

Answer: nitrogen; vanadium

Defects in Ionic and Molecular Crystals

All the defects and impurities described for metals are seen in ionic and molecular compounds as well. Because ionic compounds contain both cations and anions rather than only neutral atoms, however, they exhibit additional types of defects that are not possible in metals.

The most straightforward variant is a substitutional impurity in which a cation or an anion is replaced by another of similar charge and size. For example, Br can substitute for Cl, so tiny amounts of Br are usually present in a chloride salt such as CaCl2 or BaCl2. If the substitutional impurity and the host have different charges, however, the situation becomes more complicated. Suppose, for example, that Sr2+ (ionic radius = 118 pm) substitutes for K+ (ionic radius = 138 pm) in KCl. Because the ions are approximately the same size, Sr2+ should fit nicely into the face-centered cubic (fcc) lattice of KCl. The difference in charge, however, must somehow be compensated for so that electrical neutrality is preserved. The simplest way is for a second K+ ion to be lost elsewhere in the crystal, producing a vacancy. Thus substitution of K+ by Sr2+ in KCl results in the introduction of two defects: a site in which an Sr2+ ion occupies a K+ position and a vacant cation site. Substitutional impurities whose charges do not match the host’s are often introduced intentionally to produce compounds with specific properties (see ).

Virtually all the colored gems used in jewelry are due to substitutional impurities in simple oxide structures. For example, α-Al2O3, a hard white solid called corundum that is used as an abrasive in fine sandpaper, is the primary component, or matrix, of a wide variety of gems. Because many trivalent transition metal ions have ionic radii only a little larger than the radius of Al3+ (ionic radius = 53.5 pm), they can replace Al3+ in the octahedral holes of the oxide lattice. Substituting small amounts of Cr3+ ions (ionic radius = 75 pm) for Al3+ gives the deep red color of ruby, and a mixture of impurities (Fe2+, Fe3+, and Ti4+) gives the deep blue of sapphire. True amethyst contains small amounts of Fe3+ in an SiO2 (quartz) matrix. The same metal ion substituted into different mineral lattices can produce very different colors. For example, Fe3+ ions are responsible for the yellow color of topaz and the violet color of amethyst. The distinct environments cause differences in d orbital energies, enabling the Fe3+ ions to absorb light of different frequencies, a topic we describe in more detail in .

The same cation in different environments. An Fe3+ substitutional impurity produces substances with strikingly different colors.

Substitutional impurities are also observed in molecular crystals if the structure of the impurity is similar to the host, and they can have major effects on the properties of the crystal. Pure anthracene, for example, is an electrical conductor, but the transfer of electrons through the molecule is much slower if the anthracene crystal contains even very small amounts of tetracene despite their strong structural similarities.

If a cation or an anion is simply missing, leaving a vacant site in an ionic crystal, then for the crystal to be electrically neutral, there must be a corresponding vacancy of the ion with the opposite charge somewhere in the crystal. In compounds such as KCl, the charges are equal but opposite, so one anion vacancy is sufficient to compensate for each cation vacancy. In compounds such as CaCl2, however, two Cl anion sites must be vacant to compensate for each missing Ca2+ cation. These pairs (or sets) of vacancies are called Schottky defectsA coupled pair of vacancies—one cation and one anion—that maintains the electrical neutrality of an ionic solid. and are particularly common in simple alkali metal halides such as KCl (part (a) in ). Many microwave diodes, which are devices that allow a current to flow in a single direction, are composed of materials with Schottky defects.

Figure 12.18 The Two Most Common Defects in Ionic Solids

(a) A Schottky defect in KCl shows the missing cation/anion pair. (b) A Frenkel defect in AgI shows a misplaced Ag+ cation.

Occasionally one of the ions in an ionic lattice is simply in the wrong position. An example of this phenomenon, called a Frenkel defectA defect in an ionic lattice that occurs when one of the ions is in the wrong position., is a cation that occupies a tetrahedral hole rather than an octahedral hole in the anion lattice (part (b) in ). Frenkel defects are most common in salts that have a large anion and a relatively small cation. To preserve electrical neutrality, one of the normal cation sites, usually octahedral, must be vacant.

Frenkel defects are particularly common in the silver halides AgCl, AgBr, and AgI, which combine a rather small cation (Ag+, ionic radius = 115 pm) with large, polarizable anions. Certain more complex salts with a second cation in addition to Ag+ and Br or I have so many Ag+ ions in tetrahedral holes that they are good electrical conductors in the solid state; hence they are called solid electrolytesA solid material with a very high electrical conductivity.. (As you learned in , most ionic compounds do not conduct electricity in the solid state, although they do conduct electricity when molten or dissolved in a solvent that separates the ions, allowing them to migrate in response to an applied electric field.) In response to an applied voltage, the cations in solid electrolytes can diffuse rapidly through the lattice via octahedral holes, creating Frenkel defects as the cations migrate. Sodium–sulfur batteries use a solid Al2O3 electrolyte with small amounts of solid Na2O. Because the electrolyte cannot leak, it cannot cause corrosion, which gives a battery that uses a solid electrolyte a significant advantage over one with a liquid electrolyte.

Example 6

In a sample of NaCl, one of every 10,000 sites normally occupied by Na+ is occupied instead by Ca2+. Assuming that all of the Cl sites are fully occupied, what is the stoichiometry of the sample?

Given: ionic solid and number and type of defect

Asked for: stoichiometry

Strategy:

A Identify the unit cell of the host compound. Compute the stoichiometry if 0.01% of the Na+ sites are occupied by Ca2+. If the overall charge is greater than 0, then the stoichiometry must be incorrect.

B If incorrect, adjust the stoichiometry of the Na+ ion to compensate for the additional charge.

Solution:

A Pure NaCl has a 1:1 ratio of Na+ and Cl ions arranged in an fcc lattice (the sodium chloride structure). If all the anion sites are occupied by Cl, the negative charge is −1.00 per formula unit. If 0.01% of the Na+ sites are occupied by Ca2+ ions, the cation stoichiometry is Na0.99Ca0.01. This results in a positive charge of (0.99)(+1) + (0.01)(+2) = +1.01 per formula unit, for a net charge in the crystal of +1.01 + (−1.00) = +0.01 per formula unit. Because the overall charge is greater than 0, this stoichiometry must be incorrect.

B The most plausible way for the solid to adjust its composition to become electrically neutral is for some of the Na+ sites to be vacant. If one Na+ site is vacant for each site that has a Ca2+ cation, then the cation stoichiometry is Na0.98Ca0.01. This results in a positive charge of (0.98)(+1) + (0.01)(+2) = +1.00 per formula unit, which exactly neutralizes the negative charge. The stoichiometry of the solid is thus Na0.98Ca0.01Cl1.00.

Exercise

In a sample of MgO that has the sodium chloride structure, 0.02% of the Mg2+ ions are replaced by Na+ ions. Assuming that all of the cation sites are fully occupied, what is the stoichiometry of the sample?

Answer: If the formula of the compound is Mg0.98Na0.02O1−x, then x must equal 0.01 to preserve electrical neutrality. The formula is thus Mg0.98Na0.02O0.99.

Nonstoichiometric Compounds

The law of multiple proportions (see ), states that chemical compounds contain fixed integral ratios of atoms. In fact, nonstoichiometric compoundsA solid that has intrinsically variable stoichiometries without affecting the fundamental structure of the crystal. contain large numbers of defects, usually vacancies, which give rise to stoichiometries that can depart significantly from simple integral ratios without affecting the fundamental structure of the crystal. Nonstoichiometric compounds frequently consist of transition metals, lanthanides, and actinides, with polarizable anions such as oxide (O2−) and sulfide (S2−). Some common examples are listed in , along with their basic structure type. These compounds are nonstoichiometric because their constituent metals can exist in multiple oxidation states in the solid, which in combination preserve electrical neutrality.

Table 12.4 Some Nonstoichiometric Compounds

Compound Observed Range of x
Oxides*
FexO 0.85–0.95
NixO 0.97–1.00
TiOx 0.75–1.45
VOx 0.9–1.20
NbOx 0.9–1.04
Sulfides
CuxS 1.77–2.0
FexS 0.80–0.99
ZrSx 0.9–1.0
*All the oxides listed have the sodium chloride structure.

One example is iron(II) oxide (ferrous oxide), which produces the black color in clays and is used as an abrasive. Its stoichiometry is not FeO because it always contains less than 1.00 Fe per O2− (typically 0.90–0.95). This is possible because Fe can exist in both the +2 and +3 oxidation states. Thus the difference in charge created by a vacant Fe2+ site can be balanced by two Fe2+ sites that have Fe3+ ions instead [+2 vacancy = (3 − 2) + (3 − 2)]. The crystal lattice is able to accommodate this relatively high fraction of substitutions and vacancies with no significant change in structure.

Note the Pattern

Because a crystal must be electrically neutral, any defect that affects the number or charge of the cations must be compensated by a corresponding defect in the number or charge of the anions.

Summary

Real crystals contain large numbers of defects. Defects may affect only a single point in the lattice (a point defect), a row of lattice points (a line defect), or a plane of atoms (a plane defect). A point defect can be an atom missing from a site in the crystal (a vacancy) or an impurity atom that occupies either a normal lattice site (a substitutional impurity) or a hole in the lattice between atoms (an interstitial impurity). In an edge dislocation, an extra plane of atoms is inserted into part of the crystal lattice. Multiple defects can be introduced into materials so that the presence of one defect prevents the motion of another, in a process called pinning. Because defect motion tends to stop at grain boundaries, controlling the size of the grains in a material controls its mechanical properties. In addition, a process called work hardening introduces defects to toughen metals. Schottky defects are a coupled pair of vacancies—one cation and one anion—that maintains electrical neutrality. A Frenkel defect is an ion that occupies an incorrect site in the lattice. Cations in such compounds are often able to move rapidly from one site in the crystal to another, resulting in high electrical conductivity in the solid material. Such compounds are called solid electrolytes. Nonstoichiometric compounds have variable stoichiometries over a given range with no dramatic change in crystal structure. This behavior is due to a large number of vacancies or substitutions of one ion by another ion with a different charge.

Key Takeaway

  • Defects determine the behavior of solids, but because ionic compounds contain both cations and anions, they exhibit additional types of defects that are not found in metals.

Conceptual Problems

  1. How are defects and impurities in a solid related? Can a pure, crystalline compound be free of defects? How can a substitutional impurity produce a vacancy?

  2. Why does applying a mechanical stress to a covalent solid cause it to fracture? Use an atomic level description to explain why a metal is ductile under conditions that cause a covalent solid to fracture.

  3. How does work hardening increase the strength of a metal? How does work hardening affect the physical properties of a metal?

  4. Work-hardened metals and covalent solids such as diamonds are both susceptible to cracking when stressed. Explain how such different materials can both exhibit this property.

  5. Suppose you want to produce a ductile material with improved properties. Would impurity atoms of similar or dissimilar atomic size be better at maintaining the ductility of a metal? Why? How would introducing an impurity that forms polar covalent bonds with the metal atoms affect the ductility of the metal? Explain your reasoning.

  6. Substitutional impurities are often used to tune the properties of material. Why are substitutional impurities generally more effective at high concentrations, whereas interstitial impurities are usually effective at low concentrations?

  7. If an O2− ion (ionic radius = 132 pm) is substituted for an F ion (ionic radius = 133 pm) in an ionic crystal, what structural changes in the ionic lattice will result?

  8. How will the introduction of a metal ion with a different charge as an impurity induce the formation of oxygen vacancies in an ionic metal-oxide crystal?

  9. Many nonstoichiometric compounds are transition metal compounds. How can such compounds exist, given that their nonintegral cation:anion ratios apparently contradict one of the basic tenets of Dalton’s atomic theory of matter?

  10. If you wanted to induce the formation of oxygen vacancies in an ionic crystal, which would you introduce as substitutional impurities—cations with a higher positive charge or a lower positive charge than the cations in the parent structure? Explain your reasoning.

Answers

  1. Impurity atoms of similar size and with similar chemical properties would be most likely to maintain the ductility of the metal, because they are unlikely to have a large effect on the ease with which one layer of atoms can move past another under mechanical stress. Larger impurity atoms are likely to form “bumps” or kinks that will make it harder for layers of atoms to move across one another. Interstitial atoms that form polar covalent bonds with the metal atoms tend to occupy spaces between the layers; they act as a “glue” that holds layers of metal atoms together, which greatly decreases the ductility.

  2. Since O2− and F are both very similar in size, substitution is possible without disruption of the ionic packing. The difference in charge, however, requires the formation of a vacancy on another F site to maintain charge neutrality.

  3. Most transition metals form at least two cations that differ by only one electron. Consequently, nonstoichiometric compounds containing transition metals can maintain electrical neutrality by gaining electrons to compensate for the absence of anions or the presence of additional metal ions. Conversely, such compounds can lose electrons to compensate for the presence of additional anions or the absence of metal ions. In both cases, the positive charge on the transition metal is adjusted to maintain electrical neutrality.

Numerical Problems

  1. The ionic radius of K+ is 133 pm, whereas that of Na+ is 98 pm. Do you expect K+ to be a common substitutional impurity in compounds containing Na+? Why or why not?

  2. Given Cs (262 pm), Tl (171 pm), and B (88 pm) with their noted atomic radii, which atom is most likely to act as an interstitial impurity in an Sn lattice (Sn atomic radius = 141 pm)? Why?

  3. After aluminum, iron is the second most abundant metal in Earth’s crust. The silvery-white, ductile metal has a body-centered cubic (bcc) unit cell with an edge length of 286.65 pm.

    1. Use this information to calculate the density of iron.
    2. What would the density of iron be if 0.15% of the iron sites were vacant?
    3. How does the mass of 1.00 cm3 of iron without defects compare with the mass of 1.00 cm3 of iron with 0.15% vacancies?
  4. Certain ceramic materials are good electrical conductors due to high mobility of oxide ions resulting from the presence of oxygen vacancies. Zirconia (ZrO2) can be doped with yttrium by adding Y2O3. If 0.35 g of Y2O3 can be incorporated into 25.0 g of ZrO2 while maintaining the zirconia structure, what is the percentage of oxygen vacancies in the structure?

  5. Which of the following ions is most effective at inducing an O2− vacancy in crystal of CaO? The ionic radii are O2−, 132 pm; Ca2+, 100 pm; Sr2+, 127 pm; F, 133 pm; La3+, 104 pm; and K+, 133 pm. Explain your reasoning.

Answers

  1. No. The potassium is much larger than the sodium ion.

    1. 7.8744 g/cm3
    2. 7.86 g/cm3
    3. Without defects, the mass is 0.15% greater.
  2. The lower charge of K+ makes it the best candidate for inducing an oxide vacancy, even though its ionic radius is substantially larger than that of Ca2+. Substituting two K+ ions for two Ca2+ ions will decrease the total positive charge by two, and an oxide vacancy will maintain electrical neutrality. For example, if 10% of the Ca2+ ions are replaced by K+, we can represent the change as going from Ca20O20 to K2Ca18O20, which has a net charge of +2. Loss of one oxide ion would give a composition of K2Ca18O19, which is electrically neutral.

12.5 Correlation between Bonding and the Properties of Solids

Learning Objective

  1. To understand the correlation between bonding and the properties of solids.

Based on the nature of the forces that hold the component atoms, molecules, or ions together, solids may be formally classified as ionic, molecular, covalent (network), or metallic. The variation in the relative strengths of these four types of interactions correlates nicely with their wide variation in properties.

Ionic Solids

You learned in that an ionic solidA solid that consists of positively and negatively charged ions held together by electrostatic forces. consists of positively and negatively charged ions held together by electrostatic forces. (For more information about ionic solids, see , .) The strength of the attractive forces depends on the charge and size of the ions that compose the lattice and determines many of the physical properties of the crystal.

The lattice energy, the energy required to separate 1 mol of a crystalline ionic solid into its component ions in the gas phase, is directly proportional to the product of the ionic charges and inversely proportional to the sum of the radii of the ions. For example, NaF and CaO both crystallize in the face-centered cubic (fcc) sodium chloride structure, and the sizes of their component ions are about the same: Na+ (102 pm) versus Ca2+ (100 pm), and F (133 pm) versus O2− (140 pm). Because of the higher charge on the ions in CaO, however, the lattice energy of CaO is almost four times greater than that of NaF (3401 kJ/mol versus 923 kJ/mol). The forces that hold Ca and O together in CaO are much stronger than those that hold Na and F together in NaF, so the heat of fusion of CaO is almost twice that of NaF (59 kJ/mol versus 33.4 kJ/mol), and the melting point of CaO is 2927°C versus 996°C for NaF. In both cases, however, the values are large; that is, simple ionic compounds have high melting points and are relatively hard (and brittle) solids.

Molecular Solids

Molecular solidsA solid that consists of molecules held together by relatively weak forces, such as dipole-dipole interactions, hydrogen bonds, and London dispersion forces. consist of atoms or molecules held to each other by dipole–dipole interactions, London dispersion forces, or hydrogen bonds, or any combination of these, which were discussed in . The arrangement of the molecules in solid benzene is as follows:

The structure of solid benzene. In solid benzene, the molecules are not arranged with their planes parallel to one another but at 90° angles.

Because the intermolecular interactions in a molecular solid are relatively weak compared with ionic and covalent bonds, molecular solids tend to be soft, low melting, and easily vaporized (ΔHfus and ΔHvap are low). For similar substances, the strength of the London dispersion forces increases smoothly with increasing molecular mass. For example, the melting points of benzene (C6H6), naphthalene (C10H8), and anthracene (C14H10), with one, two, and three fused aromatic rings, are 5.5°C, 80.2°C, and 215°C, respectively. The enthalpies of fusion also increase smoothly within the series: benzene (9.95 kJ/mol) < naphthalene (19.1 kJ/mol) < anthracene (28.8 kJ/mol). If the molecules have shapes that cannot pack together efficiently in the crystal, however, then the melting points and the enthalpies of fusion tend to be unexpectedly low because the molecules are unable to arrange themselves to optimize intermolecular interactions. Thus toluene (C6H5CH3) and m-xylene [m-C6H4(CH3)2] have melting points of −95°C and −48°C, respectively, which are significantly lower than the melting point of the lighter but more symmetrical analog, benzene.

Self-healing rubber is an example of a molecular solid with the potential for significant commercial applications. The material can stretch, but when snapped into pieces it can bond back together again through reestablishment of its hydrogen-bonding network without showing any sign of weakness. Among other applications, it is being studied for its use in adhesives and bicycle tires that will self-heal.

Toluene and m-xylene. The methyl groups attached to the phenyl ring in toluene and m-xylene prevent the rings from packing together as in solid benzene.

Covalent Solids

Covalent solidsA solid that consists of two- or three-dimensional networks of atoms held together by covalent bonds. are formed by networks or chains of atoms or molecules held together by covalent bonds. A perfect single crystal of a covalent solid is therefore a single giant molecule. For example, the structure of diamond, shown in part (a) in , consists of sp3 hybridized carbon atoms, each bonded to four other carbon atoms in a tetrahedral array to create a giant network. The carbon atoms form six-membered rings.

Figure 12.19 The Structures of Diamond and Graphite

(a) Diamond consists of sp3 hybridized carbon atoms, each bonded to four other carbon atoms. The tetrahedral array forms a giant network in which carbon atoms form six-membered rings. (b) These side (left) and top (right) views of the graphite structure show the layers of fused six-membered rings and the arrangement of atoms in alternate layers of graphite. The rings in alternate layers are staggered, such that every other carbon atom in one layer lies directly under (and above) the center of a six-membered ring in an adjacent layer.

The unit cell of diamond can be described as an fcc array of carbon atoms with four additional carbon atoms inserted into four of the tetrahedral holes. It thus has the zinc blende structure described in , except that in zinc blende the atoms that compose the fcc array are sulfur and the atoms in the tetrahedral holes are zinc. Elemental silicon has the same structure, as does silicon carbide (SiC), which has alternating C and Si atoms. The structure of crystalline quartz (SiO2), shown in , can be viewed as being derived from the structure of silicon by inserting an oxygen atom between each pair of silicon atoms.

All compounds with the diamond and related structures are hard, high-melting-point solids that are not easily deformed. Instead, they tend to shatter when subjected to large stresses, and they usually do not conduct electricity very well. In fact, diamond (melting point = 3500°C at 63.5 atm) is one of the hardest substances known, and silicon carbide (melting point = 2986°C) is used commercially as an abrasive in sandpaper and grinding wheels. It is difficult to deform or melt these and related compounds because strong covalent (C–C or Si–Si) or polar covalent (Si–C or Si–O) bonds must be broken, which requires a large input of energy.

Other covalent solids have very different structures. For example, graphite, the other common allotrope of carbon, has the structure shown in part (b) in . It contains planar networks of six-membered rings of sp2 hybridized carbon atoms in which each carbon is bonded to three others. This leaves a single electron in an unhybridized 2pz orbital that can be used to form C=C double bonds, resulting in a ring with alternating double and single bonds. Because of its resonance structures, the bonding in graphite is best viewed as consisting of a network of C–C single bonds with one-third of a π bond holding the carbons together, similar to the bonding in benzene.

To completely describe the bonding in graphite, we need a molecular orbital approach similar to the one used for benzene in . In fact, the C–C distance in graphite (141.5 pm) is slightly longer than the distance in benzene (139.5 pm), consistent with a net carbon–carbon bond order of 1.33. In graphite, the two-dimensional planes of carbon atoms are stacked to form a three-dimensional solid; only London dispersion forces hold the layers together. As a result, graphite exhibits properties typical of both covalent and molecular solids. Due to strong covalent bonding within the layers, graphite has a very high melting point, as expected for a covalent solid (it actually sublimes at about 3915°C). It is also very soft; the layers can easily slide past one another because of the weak interlayer interactions. Consequently, graphite is used as a lubricant and as the “lead” in pencils; the friction between graphite and a piece of paper is sufficient to leave a thin layer of carbon on the paper. Graphite is unusual among covalent solids in that its electrical conductivity is very high parallel to the planes of carbon atoms because of delocalized C–C π bonding. Finally, graphite is black because it contains an immense number of alternating double bonds, which results in a very small energy difference between the individual molecular orbitals. Thus light of virtually all wavelengths is absorbed. Diamond, on the other hand, is colorless when pure because it has no delocalized electrons.

compares the strengths of the intermolecular and intramolecular interactions for three covalent solids, showing the comparative weakness of the interlayer interactions.

Table 12.5 A Comparison of Intermolecular (ΔHsub) and Intramolecular Interactions

Substance ΔHsub (kJ/mol) Average Bond Energy (kJ/mol)
phosphorus (s) 58.98 201
sulfur (s) 64.22 226
iodine (s) 62.42 149

Metallic Solids

Metals are characterized by their ability to reflect light, called lusterThe ability to reflect light. Metals, for instance, have a shiny surface that reflects light (metals are lustrous), whereas nonmetals do not., their high electrical and thermal conductivity, their high heat capacity, and their malleability and ductility. Every lattice point in a pure metallic element is occupied by an atom of the same metal. The packing efficiency in metallic crystals tends to be high, so the resulting metallic solidsA solid that consists of metal atoms held together by metallic bonds. are dense, with each atom having as many as 12 nearest neighbors.

Bonding in metallic solids is quite different from the bonding in the other kinds of solids we have discussed. Because all the atoms are the same, there can be no ionic bonding, yet metals always contain too few electrons or valence orbitals to form covalent bonds with each of their neighbors. Instead, the valence electrons are delocalized throughout the crystal, providing a strong cohesive force that holds the metal atoms together.

Note the Pattern

Valence electrons in a metallic solid are delocalized, providing a strong cohesive force that holds the atoms together.

The strength of metallic bonds varies dramatically. For example, cesium melts at 28.4°C, and mercury is a liquid at room temperature, whereas tungsten melts at 3680°C. Metallic bonds tend to be weakest for elements that have nearly empty (as in Cs) or nearly full (Hg) valence subshells, and strongest for elements with approximately half-filled valence shells (as in W). As a result, the melting points of the metals increase to a maximum around group 6 and then decrease again from left to right across the d block. Other properties related to the strength of metallic bonds, such as enthalpies of fusion, boiling points, and hardness, have similar periodic trends.

A somewhat oversimplified way to describe the bonding in a metallic crystal is to depict the crystal as consisting of positively charged nuclei in an electron seaValence electrons that are delocalized throughout a metallic solid. (). In this model, the valence electrons are not tightly bound to any one atom but are distributed uniformly throughout the structure. Very little energy is needed to remove electrons from a solid metal because they are not bound to a single nucleus. When an electrical potential is applied, the electrons can migrate through the solid toward the positive electrode, thus producing high electrical conductivity. The ease with which metals can be deformed under pressure is attributed to the ability of the metal ions to change positions within the electron sea without breaking any specific bonds. The transfer of energy through the solid by successive collisions between the metal ions also explains the high thermal conductivity of metals. This model does not, however, explain many of the other properties of metals, such as their metallic luster and the observed trends in bond strength as reflected in melting points or enthalpies of fusion. A more complete description of metallic bonding is presented in .

Figure 12.20 The Electron-Sea Model of Bonding in Metals

Fixed, positively charged metal nuclei from group 1 (a) or group 2 (b) are surrounded by a “sea” of mobile valence electrons. Because a group 2 metal has twice the number of valence electrons as a group 1 metal, it should have a higher melting point.

Substitutional Alloys

An alloyA solid solution of two or more metals whose properties differ from those of the constituent elements. is a mixture of metals with metallic properties that differ from those of its constituent elements. Brass (Cu and Zn in a 2:1 ratio) and bronze (Cu and Sn in a 4:1 ratio) are examples of substitutional alloysAn alloy formed by the substitution of one metal atom for another of similar size in the lattice., which are metallic solids with large numbers of substitutional impurities. In contrast, small numbers of interstitial impurities, such as carbon in the iron lattice of steel, give an interstitial alloyAn alloy formed by inserting smaller atoms into holes in the metal lattice.. Because scientists can combine two or more metals in varying proportions to tailor the properties of a material for particular applications, most of the metallic substances we encounter are actually alloys. Examples include the low-melting-point alloys used in solder (Pb and Sn in a 2:1 ratio) and in fuses and fire sprinklers (Bi, Pb, Sn, and Cd in a 4:2:1:1 ratio).

The compositions of most alloys can vary over wide ranges. In contrast, intermetallic compoundsAn alloy that consists of certain metals that combine in only specific proportions and whose properties are frequently quite different from those of their constituent elements. consist of certain metals that combine in only specific proportions. Their compositions are largely determined by the relative sizes of their component atoms and the ratio of the total number of valence electrons to the number of atoms present (the valence electron density). The structures and physical properties of intermetallic compounds are frequently quite different from those of their constituent elements, but they may be similar to elements with a similar valence electron density. For example, Cr3Pt is an intermetallic compound used to coat razor blades advertised as “platinum coated”; it is very hard and dramatically lengthens the useful life of the razor blade. With similar valence electron densities, Cu and PdZn have been found to be virtually identical in their catalytic properties.

Some general properties of the four major classes of solids are summarized in .

Table 12.6 Properties of the Major Classes of Solids

Ionic Solids Molecular Solids Covalent Solids Metallic Solids
poor conductors of heat and electricity poor conductors of heat and electricity poor conductors of heat and electricity* good conductors of heat and electricity
relatively high melting point low melting point high melting point melting points depend strongly on electron configuration
hard but brittle; shatter under stress soft very hard and brittle easily deformed under stress; ductile and malleable
relatively dense low density low density usually high density
dull surface dull surface dull surface lustrous
*Many exceptions exist. For example, graphite has a relatively high electrical conductivity within the carbon planes, and diamond has the highest thermal conductivity of any known substance.

Note the Pattern

The general order of increasing strength of interactions in a solid is molecular solids < ionic solids ≈ metallic solids < covalent solids.

Example 7

Classify Ge, RbI, C6(CH3)6, and Zn as ionic, molecular, covalent, or metallic solids and arrange them in order of increasing melting points.

Given: compounds

Asked for: classification and order of melting points

Strategy:

A Locate the component element(s) in the periodic table. Based on their positions, predict whether each solid is ionic, molecular, covalent, or metallic.

B Arrange the solids in order of increasing melting points based on your classification, beginning with molecular solids.

Solution:

A Germanium lies in the p block just under Si, along the diagonal line of semimetallic elements, which suggests that elemental Ge is likely to have the same structure as Si (the diamond structure). Thus Ge is probably a covalent solid. RbI contains a metal from group 1 and a nonmetal from group 17, so it is an ionic solid containing Rb+ and I ions. The compound C6(CH3)6 is a hydrocarbon (hexamethylbenzene), which consists of isolated molecules that stack to form a molecular solid with no covalent bonds between them. Zn is a d-block element, so it is a metallic solid.

B Arranging these substances in order of increasing melting points is straightforward, with one exception. We expect C6(CH3)6 to have the lowest melting point and Ge to have the highest melting point, with RbI somewhere in between. The melting points of metals, however, are difficult to predict based on the models presented thus far. Because Zn has a filled valence shell, it should not have a particularly high melting point, so a reasonable guess is C6(CH3)6 < Zn ~ RbI < Ge. The actual melting points are C6(CH3)6, 166°C; Zn, 419°C; RbI, 642°C; and Ge, 938°C. This agrees with our prediction.

Exercise

Classify C60, BaBr2, GaAs, and AgZn as ionic, covalent, molecular, or metallic solids and then arrange them in order of increasing melting points.

Answer: C60 (molecular) < AgZn (metallic) ~ BaBr2 (ionic) < GaAs (covalent). The actual melting points are C60, about 300°C; AgZn, about 700°C; BaBr2, 856°C; and GaAs, 1238°C.

Summary

The major types of solids are ionic, molecular, covalent, and metallic. Ionic solids consist of positively and negatively charged ions held together by electrostatic forces; the strength of the bonding is reflected in the lattice energy. Ionic solids tend to have high melting points and are rather hard. Molecular solids are held together by relatively weak forces, such as dipole–dipole interactions, hydrogen bonds, and London dispersion forces. As a result, they tend to be rather soft and have low melting points, which depend on their molecular structure. Covalent solids consist of two- or three-dimensional networks of atoms held together by covalent bonds; they tend to be very hard and have high melting points. Metallic solids have unusual properties: in addition to having high thermal and electrical conductivity and being malleable and ductile, they exhibit luster, a shiny surface that reflects light. An alloy is a mixture of metals that has bulk metallic properties different from those of its constituent elements. Alloys can be formed by substituting one metal atom for another of similar size in the lattice (substitutional alloys), by inserting smaller atoms into holes in the metal lattice (interstitial alloys), or by a combination of both. Although the elemental composition of most alloys can vary over wide ranges, certain metals combine in only fixed proportions to form intermetallic compounds with unique properties.

Key Takeaway

  • Solids can be classified as ionic, molecular, covalent (network), or metallic, where the general order of increasing strength of interactions is molecular < ionic ≈ metallic < covalent.

Conceptual Problems

  1. Four vials labeled A–D contain sucrose, zinc, quartz, and sodium chloride, although not necessarily in that order. The following table summarizes the results of the series of analyses you have performed on the contents:

    A B C D
    Melting Point high high high low
    Thermal Conductivity poor poor good poor
    Electrical Conductivity in Solid State moderate poor high poor
    Electrical Conductivity in Liquid State high poor high poor
    Hardness hard hard soft soft
    Luster none none high none

    Match each vial with its contents.

  2. Do ionic solids generally have higher or lower melting points than molecular solids? Why? Do ionic solids generally have higher or lower melting points than covalent solids? Explain your reasoning.

  3. The strength of London dispersion forces in molecular solids tends to increase with molecular mass, causing a smooth increase in melting points. Some molecular solids, however, have significantly lower melting points than predicted by their molecular masses. Why?

  4. Suppose you want to synthesize a solid that is both heat resistant and a good electrical conductor. What specific types of bonding and molecular interactions would you want in your starting materials?

  5. Explain the differences between an interstitial alloy and a substitutional alloy. Given an alloy in which the identity of one metallic element is known, how could you determine whether it is a substitutional alloy or an interstitial alloy?

  6. How are intermetallic compounds different from interstitial alloys or substitutional alloys?

Answers

    1. NaCl, ionic solid
    2. quartz, covalent solid
    3. zinc, metal
    4. sucrose, molecular solid
  1. In a substitutional alloy, the impurity atoms are similar in size and chemical properties to the atoms of the host lattice; consequently, they simply replace some of the metal atoms in the normal lattice and do not greatly perturb the structure and physical properties. In an interstitial alloy, the impurity atoms are generally much smaller, have very different chemical properties, and occupy holes between the larger metal atoms. Because interstitial impurities form covalent bonds to the metal atoms in the host lattice, they tend to have a large effect on the mechanical properties of the metal, making it harder, less ductile, and more brittle. Comparing the mechanical properties of an alloy with those of the parent metal could be used to decide whether the alloy were a substitutional or interstitial alloy.

Numerical Problems

  1. Will the melting point of lanthanum(III) oxide be higher or lower than that of ferrous bromide? The relevant ionic radii are as follows: La3+, 104 pm; O2−, 132 pm; Fe2+, 83 pm; and Br, 196 pm. Explain your reasoning.

  2. Draw a graph showing the relationship between the electrical conductivity of metallic silver and temperature.

  3. Which has the higher melting point? Explain your reasoning in each case.

    1. Os or Hf
    2. SnO2 or ZrO2
    3. Al2O3 or SiO2
  4. Draw a graph showing the relationship between the electrical conductivity of a typical semiconductor and temperature.

Answer

    1. Osmium has a higher melting point, due to more valence electrons for metallic bonding.
    2. Zirconium oxide has a higher melting point, because it has greater ionic character.
    3. Aluminum oxide has a higher melting point, again because it has greater ionic character.

12.6 Bonding in Metals and Semiconductors

Learning Objective

  1. To describe the electrical properties of a solid using band theory.

To explain the observed properties of metals, a more sophisticated approach is needed than the electron-sea model described in . The molecular orbital theory we used in to explain the delocalized π bonding in polyatomic ions and molecules such as NO2, ozone, and 1,3-butadiene can be adapted to accommodate the much higher number of atomic orbitals that interact with one another simultaneously in metals.

Band Theory

In a 1 mol sample of a metal, there can be more than 1024 orbital interactions to consider. In our molecular orbital description of metals, however, we begin by considering a simple one-dimensional example: a linear arrangement of n metal atoms, each containing a single electron in an s orbital. We use this example to describe an approach to metallic bonding called band theoryA theory used to describe the bonding in metals and semiconductors., which assumes that the valence orbitals of the atoms in a solid interact, generating a set of molecular orbitals that extend throughout the solid.

One-Dimensional Systems

If the distance between the metal atoms is short enough for the orbitals to interact, they produce bonding, antibonding, and nonbonding molecular orbitals. The left portion of , which is the same as the molecular orbital diagram in , shows the pattern of molecular orbitals that results from the interaction of ns orbitals as n increases from 2 to 5.

Figure 12.21 The Molecular Orbital Energy-Level Diagram for a Linear Arrangement of n Atoms, Each of Which Contains a Singly Occupied s Orbital

This is the same diagram as , with the addition of the far right-hand portion, corresponding to n = 30 and n = ∞. As n becomes very large, the energy separation between adjacent levels becomes so small that a single continuous band of allowed energy levels results. The lowest-energy molecular orbital corresponds to positive overlap between all the atomic orbitals to give a totally bonding combination, whereas the highest-energy molecular orbital contains a node between each pair of atoms and is thus totally antibonding.

As we saw in , the lowest-energy orbital is the completely bonding molecular orbital, whereas the highest-energy orbital is the completely antibonding molecular orbital. Molecular orbitals of intermediate energy have fewer nodes than the totally antibonding molecular orbital. The energy separation between adjacent orbitals decreases as the number of interacting orbitals increases. For n = 30, there are still discrete, well-resolved energy levels, but as n increases from 30 to a number close to Avogadro’s number, the spacing between adjacent energy levels becomes almost infinitely small. The result is essentially a continuum of energy levels, as shown on the right in , each of which corresponds to a particular molecular orbital extending throughout the linear array of metal atoms. The levels that are lowest in energy correspond to mostly bonding combinations of atomic orbitals, those highest in energy correspond to mostly antibonding combinations, and those in the middle correspond to essentially nonbonding combinations.

The continuous set of allowed energy levels shown on the right in is called an energy bandThe continuous set of allowed energy levels generated in band theory when the valence orbitals of the atoms in a solid interact with one another, thus creating a set of molecular orbitals that extend throughout the solid.. The difference in energy between the highest and lowest energy levels is the bandwidthThe difference in energy between the highest and lowest energy levels in an energy band. and is proportional to the strength of the interaction between orbitals on adjacent atoms: the stronger the interaction, the larger the bandwidth. Because the band contains as many energy levels as molecular orbitals, and the number of molecular orbitals is the same as the number of interacting atomic orbitals, the band in contains n energy levels corresponding to the combining of s orbitals from n metal atoms. Each of the original s orbitals could contain a maximum of two electrons, so the band can accommodate a total of 2n electrons. Recall, however, that each of the metal atoms we started with contained only a single electron in each s orbital, so there are only n electrons to place in the band. Just as with atomic orbitals or molecular orbitals, the electrons occupy the lowest energy levels available. Consequently, only the lower half of the band is filled. This corresponds to filling all of the bonding molecular orbitals in the linear array of metal atoms and results in the strongest possible bonding.

Multidimensional Systems

The previous example was a one-dimensional array of atoms that had only s orbitals. To extrapolate to two- or three-dimensional systems and atoms with electrons in p and d orbitals is straightforward in principle, even though in practice the mathematics becomes more complex, and the resulting molecular orbitals are more difficult to visualize. The resulting energy-level diagrams are essentially the same as the diagram of the one-dimensional example in , with the following exception: they contain as many bands as there are different types of interacting orbitals. Because different atomic orbitals interact differently, each band will have a different bandwidth and will be centered at a different energy, corresponding to the energy of the parent atomic orbital of an isolated atom.

Band Gap

Because the 1s, 2s, and 2p orbitals of a period 3 atom are filled core levels, they do not interact strongly with the corresponding orbitals on adjacent atoms. Hence they form rather narrow bands that are well separated in energy (). These bands are completely filled (both the bonding and antibonding levels are completely populated), so they do not make a net contribution to bonding in the solid. The energy difference between the highest level of one band and the lowest level of the next is the band gapThe difference in energy between the highest level of one energy band and the lowest level of the band above it, which represents a set of forbidden energies that do not correspond to any allowed combinations of atomic orbitals.. It represents a set of forbidden energies that do not correspond to any allowed combinations of atomic orbitals.

Figure 12.22 The Band Structures of the Period 3 Metals Na, Mg, and Al

The 3s and 3p valence bands overlap in energy to form a continuous set of energy levels that can hold a maximum of eight electrons per atom.

Because they extend farther from the nucleus, the valence orbitals of adjacent atoms (3s and 3p in ) interact much more strongly with one another than do the filled core levels; as a result, the valence bands have a larger bandwidth. In fact, the bands derived from the 3s and 3p atomic orbitals are wider than the energy gap between them, so the result is overlapping bandsMolecular orbitals derived from two or more different kinds of valence electrons that have similar energies.. These have molecular orbitals derived from two or more valence orbitals with similar energies. As the valence band is filled with one, two, or three electrons per atom for Na, Mg, and Al, respectively, the combined band that arises from the overlap of the 3s and 3p bands is also filling up; it has a total capacity of eight electrons per atom (two electrons for each 3s orbital and six electrons for each set of 3p orbitals). With Na, therefore, which has one valence electron, the combined valence band is one-eighth filled; with Mg (two valence electrons), it is one-fourth filled; and with Al, it is three-eighths filled, as indicated in . The partially filled valence band is absolutely crucial for explaining metallic behavior because it guarantees that there are unoccupied energy levels at an infinitesimally small energy above the highest occupied level.

Band theory can explain virtually all the properties of metals. Metals conduct electricity, for example, because only a very small amount of energy is required to excite an electron from a filled level to an empty one, where it is free to migrate rapidly throughout the crystal in response to an applied electric field. Similarly, metals have high heat capacities (as you no doubt remember from the last time a doctor or a nurse placed a stethoscope on your skin) because the electrons in the valence band can absorb thermal energy by being excited to the low-lying empty energy levels. Finally, metals are lustrous because light of various wavelengths can be absorbed, causing the valence electrons to be excited into any of the empty energy levels above the highest occupied level. When the electrons decay back to low-lying empty levels, they emit light of different wavelengths. Because electrons can be excited from many different filled levels in a metallic solid and can then decay back to any of many empty levels, light of varying wavelengths is absorbed and reemitted, which results in the characteristic shiny appearance that we associate with metals.

Requirements for Metallic Behavior

For a solid to exhibit metallic behavior, it must have a set of delocalized orbitals forming a band of allowed energy levels, and the resulting band must be only partially filled (10%–90%) with electrons. Without a set of delocalized orbitals, there is no pathway by which electrons can move through the solid.

Note the Pattern

Metallic behavior requires a set of delocalized orbitals and a band of allowed energy levels that is partially occupied.

Band theory explains the correlation between the valence electron configuration of a metal and the strength of metallic bonding. The valence electrons of transition metals occupy either their valence ns, (n − 1)d, and np orbitals (with a total capacity of 18 electrons per metal atom) or their ns and (n − 1)d orbitals (a total capacity of 12 electrons per metal atom). These atomic orbitals are close enough in energy that the derived bands overlap, so the valence electrons are not confined to a specific orbital. Metals with 6 to 9 valence electrons (which correspond to groups 6–9) are those most likely to fill the valence bands approximately halfway. Those electrons therefore occupy the highest possible number of bonding levels, while the number of antibonding levels occupied is minimal. Not coincidentally, the elements of these groups exhibit physical properties consistent with the presence of the strongest metallic bonding, such as very high melting points.

Insulators

In contrast to metals, electrical insulatorsA material that conducts electricity poorly because its valence bands are full. are materials that conduct electricity poorly because their valence bands are full. The energy gap between the highest filled levels and the lowest empty levels is so large that the empty levels are inaccessible: thermal energy cannot excite an electron from a filled level to an empty one. The valence-band structure of diamond, for example, is shown in part (a) in . Because diamond has only 4 bonded neighbors rather than the 6 to 12 typical of metals, the carbon 2s and 2p orbitals combine to form two bands in the solid, with the one at lower energy representing bonding molecular orbitals and the one at higher energy representing antibonding molecular orbitals. Each band can accommodate four electrons per atom, so only the lower band is occupied. Because the energy gap between the filled band and the empty band is very large (530 kJ/mol), at normal temperatures thermal energy cannot excite electrons from the filled level into the empty band. Thus there is no pathway by which electrons can move through the solid, so diamond has one of the lowest electrical conductivities known.

Figure 12.23 Energy-Band Diagrams for Diamond, Silicon, and Germanium

The band gap gets smaller from C to Ge.

Semiconductors

What if the difference in energy between the highest occupied level and the lowest empty level is intermediate between those of electrical conductors and insulators? This is the case for silicon and germanium, which have the same structure as diamond. Because Si–Si and Ge–Ge bonds are substantially weaker than C–C bonds, the energy gap between the filled and empty bands becomes much smaller as we go down group 14 (part (b) and part (c) of ). (For more information on bond strengths, see , .) Consequently, thermal energy is able to excite a small number of electrons from the filled valence band of Si and Ge into the empty band above it, which is called the conduction bandThe band of empty molecular orbitals in a semiconductor..

Exciting electrons from the filled valence band to the empty conduction band causes an increase in electrical conductivity for two reasons:

  1. The electrons in the previously vacant conduction band are free to migrate through the crystal in response to an applied electric field.
  2. Excitation of an electron from the valence band produces a “hole” in the valence band that is equivalent to a positive charge. The hole in the valence band can migrate through the crystal in the direction opposite that of the electron in the conduction band by means of a “bucket brigade” mechanism in which an adjacent electron fills the hole, thus generating a hole where the second electron had been, and so forth.

Consequently, Si is a much better electrical conductor than diamond, and Ge is even better, although both are still much poorer conductors than a typical metal (). Substances such as Si and Ge that have conductivities between those of metals and insulators are called semiconductorsA substance such as Si and Ge that has a conductivity between that of metals and insulators.. Many binary compounds of the main group elements exhibit semiconducting behavior similar to that of Si and Ge. For example, gallium arsenide (GaAs) is isoelectronic with Ge and has the same crystalline structure, with alternating Ga and As atoms; not surprisingly, it is also a semiconductor. The electronic structure of semiconductors is compared with the structures of metals and insulators in .

Figure 12.24 A Logarithmic Scale Illustrating the Enormous Range of Electrical Conductivities of Solids

Figure 12.25 A Comparison of the Key Features of the Band Structures of Metals, Semiconductors, and Insulators

Metallic behavior can arise either from the presence of a single partially filled band or from two overlapping bands (one full and one empty).

Temperature and Conductivity

Because thermal energy can excite electrons across the band gap in a semiconductor, increasing the temperature increases the number of electrons that have sufficient kinetic energy to be promoted into the conduction band. The electrical conductivity of a semiconductor therefore increases rapidly with increasing temperature, in contrast to the behavior of a purely metallic crystal. In a metal, as an electron travels through the crystal in response to an applied electrical potential, it cannot travel very far before it encounters and collides with a metal nucleus. The more often such encounters occur, the slower the net motion of the electron through the crystal, and the lower the conductivity. As the temperature of the solid increases, the metal atoms in the lattice acquire more and more kinetic energy. Because their positions are fixed in the lattice, however, the increased kinetic energy increases only the extent to which they vibrate about their fixed positions. At higher temperatures, therefore, the metal nuclei collide with the mobile electrons more frequently and with greater energy, thus decreasing the conductivity. This effect is, however, substantially smaller than the increase in conductivity with temperature exhibited by semiconductors. For example, the conductivity of a tungsten wire decreases by a factor of only about two over the temperature range 750–1500 K, whereas the conductivity of silicon increases approximately 100-fold over the same temperature range. These trends are illustrated in .

Figure 12.26 The Temperature Dependence of the Electrical Conductivity of a Metal versus a Semiconductor

The conductivity of the metal (tungsten) decreases relatively slowly with increasing temperature, whereas the conductivity of the semiconductor (silicon) increases much more rapidly.

Note the Pattern

The electrical conductivity of a semiconductor increases with increasing temperature, whereas the electrical conductivity of a metal decreases with increasing temperature.

n- and p-Type Semiconductors

DopingThe process of deliberately introducing small amounts of impurities into commercial semiconductors to tune their electrical properties for specific applications. is a process used to tune the electrical properties of commercial semiconductors by deliberately introducing small amounts of impurities. If an impurity contains more valence electrons than the atoms of the host lattice (e.g., when small amounts of a group 15 atom are introduced into a crystal of a group 14 element), then the doped solid has more electrons available to conduct current than the pure host has. As shown in part (a) in , adding an impurity such as phosphorus to a silicon crystal creates occasional electron-rich sites in the lattice. The electronic energy of these sites lies between those of the filled valence band and the empty conduction band but closer to the conduction band. Because the atoms that were introduced are surrounded by host atoms, and the electrons associated with the impurity are close in energy to the conduction band, those extra electrons are relatively easily excited into the empty conduction band of the host. Such a substance is called an n-type semiconductorA semiconductor that has been doped with an impurity that has more valence electrons than the atoms of the host lattice., with the n indicating that the added charge carriers are negative (they are electrons).

Figure 12.27 Structures and Band Diagrams of n-Type and p-Type Semiconductors

(a) Doping silicon with a group 15 element results in a new filled level between the valence and conduction bands of the host. (b) Doping silicon with a group 13 element results in a new empty level between the valence and conduction bands of the host. In both cases, the effective band gap is substantially decreased, and the electrical conductivity at a given temperature increases dramatically.

If the impurity atoms contain fewer valence electrons than the atoms of the host (e.g., when small amounts of a group 13 atom are introduced into a crystal of a group 14 element), then the doped solid has fewer electrons than the pure host. Perhaps unexpectedly, this also results in increased conductivity because the impurity atoms generate holes in the valence band. As shown in part (b) in , adding an impurity such as gallium to a silicon crystal creates isolated electron-deficient sites in the host lattice. The electronic energy of these empty sites also lies between those of the filled valence band and the empty conduction band of the host but much closer to the filled valence band. It is therefore relatively easy to excite electrons from the valence band of the host to the isolated impurity atoms, thus forming holes in the valence band. This kind of substance is called a p-type semiconductorA semiconductor that has been doped with an impurity that has fewer valence electrons than the atoms of the host lattice., with the p standing for positive charge carrier (i.e., a hole). Holes in what was a filled band are just as effective as electrons in an empty band at conducting electricity.

Note the Pattern

n-Type semiconductors are negative charge carriers; the impurity has more valence electrons than the host. p-Type semiconductors are positive charge carriers; the impurity has fewer valence electrons than the host.

The electrical conductivity of a semiconductor is roughly proportional to the number of charge carriers, so doping is a precise way to adjust the conductivity of a semiconductor over a wide range. The entire semiconductor industry is built on methods for preparing samples of Si, Ge, or GaAs doped with precise amounts of desired impurities and assembling silicon chips and other complex devices with junctions between n- and p-type semiconductors in varying numbers and arrangements.

Because silicon does not stand up well to temperatures above approximately 100°C, scientists have been interested in developing semiconductors made from diamonds, a more thermally stable material. A new method has been developed based on vapor deposition, in which a gaseous mixture is heated to a high temperature to produce carbon that then condenses on a diamond kernel. This is the same method now used to create cultured diamonds, which are indistinguishable from natural diamonds. The diamonds are heated to more than 2000°C under high pressure to harden them even further. Doping the diamonds with boron has produced p-type semiconductors, whereas doping them with boron and deuterium achieves n-type behavior. Because of their thermal stability, diamond semiconductors have potential uses as microprocessors in high-voltage applications.

Example 8

A crystalline solid has the following band structure, with the purple areas representing regions occupied by electrons. The lower band is completely occupied by electrons, and the upper level is about one-third filled with electrons.

  1. Predict the electrical properties of this solid.
  2. What would happen to the electrical properties if all of the electrons were removed from the upper band? Would you use a chemical oxidant or reductant to effect this change?
  3. What would happen to the electrical properties if enough electrons were added to completely fill the upper band? Would you use a chemical oxidant or reductant to effect this change?

Given: band structure

Asked for: variations in electrical properties with conditions

Strategy:

A Based on the occupancy of the lower and upper bands, predict whether the substance will be an electrical conductor. Then predict how its conductivity will change with temperature.

B After all the electrons are removed from the upper band, predict how the band gap would affect the electrical properties of the material. Determine whether you would use a chemical oxidant or reductant to remove electrons from the upper band.

C Predict the effect of a filled upper band on the electrical properties of the solid. Then decide whether you would use an oxidant or a reductant to fill the upper band.

Solution:

  1. A The material has a partially filled band, which is critical for metallic behavior. The solid will therefore behave like a metal, with high electrical conductivity that decreases slightly with increasing temperature.
  2. B Removing all of the electrons from the partially filled upper band would create a solid with a filled lower band and an empty upper band, separated by an energy gap. If the band gap is large, the material will be an electrical insulator. If the gap is relatively small, the substance will be a semiconductor whose electrical conductivity increases rapidly with increasing temperature. Removing the electrons would require an oxidant because oxidants accept electrons.
  3. C Adding enough electrons to completely fill the upper band would produce an electrical insulator. Without another empty band relatively close in energy above the filled band, semiconductor behavior would be impossible. Adding electrons to the solid would require a reductant because reductants are electron donors.

Exercise

A substance has the following band structure, in which the lower band is half-filled with electrons (purple area) and the upper band is empty.

  1. Predict the electrical properties of the solid.
  2. What would happen to the electrical properties if all of the electrons were removed from the lower band? Would you use a chemical oxidant or reductant to effect this change?
  3. What would happen to the electrical properties if enough electrons were added to completely fill the lower band? Would you use a chemical oxidant or reductant to effect this change?

Answer:

  1. The solid has a partially filled band, so it has the electrical properties of a conductor.
  2. Removing all of the electrons from the lower band would produce an electrical insulator with two empty bands. An oxidant is required.
  3. Adding enough electrons to completely fill the lower level would result in an electrical insulator if the energy gap between the upper and lower bands is relatively large, or a semiconductor if the band gap is relatively small. A reductant is required.

Summary

Band theory assumes that the valence orbitals of the atoms in a solid interact to generate a set of molecular orbitals that extend throughout the solid; the continuous set of allowed energy levels is an energy band. The difference in energy between the highest and lowest allowed levels within a given band is the bandwidth, and the difference in energy between the highest level of one band and the lowest level of the band above it is the band gap. If the width of adjacent bands is larger than the energy gap between them, overlapping bands result, in which molecular orbitals derived from two or more kinds of valence orbitals have similar energies. Metallic properties depend on a partially occupied band corresponding to a set of molecular orbitals that extend throughout the solid to form a band of energy levels. If a solid has a filled valence band with a relatively low-lying empty band above it (a conduction band), then electrons can be excited by thermal energy from the filled band into the vacant band where they can then migrate through the crystal, resulting in electrical conductivity. Electrical insulators are poor conductors because their valence bands are full. Semiconductors have electrical conductivities intermediate between those of insulators and metals. The electrical conductivity of semiconductors increases rapidly with increasing temperature, whereas the electrical conductivity of metals decreases slowly with increasing temperature. The properties of semiconductors can be modified by doping, or introducing impurities. Adding an element with more valence electrons than the atoms of the host populates the conduction band, resulting in an n-type semiconductor with increased electrical conductivity. Adding an element with fewer valence electrons than the atoms of the host generates holes in the valence band, resulting in a p-type semiconductor that also exhibits increased electrical conductivity.

Key Takeaway

  • Bonding in metals and semiconductors can be described using band theory, in which a set of molecular orbitals is generated that extends throughout the solid.

Conceptual Problems

  1. Can band theory be applied to metals with two electrons in their valence s orbitals? with no electrons in their valence s orbitals? Why or why not?

  2. Given a sample of a metal with 1020 atoms, how does the width of the band arising from p orbital interactions compare with the width of the band arising from s orbital interactions? from d orbital interactions?

  3. Diamond has one of the lowest electrical conductivities known. Based on this fact, do you expect diamond to be colored? Why? How do you account for the fact that some diamonds are colored (such as “pink” diamond or “green” diamond)?

  4. Why do silver halides, used in the photographic industry, have band gaps typical of semiconducting materials, whereas alkali metal halides have very large band gaps?

  5. As the ionic character of a compound increases, does its band gap increase or decrease? Why?

  6. Why is silicon, rather than carbon or germanium, used in the semiconductor industry?

  7. Carbon is an insulator, and silicon and germanium are semiconductors. Explain the relationship between the valence electron configuration of each element and their band structures. Which will have the higher electrical conductivity at room temperature—silicon or germanium?

  8. How does doping affect the electrical conductivity of a semiconductor? Draw the effect of doping on the energy levels of the valence band and the conduction band for both an n-type and a p-type semiconductor.

Answers

  1. The low electrical conductivity of diamond implies a very large band gap, corresponding to the energy of a photon of ultraviolet light rather than visible light. Consequently, diamond should be colorless. Pink or green diamonds contain small amounts of highly colored impurities that are responsible for their color.

  2. As the ionic character of a compound increases, the band gap will also increase due to a decrease in orbital overlap. Remember that overlap is greatest for orbitals of the same energy, and that the difference in energy between orbitals on adjacent atoms increases as the difference in electronegativity between the atoms increases. Thus, large differences in electronegativity increase the ionic character, decrease the orbital overlap, and increase the band gap.

Numerical Problems

  1. Of Ca, N, B, and Ge, which will convert pure silicon into a p-type semiconductor when doping? Explain your reasoning.

  2. Of Ga, Si, Br, and P, which will convert pure germanium into an n-type semiconductor when doping? Explain your reasoning.

12.7 Superconductors

Learning Objective

  1. To become familiar with the properties of superconductors.

The phenomenon of superconductivity was discovered by the Danish physicist H. Kamerlingh Onnes (1853–1926; Nobel Prize in Physics, 1913), who found a way to liquefy helium, which boils at 4.2 K and 1 atm pressure. To exploit the very low temperatures made possible by this new cryogenic fluid, he began a systematic study of the properties of metals, especially their electrical properties. Because the electrical resistance of a sample is technically easier to measure than its conductivity, Onnes measured the resistivity of his samples. The resistivity and conductivity of a material are inversely proportional:

Equation 12.2

conductivity=1resistivity

In 1911, Onnes discovered that at about 4 K, the resistivity of metallic mercury (melting point = 234 K) decreased suddenly to essentially zero, rather than continuing to decrease only slowly with decreasing temperature as expected (). He called this phenomenon superconductivityThe phenomenon in which a solid at low temperatures exhibits zero resistance to the flow of electrical current. because a resistivity of zero means that an electrical current can flow forever. Onnes soon discovered that many other metallic elements exhibit superconductivity at very low temperatures. Each of these superconductorsA solid that at low temperatures exhibits zero resistance to the flow of electrical current. has a characteristic superconducting transition temperature (Tc)The temperature at which the electrical resistance of a material drops to zero. at which its resistivity drops to zero. At temperatures less than their Tc, superconductors also completely expel a magnetic field from their interior (part (a) in ). This phenomenon is called the Meissner effectThe phenomenon in which a superconductor completely expels a magnetic field from its interior. after one of its discoverers, the German physicist Walther Meissner, who described the phenomenon in 1933. Due to the Meissner effect, a superconductor will actually “float” over a magnet, as shown in part (b) in .

Figure 12.28 The Temperature Dependence of the Electrical Resistivity of a Normal Metal and a Superconductor

The superconducting transition temperature (Tc) is the temperature at which the resistivity of a superconductor drops to zero.

Figure 12.29 The Meissner Effect

(a) Below its Tc, a superconductor completely expels magnetic lines of force from its interior. (b) In magnetic levitation, a small magnet “floats” over a disk of a high-temperature superconducting material (YBa2Cu3O7−x) cooled in liquid nitrogen.

BCS Theory

For many years, the phenomenon of superconductivity could not be satisfactorily explained by the laws of conventional physics. In the early 1950s, however, American physicists John Bardeen, Leon Cooper, and John Schrieffer formulated a theory for superconductivity that earned them the Nobel Prize in Physics in 1972. According to the BCS theoryA theory used to explain the phenomenon of superconductivity. (named for the initials of their last names), electrons are able to travel through a solid with zero resistance because of attractive interactions involving two electrons that are at some distance from each other. As one electron moves through the lattice, the surrounding nuclei are attracted to it. The motion of the nuclei can create a transient (short-lived) hole that pulls the second electron in the same direction as the first. The nuclei then return to their original positions to avoid colliding with the second electron as it approaches. The pairs of electrons, called Cooper pairsPairs of electrons that migrate through a superconducting material as a unit., migrate through the crystal as a unit. The electrons in Cooper pairs change partners frequently, like dancers in a ballet.

According to the BCS theory, as the temperature of the solid increases, the vibrations of the atoms in the lattice increase continuously, until eventually the electrons cannot avoid colliding with them. The collisions result in the loss of superconductivity at higher temperatures.

The phenomenon of superconductivity suggested many exciting technological applications. For example, using superconducting wires in power cables would result in zero power losses, even over distances of hundreds of miles. Additionally, because superconductors expel magnetic fields, a combination of magnetic rails and superconducting wheels (or vice versa) could be used to produce magnetic levitation of, for example, a train over the track, resulting in friction-free transportation.

Unfortunately, for many years the only superconductors known had serious limitations, especially the need for very low temperatures, which required the use of expensive cryogenic fluids such as liquid He. In addition, the superconducting properties of many substances are destroyed by large electrical currents or even moderately large magnetic fields, making them useless for applications in power cables or high-field magnets. The ability of materials such as NbTi, NbSn, Nb3Si, and Nb3Ge to tolerate rather high magnetic fields, however, has led to a number of commercial applications of superconductors, including high-field magnets for nuclear magnetic resonance (NMR) spectrometers and magnetic resonance imaging (MRI) instruments in medicine, which, unlike x-rays, can detect small changes in soft tissues in the body.

High-Temperature Superconductors

Because of these limitations, scientists continued to search for materials that exhibited superconductivity at temperatures greater than 77 K (the temperature of liquid nitrogen, the least expensive cryogenic fluid). In 1986, Johannes G. Bednorz and Karl A. Müller, working for IBM in Zurich, showed that certain mixed-metal oxides containing La, Ba, and Cu exhibited superconductivity above 30 K. These compounds had been prepared by French workers as potential solid catalysts some years earlier, but their electrical properties had never been examined at low temperatures. Although initially the scientific community was extremely skeptical, the compounds were so easy to prepare that the results were confirmed within a few weeks. These high-temperature superconductorsA material that becomes a superconductor at temperatures greater than 30 K. earned Bednorz and Müller the Nobel Prize in Physics in 1987. Subsequent research has produced new compounds with related structures that are superconducting at temperatures as high as 135 K. The best known of these was discovered by Paul Chu and Maw-Kuen Wu Jr. and is called the “Chu–Wu phase” or the 1-2-3 superconductor.

The formula for the 1-2-3 superconductor is YBa2Cu3O7−x, where x is about 0.1 for samples that superconduct at about 95 K. If x ≈ 1.0, giving a formula of YBa2Cu3O6, the material is an electrical insulator. The superconducting phase is thus a nonstoichiometric compound, with a fixed ratio of metal atoms but a variable oxygen content. The overall equation for the synthesis of this material is as follows:

Equation 12.3

Y2O3(s)+4BaCO3(s)+6CuO(s)+12O2(g)Δ2YBa2Cu3O7(s)+4CO2(g)

If we assume that the superconducting phase is really stoichiometric YBa2Cu3O7, then the average oxidation states of O, Y, Ba, and Cu are −2, +3, +2, and +73, respectively. The simplest way to view the average oxidation state of Cu is to assume that two Cu atoms per formula unit are present as Cu2+ and one is present as the rather unusual Cu3+. In YBa2Cu3O6, the insulating form, the oxidation state of Cu is +53, so there are two Cu2+ and one Cu+ per formula unit.

As shown in , the unit cell of the 1-2-3 superconductor is related to the unit cell of the simple perovskite structure (part (b) in ). The only difference between the superconducting and insulating forms of the compound is that an O atom has been removed from between the Cu3+ ions, which destroys the chains of Cu atoms and leaves the Cu in the center of the unit cell as Cu+. The chains of Cu atoms are crucial to the formation of the superconducting state.

Figure 12.30 The Relationship of the Structure of a Superconductor Consisting of Y-Ba-Cu-O to a Simple Perovskite Structure

(a) Stacking three unit cells of the Ca-centered CaTiO3 perovskite structure (part (b) in ) together with (b) replacement of all Ti atoms by Cu, replacement of Ca in the top and bottom cubes by Ba, and replacement of Ca in the central cube by Y gives a YBa2Cu3O9 stoichiometry. (c) The removal of two oxygen atoms per unit cell gives the nominal YBa2Cu3O7 stoichiometry of the superconducting material.

lists the ideal compositions of some of the known high-temperature superconductors that have been discovered in recent years. Engineers have learned how to process the brittle polycrystalline 1-2-3 and related compounds into wires, tapes, and films that can carry enormous electrical currents. Commercial applications include their use in infrared sensors and in analog signal processing and microwave devices.

Table 12.7 The Composition of Various Superconductors

Compound Tc (K)
Ba(Pb1−xBix)O3 13.5
(La2−xSrx)CuO4 35
YBa2Cu3O7−x 95
Bi2(Sr2−xCax)CuO6* 80
Bi2Ca2Sr2Cu3O10* 110
Tl2Ba2Ca2Cu3O10* 125
HgBa2Ca2Cu3O8* 133
K3C60 18
Rb3C60 30
*Nominal compositions only. Oxygen deficiencies or excesses are common in these compounds.

Example 9

Calculate the average oxidation state of Cu in a sample of YBa2Cu3O7−x with x = 0.5. How do you expect its structure to differ from those shown in for YBa2Cu3O9 and YBa2Cu3O7?

Given: stoichiometry

Asked for: average oxidation state and structure

Strategy:

A Based on the oxidation states of the other component atoms, calculate the average oxidation state of Cu that would make an electrically neutral compound.

B Compare the stoichiometry of the structures shown in with the stoichiometry of the given compound to predict how the structures differ.

Solution:

A The net negative charge from oxygen is (7.0 − 0.5) (−2) = −13, and the sum of the charges on the Y and Ba atoms is [1 × (+3)] + [2 × (+2)] = +7. This leaves a net charge of −6 per unit cell, which must be compensated for by the three Cu atoms, for a net charge of +63=+2 per Cu.

B The most likely structure would be one in which every other O atom between the Cu atoms in the Cu chains of YBa2Cu3O7 has been removed.

Exercise

Calculate the average oxidation state of Cu in a sample of HgBa2Ca2Cu3O8. Assume that Hg is present as Hg2+.

Answer: +2

Summary

Superconductors are solids that at low temperatures exhibit zero resistance to the flow of electrical current, a phenomenon known as superconductivity. The temperature at which the electrical resistance of a substance drops to zero is its superconducting transition temperature (Tc). Superconductors also expel a magnetic field from their interior, a phenomenon known as the Meissner effect. Superconductivity can be explained by the BCS theory, which says that electrons are able to travel through a solid with no resistance because they couple to form pairs of electrons (Cooper pairs). High-temperature superconductors have Tc values greater than 30 K.

Key Takeaway

  • Superconductivity can be described using the BCS theory, in which Cooper pairs of electrons migrate through the crystal as a unit.

Conceptual Problems

  1. Why does the BCS theory predict that superconductivity is not possible at temperatures above approximately 30 K?

  2. How does the formation of Cooper pairs lead to superconductivity?

Answer

  1. According to BCS theory, the interactions that lead to formation of Cooper pairs of electrons are so weak that they should be disrupted by thermal vibrations of lattice atoms above about 30 K.

12.8 Polymeric Solids

Learning Objective

  1. To understand the differences between synthetic and biological polymers.

Most of the solids discussed so far have been molecules or ions with low molecular masses, ranging from tens to hundreds of atomic mass units. Many of the molecular materials in consumer goods today, however, have very high molecular masses, ranging from thousands to millions of atomic mass units, and are formed from a carefully controlled series of reactions that produce giant molecules called polymersA giant molecule that consists of many basic structural units (monomers) connected in a chain or network by covalent bonds. (from the Greek poly and meros, meaning “many parts”). Polymers are used in corrective eye lenses, plastic containers, clothing and textiles, and medical implant devices, among many other uses. They consist of basic structural units called monomersThe basic structural unit of a polymer., which are repeated many times in each molecule. As shown schematically in , polymerizationA process by which monomers are connected into chains or networks by covalent bonds. is the process by which monomers are connected into chains or networks by covalent bonds. Polymers can form via a condensation reaction, in which two monomer molecules are joined by a new covalent bond and a small molecule such as water is eliminated, or by an addition reaction, a variant of a condensation reaction in which the components of a species AB are added to adjacent atoms of a multiple bond. (For more information about condensation and addition reactions, see , .) Many people confuse the terms plastics and polymers. PlasticThe property of a material that allows it to be molded into almost any shape. is the property of a material that allows it to be molded into almost any shape. Although many plastics are polymers, many polymers are not plastics. In this section, we introduce the reactions that produce naturally occurring and synthetic polymers.

Figure 12.31 Polymer Formation

During a polymerization reaction, a large number of monomers become connected by covalent bonds to form a single long molecule, a polymer.

Note the Pattern

Polymers are formed via condensation or addition reactions.

Naturally Occurring Polymers: Peptides and Proteins

Polymers that occur naturally are crucial components of all organisms and form the fabric of our lives. Hair, silk, skin, feathers, muscle, and connective tissue are all primarily composed of proteins, the most familiar kind of naturally occurring, or biological, polymer. The monomers of many biological polymers are the amino acids introduced in , , each called an amino acid residue. The residues are linked together by amide bonds, also called peptide bonds, via a condensation reaction where H2O is eliminated:

In the above equation, R represents an alkyl or aryl group, or hydrogen, depending on the amino acid. We write the structural formula of the product with the free amino group on the left (the N-terminus) and the free carboxylate group on the right (the C-terminus). For example, the structural formula for the product formed from the amino acids glycine and valine (glycyl-valine) is as follows:

The most important difference between synthetic and naturally occurring polymers is that the former usually contain very few different monomers, whereas biological polymers can have as many as 20 different kinds of amino acid residues arranged in many different orders. Chains with less than about 50 amino acid residues are called peptidesBiological polymers with less than about 50 amino acid residues., whereas those with more than about 50 amino acid residues are called proteinsBiological polymers with more than 50 amino acid residues linked together by amide bonds.. Many proteins are enzymesCatalysts that occur naturally in living organisms and that catalyze biological reactions., which are catalysts that increase the rate of a biological reaction.

Note the Pattern

Synthetic polymers usually contain only a few different monomers, whereas biological polymers can have many kinds of monomers, such as amino acids arranged in different orders.

Many small peptides have potent physiological activities. The endorphins, for example, are powerful, naturally occurring painkillers found in the brain. Other important peptides are the hormones vasopressin and oxytocin. Although their structures and amino acid sequences are similar, vasopressin is a blood pressure regulator, whereas oxytocin induces labor in pregnant women and milk production in nursing mothers. Oxytocin was the first biologically active peptide to be prepared in the laboratory by Vincent du Vigneaud (1901–1978), who was awarded the Nobel Prize in Chemistry in 1955.

Synthetic Polymers

Many of the synthetic polymers we use, such as plastics and rubbers, have commercial advantages over naturally occurring polymers because they can be produced inexpensively. Moreover, many synthetic polymers are actually more desirable than their natural counterparts because scientists can select monomer units to tailor the physical properties of the resulting polymer for particular purposes. For example, in many applications, wood has been replaced by plastics that are more durable, lighter, and easier to shape and maintain. Polymers are also increasingly used in engineering applications where weight reduction and corrosion resistance are required. Steel rods used to support concrete structures, for example, are often coated with a polymeric material when the structures are near ocean environments where steel is vulnerable to corrosion (For more information on corrosion, see , .) In fact, the use of polymers in engineering applications is a very active area of research.

Probably the best-known example of a synthetic polymer is nylon (). Its monomers are linked by amide bonds (which are called peptide bonds in biological polymers), so its physical properties are similar to those of some proteins because of their common structural unit—the amide group. Nylon is easily drawn into silky fibersA particle of a synthetic polymer that is more than 100 times longer than it is wide. that are more than a hundred times longer than they are wide and can be woven into fabrics. Nylon fibers are so light and strong that during World War II, all available nylon was commandeered for use in parachutes, ropes, and other military items. With polymer chains that are fully extended and run parallel to the fiber axis, nylon fibers resist stretching, just like naturally occurring silk fibers, although the structures of nylon and silk are otherwise different. Replacing the flexible –CH2– units in nylon by aromatic rings produces a stiffer and stronger polymer, such as the very strong polymer known as Kevlar. Kevlar fibers are so strong and rigid that they are used in lightweight army helmets, bulletproof vests, and even sailboat and canoe hulls, all of which contain multiple layers of Kevlar fabric.

Figure 12.32 The Synthesis of Nylon

Nylon is a synthetic condensation polymer created by the reaction of a dicarboxylic acid and a diamine to form amide bonds and water.

A fiberglass mat (left) and a Kevlar vest (right).

Not all synthetic polymers are linked by amide bonds—for example, polyesters contain monomers that are linked by ester bonds. Polyesters are sold under trade names such as Dacron, Kodel, and Fortrel, which are used in clothing, and Mylar, which is used in magnetic tape, helium-filled balloons, and high-tech sails for sailboats. Although the fibers are flexible, properly prepared Mylar films are almost as strong as steel.

Polymers based on skeletons with only carbon are all synthetic. Most of these are formed from ethylene (CH2=CH2), a two-carbon building block, and its derivatives. The relative lengths of the chains and any branches control the properties of polyethylene. For example, higher numbers of branches produce a softer, more flexible, lower-melting-point polymer called low-density polyethylene (LDPE), whereas high-density polyethylene (HDPE) contains few branches. Substances such as glass that melt at relatively low temperatures can also be formed into fibers, producing fiberglass.

Because most synthetic fibers are neither soluble nor low melting, multistep processes are required to manufacture them and form them into objects. Graphite fibers are formed by heating a precursor polymer at high temperatures to decompose it, a process called pyrolysisA high-temperature decomposition reaction that can be used to form fibers of synthetic polymers.. The usual precursor for graphite is polyacrylonitrile, better known by its trade name—Orlon. A similar approach is used to prepare fibers of silicon carbide using an organosilicon precursor such as polydimethylsilane {[–(CH3)2Si–]n}. A new type of fiber consisting of carbon nanotubes, hollow cylinders of carbon just one atom thick, is lightweight, strong, and impact resistant. Its performance has been compared to that of Kevlar, and it is being considered for use in body armor, flexible solar panels, and bombproof trash bins, among other uses.

Because there are no good polymer precursors for elemental boron or boron nitride, these fibers have to be prepared by time-consuming and costly indirect methods. Even though boron fibers are about eight times stronger than metallic aluminum and 10% lighter, they are significantly more expensive. Consequently, unless an application requires boron’s greater resistance to oxidation, these fibers cannot compete with less costly graphite fibers.

Example 10

Polyethylene is used in a wide variety of products, including beach balls and the hard plastic bottles used to store solutions in a chemistry laboratory. Which of these products is formed from the more highly branched polyethylene?

Given: type of polymer

Asked for: application

Strategy:

Determine whether the polymer is LDPE, which is used in applications that require flexibility, or HDPE, which is used for its strength and rigidity.

Solution:

A highly branched polymer is less dense and less rigid than a relatively unbranched polymer. Thus hard, strong polyethylene objects such as bottles are made of HDPE with relatively few branches. In contrast, a beach ball must be flexible so it can be inflated. It is therefore made of highly branched LDPE.

Exercise

Which products are manufactured from LDPE and which from HPDE?

  1. lawn chair frames
  2. rope
  3. disposable syringes
  4. automobile protective covers

Answer:

  1. HDPE
  2. LDPE
  3. HDPE
  4. LDPE

Summary

Polymers are giant molecules that consist of long chains of units called monomers connected by covalent bonds. Polymerization is the process of linking monomers together to form a polymer. Plastic is the property of a material that allows it to be molded. Biological polymers formed from amino acid residues are called peptides or proteins, depending on their size. Enzymes are proteins that catalyze a biological reaction. A particle that is more than a hundred times longer than it is wide is a fiber, which can be formed by a high-temperature decomposition reaction called pyrolysis.

Key Takeaway

  • Polymers are giant molecules formed from addition or condensation reactions and can be classified as either biological or synthetic polymers.

Conceptual Problems

  1. How are amino acids and proteins related to monomers and polymers? Draw the general structure of an amide bond linking two amino acid residues.

  2. Although proteins and synthetic polymers (such as nylon) both contain amide bonds, different terms are used to describe the two types of polymer. Compare and contrast the terminology used for the

    1. smallest repeating unit.
    2. covalent bond connecting the units.

12.9 Contemporary Materials

Learning Objective

  1. To become familiar with the properties of some contemporary materials.

In addition to polymers, other materials, such as ceramics, high-strength alloys, and composites, play a major role in almost every aspect of our lives. Until relatively recently, steel was used for any application that required an especially strong and durable material, such as bridges, automobiles, airplanes, golf clubs, and tennis rackets. In the last 15 to 20 years, however, graphite or boron fiber golf clubs and tennis rackets have made wood and steel obsolete for these items. Likewise, a modern jet engine now is largely composed of Ti and Ni by weight rather than steel (). The percentage of iron in wings and fuselages is similarly low, which indicates the extent to which other materials have supplanted steel. The Chevrolet Corvette introduced in 1953 was considered unusual because its body was constructed of fiberglass, a composite material, rather than steel; by 1992, Jaguar fabricated an all-aluminum limited-edition vehicle. In fact, the current models of many automobiles have engines that are made mostly of aluminum rather than steel. In this section, we describe some of the chemistry behind three classes of contemporary materials: ceramics, superalloys, and composites.

Table 12.8 The Approximate Elemental Composition of a Modern Jet Engine

Element Percentage by Mass
titanium 38
nickel 37
chromium 12
cobalt 6
aluminum 3
niobium 1
tantalum 0.025

Ceramics

A ceramicAny nonmetallic inorganic solid that is strong eneough to be used in structural applications. is any nonmetallic, inorganic solid that is strong enough for use in structural applications. Traditional ceramics, which are based on metal silicates or aluminosilicates, are the materials used to make pottery, china, bricks, and concrete. Modern ceramics contain a much wider range of components and can be classified as either ceramic oxides, which are based on metal oxides such as alumina (Al2O3), zirconia (ZrO2), and beryllia (BeO), or nonoxide ceramics, which are based on metal carbides such as silicon carbide (carborundum, SiC) and tungsten carbide (WC), or nitrides like silicon nitride (Si3N4) and boron nitride (BN).

All modern ceramics are hard, lightweight, and stable at very high temperatures. Unfortunately, however, they are also rather brittle, tending to crack or break under stresses that would cause metals to bend or dent. Thus a major challenge for materials scientists is to take advantage of the desirable properties of ceramics, such as their thermal and oxidative stability, chemical inertness, and toughness, while finding ways to decrease their brittleness to use them in new applications. Few metals can be used in jet engines, for example, because most lose mechanical strength and react with oxygen at the very high operating temperatures inside the engines (approximately 2000°C). In contrast, ceramic oxides such as Al2O3 cannot react with oxygen regardless of the temperature because aluminum is already in its highest possible oxidation state (Al3+). Even nonoxide ceramics such as silicon and boron nitrides and silicon carbide are essentially unreactive in air up to about 1500°C. Producing a high-strength ceramic for service use involves a process called sinteringA process that fuses the grains of a ceramic into a dense, strong material. Sintering is used to produce high-strength ceramics., which fuses the grains into a dense and strong material ().

Figure 12.33 Sintering

These photos show the effects of sintering magnesium oxide grains: (a) the microstructure before sintering; (b) the microstructure of the ceramic after sintering for two hours at 1250°C; and (c) the microstructure after sintering for two hours at 1450°C. During the sintering process, the grains fuse, forming a dense and strong material.

Note the Pattern

Ceramics are hard, lightweight, and able to withstand high temperatures, but they are also brittle.

One of the most widely used raw materials for making ceramics is clay. Clay minerals consist of hydrated alumina (Al2O3) and silica (SiO2) that have a broad range of impurities, including barium, calcium, sodium, potassium, and iron. Although the structures of clay minerals are complicated, they all contain layers of metal atoms linked by oxygen atoms. Water molecules fit between the layers to form a thin film of water. When hydrated, clays can be easily molded, but during high-temperature heat treatment, called firing, a dense and strong ceramic is produced.

Because ceramics are so hard, they are easily contaminated by the material used to grind them. In fact, the ceramic often grinds the metal surface of the mill almost as fast as the mill grinds the ceramic! The sol-gel processA process used to manufacture ceramics by producing fine powders of ceramic oxides with uniformly sized particles. was developed to address this problem. In this process, a water-soluble precursor species, usually a metal or semimetal alkoxide [M(OR)n] undergoes a hydrolysis reaction to form a cloudy aqueous dispersion called a sol. The sol contains particles of the metal or semimetal hydroxide [M(OH)n], which are typically 1–100 nm in diameter. As the reaction proceeds, molecules of water are eliminated from between the M(OH)n units in a condensation reaction, and the particles fuse together, producing oxide bridges, M–O–M. Eventually, the particles become linked in a three-dimensional network that causes the solution to form a gel, similar to a gelatin dessert. Heating the gel to 200°C–500°C causes more water to be eliminated, thus forming small particles of metal oxide that can be amazingly uniform in size. This chemistry starts with highly pure SiCl4 and proceeds via the following reactions:

Equation 12.4

SiCl4(s)+4CH3CH2OH(l)+4NH3(g)Si(OCH2CH3)4(s)+4NH4Cl(s)                alkoxide formation

Equation 12.5

Si(OCH2CH3)4(s)+4H2O(l)(ΗΟ)3SiOH(s)+4CH3CH2OH(aq)hydroylysis of the alkoxide (sol)

Equation 12.6

(HO)3SiOH(s)+nHOSi(OH)3(s)(ΗΟ)3Si(OSi(OH)3)n(s)+nH2O(l)condensation

Nature uses the same process to create opal gemstones.

Superalloys

SuperalloysA high-strength alloy based on cobalt, nickel, and iron, often of complex composition, that is used in applications that require mechanical strength, high surface stability, and resistance to high temperatures. are high-strength alloys, often with a complex composition, that are used in systems requiring mechanical strength, high surface stability (minimal flaking or pitting), and resistance to high temperatures. The aerospace industry, for example, requires materials that have high strength-to-weight ratios to improve the fuel efficiency of advanced propulsion systems, and these systems must operate safely at temperatures greater than 1000°C.

Note the Pattern

Superalloys are used in systems requiring mechanical strength, minimal flaking or pitting, and high-temperature resistance.

Although most superalloys are based on nickel, cobalt, or iron, other metals are used as well. Pure nickel or cobalt is relatively easily oxidized, but adding small amounts of other metals (Al, Co, Cr, Mo, Nb, Ti, and W) results in an alloy that has superior properties. Consequently, most of the internal parts of modern gas turbine jet engines are now made of superalloys based on either nickel (used in blades and disks) or cobalt (used in vanes, combustion chamber liners, and afterburners). The cobalt-based superalloys are not as strong as the nickel-based ones, but they have excellent corrosion resistance at high temperatures.

Other alloys, such as aluminum–lithium and alloys based on titanium, also have applications in the aerospace industry. Because aluminum–lithium alloys are lighter, stiffer, and more resistant to fatigue at high temperatures than aluminum itself, they are used in engine parts and in the metal “skins” that cover wings and bodies. Titanium’s high strength, corrosion resistance, and lightweight properties are equally desirable for applications where minimizing weight is important (as in airplanes). Unfortunately, however, metallic titanium reacts rapidly with air at high temperatures to form TiN and TiO2. The welding of titanium or any similar processes must therefore be carried out in an argon or inert gas atmosphere, which adds significantly to the cost. Initially, titanium and its alloys were primarily used in military applications, but more recently, they have been used as components of the airframes of commercial planes, in ship structures, and in biological implants.

Composite Materials

Composite materialsA material that consists of at least two distinct phases: the matrix (which constitutes the bulk of the material) and fibers or granules that are embedded within the matrix. have at least two distinct components: the matrix (which constitutes the bulk of the material) and fibers or granules that are embedded within the matrix and limit the growth of cracks by pinning defects in the bulk material (). The resulting material is stronger, tougher, stiffer, and more resistant to corrosion than either component alone. Composites are thus the nanometer-scale equivalent of reinforced concrete, in which steel rods greatly increase the mechanical strength of the cement matrix, and are extensively used in the aircraft industry, among others. For example, the Boeing 777 is 9% composites by weight, whereas the newly developed Boeing 787 is 50% composites by weight. Not only does the use of composite materials reduce the weight of the aircraft, and therefore its fuel consumption, but it also allows new design concepts because composites can be molded. Moreover, by using composites in the Boeing 787 multiple functions can be integrated into a single system, such as acoustic damping, thermal regulation, and the electrical system.

Three distinct types of composite material are generally recognized, distinguished by the nature of the matrix. These are polymer-matrix composites, metal-matrix composites, and ceramic-matrix composites.

Figure 12.34 Some Possible Arrangements of Fibers in Fiber-Reinforced Composite Materials

The arrangements shown range from discontinuous and randomly oriented to continuous and aligned. The fibers limit the growth of cracks by pinning defects within the matrix.

Note the Pattern

Composites are stronger, tougher, stiffer, and more resistant to corrosion than their components alone.

Fiberglass is a polymer-matrix compositeA compositie that consists of reinforcing fibers embedded in a polymer matrix. that consists of glass fibers embedded in a polymer, forming tapes that are then arranged in layers impregnated with epoxy. The result is a strong, stiff, lightweight material that is resistant to chemical degradation. It is not strong enough, however, to resist cracking or puncturing on impact. Stronger, stiffer polymer-matrix composites contain fibers of carbon (graphite), boron, or polyamides such as Kevlar. High-tech tennis rackets and golf clubs as well as the skins of modern military aircraft such as the “stealth” F-117A fighters and B-2 bombers are made from both carbon fiber–epoxy and boron fiber–epoxy composites. Compared with metals, these materials are 25%–50% lighter and thus reduce operating costs. Similarly, the space shuttle payload bay doors and panels are made of a carbon fiber–epoxy composite. The structure of the Boeing 787 has been described as essentially one giant macromolecule, where everything is fastened through cross-linked chemical bonds reinforced with carbon fiber.

Metal-matrix compositesA composite that consists of reinforcing fibers embedded in a metal or a metal alloy matrix. consist of metals or metal alloys reinforced with fibers. They offer significant advantages for high-temperature applications but pose major manufacturing challenges. For example, obtaining a uniform distribution and alignment of the reinforcing fibers can be difficult, and because organic polymers cannot survive the high temperatures of molten metals, only fibers composed of boron, carbon, or ceramic (such as silicon carbide) can be used. Aluminum alloys reinforced with boron fibers are used in the aerospace industry, where their strength and lightweight properties make up for their relatively high cost. The skins of hypersonic aircraft and structural units in the space shuttle are made of metal-matrix composites.

Ceramic-matrix compositesA composite consisting of reinforcing fibers embedded in a ceramic matrix. contain ceramic fibers in a ceramic matrix material. A typical example is alumina reinforced with silicon carbide fibers. Combining the two very high-melting-point materials results in a composite that has excellent thermal stability, great strength, and corrosion resistance, while the SiC fibers reduce brittleness and cracking. Consequently, these materials are used in very high-temperature applications, such as the leading edge of wings of hypersonic airplanes and jet engine parts. They are also used in the protective ceramic tiles on the space shuttle, which contain short fibers of pure SiO2 mixed with fibers of an aluminum–boron–silicate ceramic. These tiles are excellent thermal insulators and extremely light (their density is only about 0.2 g/cm3). Although their surface reaches a temperature of about 1250°C during reentry into Earth’s atmosphere, the temperature of the underlying aluminum alloy skin stays below 200°C.

Example 11

An engineer is tasked with designing a jet ski hull. What material is most suited to this application? Why?

Given: design objective

Asked for: most suitable material

Strategy:

Determine under what conditions the design will be used. Then decide what type of material is most appropriate.

Solution:

A jet ski hull must be lightweight to maximize speed and fuel efficiency. Because of its use in a marine environment, it must also be resistant to impact and corrosion. A ceramic material provides rigidity but is brittle and therefore tends to break or crack under stress, such as when it impacts waves at high speeds. Superalloys provide strength and stability, but a superalloy is probably too heavy for this application. Depending on the selection of metals, it might not be resistant to corrosion in a marine environment either. Composite materials, however, provide strength, stiffness, and corrosion resistance; they are also lightweight materials. This is not a high-temperature application, so we do not need a metal-matrix composite or a ceramic-matrix composite. The best choice of material is a polymer-matrix composite with Kevlar fibers to increase the strength of the composite on impact.

Exercise

In designing a new generation of space shuttle, National Aeronautics and Space Administration (NASA) engineers are considering thermal-protection devices to protect the skin of the craft. Among the materials being considered are titanium- or nickel-based alloys and silicon-carbide ceramic reinforced with carbon fibers. Why are these materials suitable for this application?

Answer: Ti- or Ni-based alloys have a high strength-to-weight ratio, resist corrosion, and are safe at high temperatures. Reinforced ceramic is lightweight; has high thermal and oxidative stability; and is chemically inert, tough, and impact resistant.

Summary

Ceramics are nonmetallic, inorganic solids that are typically strong; they have high melting points but are brittle. The two major classes of modern ceramics are ceramic oxides and nonoxide ceramics, which are composed of nonmetal carbides or nitrides. The production of ceramics generally involves pressing a powder of the material into the desired shape and sintering at a temperature just below its melting point. The necessary fine powders of ceramic oxides with uniformly sized particles can be produced by the sol-gel process. Superalloys are new metal phases based on cobalt, nickel, or iron that exhibit unusually high temperature stability and resistance to oxidation. Composite materials consist of at least two phases: a matrix that constitutes the bulk of the material and fibers or granules that act as a reinforcement. Polymer-matrix composites have reinforcing fibers embedded in a polymer matrix. Metal-matrix composites have a metal matrix and fibers of boron, graphite, or ceramic. Ceramic-matrix composites use reinforcing fibers, usually also ceramic, to make the matrix phase less brittle.

Key Takeaway

  • Materials that have contemporary applications include ceramics, high-strength alloys, and composites, whose properties can be modified as needed.

Conceptual Problems

  1. Can a compound based on titanium oxide qualify as a ceramic material? Explain your answer.

  2. What features make ceramic materials attractive for use under extreme conditions? What are some potential drawbacks of ceramics?

  3. How do composite materials differ from the other classes of materials discussed in this chapter? What advantages do composites have versus other materials?

  4. How does the matrix control the properties of a composite material? What is the role of an additive in determining the properties of a composite material?

12.10 End-of-Chapter Material

Application Problems

    Problems marked with a ♦ involve multiple concepts.

  1. ♦ Cadmium selenide (CdSe) is a semiconductor used in photoconductors and photoelectric cells that conduct electricity when illuminated. In a related process, a CdSe crystal can absorb enough energy to excite electrons from the valence band to the conduction band, and the excited electrons can return to the valence band by emitting light. The relative intensity and peak wavelength of the emitted light in one experiment are shown in the following table:

    Relative Intensity (%) Wavelength (nm) Temperature (°C)
    100 720 23
    50 725 45
    10 730 75
    1. Explain why the emitted light shifts to longer wavelength at higher temperatures. (Hint: consider the expansion of the crystal and the resulting changes in orbital interactions when heated.)
    2. Why does the relative intensity of the emitted light decrease as the temperature increases?
  2. A large fraction of electrical energy is currently lost as heat during transmission due to the electrical resistance of transmission wires. How could superconducting technology improve the transmission of electrical power? What are some potential drawbacks of this technology?

  3. Light-emitting diodes (LEDs) are semiconductor-based devices that are used in consumer electronics products ranging from digital clocks to fiber-optic telephone transmission lines. The color of the emitted light is determined in part by the band gap of the semiconductor. Electrons can be promoted to the conduction band and return to the valence band by emitting light or by increasing the magnitude of atomic vibrations in the crystal, which increases its temperature. If you wanted to increase the efficiency of an LED display, and thereby the intensity of the emitted light, would you increase or decrease the operating temperature of the LED? Explain your answer.

  4. ♦ Strips of pure Au and Al are often used in close proximity to each other on circuit boards. As the boards become warm during use, however, the metals can diffuse, forming a purple alloy known as “the purple plague” between the strips. Because the alloy is electrically conductive, the board short-circuits. A structural analysis of the purple alloy showed that its structure contained a face-centered cubic (fcc) lattice of atoms of one element, with atoms of the other element occupying tetrahedral holes. What type of alloy is this? Which element is most likely to form the fcc lattice? Which element is most likely to occupy the tetrahedral holes? Explain your answers. What is the empirical formula of the “purple plague”?

  5. ♦ Glasses are mixtures of oxides, the main component of which is silica (SiO2). Silica is called the glass former, while additives are referred to as glass modifiers. The crystalline lattice of the glass former breaks down during heating, producing the random atomic arrangements typical of a liquid. Adding a modifier and cooling the melt rapidly produces a glass. How does the three-dimensional structure of the glass differ from that of the crystalline glass former? Would you expect the melting point of a glass to be higher or lower than that of pure SiO2? Lead glass, a particular favorite of the Romans, was formed by adding lead oxide as the modifier. Would you expect lead glass to be more or less dense than soda-lime glass formed by adding sodium and potassium salts as modifiers?

  6. Many glasses eventually crystallize, rendering them brittle and opaque. Modifying agents such as TiO2 are frequently added to molten glass to reduce their tendency to crystallize. Why does the addition of small amounts of TiO2 stabilize the amorphous structure of glass?

  7. ♦ The carbon–carbon bond distances in polyacetylene (–CH=CH–)n alternate between short and long, resulting in the following band structure:

    1. Is polyacetylene a metal, a semiconductor, or an insulator?
    2. Based on its band structure, how would treating polyacetylene with a potent oxidant affect its electrical conductivity? What would be the effect of treating polyacetylene with small amounts of a powerful reductant? Explain your answers.
  8. Enkephalins are pentapeptides, short biopolymers that are synthesized by humans to control pain. Enkephalins bind to certain receptors in brain cells, which are also known to bind morphine and heroin. One enkephalin has the structure tyrosine–glycine–glycine–phenylalanine–methionine. Draw its structure.

  9. A polymerization reaction is used to synthesize Saran, a flexible material used in packaging film and seat covers. The monomeric unit for Saran is 1,1-dichloroethylene (CH2=CCl2), also known as vinylidene chloride. Draw a reasonable structure for the polymer. Why do pieces of Saran “cling” to one another when they are brought in contact?

  10. Polymers are often amorphous solids. Like other materials, polymers can also undergo phase changes. For example, many polymers are flexible above a certain temperature, called the glass-transition temperature (Tg). Below the glass transition temperature, the polymer becomes hard and brittle. Biomedical devices that replace or augment parts of the human body often contain a wide variety of materials whose properties must be carefully controlled.

    1. Polydimethylsiloxane has a Tg of −123°C, whereas poly(methylmethacrylate) has a Tg of 105°C. Which of these polymers is likely to be used in dentures, and which is likely to be used for soft-tissue replacement?
    2. If you were designing biomedical devices, which class of biomaterials (alloys, ceramics, or polymers) would you consider for finger joint replacements, eyeball replacements, windpipe replacements, shoulder joint replacements, and bridging bone fractures? Explain your answers.

Chapter 13 Solutions

We explored the general properties of gases, liquids, and solids in , , and , respectively. Most of the discussion focused on pure substances containing a single kind of atom, molecule, or cation–anion pair. The substances we encounter in our daily lives, however, are usually mixtures rather than pure substances. Some are heterogeneous mixtures, which consist of at least two phases that are not uniformly dispersed on a microscopic scale; others are homogeneous mixtures, consisting of a single phase in which the components are uniformly distributed. (For more information about homogeneous mixtures, see , .) Homogeneous mixtures are also called solutionsA homogeneous mixture of two or more substances in which the substances present in lesser amounts (the solutes) are dispersed uniformly throughout the substance present in greater amount (the solvent).; they include the air we breathe, the gas we use to cook and heat our homes, the water we drink, the gasoline or diesel fuel that powers engines, and the gold and silver jewelry we wear.

Beads of oil in water. When a nonpolar liquid such as oil is dispersed in a polar solvent such as water, it does not dissolve, but forms spherical beads. Oil is insoluble in water because the intermolecular interactions within the solute (oil) and the solvent (water) are stronger than the intermolecular interactions between the solute and the solvent.

Many of the concepts that we will use in our discussion of solutions were introduced in earlier chapters. In , for example, we described reactions that occur in aqueous solution and how to use molarity to describe concentrations. In , , and , we introduced the principles that govern ion–ion and molecule–molecule interactions in pure substances; similar interactions also occur in solutions. Now we use the principles developed in those chapters to understand the factors that determine how much of one substance can dissolve in another, and how the properties of a solution differ from those of its components.

The properties of mixtures of gases were described in , and the properties of certain types of solid solutions, such as alloys and doped semiconductors, were discussed in . This chapter focuses on liquid solutions, aqueous or otherwise. By the end of this chapter, your understanding of solutions will enable you to explain why the radiator in your car must contain ethylene glycol to avoid damage to the engine on cold winter nights, why salt is spread on icy roads in the winter (and why it isn’t effective when the temperature is too low), why certain vitamins accumulate in your body at toxic levels while others are rapidly excreted, and how salt can be removed from seawater to provide drinking water.

13.1 Factors Affecting Solution Formation

Learning Objective

  1. To understand how enthalpy and entropy changes affect solution formation.

In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent. lists some common examples of gaseous, liquid, and solid solutions and identifies the physical states of the solute and solvent in each.

Table 13.1 Types of Solutions

Solution Solute Solvent Examples
gas gas gas air, natural gas
liquid gas liquid seltzer water (CO2 gas in water)
liquid liquid liquid alcoholic beverage (ethanol in water), gasoline
liquid solid liquid tea, salt water
solid gas solid H2 in Pd (used for H2 storage)
solid solid liquid mercury in silver or gold (amalgam often used in dentistry)

Forming a Solution

The formation of a solution from a solute and a solvent is a physical process, not a chemical one. That is, both solute and solvent can be recovered in chemically unchanged forms using appropriate separation methods. For example, solid zinc nitrate dissolves in water to form an aqueous solution of zinc nitrate:

Equation 13.1

Zn(NO3)(s)H2O(l)Zn2+(aq)+2NO3 –(aq)

Because Zn(NO3)2 can be recovered easily by evaporating the water, this is a physical process. In contrast, metallic zinc appears to dissolve in aqueous hydrochloric acid. In fact, the two substances undergo a chemical reaction to form an aqueous solution of zinc chloride with evolution of hydrogen gas:

Equation 13.2

Zn(s) + 2H+(aq) + 2Cl(aq) → Zn2+(aq) + 2Cl(aq) + H2(g)

Note the Pattern

Dissolution of a solute in a solvent to form a solution does not involve a chemical transformation.

When the solution evaporates, we do not recover metallic zinc, so we cannot say that metallic zinc is soluble in aqueous hydrochloric acid because it is chemically transformed when it dissolves. The dissolution of a solute in a solvent to form a solution does not involve a chemical transformation.

Substances that form a single homogeneous phase in all proportions are said to be completely miscibleCapable of forming a single homogeneous phase, regardless of the proportions with which the substances are mixed. in one another. Ethanol and water are miscible, just as mixtures of gases are miscible. If two substances are essentially insoluble in each other, such as oil and water, they are immiscible. Examples of gaseous solutions that we have already discussed include Earth’s atmosphere (see ) and natural gas (see ).

The Role of Enthalpy in Solution Formation

As we saw in , energy is required to overcome the intermolecular interactions in a solute. This energy can be supplied only by the new interactions that occur in the solution, when each solute particle is surrounded by particles of the solvent in a process called solvationThe process of surrounding each solute particle with particles of solvent., or hydrationThe process of surrounding solute particles with water molecules. when the solvent is water. Thus all of the solute–solute interactions and many of the solvent–solvent interactions must be disrupted for a solution to form. In this section, we describe the role of enthalpy in this process.

Because enthalpy is a state function, we can use the same type of thermochemical cycle described in to analyze the energetics of solution formation. (For more information about state functions, see , .) The process occurs in three discrete steps, indicated by ΔH1, ΔH2, and ΔH3 in . The overall enthalpy change in the formation of the solution (ΔHsoln) is the sum of the enthalpy changes in the three steps:

Equation 13.3

ΔHsoln = ΔH1 + ΔH2 + ΔH3

When a solvent is added to a solution, steps 1 and 2 are both endothermic because energy is required to overcome the intermolecular interactions in the solvent (ΔH1) and the solute (ΔH2). Because ΔH is positive for both steps 1 and 2, the solute–solvent interactions (ΔH3) must be stronger than the solute–solute and solvent–solvent interactions they replace in order for the dissolution process to be exothermic (ΔHsoln < 0). When the solute is an ionic solid, ΔH2 corresponds to the lattice energy that must be overcome to form a solution. As you learned in , the higher the charge of the ions in an ionic solid, the higher the lattice energy. Consequently, solids that have very high lattice energies, such as MgO (−3791 kJ/mol), are generally insoluble in all solvents.

Figure 13.1 Enthalpy Changes That Accompany the Formation of a Solution

Solvation can be an exothermic or endothermic process depending on the nature of the solute and solvent. In both cases, step 1, separation of the solvent particles, is energetically uphill (ΔH1 > 0), as is step 2, separation of the solute particles (ΔH2 > 0). In contrast, energy is released in step 3 (ΔH3 < 0) because of interactions between the solute and solvent. (a) When ΔH3 is larger in magnitude than the sum of ΔH1 and ΔH2, the overall process is exothermic (ΔHsoln < 0), as shown in the thermochemical cycle. (b) When ΔH3 is smaller in magnitude than the sum of ΔH1 and ΔH2, the overall process is endothermic (ΔHsoln > 0).

As you will see in , a positive value for ΔHsoln does not mean that a solution will not form. Whether a given process, including formation of a solution, occurs spontaneously depends on whether the total energy of the system is lowered as a result. Enthalpy is only one of the contributing factors. A high ΔHsoln is usually an indication that the substance is not very soluble. Instant cold packs used to treat athletic injuries, for example, take advantage of the large positive ΔHsoln of ammonium nitrate during dissolution (+25.7 kJ/mol), which produces temperatures less than 0°C ().

Figure 13.2 Commercial Cold Packs for Treating Injuries

These packs contain solid NH4NO3 and water in separate compartments. When the seal between the compartments is broken, the NH4NO3 dissolves in the water. Because ΔHsoln for NH4NO3 is much greater than zero, heat is absorbed by the cold pack during the dissolution process, producing local temperatures less than 0°C.

Entropy and Solution Formation

The enthalpy change that accompanies a process is important because processes that release substantial amounts of energy tend to occur spontaneously. A second property of any system, its entropy, is also important in helping us determine whether a given process occurs spontaneously. We will discuss entropy in more detail in , but for now we can state that entropy(S)The degree of disorder in a thermodynamic system. The greater the number of possible microstates for a system, the higher the entropy. is a thermodynamic property of all substances that is proportional to their degree of disorder. A perfect crystal at 0 K, whose atoms are regularly arranged in a perfect lattice and are motionless, is arbitrarily assigned an entropy of zero. In contrast, gases have large positive entropies because their molecules are highly disordered and in constant motion at high speeds.

The formation of a solution disperses molecules, atoms, or ions of one kind throughout a second substance, which generally increases the disorder and results in an increase in the entropy of the system. Thus entropic factors almost always favor formation of a solution. In contrast, a change in enthalpy may or may not favor solution formation. The London dispersion forces that hold cyclohexane and n-hexane together in pure liquids, for example, are similar in nature and strength. Consequently, ΔHsoln should be approximately zero, as is observed experimentally. Mixing equal amounts of the two liquids, however, produces a solution in which the n-hexane and cyclohexane molecules are uniformly distributed over approximately twice the initial volume. In this case, the driving force for solution formation is not a negative ΔHsoln but rather the increase in entropy due to the increased disorder in the mixture. All spontaneous processes with ΔH ≥ 0 are characterized by an increase in entropy. In other cases, such as mixing oil with water, salt with gasoline, or sugar with hexane, the enthalpy of solution is large and positive, and the increase in entropy resulting from solution formation is not enough to overcome it. Thus in these cases a solution does not form.

Note the Pattern

All spontaneous processes with ΔH ≥ 0 are characterized by an increase in entropy.

summarizes how enthalpic factors affect solution formation for four general cases. The column on the far right uses the relative magnitudes of the enthalpic contributions to predict whether a solution will form from each of the four. Keep in mind that in each case entropy favors solution formation. In two of the cases the enthalpy of solution is expected to be relatively small and can be either positive or negative. Thus the entropic contribution dominates, and we expect a solution to form readily. In the other two cases the enthalpy of solution is expected to be large and positive. The entropic contribution, though favorable, is usually too small to overcome the unfavorable enthalpy term. Hence we expect that a solution will not form readily.

Table 13.2 Relative Changes in Enthalpies for Different Solute–Solvent Combinations*

ΔH1 (separation of solvent molecules) ΔH2 (separation of solute particles) ΔH3 (solute–solvent interactions) ΔHsolnH1 + ΔH2H3) Result of Mixing Solute and Solvent
large; positive large; positive large; negative small; positive or negative solution will usually form
small; positive large; positive small; negative large; positive solution will not form
large; positive small; positive small; negative large; positive solution will not form
small; positive small; positive small; negative small; positive or negative solution will usually form
H1, ΔH2, and ΔH3 refer to the processes indicated in the thermochemical cycle shown in .
In all four cases, entropy increases.

In contrast to liquid solutions, the intermolecular interactions in gases are weak (they are considered to be nonexistent in ideal gases). Hence mixing gases is usually a thermally neutral process (ΔHsoln ≈ 0), and the entropic factor due to the increase in disorder is dominant (). Consequently, all gases dissolve readily in one another in all proportions to form solutions. We will return to a discussion of enthalpy and entropy in , where we treat their relationship quantitatively.

Figure 13.3 Formation of a Solution of Two Gases

(top) Pure samples of two different gases are in separate bulbs. (bottom) When the connecting stopcock is opened, diffusion causes the two gases to mix together and form a solution. Even though ΔHsoln is zero for the process, the increased entropy of the solution (the increased disorder) versus that of the separate gases favors solution formation.

Example 1

Considering LiCl, benzoic acid (C6H5CO2H), and naphthalene, which will be most soluble and which will be least soluble in water?

Given: three compounds

Asked for: relative solubilities in water

Strategy:

Assess the relative magnitude of the enthalpy change for each step in the process shown in . Then use to predict the solubility of each compound in water and arrange them in order of decreasing solubility.

Solution:

The first substance, LiCl, is an ionic compound, so a great deal of energy is required to separate its anions and cations and overcome the lattice energy (ΔH2 is far greater than zero in ). Because water is a polar substance, the interactions between both Li+ and Cl ions and water should be favorable and strong. Thus we expect ΔH3 to be far less than zero, making LiCl soluble in water. In contrast, naphthalene is a nonpolar compound, with only London dispersion forces holding the molecules together in the solid state. We therefore expect ΔH2 to be small and positive. We also expect the interaction between polar water molecules and nonpolar naphthalene molecules to be weak ΔH3 ≈ 0. Hence we do not expect naphthalene to be very soluble in water, if at all. Benzoic acid has a polar carboxylic acid group and a nonpolar aromatic ring. We therefore expect that the energy required to separate solute molecules (ΔH2) will be greater than for naphthalene and less than for LiCl. The strength of the interaction of benzoic acid with water should also be intermediate between those of LiCl and naphthalene. Hence benzoic acid is expected to be more soluble in water than naphthalene but less soluble than LiCl. We thus predict LiCl to be the most soluble in water and naphthalene to be the least soluble.

Exercise

Considering ammonium chloride, cyclohexane, and ethylene glycol (HOCH2CH2OH), which will be most soluble and which will be least soluble in benzene?

Answer: The most soluble is cyclohexane; the least soluble is ammonium chloride.

Summary

Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. Solvation is the process in which solute particles are surrounded by solvent molecules. When the solvent is water, the process is called hydration. The overall enthalpy change that accompanies the formation of a solution, ΔHsoln, is the sum of the enthalpy change for breaking the intermolecular interactions in both the solvent and the solute and the enthalpy change for the formation of new solute–solvent interactions. Exothermic (ΔHsoln < 0) processes favor solution formation. In addition, the change in entropy, the degree of disorder of the system, must be considered when predicting whether a solution will form. An increase in entropy (a decrease in order) favors dissolution.

Key Takeaway

  • The magnitude of the changes in both enthalpy and entropy must be considered when predicting whether a given solute–solvent combination will spontaneously form a solution.

Conceptual Problems

  1. Classify each of the following as a heterogeneous mixture or homogeneous mixture. Explain your rationale in each case.

    1. aqueous ammonia
    2. liquid decongestant
    3. vinegar
    4. seawater
    5. gasoline
    6. fog
  2. Solutions and heterogeneous mixtures are at the extreme ends of the solubility scale. Name one type of mixture that is intermediate on this scale. How are the properties of the mixture you have chosen different from those of a solution or a heterogeneous mixture?

  3. Classify each process as simple dissolution or a chemical reaction.

    1. a naphthalene mothball dissolving in benzene
    2. a sample of a common drain cleaner that has a mixture of NaOH crystals and Al chunks dissolving in water to give H2 gas and an aqueous solution of Na+, OH, and Al3+ ions
    3. an iron ship anchor slowly dissolving in seawater
    4. sodium metal dissolving in liquid ammonia
  4. Classify each process as simple dissolution or a chemical reaction.

    1. a sugar cube dissolving in a cup of hot tea
    2. SO3 gas dissolving in water to produce sulfuric acid
    3. calcium oxide dissolving in water to produce a basic solution
    4. metallic gold dissolving in a small quantity of liquid mercury
  5. You notice that a gas is evolved as you are dissolving a solid in a liquid. Will you be able to recover your original solid by evaporation? Why or why not?

  6. Why is heat evolved when sodium hydroxide pellets are dissolved in water? Does this process correspond to simple dissolution or a chemical reaction? Justify your answer.

  7. Which process(es) is the simple formation of a solution, and which process(es) involves a chemical reaction?

    1. mixing an aqueous solution of NaOH with an aqueous solution of HCl
    2. bubbling HCl gas through water
    3. adding iodine crystals to CCl4
    4. adding sodium metal to ethanol to produce sodium ethoxide (C2H5ONa+) and hydrogen gas
  8. Using thermochemical arguments, explain why some substances that do not form a solution at room temperature will form a solution when heated. Explain why a solution can form even when ΔHsoln is positive.

  9. If you wanted to formulate a new compound that could be used in an instant cold pack, would you select a compound with a positive or negative value of ΔHsoln in water? Justify your answer.

  10. Why is entropy the dominant factor in the formation of solutions of two or more gases? Is it possible for two gases to be immiscible? Why or why not?

Answers

  1. Homogeneous mixtures: aqueous ammonia, liquid decongestant, vinegar, and gasoline. Heterogeneous mixtures: seawater and fog.

  2. All are chemical reactions except dissolving iodine crystals in CCl4.

13.2 Solubility and Molecular Structure

Learning Objective

  1. To understand the relationship between solubility and molecular structure.

When a solute dissolves, its individual atoms, molecules, or ions interact with the solvent, become solvated, and are able to diffuse independently throughout the solution (part (a) in ). This is not, however, a unidirectional process. If the molecule or ion happens to collide with the surface of a particle of the undissolved solute, it may adhere to the particle in a process called crystallization. Dissolution and crystallization continue as long as excess solid is present, resulting in a dynamic equilibrium analogous to the equilibrium that maintains the vapor pressure of a liquid. (For more information about vapor pressure, see , .) We can represent these opposing processes as follows:

Equation 13.4

solute + solventcrystallizationdissolutionsolution

Although the terms precipitation and crystallization are both used to describe the separation of solid solute from a solution, crystallization refers to the formation of a solid with a well-defined crystalline structure, whereas precipitation refers to the formation of any solid phase, often one with very small particles.

Figure 13.4 Dissolution and Precipitation

(a) When a solid is added to a solvent in which it is soluble, solute particles leave the surface of the solid and become solvated by the solvent, initially forming an unsaturated solution. (b) When the maximum possible amount of solute has dissolved, the solution becomes saturated. If excess solute is present, the rate at which solute particles leave the surface of the solid equals the rate at which they return to the surface of the solid. (c) A supersaturated solution can usually be formed from a saturated solution by filtering off the excess solute and lowering the temperature. (d) When a seed crystal of the solute is added to a supersaturated solution, solute particles leave the solution and form a crystalline precipitate.

Factors Affecting Solubility

The maximum amount of a solute that can dissolve in a solvent at a specified temperature and pressure is its solubilityA measure of the how much of a solid substance remains dissolved in a given amount of a specified liquid at a specified temperature and pressure.. Solubility is often expressed as the mass of solute per volume (g/L) or mass of solute per mass of solvent (g/g), or as the moles of solute per volume (mol/L). Even for very soluble substances, however, there is usually a limit to how much solute can dissolve in a given quantity of solvent. In general, the solubility of a substance depends on not only the energetic factors we have discussed but also the temperature and, for gases, the pressure. At 20°C, for example, 177 g of NaI, 91.2 g of NaBr, 35.9 g of NaCl, and only 4.1 g of NaF dissolve in 100 g of water. At 70°C, however, the solubilities increase to 295 g of NaI, 119 g of NaBr, 37.5 g of NaCl, and 4.8 g of NaF. As you learned in , the lattice energies of the sodium halides increase from NaI to NaF. The fact that the solubilities decrease as the lattice energy increases suggests that the ΔH2 term in dominates for this series of compounds.

A solution with the maximum possible amount of solute is saturatedA solution with the maximum possible amount of a solute under a given set of conditions.. If a solution contains less than the maximum amount of solute, it is unsaturated. When a solution is saturated and excess solute is present, the rate of dissolution is exactly equal to the rate of crystallization (part (b) in ). Using the value just stated, a saturated aqueous solution of NaCl, for example, contains 35.9 g of NaCl per 100 mL of water at 20°C. We can prepare a homogeneous saturated solution by adding excess solute (in this case, greater than 35.9 g of NaCl) to the solvent (water), stirring until the maximum possible amount of solute has dissolved, and then removing undissolved solute by filtration.

Note the Pattern

The solubility of most solids increases with increasing temperature.

Because the solubility of most solids increases with increasing temperature, a saturated solution that was prepared at a higher temperature usually contains more dissolved solute than it would contain at a lower temperature. When the solution is cooled, it can therefore become supersaturatedAn unstable solution with more dissolved solute than it would normally contain under the given set of conditions. (part (c) in ). Like a supercooled or superheated liquid (see ), a supersaturated solution is unstable. Consequently, adding a small particle of the solute, a seed crystalA solid sample of a substance that can be added to a supercooled liquid or a supersaturated solution to help induce crystallization., will usually cause the excess solute to rapidly precipitate or crystallize, sometimes with spectacular results, as was shown in . The rate of crystallization in is greater than the rate of dissolution, so crystals or a precipitate form (part (d) in ). In contrast, adding a seed crystal to a saturated solution reestablishes the dynamic equilibrium, and the net quantity of dissolved solute no longer changes.

Because crystallization is the reverse of dissolution, a substance that requires an input of heat to form a solution (ΔHsoln > 0) releases that heat when it crystallizes from solution (ΔHcrys < 0). The amount of heat released is proportional to the amount of solute that exceeds its solubility. Two substances that have a positive enthalpy of solution are sodium thiosulfate (Na2S2O3) and sodium acetate (CH3CO2Na), both of which are used in commercial hot packs, small bags of supersaturated solutions used to warm hands (see ).

Interactions in Liquid Solutions

The interactions that determine the solubility of a substance in a liquid depend largely on the chemical nature of the solute (such as whether it is ionic or molecular) rather than on its physical state (solid, liquid, or gas). We will first describe the general case of forming a solution of a molecular species in a liquid solvent and then describe the formation of a solution of an ionic compound.

Solutions of Molecular Substances in Liquids

The London dispersion forces, dipole–dipole interactions, and hydrogen bonds that hold molecules to other molecules are generally weak. Even so, energy is required to disrupt these interactions. As we described in , unless some of that energy is recovered in the formation of new, favorable solute–solvent interactions, the increase in entropy on solution formation is not enough for a solution to form.

For solutions of gases in liquids, we can safely ignore the energy required to separate the solute molecules (ΔH2 = 0) because the molecules are already separated. Thus we need to consider only the energy required to separate the solvent molecules (ΔH1) and the energy released by new solute–solvent interactions (ΔH3). Nonpolar gases such as N2, O2, and Ar have no dipole moment and cannot engage in dipole–dipole interactions or hydrogen bonding. Consequently, the only way they can interact with a solvent is by means of London dispersion forces, which may be weaker than the solvent–solvent interactions in a polar solvent. It is not surprising, then, that nonpolar gases are most soluble in nonpolar solvents. In this case, ΔH1 and ΔH3 are both small and of similar magnitude. In contrast, for a solution of a nonpolar gas in a polar solvent, ΔH1 is far greater than ΔH3. As a result, nonpolar gases are less soluble in polar solvents than in nonpolar solvents. For example, the concentration of N2 in a saturated solution of N2 in water, a polar solvent, is only 7.07 × 10−4 M compared with 4.5 × 10−3 M for a saturated solution of N2 in benzene, a nonpolar solvent.

The solubilities of nonpolar gases in water generally increase as the molecular mass of the gas increases, as shown in . This is precisely the trend expected: as the gas molecules become larger, the strength of the solvent–solute interactions due to London dispersion forces increases, approaching the strength of the solvent–solvent interactions.

Table 13.3 Solubilities of Selected Gases in Water at 20°C and 1 atm Pressure

Gas Solubility (M) × 10−4
He 3.90
Ne 4.65
Ar 15.2
Kr 27.9
Xe 50.2
H2 8.06
N2 7.07
CO 10.6
O2 13.9
N2O 281
CH4 15.5

Virtually all common organic liquids, whether polar or not, are miscible. The strengths of the intermolecular attractions are comparable; thus the enthalpy of solution is expected to be small (ΔHsoln ≈ 0), and the increase in entropy drives the formation of a solution. If the predominant intermolecular interactions in two liquids are very different from one another, however, they may be immiscible. For example, organic liquids such as benzene, hexane, CCl4, and CS2 (S=C=S) are nonpolar and have no ability to act as hydrogen bond donors or acceptors with hydrogen-bonding solvents such as H2O, HF, and NH3; hence they are immiscible in these solvents. When shaken with water, they form separate phases or layers separated by an interface (), the region between the two layers. Just because two liquids are immiscible, however, does not mean that they are completely insoluble in each other. For example, 188 mg of benzene dissolves in 100 mL of water at 23.5°C. Adding more benzene results in the separation of an upper layer consisting of benzene with a small amount of dissolved water (the solubility of water in benzene is only 178 mg/100 mL of benzene).

Figure 13.5 Immiscible Liquids

Water is immiscible with both CCl4 and hexane. When all three liquids are mixed, they separate into three distinct layers. Because water is less dense than CCl4, the water layer floats on the CCl4. In contrast, hexane is less dense than water, so the hexane floats on the water layer. Because I2 is intensely purple and quite soluble in both CCl4 and hexane, but insoluble in water, a small amount of I2 has been added to help identify the hexane and CCl4 layers.

The solubilities of simple alcohols in water are given in . Only the three lightest alcohols (methanol, ethanol, and n-propanol) are completely miscible with water. As the molecular mass of the alcohol increases, so does the proportion of hydrocarbon in the molecule. Correspondingly, the importance of hydrogen bonding and dipole–dipole interactions in the pure alcohol decreases, while the importance of London dispersion forces increases, which leads to progressively fewer favorable electrostatic interactions with water. Organic liquids such as acetone, ethanol, and tetrahydrofuran are sufficiently polar to be completely miscible with water yet sufficiently nonpolar to be completely miscible with all organic solvents.

Table 13.4 Solubilities of Straight-Chain Organic Alcohols in Water at 20°C

Alcohol Solubility (mol/100 g of H2O)
methanol completely miscible
ethanol completely miscible
n-propanol completely miscible
n-butanol 0.11
n-pentanol 0.030
n-hexanol 0.0058
n-heptanol 0.0008

The same principles govern the solubilities of molecular solids in liquids. For example, elemental sulfur is a solid consisting of cyclic S8 molecules that have no dipole moment. Because the S8 rings in solid sulfur are held to other rings by London dispersion forces, elemental sulfur is insoluble in water. It is, however, soluble in nonpolar solvents that have comparable London dispersion forces, such as CS2 (23 g/100 mL). In contrast, glucose contains five –OH groups that can form hydrogen bonds. Consequently, glucose is very soluble in water (91 g/120 mL of water) but essentially insoluble in nonpolar solvents such as benzene. The structure of one isomer of glucose is shown here.

Low-molecular-mass hydrocarbons with highly electronegative and polarizable halogen atoms, such as chloroform (CHCl3) and methylene chloride (CH2Cl2), have both significant dipole moments and relatively strong London dispersion forces. These hydrocarbons are therefore powerful solvents for a wide range of polar and nonpolar compounds. Naphthalene, which is nonpolar, and phenol (C6H5OH), which is polar, are very soluble in chloroform. In contrast, the solubility of ionic compounds is largely determined not by the polarity of the solvent but rather by its dielectric constant, a measure of its ability to separate ions in solution, as you will soon see.

Example 2

Identify the most important solute–solvent interactions in each solution.

  1. iodine in benzene
  2. aniline (C6H5NH2) in dichloromethane (CH2Cl2)

  3. iodine in water

Given: components of solutions

Asked for: predominant solute–solvent interactions

Strategy:

Identify all possible intermolecular interactions for both the solute and the solvent: London dispersion forces, dipole–dipole interactions, or hydrogen bonding. Determine which is likely to be the most important factor in solution formation.

Solution:

  1. Benzene and I2 are both nonpolar molecules. The only possible attractive forces are London dispersion forces.
  2. Aniline is a polar molecule with an –NH2 group, which can act as a hydrogen bond donor. Dichloromethane is also polar, but it has no obvious hydrogen bond acceptor. Therefore, the most important interactions between aniline and CH2Cl2 are likely to be London interactions.
  3. Water is a highly polar molecule that engages in extensive hydrogen bonding, whereas I2 is a nonpolar molecule that cannot act as a hydrogen bond donor or acceptor. The slight solubility of I2 in water (1.3 × 10−3 mol/L at 25°C) is due to London dispersion forces.

Exercise

Identify the most important interactions in each solution:

  1. ethylene glycol (HOCH2CH2OH) in acetone
  2. acetonitrile (CH3C≡N) in acetone
  3. n-hexane in benzene

Answer:

  1. hydrogen bonding
  2. London interactions
  3. London dispersion forces

Hydrophilic and Hydrophobic Solutes

A solute can be classified as hydrophilicA substance attracted to water. Hydrophilic substances are polar and can form hydrogen bonds to water. (literally, “water loving”), meaning that it has an electrostatic attraction to water, or hydrophobicA substance that repels water. Hydrophobic substances do not interact favorably with water. (“water fearing”), meaning that it repels water. A hydrophilic substance is polar and often contains O–H or N–H groups that can form hydrogen bonds to water. For example, glucose with its five O–H groups is hydrophilic. In contrast, a hydrophobic substance may be polar but usually contains C–H bonds that do not interact favorably with water, as is the case with naphthalene and n-octane. Hydrophilic substances tend to be very soluble in water and other strongly polar solvents, whereas hydrophobic substances are essentially insoluble in water and soluble in nonpolar solvents such as benzene and cyclohexane.

The difference between hydrophilic and hydrophobic substances has substantial consequences in biological systems. For example, vitamins can be classified as either fat soluble or water soluble. Fat-soluble vitamins, such as vitamin A, are mostly nonpolar, hydrophobic molecules. As a result, they tend to be absorbed into fatty tissues and stored there. In contrast, water-soluble vitamins, such as vitamin C, are polar, hydrophilic molecules that circulate in the blood and intracellular fluids, which are primarily aqueous. Water-soluble vitamins are therefore excreted much more rapidly from the body and must be replenished in our daily diet. A comparison of the chemical structures of vitamin A and vitamin C quickly reveals why one is hydrophobic and the other hydrophilic.

Because water-soluble vitamins are rapidly excreted, the risk of consuming them in excess is relatively small. Eating a dozen oranges a day is likely to make you tired of oranges long before you suffer any ill effects due to their high vitamin C content. In contrast, fat-soluble vitamins constitute a significant health hazard when consumed in large amounts. For example, the livers of polar bears and other large animals that live in cold climates contain large amounts of vitamin A, which have occasionally proven fatal to humans who have eaten them.

Example 3

The following substances are essential components of the human diet:

Using what you know of hydrophilic and hydrophobic solutes, classify each as water soluble or fat soluble and predict which are likely to be required in the diet on a daily basis.

  1. arginine
  2. pantothenic acid
  3. oleic acid

Given: chemical structures

Asked for: classification as water soluble or fat soluble; dietary requirement

Strategy:

Based on the structure of each compound, decide whether it is hydrophilic or hydrophobic. If it is hydrophilic, it is likely to be required on a daily basis.

Solution:

  1. Arginine is a highly polar molecule with two positively charged groups and one negatively charged group, all of which can form hydrogen bonds with water. As a result, it is hydrophilic and required in our daily diet.
  2. Although pantothenic acid contains a hydrophobic hydrocarbon portion, it also contains several polar functional groups (–OH and –CO2H) that should interact strongly with water. It is therefore likely to be water soluble and required in the diet. (In fact, pantothenic acid is almost always a component of multiple-vitamin tablets.)
  3. Oleic acid is a hydrophobic molecule with a single polar group at one end. It should be fat soluble and not required daily.

Exercise

These compounds are consumed by humans: caffeine, acetaminophen, and vitamin D. Identify each as primarily hydrophilic (water soluble) or hydrophobic (fat soluble), and predict whether each is likely to be excreted from the body rapidly or slowly.

Answer: Caffeine and acetaminophen are water soluble and rapidly excreted, whereas vitamin D is fat soluble and slowly excreted.

Solid Solutions

Solutions are not limited to gases and liquids; solid solutions also exist. For example, amalgamsA solution (usually a solid solution) of a metal in liquid mercury., which are usually solids, are solutions of metals in liquid mercury. Because most metals are soluble in mercury, amalgams are used in gold mining, dentistry, and many other applications. A major difficulty when mining gold is separating very small particles of pure gold from tons of crushed rock. One way to accomplish this is to agitate a suspension of the crushed rock with liquid mercury, which dissolves the gold (as well as any metallic silver that might be present). The very dense liquid gold–mercury amalgam is then isolated and the mercury distilled away.

An alloy is a solid or liquid solution that consists of one or more elements in a metallic matrix. A solid alloy has a single homogeneous phase in which the crystal structure of the solvent remains unchanged by the presence of the solute. Thus the microstructure of the alloy is uniform throughout the sample. Examples are substitutional and interstitial alloys such as brass or solder. (For more information about alloys, see , ) Liquid alloys include sodium/potassium and gold/mercury. In contrast, a partial alloy solution has two or more phases that can be homogeneous in the distribution of the components, but the microstructures of the two phases are not the same. As a liquid solution of lead and tin is cooled, for example, different crystalline phases form at different cooling temperatures. As you learned in , alloys usually have properties that differ from those of the component elements.

Network solids such as diamond, graphite, and SiO2 are insoluble in all solvents with which they do not react chemically. The covalent bonds that hold the network or lattice together are simply too strong to be broken under normal conditions. They are certainly much stronger than any conceivable combination of intermolecular interactions that might occur in solution. Most metals are insoluble in virtually all solvents for the same reason: the delocalized metallic bonding is much stronger than any favorable metal atom–solvent interactions. Many metals react with solutions such as aqueous acids or bases to produce a solution. However, as we saw in , in these instances the metal undergoes a chemical transformation that cannot be reversed by simply removing the solvent.

Note the Pattern

Solids with very strong intermolecular bonding tend to be insoluble.

Solubilities of Ionic Substances in Liquids

introduced you to guidelines for predicting the solubility of ionic compounds in water. Ionic substances are generally most soluble in polar solvents; the higher the lattice energy, the more polar the solvent must be to overcome the lattice energy and dissolve the substance. Because of its high polarity, water is the most common solvent for ionic compounds. Many ionic compounds are soluble in other polar solvents, however, such as liquid ammonia, liquid hydrogen fluoride, and methanol. Because all these solvents consist of molecules that have relatively large dipole moments, they can interact favorably with the dissolved ions.

The interaction of water with Na+ and Cl ions in an aqueous solution of NaCl was illustrated in . The ion–dipole interactions between Li+ ions and acetone molecules in a solution of LiCl in acetone are shown in . The energetically favorable Li+–acetone interactions make ΔH3 in sufficiently negative to overcome the positive ΔH1 and ΔH2. Because the dipole moment of acetone (2.88 D), and thus its polarity, is actually larger than that of water (1.85 D), one might even expect that LiCl would be more soluble in acetone than in water. In fact, the opposite is true: 83 g of LiCl dissolve in 100 mL of water at 20°C, but only about 4.1 g of LiCl dissolve in 100 mL of acetone. This apparent contradiction arises from the fact that the dipole moment is a property of a single molecule in the gas phase. A more useful measure of the ability of a solvent to dissolve ionic compounds is its dielectric constant (ε)A constant that expresses the ability of a bulk substance to decrease the electrostatic forces between two charged particles., which is the ability of a bulk substance to decrease the electrostatic forces between two charged particles. By definition, the dielectric constant of a vacuum is 1. In essence, a solvent with a high dielectric constant causes the charged particles to behave as if they have been moved farther apart. At 25°C, the dielectric constant of water is 80.1, one of the highest known, and that of acetone is only 21.0. Hence water is better able to decrease the electrostatic attraction between Li+ and Cl ions, so LiCl is more soluble in water than in acetone. This behavior is in contrast to that of molecular substances, for which polarity is the dominant factor governing solubility.

Note the Pattern

A solvent’s dielectric constant is the most useful measure of its ability to dissolve ionic compounds. A solvent’s polarity is the dominant factor in dissolving molecular substances.

Figure 13.6 Ion–Dipole Interactions in the Solvation of Li+ Ions by Acetone, a Polar Solvent

It is also possible to dissolve ionic compounds in organic solvents using crown ethersCyclic polyether with four or more oxygen atoms separated by two or three carbon atoms. All crown ethers have a central cavity that can accommodate a metal ion coordinated to the ring of oxygen atoms., cyclic compounds with the general formula (OCH2CH2)n. Crown ethers are named using both the total number of atoms in the ring and the number of oxygen atoms. Thus 18-crown-6 is an 18-membered ring with six oxygen atoms (part (a) in ). The cavity in the center of the crown ether molecule is lined with oxygen atoms and is large enough to be occupied by a cation, such as K+. The cation is stabilized by interacting with lone pairs of electrons on the surrounding oxygen atoms. Thus crown ethers solvate cations inside a hydrophilic cavity, whereas the outer shell, consisting of C–H bonds, is hydrophobic. Crown ethers are useful for dissolving ionic substances such as KMnO4 in organic solvents such as isopropanol [(CH3)2CHOH] (). The availability of crown ethers with cavities of different sizes allows specific cations to be solvated with a high degree of selectivity.

Figure 13.7 Crown Ethers and Cryptands

(a) The potassium complex of the crown ether 18-crown-6. Note how the cation is nestled within the central cavity of the molecule and interacts with lone pairs of electrons on the oxygen atoms. (b) The potassium complex of 2,2,2-cryptand, showing how the cation is almost hidden by the cryptand. Cryptands solvate cations via lone pairs of electrons on both oxygen and nitrogen atoms.

Figure 13.8 Effect of a Crown Ether on the Solubility of KMnO4 in Isopropanol (2-Propanol)

(a) Normally KMnO4, which is intensely purple, is completely insoluble in isopropanol, which has a relatively low dielectric constant. (b) In the presence of a small amount of 18-crown-6, KMnO4 dissolves in isopropanol, as shown by the reddish-purple color caused by permanganate ions in solution.

CryptandsConsisting of three (–OCH2CH2O–)n chains connected by two nitrogen atoms, cryptands have a central cavity that can encapsulate a metal ion coordinated to the oxygen and nitrogen atoms. (from the Greek kryptós, meaning “hidden”) are compounds that can completely surround a cation with lone pairs of electrons on oxygen and nitrogen atoms (part (b) in ). The number in the name of the cryptand is the number of oxygen atoms in each strand of the molecule. Like crown ethers, cryptands can be used to prepare solutions of ionic compounds in solvents that are otherwise too nonpolar to dissolve them.

Summary

The solubility of a substance is the maximum amount of a solute that can dissolve in a given quantity of solvent; it depends on the chemical nature of both the solute and the solvent and on the temperature and pressure. When a solution contains the maximum amount of solute that can dissolve under a given set of conditions, it is a saturated solution. Otherwise, it is unsaturated. Supersaturated solutions, which contain more dissolved solute than allowed under particular conditions, are not stable; the addition of a seed crystal, a small particle of solute, will usually cause the excess solute to crystallize. A system in which crystallization and dissolution occur at the same rate is in dynamic equilibrium. The solubility of a substance in a liquid is determined by intermolecular interactions, which also determine whether two liquids are miscible. Solutes can be classified as hydrophilic (water loving) or hydrophobic (water fearing). Vitamins with hydrophilic structures are water soluble, whereas those with hydrophobic structures are fat soluble. Many metals dissolve in liquid mercury to form amalgams. Covalent network solids and most metals are insoluble in nearly all solvents. The solubility of ionic compounds is largely determined by the dielectric constant (ε) of the solvent, a measure of its ability to decrease the electrostatic forces between charged particles. Solutions of many ionic compounds in organic solvents can be dissolved using crown ethers, cyclic polyethers large enough to accommodate a metal ion in the center, or cryptands, compounds that completely surround a cation.

Key Takeaway

  • The strength of intramolecular bonding determines the solubility of a solute in a given solvent.

Conceptual Problems

  1. If a compound is only slightly soluble in a particular solvent, what are the relative strengths of the solvent–solvent and solute–solute interactions versus the solute–solvent interactions?

  2. Predict whether each of the following sets of conditions favors formation of a solution:

    Intermolecular Attractive Forces (Solute) Intermolecular Attractive Forces (Solvent) ΔHsoln
    London dispersion hydrogen bonding slightly positive
    dipole–dipole hydrogen bonding very negative
    ionic dipole–dipole slightly positive
    ionic London dispersion positive
  3. Arrange the following liquids in order of increasing solubility in water: t-butanol [(CH3)3COH], benzene, ammonia, and heptane. Justify your answer.

  4. Which compound in each pair will be more soluble in water? Explain your reasoning in each case.

    1. toluene (C7H8) or ethyl ether (C2H5OC2H5)
    2. chloroform (CHCl3) or acetone (CH3COCH3)
    3. carbon tetrachloride (CCl4) or tetrahydrofuran (C4H8O)
    4. CaCl2 or CH2Cl2
  5. Which compound in each pair will be more soluble in benzene? Explain your reasoning in each case.

    1. cyclohexane or methanol
    2. I2 or MgCl2
    3. methylene chloride (CH2Cl2) or acetic acid
  6. Two water-insoluble compounds—n-decylamine [CH3(CH2)9NH2] and n-decane—can be separated by the following procedure: The compounds are dissolved in a solvent such as toluene that is immiscible with water. When adding an aqueous HCl solution to the mixture and stirring vigorously, the HCl reacts with one of the compounds to produce a salt. When the stirring is stopped and the mixture is allowed to stand, two layers are formed. At this point, each layer contains only one of the two original compounds. After the layers are separated, adding aqueous NaOH to the aqueous layer liberates one of the original compounds, which can then be removed by stirring with a second portion of toluene to extract it from the water.

    1. Identify the compound that is present in each layer following the addition of HCl. Explain your reasoning.
    2. How can the original compounds be recovered from the toluene solution?
  7. Bromine and iodine are both soluble in CCl4, but bromine is much more soluble. Why?

  8. A solution is made by mixing 50.0 mL of liquid A with 75.0 mL of liquid B. Which is the solute, and which is the solvent? Is it valid to assume that the volume of the resulting solution will be 125 mL? Explain your answer.

  9. The compounds NaI, NaBr, and NaCl are far more soluble in water than NaF, a substance that is used to fluoridate drinking water. In fact, at 25°C the solubility of NaI is 184 g/100 mL of water, versus only 4.2 g/100 mL of water for NaF. Why is sodium iodide so much more soluble in water? Do you expect KCl to be more soluble or less soluble in water than NaCl?

  10. When water is mixed with a solvent with which it is immiscible, the two liquids usually form two separate layers. If the density of the nonaqueous solvent is 1.75 g/mL at room temperature, sketch the appearance of the heterogeneous mixture in a beaker and label which layer is which. If you were not sure of the density and the identity of the other liquid, how might you be able to identify which is the aqueous layer?

  11. When two liquids are immiscible, the addition of a third liquid can occasionally be used to induce the formation of a homogeneous solution containing all three.

    1. Ethylene glycol (HOCH2CH2OH) and hexane are immiscible, but adding acetone [(CH3)2CO] produces a homogeneous solution. Why does adding a third solvent produce a homogeneous solution?
    2. Methanol and n-hexane are immiscible. Which of the following solvents would you add to create a homogeneous solution—water, n-butanol, or cyclohexane? Justify your choice.
  12. Some proponents of vitamin therapy for combating illness encourage the consumption of large amounts of fat-soluble vitamins. Why can this be dangerous? Would it be as dangerous to consume large amounts of water-soluble vitamins? Why or why not?

  13. Why are most metals insoluble in virtually all solvents?

  14. Because sodium reacts violently with water, it is difficult to weigh out small quantities of sodium metal for a reaction due to its rapid reaction with small amounts of moisture in the air. Would a Na/Hg amalgam be as sensitive to moisture as metallic sodium? Why or why not? A Na/K alloy is a liquid at room temperature. Will it be more or less sensitive to moisture than solid Na or K?

  15. Dental amalgams often contain high concentrations of Hg, which is highly toxic. Why isn’t dental amalgam toxic?

  16. Arrange 2,2,3-trimethylpentane, 1-propanol, toluene (C7H8), and dimethyl sulfoxide [(CH3)2S=O] in order of increasing dipole moment. Explain your reasoning.

  17. Arrange acetone, chloroform, cyclohexane, and 2-butanol in order of increasing dielectric constant. Explain your reasoning.

  18. Dissolving a white crystalline compound in ethanol gave a blue solution. Evaporating the ethanol from the solution gave a bluish-crystalline product, which slowly transformed into the original white solid on standing in the air for several days. Explain what happened. How does the mass of the initial bluish solid compare with the mass of the white solid finally recovered?

  19. You have been asked to develop a new drug that could be used to bind Fe3+ ions in patients who suffer from iron toxicity, allowing the bound iron to be excreted in the urine. Would you consider a crown ether or a cryptand to be a reasonable candidate for such a drug? Explain your answer.

  20. Describe two different situations in which fractional crystallization will not work as a separation technique when attempting to isolate a single compound from a mixture.

  21. You have been given a mixture of two compounds—A and B—and have been told to isolate pure A. You know that pure A has a lower solubility than pure B and that the solubilities of both A and B increase with temperature. Outline a procedure to isolate pure A. If B had the lower solubility, could you use the same procedure to isolate pure A? Why or why not?

Answers

  1. London dispersion forces increase with increasing atomic mass. Iodine is a solid while bromine is a liquid due to the greater intermolecular interactions between the heavier iodine atoms. Iodine is less soluble than bromine in virtually all solvents because it requires more energy to separate I2 molecules than Br2 molecules.

    1. A third solvent with intermediate polarity and/or dielectric constant can effectively dissolve both of the immiscible solvents, creating a single liquid phase.
    2. n-butanol—it is intermediate in polarity between methanol and n-hexane, while water is more polar than either and cyclohexane is comparable to n-hexane.
  2. In dental amalgam, the mercury atoms are locked in a solid phase that does not undergo corrosion under physiological conditions; hence, the mercury atoms cannot readily diffuse to the surface where they could vaporize or undergo chemical reaction.

  3. Dissolve the mixture of A and B in a solvent in which they are both soluble when hot and relatively insoluble when cold, filter off any undissolved B, and cool slowly. Pure A should crystallize, while B stays in solution. If B were less soluble, it would be impossible to obtain pure A by this method in a single step, because some of the less soluble compound (B) will always be present in the solid that crystallizes from solution.

13.3 Units of Concentration

Learning Objective

  1. To describe the concentration of a solution in the way that is most appropriate for a particular problem or application.

There are several different ways to quantitatively describe the concentrationThe quantity of solute that is dissolved in a particular quantity of solvent or solution. of a solution. For example, molarity was introduced in as a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions, introduced in , are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example 4 reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known.

Example 4

Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL.

  1. What is its molarity?
  2. What is its mole fraction?

Given: mass of substance and mass and density of solution

Asked for: molarity and mole fraction

Strategy:

A Calculate the number of moles of acetic acid in the sample. Then calculate the number of liters of solution from its mass and density. Use these results to determine the molarity of the solution.

B Determine the mass of the water in the sample and calculate the number of moles of water. Then determine the mole fraction of acetic acid by dividing the number of moles of acetic acid by the total number of moles of substances in the sample.

Solution:

  1. A The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass. The volume of the solution equals its mass divided by its density. The calculations follow:

    moles CH3CO2H=3.78 g CH3CO2H60.05 g/mol=0.0629 mol volume=massdensity=100.0 g solution1.00 g/mL=100 mL molarity of CH3CO2H=moles CH3CO2Hliter solution=0.0629 mol CH3CO2H(100 mL)(1 L/1000 mL)=0.629 M CH3CO2H

    This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than 12 mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL.

  2. B To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have

    moles H2O=96.2 g H2O18.02 g/mol=5.34 mol H2O

    The mole fraction X of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present:

    XCH3CO2H=moles CH3CO2Hmoles CH3CO2+ moles H2O=0.0629 mol0.0629 mol+5.34 mol=0.0116=1.16×102

    This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01.

Exercise

A solution of HCl gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of HCl per 100.0 g of solution, and its density is 1.10 g/mL.

  1. What is its molarity?
  2. What is its mole fraction?

Answer:

  1. 6.10 M HCl
  2. XHCl = 0.111

The concentration of a solution can also be described by its molality (m)The number of moles of solute present in exactly 1 kg of solvent., the number of moles of solute per kilogram of solvent:

Equation 13.5

molality(m)=moles solutekilogram solvent

Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of H2O under these conditions is very close to 1.0 L, and a 0.50 M solution of KBr in water, for example, has approximately the same concentration as a 0.50 m solution.

Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentageThe ratio of the total mass of the solute to the total mass of the solution., parts per million (ppm)Milligrams of solute per kilogram of solvent., or parts per billion (ppb)Micrograms of solute per kilogram of solvent.:

Equation 13.6

mass percentage=(mass of solutemass of solution)(100)

Equation 13.7

parts per million (ppm)=(mass of solutemass of solution)(106)

Equation 13.8

parts per billion (ppb)=(mass of solutemass of solution)109

In the health sciences, the concentration of a solution is typically expressed as parts per thousand (ppt)Grams of solute per kilogram of solvent, primarily used in the health sciences., indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution.

The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of H2SO4 per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL).

Example 5

Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL.

  1. What is the molarity of the solution?
  2. What is the mass of benzene in the sample?

Given: volume of sample, solute concentration, and density of solution

Asked for: molarity of solute and mass of solute in 250 mL

Strategy:

A Use the concentration of the solute in parts per million to calculate the molarity.

B Use the concentration of the solute in parts per million to calculate the mass of the solute in the specified volume of solution.

Solution:

  1. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore

    molarity=molesliter solution=(12.7 mg)(1 g1000 mg)(1 mol78.114 g)1.00 L=1.63×104 M
  2. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is

    mass of benzene=(12.7 mg benzene)(250 mL)1000 mL=3.18 mg=3.18×103 g benzene

Exercise

The maximum allowable concentration of lead in drinking water is 9.0 ppb. What is the molarity of Pb2+ in a 9.0 ppb aqueous solution? Use your calculated concentration to determine how many grams of Pb2+ are in an 8 oz glass of water.

Answer: 4.3 × 10−8 M; 2 × 10−6 g

How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature.

Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as we saw in . As you will learn in , mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies.

summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example 6.

Table 13.5 Different Units for Expressing the Concentrations of Solutions*

Unit Definition Application
molarity (M) moles of solute/liter of solution (mol/L) Used for quantitative reactions in solution and titrations; mass and molecular mass of solute and volume of solution are known.
mole fraction (X) moles of solute/total moles present (mol/mol) Used for partial pressures of gases and vapor pressures of some solutions; mass and molecular mass of each component are known.
molality (m) moles of solute/kg of solvent (mol/kg) Used in determining how colligative properties vary with solute concentration; masses and molecular mass of solute are known.
mass percentage (%) [mass of solute (g)/mass of solution (g)] × 100 Useful when masses are known but molecular masses are unknown.
parts per thousand (ppt) [mass of solute/mass of solution] × 103 (g solute/kg solution) Used in the health sciences, ratio solutions are typically expressed as a proportion, such as 1:1000.
parts per million (ppm) [mass of solute/mass of solution] × 106 (mg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
parts per billion (ppb) [mass of solute/mass of solution] × 109 (µg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
*The molarity of a solution is temperature dependent, but the other units shown in this table are independent of temperature.

Example 6

Vodka is essentially a solution of pure ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka.

  1. the mass percentage
  2. the mole fraction
  3. the molarity
  4. the molality

Given: volume percent and density

Asked for: mass percentage, mole fraction, molarity, and molality

Strategy:

A Use the density of the solute to calculate the mass of the solute in 100.0 mL of solution. Calculate the mass of water in 100.0 mL of solution.

B Determine the mass percentage of solute by dividing the mass of ethanol by the mass of the solution and multiplying by 100.

C Convert grams of solute and solvent to moles of solute and solvent. Calculate the mole fraction of solute by dividing the moles of solute by the total number of moles of substances present in solution.

D Calculate the molarity of the solution: moles of solute per liter of solution. Determine the molality of the solution by dividing the number of moles of solute by the kilograms of solvent.

Solution:

The key to this problem is to use the density of pure ethanol to determine the mass of ethanol (CH3CH2OH), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units. A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density:

mass of EtOH=(40.0 mL)(0.789 gmL)=31.6 g EtOH

If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution.

  1. B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage:

    %EtOH=(mass of EtOHmass of solution)(100)=(31.6 g EtOH31.6 g EtOH+60.0 g Η2O)(100)=34.5%
  2. C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL:

    moles EtOH=(31.6 g EtOH)(1 mol46.07 g EtOH)=0.686 mol CH3CH2OH

    Similarly, the number of moles of water is

    moles H2Ο=(60.0 g H2O)(1 mol H2O18.02 g H2O)=3.33 mol H2O

    The mole fraction of ethanol is thus

    XEtOH=0.686 mol0.686 mol+3.33 mol=0.171
  3. D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is

    MEtOH=(0.686 mol100 mL)(1000 mLL)=6.86 M
  4. The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward:

    mEtOH=(0.686 mol EtOH60.0 g H2O)(1000 gkg)=11.4 mol EtOHkg H2O=11.4 m

Exercise

A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene.

  1. mass percentage
  2. mole fraction
  3. molarity
  4. molality

Answer:

  1. mass percentage toluene = 24.8%
  2. Xtoluene = 0.219
  3. 2.35 M toluene
  4. 3.59 m toluene

Summary

The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb), micrograms of solute per kilogram of solution; and molality (m), the number of moles of solute per kilogram of solvent.

Key Takeaway

  • Different units are used to express the concentrations of a solution depending on the application.

Key Equations

molality

:   molality (m)=moles solutekilogram solvent

mass percentage

:   mass percentage=(mass of solutemass of solution)(100)

parts per million

:   ppm=(mass of solutemass of solution)(106)

parts per billion

:   ppb=(mass of solutemass of solution)(109)

Conceptual Problems

  1. Does the molality have the same numerical value as the molarity for a highly concentrated aqueous solution of fructose (C6H12O6) (approximately 3.2 M)? Why or why not?

  2. Explain why the molality and molarity of an aqueous solution are not always numerically identical. Will the difference between the two be greater for a dilute or a concentrated solution? Explain your answer.

  3. Under what conditions are molality and molarity likely to be equal? Is the difference between the two greater when water is the solvent or when the solvent is not water? Why?

  4. What is the key difference between using mole fraction or molality versus molarity to describe the concentration of a solution? Which unit(s) of concentration is most appropriate for experiments that must be carried out at several different temperatures?

  5. An experiment that relies on very strict control of the reaction stoichiometry calls for adding 50.0 mL of a 0.95 M solution of A to 225 mL of a 1.01 M solution of B, followed by heating for 1 h at 60°C. To save time, a student decided to heat solution B to 60°C before measuring out 225 mL of solution B, transferring it to the flask containing solution A, and proceeding normally. This change in procedure caused the yield of product to be less than usual. How could such an apparently minor change in procedure have resulted in a decrease in the yield?

Numerical Problems

  1. Complete the following table for aqueous solutions of the compounds given.

    Compound Molarity (M) Solution Density (g/mL) Mole Fraction (X)
    H2SO4 18.0 1.84
    CH3COOH 1.00 7.21 × 10−3
    KOH 3.60 1.16
  2. Complete the following table for each compound given.

    Compound Mass (g) Volume of Solution (mL) Molarity (M)
    Na2SO4 7.80 225
    KNO3 125 1.27
    NaO2CCH3 18.64 0.95
  3. How would you prepare 100.0 mL of an aqueous solution with 0.40 M KI? a solution with 0.65 M NaCN?

  4. Calculate the molality of a solution with 775 mg of NaCl in 500.0 g of water. Do you expect the molarity to be the same as the molality? Why or why not?

  5. What is the molarity of each solution?

    1. 12.8 g of glucose (C6H12O6) in water, total volume 150.0 mL
    2. 9.2 g of Na3PO4 in water, total volume 200.0 mL
    3. 843 mg of I2 in EtOH, total volume 150.0 mL
  6. A medication used to treat abnormal heart rhythms is labeled “Procainamide 0.5 g/250 cc.” Express this concentration in parts per thousand.

  7. Meperidine is a medication used for pain relief. A bottle of meperidine is labeled as 50 mg/mL. Express this concentration in parts per thousand.

  8. An aqueous solution that is 4.61% NaOH by mass has a density of 1.06 g/mL. Calculate the molarity of the solution, the mole fraction of NaOH, and the molality of the solution.

  9. A solution of concentrated phosphoric acid contains 85.0% H3PO4 by mass and has a density of 1.684 g/mL. Calculate the following.

    1. the molarity of the solution
    2. the mole fraction of H3PO4
    3. the molality of the solution
  10. A solution of commercial concentrated nitric acid is 16 M HNO3 and has a density of 1.42 g/mL. What is the percentage of HNO3 in the solution by mass? What is the molality?

  11. A commercial aqueous ammonia solution contains 28.0% NH3 by massand has a density of 0.899 g/mL. Calculate the following.

    1. the molarity
    2. the mole fraction
  12. Concentrated, or glacial, acetic acid is pure acetic acid and has a density of 1.053 g/mL. It is widely used in organic syntheses, in the manufacture of rayon and plastics, as a preservative in foods, and occasionally to treat warts. What volume of glacial acetic acid is required to prepare 5.00 L of a 1.75 M solution of acetic acid in ethanol?

  13. Solutions of sodium carbonate decahydrate, also known as washing soda, are used as skin cleansers. The solubility of this compound in cold water is 21.52 g/100 mL. If a saturated solution has a density of 1.20 g/mL, what is its molarity? What is the mole fraction of sodium carbonate decahydrate in this solution?

  14. Hydrogen peroxide (H2O2) is usually sold over the counter as an aqueous solution that is 3% by mass. Assuming a solution density of 1.01 g/mL, what is the molarity of hydrogen peroxide? What is the molar concentration of a solution that is 30% hydrogen peroxide by mass (density = 1.112 g/mL)? How would you prepare 100.0 mL of a 3% solution from the 30% solution?

  15. Determine the concentration of a solution with 825 mg of Na2HPO4 dissolved in 450.0 mL of H2O at 20°C in molarity, molality, mole fraction, and parts per million. Assume that the density of the solution is the same as that of water. Which unit of concentration is most convenient for calculating vapor pressure changes? Why?

  16. How many moles of Cl are there in 25.0 mL of a 0.15 M CaCl2 solution?

  17. How many moles of Na+ are there in 25.0 g of a 1.33 × 10−3m Na2HPO4 solution? What is the sodium concentration of this solution in ppb?

  18. How many grams of copper are there in 30.0 mL of a 0.100 M CuSO4 solution?

  19. How many grams of nitrate ion are there in 75.0 g of a 1.75 × 10−4m Pb(NO3)2 solution? What is the nitrate concentration of the solution in ppb?

  20. How many milliliters of a 0.750 M solution of K2CrO4 are required to deliver 250 mg of chromate ion?

  21. How many milliliters of a 1.95 × 10−6 M solution of Ag3PO4 are required to deliver 14.0 mg of Ag+?

  22. Iron reacts with bromine according to the following equation:

    2Fe(s) + 3Br2(aq) → 2FeBr3(aq)

    How many milliliters of a 5.0 × 10−2 M solution of bromine in water are required to react completely with 750.0 mg of iron metal?

  23. Aluminum reacts with HCl according to the following equation:

    2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

    If 25.0 mL of a solution of HCl in water is required to react completely with 1.05 g of aluminum metal, what is the molarity of the HCl solution?

  24. The precipitation of silver chloride is a diagnostic test for the presence of chloride ion. If 25.0 mL of 0.175 M AgNO3 are required to completely precipitate the chloride ions from 10.0 mL of an NaCl solution, what was the original concentration of NaCl?

  25. Barium sulfate is virtually insoluble. If a 10.0 mL solution of 0.333 M Ba(NO3)2 is stirred with 40.0 mL of a 0.100 M Na2SO4, how many grams of barium sulfate will precipitate? Which reactant is present in excess? What is its final concentration?

Answers

  1. Compound Molarity (M) Solution Density (g/mL) Mole Fraction (X)
    H2SO4 18.0 1.84 0.82
    CH3COOH 0.393 1.00 7.21 × 10−3
    KOH 3.60 1.16 6.33 × 10 −2
  2. 100.0 ml of 0.40 M KI: dissolve 6.64 g of KI in enough water to make 100.0 mL of solution; 100.0 ml of 0.65 M NaCN: dissolve 3.18 g of NaCN in enough water to make 100.0 mL of solution.

    1. 0.474 M glucose
    2. 0.28 M Na3PO4
    3. 0.0221 M I2
    1. 14.6 M
    2. X = 0.510
    3. 57.7 m
    1. 14.8 M
    2. X = 0.292
  3. The molarity is 0.745 M, and the mole fraction is 0.0134.

  4. The molarity is 0.0129 M, the molality is 0.0129 m, the mole fraction is 2.33 × 10−4, and the solution contains 1830 ppm Na2HPO4. Mole fraction is most useful for calculating vapor pressure, because Raoult’s law states that the vapor pressure of a solution containing a non-volatile solute is equal to the mole fraction of solvent times the vapor pressure of the pure solvent. The mole fraction of the solvent is just one minus the mole fraction of solute.

  5. 6.65 × 10−5 mol sodium; 6.14 × 104 ppb

  6. 1.63 × 10−3 g; 2.17 × 104 ppb

  7. 2.22 × 104 mL or 22.2 L

  8. 4.68 M HCl

  9. 0.777 g BaSO4; Na2SO4; 0.0134 M Na2SO4

13.4 Effects of Temperature and Pressure on Solubility

Learning Objective

  1. To understand the relationship among temperature, pressure, and solubility.

Experimentally it is found that the solubility of most compounds depends strongly on temperature and, if a gas, on pressure as well. As we shall see, the ability to manipulate the solubility by changing the temperature and pressure has several important consequences.

Effect of Temperature on the Solubility of Solids

shows plots of the solubilities of several organic and inorganic compounds in water as a function of temperature. Although the solubility of a solid generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. Many compounds (such as glucose and CH3CO2Na) exhibit a dramatic increase in solubility with increasing temperature. Others (such as NaCl and K2SO4) exhibit little variation, and still others (such as Li2SO4) become less soluble with increasing temperature.

Figure 13.9 Solubilities of Several Inorganic and Organic Solids in Water as a Function of Temperature

Solubility may increase or decrease with temperature; the magnitude of this temperature dependence varies widely among compounds.

Notice in particular the curves for NH4NO3 and CaCl2. The dissolution of ammonium nitrate in water is endothermic (ΔHsoln = +25.7 kJ/mol), whereas the dissolution of calcium chloride is exothermic (ΔHsoln = −68.2 kJ/mol), yet shows that the solubility of both compounds increases sharply with increasing temperature. In fact, the magnitudes of the changes in both enthalpy and entropy for dissolution are temperature dependent. Because the solubility of a compound is ultimately determined by relatively small differences between large numbers, there is generally no good way to predict how the solubility will vary with temperature.

The variation of solubility with temperature has been measured for a wide range of compounds, and the results are published in many standard reference books. Chemists are often able to use this information to separate the components of a mixture by fractional crystallizationThe separation of compounds based on their relative solubilities in a given solvent., the separation of compounds on the basis of their solubilities in a given solvent. For example, if we have a mixture of 150 g of sodium acetate (CH3CO2Na) and 50 g of KBr, we can separate the two compounds by dissolving the mixture in 100 g of water at 80°C and then cooling the solution slowly to 0°C. According to the temperature curves in , both compounds dissolve in water at 80°C, and all 50 g of KBr remains in solution at 0°C. Only about 36 g of CH3CO2Na are soluble in 100 g of water at 0°C, however, so approximately 114 g (150 g − 36 g) of CH3CO2Na crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original CH3CO2Na in essentially pure form in only one step.

Fractional crystallization is a common technique for purifying compounds as diverse as those shown in and from antibiotics to enzymes. For the technique to work properly, the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution. In addition, the impurities must be more soluble than the compound of interest (as was KBr in this example) and preferably present in relatively small amounts.

Effect of Temperature on the Solubility of Gases

The solubility of gases in liquids decreases with increasing temperature, as shown in . Attractive intermolecular interactions in the gas phase are essentially zero for most substances. When a gas dissolves, it does so because its molecules interact with solvent molecules. Because heat is released when these new attractive interactions form, dissolving most gases in liquids is an exothermic process (ΔHsoln < 0). Conversely, adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas.The phenomenon is similar to that involved in the increase in vapor pressure of a pure liquid with increasing temperature, as discussed in . In the case of vapor pressure, however, it is attractive forces between solvent molecules that are being overcome by the added thermal energy when the temperature is increased.

Figure 13.10 Solubilities of Several Common Gases in Water as a Function of Temperature at Partial Pressure of 1 atm

The solubilities of all gases decrease with increasing temperature.

The decrease in the solubilities of gases at higher temperatures has both practical and environmental implications. Anyone who routinely boils water in a teapot or electric kettle knows that a white or gray deposit builds up on the inside and must eventually be removed. The same phenomenon occurs on a much larger scale in the giant boilers used to supply hot water or steam for industrial applications, where it is called “boiler scale,” a deposit that can seriously decrease the capacity of hot water pipes (). The problem is not a uniquely modern one: aqueducts that were built by the Romans 2000 years ago to carry cold water from alpine regions to warmer, drier regions in southern France were clogged by similar deposits. The chemistry behind the formation of these deposits is moderately complex and will be described in more detail in , but the driving force is the loss of dissolved CO2 from solution. Hard water contains dissolved Ca2+ and HCO3 (bicarbonate) ions. Calcium bicarbonate [Ca(HCO3)2] is rather soluble in water, but calcium carbonate (CaCO3) is quite insoluble. A solution of bicarbonate ions can react to form carbon dioxide, carbonate ion, and water:

Equation 13.9

2HCO3(aq) → CO22−(aq) + H2O(l) + CO2(aq)

Heating the solution decreases the solubility of CO2, which escapes into the gas phase above the solution. In the presence of calcium ions, the carbonate ions precipitate as insoluble calcium carbonate, the major component of boiler scale.

Figure 13.11 Boiler Scale in a Water Pipe

Calcium carbonate (CaCO3) deposits in hot water pipes can significantly reduce pipe capacity. These deposits, called boiler scale, form when dissolved CO2 is driven into the gas phase at high temperatures.

In thermal pollution, lake or river water that is used to cool an industrial reactor or a power plant is returned to the environment at a higher temperature than normal. Because of the reduced solubility of O2 at higher temperatures (), the warmer water contains less dissolved oxygen than the water did when it entered the plant. Fish and other aquatic organisms that need dissolved oxygen to live can literally suffocate if the oxygen concentration of their habitat is too low. Because the warm, oxygen-depleted water is less dense, it tends to float on top of the cooler, denser, more oxygen-rich water in the lake or river, forming a barrier that prevents atmospheric oxygen from dissolving. Eventually even deep lakes can be suffocated if the problem is not corrected. Additionally, most fish and other nonmammalian aquatic organisms are cold-blooded, which means that their body temperature is the same as the temperature of their environment. Temperatures substantially greater than the normal range can lead to severe stress or even death. Cooling systems for power plants and other facilities must be designed to minimize any adverse effects on the temperatures of surrounding bodies of water.

A similar effect is seen in the rising temperatures of bodies of water such as the Chesapeake Bay, the largest estuary in North America, where global warming has been implicated as the cause (For more information on global warming, see , .) For each 1.5°C that the bay’s water warms, the capacity of water to dissolve oxygen decreases by about 1.1%. Many marine species that are at the southern limit of their distributions have shifted their populations farther north. In 2005, the eelgrass, which forms an important nursery habitat for fish and shellfish, disappeared from much of the bay following record high water temperatures. Presumably, decreased oxygen levels decreased populations of clams and other filter feeders, which then decreased light transmission to allow the eelsgrass to grow. The complex relationships in ecosystems such as the Chesapeake Bay are especially sensitive to temperature fluctuations that cause a deterioration of habitat quality.

Effect of Pressure on the Solubility of Gases: Henry’s Law

External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in , which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures.

Figure 13.12 A Model Depicting Why the Solubility of a Gas Increases as the Partial Pressure Increases at Constant Temperature

(a) When a gas comes in contact with a pure liquid, some of the gas molecules (purple spheres) collide with the surface of the liquid and dissolve. When the concentration of dissolved gas molecules has increased so that the rate at which gas molecules escape into the gas phase is the same as the rate at which they dissolve, a dynamic equilibrium has been established, as depicted here. This equilibrium is entirely analogous to the one that maintains the vapor pressure of a liquid. (For more information on vapor pressure, see , .) (b) Increasing the pressure of the gas increases the number of molecules of gas per unit volume, which increases the rate at which gas molecules collide with the surface of the liquid and dissolve. (c) As additional gas molecules dissolve at the higher pressure, the concentration of dissolved gas increases until a new dynamic equilibrium is established.

The relationship between pressure and the solubility of a gas is described quantitatively by Henry’s lawAn equation that quantifies the relationship between the pressure and the solubility of a gas: C=kP., which is named for its discoverer, the English physician and chemist, William Henry (1775–1836):

Equation 13.10

C = kP

where C is the concentration of dissolved gas at equilibrium, P is the partial pressure of the gas, and k is the Henry’s law constant, which must be determined experimentally for each combination of gas, solvent, and temperature. Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry’s law constant are therefore mol/(L·atm) = M/atm. Values of the Henry’s law constants for solutions of several gases in water at 20°C are listed in .

As the data in demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the elements of group 18, the Henry’s law constants increase smoothly from He to Ne to Ar. The table also shows that O2 is almost twice as soluble as N2. Although London dispersion forces are too weak to explain such a large difference, O2 is paramagnetic and hence more polarizable than N2, which explains its high solubility.

Table 13.6 Henry’s Law Constants for Selected Gases in Water at 20°C

Gas Henry’s Law Constant [mol/(L·atm)] × 10−4
He 3.9
Ne 4.7
Ar 15
H2 8.1
N2 7.1
O2 14
CO2 392

Gases that react chemically with water, such as HCl and the other hydrogen halides, H2S, and NH3, do not obey Henry’s law; all of these gases are much more soluble than predicted by Henry’s law. For example, HCl reacts with water to give H+(aq) and Cl(aq), not dissolved HCl molecules, and its dissociation into ions results in a much higher solubility than expected for a neutral molecule.

Note the Pattern

Gases that react with water do not obey Henry’s law.

Henry’s law has important applications. For example, bubbles of CO2 form as soon as a carbonated beverage is opened because the drink was bottled under CO2 at a pressure greater than 1 atm. When the bottle is opened, the pressure of CO2 above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry’s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more N2 from the air dissolves in the diver’s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of N2 to form throughout the body, a condition known as “the bends.” These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases.

Due to the low Henry’s law constant for O2 in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the O2 concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind O2 reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds O2 and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four O2 molecules. Although the concentration of dissolved O2 in blood serum at 37°C (normal body temperature) is only 0.010 mM, the total dissolved O2 concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these “blood substitutes” do not require refrigeration and have a long shelf life. Their very high Henry’s law constants for O2 result in dissolved oxygen concentrations comparable to those in normal blood.

Example 7

The Henry’s law constant for O2 in water at 25°C is 1.27 × 10−3 M/atm, and the mole fraction of O2 in the atmosphere is 0.21. Calculate the solubility of O2 in water at 25°C at an atmospheric pressure of 1.00 atm.

Given: Henry’s law constant, mole fraction of O2, and pressure

Asked for: solubility

Strategy:

A Use Dalton’s law of partial pressures to calculate the partial pressure of oxygen. (For more information about Dalton’s law of partial pressures, see , .)

B Use Henry’s law to calculate the solubility, expressed as the concentration of dissolved gas.

Solution:

A According to Dalton’s law, the partial pressure of O2 is proportional to the mole fraction of O2:

PA = XAPt = (0.21)(1.00 atm) = 0.21 atm

B From Henry’s law, the concentration of dissolved oxygen under these conditions is

CO2=kPO2=(1.27×103 M/atm)(0.21 atm)=2.7×104 M

Exercise

To understand why soft drinks “fizz” and then go “flat” after being opened, calculate the concentration of dissolved CO2 in a soft drink

  1. bottled under a pressure of 5.0 atm of CO2.
  2. in equilibrium with the normal partial pressure of CO2 in the atmosphere (approximately 3 × 10−4 atm).

The Henry’s law constant for CO2 in water at 25°C is 3.4 × 10−2 M/atm.

Answer:

  1. 0.17 M
  2. 1 × 10−5 M

Summary

The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas.

Key Takeaway

  • The solubility of a solid may increase or decrease with increasing temperature, whereas the solubility of a gas decreases with an increase in temperature and a decrease in pressure.

Conceptual Problems

  1. Use the kinetic molecular theory of gases discussed in to explain why the solubility of virtually all gases in liquids decreases with increasing temperature.

  2. An industrial plant uses water from a nearby stream to cool its reactor and returns the water to the stream after use. Over a period of time, dead fish start to appear downstream from the plant, but there is no evidence for any leaks of potentially toxic chemicals into the stream. What other factor might be causing the fish to die?

  3. One manufacturer’s instructions for setting up an aquarium specify that if boiled water is used, the water must be cooled to room temperature and allowed to stand overnight before fish are added. Why is it necessary for the water to stand?

  4. Using a carbonated beverage as an example, discuss the effect of temperature on the “fizz.” How does the “foaminess” of a carbonated beverage differ between Los Angeles, California, and Denver, Colorado?

  5. A common laboratory technique for degassing a solvent is to place it in a flask that is sealed to the atmosphere and then evacuate the flask to remove any gases above the liquid. Why is this procedure effective? Why does the temperature of the solvent usually decrease substantially during this process?

Answers

  1. When water is boiled, all of the dissolved oxygen and nitrogen are removed. When the water is cooled to room temperature, it initially contains very little dissolved oxygen. Allowing the water to stand overnight allows oxygen in the air to dissolve, so that the fish will not suffocate.

  2. Evacuating the flask to remove gases decreases the partial pressure of oxygen above the solution. According to Henry’s law, the solubility of any gas decreases as its partial pressure above the solution decreases. Consequently, dissolved oxygen escapes from solution into the gas phase, where it is removed by the vacuum pump. Filling the flask with nitrogen gas and repeating this process several times effectively removes almost all of the dissolved oxygen. The temperature of the solvent decreases because some solvent evaporates as well during this process. The heat that is required to evaporate some of the liquid is initially removed from the rest of the solvent, decreasing its temperature.

Numerical Problems

  1. The solubility of CO2 in water at 0°C and 1 atm is 0.335 g/100 g of H2O. At 20°C and 1 atm, the solubility of CO2 in water is 0.169 g/100 g of H2O.

    1. What volume of CO2 would be released by warming 750 g of water saturated with CO2 from 0°C to 20°C?
    2. What is the value of the Henry’s law constant for CO2 under each set of conditions?
  2. The solubility of O2 in 100 g of H2O at varying temperatures and a pressure of 1 atm is given in the following table:

    Solubility (g) Temperature (°C)
    0.0069 0
    0.0054 10
    0.0043 20
    1. What is the value of the Henry’s law constant at each temperature?
    2. Does Henry’s law constant increase or decrease with increasing temperature?
    3. At what partial pressure of O2 would the concentration of O2 in water at 0°C be the same as the concentration in water at 20°C at a partial pressure of 1 atm?
    4. Assuming that air is 20% O2 by volume, at what atmospheric pressure would the O2 concentration be the same at 20°C as it is at atmospheric pressure and 0°C?

Answer

    1. 0.678 L CO2
    2. k0°C = 7.61 × 10−2 M/atm, k20°C = 3.84 × 10−2 M/atm

13.5 Colligative Properties of Solutions

Learning Objective

  1. To describe the relationship between solute concentration and the physical properties of a solution.

Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode.

Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative propertiesA property of a solution that depends primarily on the number of solute particles rather than the kind of solute particles. (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties.

When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, NaCl, and CaCl2. Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both NaCl and CaCl2 are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of NaCl contains 0.01 M Na+ ions and 0.01 M Cl ions, for a total particle concentration of 0.02 M. Similarly, the CaCl2 solution contains 0.01 M Ca2+ ions and 0.02 M Cl ions, for a total particle concentration of 0.03 M.These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete. At higher concentrations (typically >1 M), especially with salts of small, highly charged ions (such as Mg2+ or Al3+), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete (see ). The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind.

Vapor Pressure of Solutions and Raoult’s Law

Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining , which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent.

Figure 13.13 A Model Depicting Why the Vapor Pressure of a Solution of Glucose Is Less Than the Vapor Pressure of Pure Water

(a) When water or any volatile solvent is in a closed container, water molecules move into and out of the liquid phase at the same rate in a dynamic equilibrium. (b) If a nonvolatile solute such as glucose is added, some fraction of the surface area is occupied by solvated solute molecules. As a result, the rate at which water molecules evaporate is decreased, although initially their rate of condensation is unchanged. (c) When the glucose solution reaches equilibrium, the concentration of water molecules in the vapor phase, and hence the vapor pressure, is less than that of pure water.

shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant.

Figure 13.14 Transfer of Water to a Beaker Containing a Glucose Solution

(top) One beaker contains an aqueous solution of glucose, and the other contains pure water. If they are placed in a sealed chamber, the lower vapor pressure of water in the glucose solution results in a net transfer of water from the beaker containing pure water to the beaker containing the glucose solution. (bottom) Eventually, all of the water is transferred to the beaker that has the glucose solution.

If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore

Equation 13.11

PA=XAPA0

where PA is the vapor pressure of component A of the solution (in this case the solvent), XA is the mole fraction of A in solution, and PA0 is the vapor pressure of pure A. is known as Raoult’s lawAn equation that quantifies the relationship between solution composition and vapor pressure: PA=XAPA0., after the French chemist who developed it. If the solution contains only a single nonvolatile solute (B), then XA + XB = 1, and we can substitute XA = 1 − XB to obtain

Equation 13.12

PA=(1XB)PA0=PA0XBPA0

Rearranging and defining ΔPA=PA0PA, we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute:

Equation 13.13

PA0PA=ΔPA=XBPA0

We can solve vapor pressure problems in either of two ways: by using to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute.

Example 8

Ethylene glycol (HOCH2CH2OH), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter.

Given: identity of solute, percentage by mass, and vapor pressure of pure solvent

Asked for: vapor pressure of solution

Strategy:

A Calculate the number of moles of ethylene glycol in an arbitrary quantity of water, and then calculate the mole fraction of water.

B Use Raoult’s law to calculate the vapor pressure of the solution.

Solution:

A A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present:

moles EG=(302 g)(1 mol62.07 g)=4.87 mol EG moles H2O=(698 g)(1 mol18.02 g)=38.7 mol H2O

The mole fraction of water is thus

XH2O=38.7 molH2O38.7 mol H2O+4.87 mol EG=0.888

B From Raoult’s law (), the vapor pressure of the solution is

PH2O=(XH2O)(PH2O0)=(0.888)(760 mmHg)=675 mmHg

Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using to calculate the resulting decrease in vapor pressure:

XEG=4.87 mol EG4.87 mol EG+38.7 mol H2O=0.112 ΔPH2O=(XEG)(PH2O0)=(0.112)(760 mmHg)=85.1 mmHg PH2O=PH2O0ΔPH2O=760 mmHg85.1 mmHg=675 mmHg

The same result is obtained using either method.

Exercise

Seawater is an approximately 3.0% aqueous solution of NaCl by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of NaCl, remembering that 1 mol of NaCl produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg.

Answer: 0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling.

Even when a solute is volatile, meaning that it has a measurable vapor pressure, we can still use Raoult’s law. In this case, we calculate the vapor pressure of each component separately. The total vapor pressure of the solution (PT) is the sum of the vapor pressures of the components:

Equation 13.14

PT=PA+PB=XAPA0+XBPB0

Because XB = 1 − XA for a two-component system,

Equation 13.15

PT=XAPA0+(1XA)PB0

Thus we need to specify the mole fraction of only one of the components in a two-component system. Consider, for example, the vapor pressure of solutions of benzene and toluene of various compositions. At 20°C, the vapor pressures of pure benzene and toluene are 74.7 and 22.3 mmHg, respectively. The vapor pressure of benzene in a benzene–toluene solution is

Equation 13.16

PC6H6=XC6H6PC6H60

and the vapor pressure of toluene in the solution is

Equation 13.17

PC6H5CH3=XC6H5CH3PC6H5CH30

and are both in the form of the equation for a straight line: y = mx + b, where b = 0. Plots of the vapor pressures of both components versus the mole fractions are therefore straight lines that pass through the origin, as shown in . Furthermore, a plot of the total vapor pressure of the solution versus the mole fraction is a straight line that represents the sum of the vapor pressures of the pure components. Thus the vapor pressure of the solution is always greater than the vapor pressure of either component.

Figure 13.15 Vapor Pressures of Benzene–Toluene Solutions

Plots of the vapor pressures of benzene (C6H6) and toluene (C6H5CH3) versus the mole fractions at 20°C are straight lines. For a solution like this, which approximates an ideal solution, the total vapor pressure of the solution (Pt) is the sum of the vapor pressures of the components.

A solution of two volatile components that behaves like the solution in is an ideal solutionA solution that obeys Raoult’s law., which is defined as a solution that obeys Raoult’s law. Like an ideal gas, an ideal solution is a hypothetical system whose properties can be described in terms of a simple model. Mixtures of benzene and toluene approximate an ideal solution because the intermolecular forces in the two pure liquids are almost identical in both kind and magnitude. Consequently, the change in enthalpy on solution formation is essentially zero (ΔHsoln ≈ 0), which is one of the defining characteristics of an ideal solution.

Note the Pattern

Ideal solutions and ideal gases are both simple models that ignore intermolecular interactions.

Most real solutions, however, do not obey Raoult’s law precisely, just as most real gases do not obey the ideal gas law exactly. Real solutions generally deviate from Raoult’s law because the intermolecular interactions between the two components A and B differ. We can distinguish between two general kinds of behavior, depending on whether the intermolecular interactions between molecules A and B are stronger or weaker than the A–A and B–B interactions in the pure components. If the A–B interactions are stronger than the A–A and B–B interactions, each component of the solution exhibits a lower vapor pressure than expected for an ideal solution, as does the solution as a whole. The favorable A–B interactions effectively stabilize the solution compared with the vapor. This kind of behavior is called a negative deviation from Raoult’s law. Systems stabilized by hydrogen bonding between two molecules, such as acetone and ethanol, exhibit negative deviations from Raoult’s law. Conversely, if the A–B interactions are weaker than the A–A and B–B interactions yet the entropy increase is enough to allow the solution to form, both A and B have an increased tendency to escape from the solution into the vapor phase. The result is a higher vapor pressure than expected for an ideal solution, producing a positive deviation from Raoult’s law. In a solution of CCl4 and methanol, for example, the nonpolar CCl4 molecules interrupt the extensive hydrogen bonding network in methanol, and the lighter methanol molecules have weaker London dispersion forces than the heavier CCl4 molecules. Consequently, solutions of CCl4 and methanol exhibit positive deviations from Raoult’s law.

Example 9

For each system, compare the intermolecular interactions in the pure liquids and in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation).

  1. cyclohexane and ethanol
  2. methanol and acetone
  3. n-hexane and isooctane (2,2,4-trimethylpentane)

Given: identity of pure liquids

Asked for: predicted deviation from Raoult’s law

Strategy:

Identify whether each liquid is polar or nonpolar, and then predict the type of intermolecular interactions that occur in solution.

Solution:

  1. Liquid ethanol contains an extensive hydrogen bonding network, and cyclohexane is nonpolar. Because the cyclohexane molecules cannot interact favorably with the polar ethanol molecules, they will disrupt the hydrogen bonding. Hence the A–B interactions will be weaker than the A–A and B–B interactions, leading to a higher vapor pressure than predicted by Raoult’s law (a positive deviation).
  2. Methanol contains an extensive hydrogen bonding network, but in this case the polar acetone molecules create A–B interactions that are stronger than the A–A or B–B interactions, leading to a negative enthalpy of solution and a lower vapor pressure than predicted by Raoult’s law (a negative deviation).
  3. Hexane and isooctane are both nonpolar molecules (isooctane actually has a very small dipole moment, but it is so small that it can be ignored). Hence the predominant intermolecular forces in both liquids are London dispersion forces. We expect the A–B interactions to be comparable in strength to the A–A and B–B interactions, leading to a vapor pressure in good agreement with that predicted by Raoult’s law (an ideal solution).

Exercise

For each system, compare the intermolecular interactions in the pure liquids with those in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation):

  1. benzene and n-hexane
  2. ethylene glycol and CCl4
  3. acetic acid and n-propanol

Answer:

  1. approximately equal
  2. positive deviation (vapor pressure greater than predicted)
  3. negative deviation (vapor pressure less than predicted)

Boiling Point Elevation

Recall from that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to P = 1 atm at a higher temperature than does the curve for pure water.

Note the Pattern

The boiling point of a solution with a nonvolatile solute is always greater than the boiling point of the pure solvent.

Figure 13.16 Phase Diagrams of Pure Water and an Aqueous Solution of a Nonvolatile Solute

The vaporization curve for the solution lies below the curve for pure water at all temperatures, which results in an increase in the boiling point and a decrease in the freezing point of the solution.

The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (). We can define the boiling point elevation (ΔTb)The difference between the boiling point of a solution and the boiling point of the pure solvent. as the difference between the boiling points of the solution and the pure solvent:

Equation 13.18

ΔTb=TbTb0

where Tb is the boiling point of the solution and Tb0 is the boiling point of the pure solvent. We can express the relationship between ΔTb and concentration as follows:

Equation 13.19

ΔTb = mKb

where m is the concentration of the solute expressed in molality, and Kb is the molal boiling point elevation constant of the solvent, which has units of °C/m. lists characteristic Kb values for several commonly used solvents.

Figure 13.17 Vapor Pressure Decrease and Boiling Point Increase as Functions of the Mole Fraction of a Nonvolatile Solute

For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration.

Table 13.7 Boiling Point Elevation Constants (Kb) and Freezing Point Depression Constants (Kf) for Some Solvents

Solvent Boiling Point (°C) Kb (°C/m) Freezing Point (°C) Kf (°C/m)
acetic acid 117.90 3.22 16.64 3.63
benzene 80.09 2.64 5.49 5.07
d-(+)-camphor 207.4 4.91 178.8 37.8
carbon disulfide 46.2 2.42 −112.1 3.74
carbon tetrachloride 76.8 5.26 −22.62 31.4
chloroform 61.17 3.80 −63.41 4.60
nitrobenzene 210.8 5.24 5.70 6.87
water 100.00 0.51 0.00 1.86

The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros.

According to , the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous NaCl solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of NaCl produces 2 mol of dissolved ions. Hence a 1.00 m NaCl solution will have a boiling point of about 101.02°C.

Example 10

In Example 8, we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution.

Given: composition of solution

Asked for: boiling point

Strategy:

Calculate the molality of ethylene glycol in the 30.2% solution. Then use to calculate the increase in boiling point.

Solution:

From Example 8, we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus

molality of ethylene glycol=(4.87 mol698 g H2O)(1000 g1 kg)=6.98 m

From , the increase in boiling point is therefore

ΔTb=mKb=(6.98 m)(0.51°C/m)=3.6°C

The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain may not be valid.

Exercise

Assume that a tablespoon (5.00 g) of NaCl is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil?

Answer: 100.04°C, or 100°C to three significant figures. (Recall that 1 mol of NaCl produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.)

Freezing Point Depression

The phase diagram in shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to P = 1 atm at a lower temperature than the curve for pure water.

We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice.

By analogy to our treatment of boiling point elevation, the freezing point depression (ΔTf)The difference between the freezing point of a pure solvent and the freezing point of the solution. is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution:

Equation 13.20

ΔTf=Tf0Tf

where Tf0 is the freezing point of the pure solvent and Tf is the freezing point of the solution. The order of the terms is reversed compared with to express the freezing point depression as a positive number. The relationship between ΔTf and the solute concentration is given by an equation analogous to :

Equation 13.21

ΔTf = mKf

where m is the molality of the solution and Kf is the molal freezing point depression constant for the solvent (in units of °C/m). Like Kb, each solvent has a characteristic value of Kf (see ). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous NaCl solution has twice as large a freezing point depression as a glucose solution of the same molality.

People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing.

Note the Pattern

The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles.

Example 11

In colder regions of the United States, NaCl or CaCl2 is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in to estimate the concentrations of two saturated solutions at 0°C, one of NaCl and one of CaCl2, and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice.

Given: solubilities of two compounds

Asked for: concentrations and freezing points

Strategy:

A Estimate the solubility of each salt in 100 g of water from . Determine the number of moles of each in 100 g and calculate the molalities.

B Determine the concentrations of the dissolved salts in the solutions. Substitute these values into to calculate the freezing point depressions of the solutions.

Solution:

A From , we can estimate the solubilities of NaCl and CaCl2 to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are

mNaCl=(36 g NaCl100 g H2O)(1 mol NaCl58.44 g NaCl)(1000 g1 kg)=6.2 mmCaCl2=(60 g CaCl2100 g H2O)(1 mol CaCl2110.98 g CaCl2)(1000 g1 kg)=5.4 m

The lower formula mass of NaCl more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than CaCl2.

B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of NaCl and CaCl2, respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for NaCl and 3 × 5.4 m = 16 m for CaCl2. The resulting freezing point depressions can be calculated using :

NaCl: ΔTf=mKf=(12m)(1.86°C/m)=22°CCaCl2: ΔTf=mKf=(16m)(1.86°C/m)=30°C

Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that CaCl2 is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, CaCl2 is the salt usually sold for home use, and it is also often used on highways.

Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer.

Exercise

Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Example 13.8 and Example 13.10.

Answer: −13.0°C

Example 12

Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m KCl, 0.1 m glucose, 0.1 m SrCl2, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m HCl.

Given: molalities of six solutions

Asked for: relative freezing points

Strategy:

A Identify each solute as a strong, weak, or nonelectrolyte, and use this information to determine the number of solute particles produced.

B Multiply this number by the concentration of the solution to obtain the effective concentration of solute particles. The solution with the highest effective concentration of solute particles has the largest freezing point depression.

Solution:

A Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. KCl, SrCl2, and HCl are strong electrolytes, producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule).

B The molalities of the solutions in terms of the total particles of solute are: KCl and HCl, 0.2 m; SrCl2, 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > HCl = KCl > SrCl2.

Exercise

Arrange these aqueous solutions in order of increasing freezing points: 0.2 m NaCl, 0.3 m acetic acid, 0.1 m CaCl2, and 0.2 m sucrose.

Answer: 0.2 m NaCl (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m CaCl2 < 0.2 m sucrose (highest freezing point)

In biological systems, freezing plant and animal tissues produces ice crystals that rip cells apart, causing severe frostbite and degrading the quality of fish or meat. How, then, can living organisms survive in freezing climates, where we might expect that exposure to freezing temperatures would be fatal? Many organisms that live in cold climates are able to survive at temperatures well below freezing by synthesizing their own chemical antifreeze in concentrations that prevent freezing. Such substances are typically small organic molecules with multiple –OH groups analogous to ethylene glycol.

Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing Kb and Kf values in , we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of Kf (37.8°C/m), d-(+)-camphor () is often used to determine the molar mass of organic compounds by this method.

Example 13

A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of CS2 to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved Sn molecules (i.e., what is the value of n?).

Given: masses of solute and solvent and freezing point

Asked for: molar mass and number of S atoms per molecule

Strategy:

A Use , the measured freezing point of the solution, and the freezing point of CS2 from to calculate the freezing point depression. Then use and the value of Kf from to calculate the molality of the solution.

B From the calculated molality, determine the number of moles of solute present.

C Use the mass and number of moles of the solute to calculate the molar mass of sulfur in solution. Divide the result by the molar mass of atomic sulfur to obtain n, the number of sulfur atoms per mole of dissolved sulfur.

Solution:

A The first step is to calculate the freezing point depression using :

ΔTf=Tf0Tf=112.1°C(113.5°C)=1.4°C

Then gives

m=ΔTfKf=1.4°C3.74°C/m=0.37m

B The total number of moles of solute present in the solution is

moles solute=(0.37 molkg)(75.0 g)(1 kg1000 g)=0.028 mol

C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus

molar mass=7.08 g0.028 mol=260 g/mol

The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of S8.

Exercise

One of the byproducts formed during the synthesis of C60 is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of CCl4 has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance?

Answer: 847 g/mol; C70

Osmotic Pressure

Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membraneA barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions., a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosisThe net flow of solvent through a semipermeable membrane. (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration.

Osmosis can be demonstrated using a U-tube like the one shown in , which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure (Π)The pressure difference between the two sides of a semipermeable membrane that separates a pure solvent from a solution prepared from the same solvent. of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution.

Figure 13.18 Osmotic Pressure

(a) A dilute solution of glucose in water is placed in the right arm of a U-tube, and the left arm is filled to the same height with pure water; a semipermeable membrane separates the two arms. Because the flow of pure solvent through the membrane from left to right (from pure water to the solution) is greater than the flow of solvent in the reverse direction, the level of liquid in the right tube rises. (b) At equilibrium, the pressure differential, equal to the osmotic pressure of the solution (Πsoln), equalizes the flow rate of solvent in both directions. (c) Applying an external pressure equal to the osmotic pressure of the original glucose solution to the liquid in the right arm reverses the flow of solvent and restores the original situation.

Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation:

Equation 13.22

Π=nRTV=MRT

where M is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), R is the ideal gas constant, and T is the absolute temperature. As shown in Example 14, osmotic pressures tend to be quite high, even for rather dilute solutions.

Example 14

When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% NaCl by mass; the solution density is 1.02 g/mL at 25°C.

  1. Calculate the osmotic pressure of a 4.0% aqueous NaCl solution at 25°C.
  2. If the normal osmotic pressure inside a yeast cell is 7.3 atm, corresponding to a total concentration of dissolved particles of 0.30 M, what concentration of glycerol must the cells synthesize to exactly balance the external osmotic pressure at 25°C?

Given: concentration, density, and temperature of NaCl solution; internal osmotic pressure of cell

Asked for: osmotic pressure of NaCl solution and concentration of glycerol needed

Strategy:

A Calculate the molarity of the NaCl solution using the formula mass of the solute and the density of the solution. Then calculate the total concentration of dissolved particles.

B Use to calculate the osmotic pressure of the solution.

C Subtract the normal osmotic pressure of the cells from the osmotic pressure of the salt solution to obtain the additional pressure needed to balance the two. Use to calculate the molarity of glycerol needed to create this osmotic pressure.

Solution:

  1. A The solution contains 4.0 g of NaCl per 100 g of solution. Using the formula mass of NaCl (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity:

    MNaCl=moles NaClliter solution=(4.0 g NaCl58.44 g/mol NaCl)(1100 g solution)(1.02 g solution1.00 mL solution)(1000 mLL)=0.70 M NaCl

    Because 1 mol of NaCl produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M.

    B Now we can use to calculate the osmotic pressure of the solution:

    Π=MRT=(1.4 mol/L)[0.0821(L·atm)/(K·mol)](298 K)=34 atm
  2. C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve for the molarity corresponding to this osmotic pressure:

    M=ΠRT=27 atm[0.0821(L·atm)/(K·mol)](298 K)=1.1 M glycerol

    In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the NaCl solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol.

Exercise

Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them.

Answer: 24 atm

Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysisA process that uses a semipermeable membrane with pores large enough to allow small solute molecules and solvent molecules to pass through but not large solute molecules., which permits both water and small molecules to pass through but not large molecules such as proteins.

The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in , which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells.

Figure 13.19 Effect on Red Blood Cells of the Surrounding Solution’s Osmotic Pressure

(a) When red blood cells are placed in a dilute salt solution having the same osmotic pressure as the intracellular fluid, the rate of flow of water into and out of the cells is the same and their shape does not change. (b) When cells are placed in distilled water whose osmotic pressure is less than that of the intracellular fluid, the rate of flow of water into the cells is greater than the rate of flow out of the cells. The cells swell and eventually burst. (c) When cells are placed in a concentrated salt solution with an osmotic pressure greater than that of the intracellular fluid, the rate of flow of water out of the cells is greater than the rate of flow into the cells. The cells shrivel and become so deformed that they cannot function.

In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. (For more information about capillary action, see , .) Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves.

Finally, a process called reverse osmosisA process that uses the application of an external pressure greater than the osmotic pressure of a solution to reverse the flow of solvent through the semipermeable membrane. can be used to produce pure water from seawater. As shown schematically in , applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats.

Figure 13.20 Desalinization of Seawater by Reverse Osmosis

(top) When the pressure applied to seawater equals its osmotic pressure (Πsoln), there is no net flow of water across the semipermeable membrane. (bottom) The application of pressure greater than the osmotic pressure of seawater forces water molecules to flow through the membrane, leaving behind a concentrated salt solution. In desalinization plants, seawater is continuously introduced under pressure and pure water is collected, so the process continues indefinitely.

Colligative Properties of Electrolyte Solutions

Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of NaCl in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of NaCl and KCl are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution.

The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van’t Hoff factor(i)The ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt. and is defined as follows:Named for Jacobus Hendricus van’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions.

Equation 13.23

i=apparent number of particles in solutionnumber of moles of solute dissolved

Note the Pattern

As the solute concentration increases the van’t Hoff factor decreases.

The van’t Hoff factor is therefore a measure of a deviation from ideal behavior. The lower the van’t Hoff factor, the greater the deviation. As the data in show, the van’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution. Instead, some of the ions exist as ion pairsA cation and anion that are in intimate contact in solution rather than separated by solvent and that migrates in solution as a single unit., a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as Mg2+, Al3+, SO42−, and PO43− have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations.

Figure 13.21 Ion Pairs

In concentrated solutions of electrolytes like NaCl, some of the ions form neutral ion pairs that are not separated by solvent and diffuse as single particles.

Table 13.8 van’t Hoff Factors for 0.0500 M Aqueous Solutions of Selected Compounds at 25°C

Compound i (measured) i (ideal)
glucose 1.0 1.0
sucrose 1.0 1.0
NaCl 1.9 2.0
HCl 1.9 2.0
MgCl2 2.7 3.0
FeCl3 3.4 4.0
Ca(NO3)2 2.5 3.0
AlCl3 3.2 4.0
MgSO4 1.4 2.0

Example 15

A 0.0500 M aqueous solution of FeCl3 has an osmotic pressure of 4.15 atm at 25°C. Calculate the van’t Hoff factor i for the solution.

Given: solute concentration, osmotic pressure, and temperature

Asked for: van’t Hoff factor

Strategy:

A Use to calculate the expected osmotic pressure of the solution based on the effective concentration of dissolved particles in the solvent.

B Calculate the ratio of the observed osmotic pressure to the expected value. Multiply this number by the number of ions of solute per formula unit, and then use to calculate the van’t Hoff factor.

Solution:

A If FeCl3 dissociated completely in aqueous solution, it would produce four ions per formula unit [Fe3+(aq) plus 3Cl(aq)] for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be

Π=MRT=(0.200 mol/L)[0.0821(L·atm)/(K·mol)](298 K)=4.89 atm

B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 particles per mole of FeCl3 dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm:

4.15 atm=M[0.0821(L·atm)/(K·mol)](298 K)0.170 mol/L=M

The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 particles per mole of FeCl3 dissolved. From , the van’t Hoff factor for the solution is

i=3.40 particles observed1 formula unit FeCl3=3.40

Exercise

Calculate the van’t Hoff factor for a 0.050 m aqueous solution of MgCl2 that has a measured freezing point of −0.25°C.

Answer: 2.7 (versus an ideal value of 3)

Summary

The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law. Solutions that obey Raoult’s law are called ideal solutions. Most real solutions exhibit positive or negative deviations from Raoult’s law. The boiling point elevation (ΔTb) and freezing point depression (ΔTf) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt.

Key Takeaway

  • The total number of nonvolatile solute particles determines the decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus the pure solvent.

Key Equations

Henry’s law

:   C = kP

Raoult’s law

:   PA=XAPA0

vapor pressure lowering

:   PA0PA=ΔP=XBPA0

vapor pressure of a system containing two volatile components

:   PT=XAPA0+(1XA)PB0

boiling point elevation

:   ΔTb = mKb

freezing point depression

:   ΔTf = mKf

osmotic pressure

:   Π=nRTV=MRT

van’t Hoff factor

:   i=apparent number of particles in solutionnumber of moles of solute dissolved

Conceptual Problems

  1. Why does the vapor pressure of a solvent decrease when adding a nonvolatile solute?

  2. Does seawater boil at the same temperature as distilled water? If not, which has the higher boiling point? Explain your answer.

  3. Which will be more soluble in benzene—O2 or HCl? Will H2S or HCl be more soluble in water? Explain your reasoning in each case.

  4. Will the vapor pressure of a solution of hexane and heptane have an ideal vapor pressure curve (i.e., obey Raoult’s law)? Explain your answer. What properties of two liquids determine whether a solution of the two will exhibit an ideal behavior?

  5. Predict whether the following mixtures will exhibit negative, zero, or positive deviations from Raoult’s law. Explain your reasoning in each case.

    1. carbon tetrachloride and heptane
    2. methanol and tetrahydrofuran (C4H8O)
    3. acetone [(CH3)2C=CO] and dichloromethane
    4. hexane and methanol
  6. Why are deviations from the ideal behavior predicted by Raoult’s law more common for solutions of liquids than are deviations from the ideal behavior predicted by the ideal gas law for solutions of gases?

  7. Boiling point elevation is proportional to the molal concentration of the solute. Is it also proportional to the molar concentration of the solution? Why or why not?

  8. Many packaged foods in sealed bags are cooked by placing the bag in boiling water. How could you reduce the time required to cook the contents of the bag using this cooking method?

  9. If the costs per kilogram of ethylene glycol and of ethanol were the same, which would be the more cost-effective antifreeze?

  10. Many people get thirsty after eating foods such as ice cream or potato chips that have a high sugar or salt content, respectively. Suggest an explanation for this phenomenon.

  11. When two aqueous solutions with identical concentrations are separated by a semipermeable membrane, no net movement of water occurs. What happens when a solute that cannot penetrate the membrane is added to one of the solutions? Why?

  12. A solution injected into blood vessels must have an electrolyte concentration nearly identical to that found in blood plasma. Why? What would happen if red blood cells were placed in distilled water? What would happen to red blood cells if they were placed in a solution that had twice the electrolyte concentration of blood plasma?

  13. If you were stranded on a desert island, why would drinking seawater lead to an increased rate of dehydration, eventually causing you to die of thirst?

  14. What is the relationship between the van’t Hoff factor for a compound and its lattice energy?

Numerical Problems

  1. Hemoglobin is the protein that is responsible for the red color of blood and for transporting oxygen from the lungs to the tissues. A solution with 11.2 mg of hemoglobin per mL has an osmotic pressure of 2.9 mmHg at 5°C. What is the molecular mass of hemoglobin?

  2. To determine the molar mass of the antifreeze protein from the Arctic right-eye flounder, the osmotic pressure of a solution containing 13.2 mg of protein per mL was measured and found to be 21.2 mmHg at 10°C. What is the molar mass of the protein?

  3. What is the osmotic pressure at 21.0°C of 13.5 mL of a solution with 1.77 g of sucrose (C12H22O11)?

  4. A solution of NaNO3 is generated by dissolving 1.25 g of NaNO3 in enough water to give a final volume of 25.0 mL. What is the osmotic pressure of this sample at 25.0°C?

  5. Which would have the lower vapor pressure—an aqueous solution that is 0.12 M in glucose or one that is 0.12 M in CaCl2? Why?

  6. What is the total particle concentration expected for each aqueous solution? Which would produce the highest osmotic pressure?

    1. 0.35 M KBr
    2. 0.11 M MgSO4
    3. 0.26 M MgCl2
    4. 0.24 M glucose (C6H12O6)
  7. The boiling point of an aqueous solution of sodium chloride is 100.37°C. What is the molality of the solution? How many grams of NaCl are present in 125 g of the solution?

  8. Calculate the boiling point of a solution of sugar prepared by dissolving 8.4 g of glucose (C6H12O6) in 250 g of water.

  9. At 37°C, the vapor pressure of 300.0 g of water was reduced from 0.062 atm to 0.058 atm by the addition of NaBr. How many grams of NaBr were added?

  10. How many grams of KCl must be added to reduce the vapor pressure of 500.0 g of H2O from 17.5 mmHg to 16.0 mmHg at 20.0°C?

  11. How much NaCl would you have to add to 2.0 L of water at a mountain lodge at an elevation of 7350 ft, where the pressure is 0.78 atm and the boiling point of water is 94°C, to get the water to boil at the same temperature as in New Orleans, Louisiana, where the pressure is 1.00 atm?

  12. You have three solutions with the following compositions: 12.5 g of KCl in 250 mL of water, 12.5 g of glucose in 400 mL of water, and 12.5 g of MgCl2 in 350 mL of water. Which will have the highest boiling point?

  13. Assuming the price per kilogram is the same, which is a better salt to use for deicing wintry roads—NaCl or MgCl2? Why? Would magnesium chloride be an effective deicer at a temperature of −8°C?

  14. How many grams of KNO3 must be added to water to produce the same boiling point elevation as a solution of 2.03 g of MgCl2 in a total volume of 120.0 mL of solution, assuming complete dissociation? If the van’t Hoff factor for MgCl2 at this concentration is 2.73, how much KNO3 would be needed?

  15. Calculate the quantity of each compound that would need to be added to lower the freezing point of 500.0 mL of water by 1.0°C: KBr, ethylene glycol, MgBr2, ethanol. Assume that the density of water is 1.00 g/cm3.

  16. The melting point depression of biphenyl (melting point = 69.0°C) can be used to determine the molecular mass of organic compounds. A mixture of 100.0 g of biphenyl and 2.67 g of naphthalene (C10H8) has a melting point of 68.50°C. If a mixture of 1.00 g of an unknown compound with 100.0 g of biphenyl has a melting point of 68.86°C, what is the molar mass of the unknown compound?

  17. Four solutions of urea in water were prepared, with concentrations of 0.32 m, 0.55 m, 1.52 m, and 3.16 m. The freezing points of these solutions were found to be −0.595°C, −1.02°C, −2.72°C, and −5.71°C, respectively. Graphically determine the freezing point depression constant for water. A fifth solution made by dissolving 6.22 g of urea in 250.0 g of water has a freezing point of −0.75°C. Use these data to determine the molar mass of urea.

  18. The term osmolarity has been used to describe the total solute concentration of a solution (generally water), where 1 osmole (Osm) is equal to 1 mol of an ideal, nonionizing molecule.

    1. What is the osmolarity of a 1.5 M solution of glucose? a 1.5 M solution of NaCl? a 1.5 M solution of CaCl2?
    2. What is the relationship between osmolarity and the concentration of water?
    3. What would be the direction of flow of water through a semipermeable membrane separating a 0.1 M solution of NaCl and a 0.1 M solution of CaCl2?
  19. At 40°C, the vapor pressures of pure CCl4 and cyclohexane are 0.2807 atm and 0.2429 atm, respectively. Assuming ideal behavior, what is the vapor pressure of a solution with a CCl4 mole fraction of 0.475? What is the mole fraction of cyclohexane in the vapor phase? The boiling points of CCl4 and cyclohexane are 76.8°C and 80.7°C, respectively.

  20. A benzene/toluene solution with a mole fraction of benzene of 0.6589 boils at 88°C at 1 atm. The vapor pressures of pure benzene and toluene at this temperature are 1.259 atm and 0.4993 atm, respectively. What is the composition of the vapor above the boiling solution at this temperature?

  21. Plot the vapor pressure of the solution versus composition for the system CCl4–CH3CN at 45°C, given the following experimental data:

    XCCl4 (liquid) 0.035 0.375 0.605 0.961
    XCCl4 (vapor) 0.180 0.543 0.594 0.800
    Total P (atm) 0.326 0.480 0.488 0.414

    Does your diagram show behavior characteristic of an ideal solution? Explain your answer.

Answers

  1. 6.7 × 104 amu

  2. 9.24 atm

  3. The CaCl2 solution will have a lower vapor pressure, because it contains three times as many particles as the glucose solution.

  4. 0.36 m NaCl, 2.6 g NaCl

  5. 60 g NaBr

  6. 700 g NaCl

  7. MgCl2 produces three particles in solution versus two for NaCl, so the same molal concentration of MgCl2 will produce a 50% greater freezing point depression than for NaCl. Nonetheless, the molar mass of MgCl2 is 95.3 g/mol versus 48.45 g/mol for NaCl. Consequently, a solution containing 1 g NaCl per 1000 g H2O will produce a freezing point depression of 0.064°C versus 0.059°C for a solution containing 1 g MgCl2 per 1000 g H2O. Thus, given equal cost per gram, NaCl is more effective. Yes, MgCl2 would be effective at −8°C; a 1.43 m solution (136 g per 1000 g H2O) would be required.

  8.  

    kf = 1.81(°C·kg)/mol, molecular mass of urea = 60.0 g/mol

13.6 Aggregate Particles in Aqueous Solution

Learning Objective

  1. To distinguish between true solutions and solutions with aggregate particles.

Suspensions and colloids are two common types of mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures. A suspensionA heterogeneous mixture of particles with diameters of about 1 µm that are distributed throughout a second phase and that separate from the dispersing phase on standing. is a heterogeneous mixture of particles with diameters of about 1 µm (1000 nm) that are distributed throughout a second phase. Common suspensions include paint, blood, and hot chocolate, which are solid particles in a liquid, and aerosol sprays, which are liquid particles in a gas. If the suspension is allowed to stand, the two phases will separate, which is why paints must be thoroughly stirred or shaken before use. A colloidA heterogeneous mixture of particles with diameters of about 2–500 nm that are distributed throughout a second phase and do not separate from the dispersing phase on standing. is also a heterogeneous mixture, but the particles of a colloid are typically smaller than those of a suspension, generally in the range of 2 to about 500 nm in diameter. Colloids include fog and clouds (liquid particles in a gas), milk (solid particles in a liquid), and butter (solid particles in a solid). Other colloids are used industrially as catalysts. Unlike in a suspension, the particles in a colloid do not separate into two phases on standing. The only combination of substances that cannot produce a suspension or a colloid is a mixture of two gases because their particles are so small that they always form true solutions. The properties of suspensions, colloids, and solutions are summarized in Table 13.9 "Properties of Liquid Solutions, Colloids, and Suspensions".

Table 13.9 Properties of Liquid Solutions, Colloids, and Suspensions

Type of Mixture Approximate Size of Particles (nm) Characteristic Properties Examples
solution < 2 not filterable; does not separate on standing; does not scatter visible light air, white wine, gasoline, salt water
colloid 2–500 scatters visible light; translucent or opaque; not filterable; does not separate on standing smoke, fog, ink, milk, butter, cheese
suspension 500–1000 cloudy or opaque; filterable; separates on standing muddy water, hot cocoa, blood, paint

Colloids and Suspensions

Colloids were first characterized in about 1860 by Thomas Graham, who also gave us Graham’s law of diffusion and effusion. Although some substances, such as starch, gelatin, and glue, appear to dissolve in water to produce solutions, Graham found that they diffuse very slowly or not at all compared with solutions of substances such as salt and sugar. Graham coined the word colloid (from the Greek kólla, meaning “glue”) to describe these substances, as well as the words solA dispersion of solid particles in a liquid or solid. and gelA semisolid sol in which all of the liquid phase has been absorbed by the solid particles. to describe certain types of colloids in which all of the solvent has been absorbed by the solid particles, thus preventing the mixture from flowing readily, as we see in Jell-O. Two other important types of colloids are aerosolsA dispersion of solid or liquid particles in a gas., which are dispersions of solid or liquid particles in a gas, and emulsions, which are dispersions of one liquid in another liquid with which it is immiscible.

Colloids share many properties with solutions. For example, the particles in both are invisible without a powerful microscope, do not settle on standing, and pass through most filters. However, the particles in a colloid scatter a beam of visible light, a phenomenon known as the Tyndall effectThe phenomenon of scattering a beam of visible light.,The effect is named after its discoverer, John Tyndall, an English physicist (1820–1893). whereas the particles of a solution do not. The Tyndall effect is responsible for the way the beams from automobile headlights are clearly visible from the side on a foggy night but cannot be seen from the side on a clear night. It is also responsible for the colored rays of light seen in many sunsets, where the sun’s light is scattered by water droplets and dust particles high in the atmosphere. An example of the Tyndall effect is shown in Figure 13.22 "Tyndall Effect, the Scattering of Light by Colloids".

Figure 13.22 Tyndall Effect, the Scattering of Light by Colloids

Both cylinders contain a solution of red food coloring in water, but a small amount of gelatin has been added to the cylinder on the right to form a colloidal suspension of gelatin particles. The beam of light goes straight through the true solution on the left, but the light beam is scattered by the colloid on the right.

Although colloids and suspensions can have particles similar in size, the two differ in stability: the particles of a colloid remain dispersed indefinitely unless the temperature or chemical composition of the dispersing medium is changed. The chemical explanation for the stability of colloids depends on whether the colloidal particles are hydrophilic or hydrophobic.

Most proteins, including those responsible for the properties of gelatin and glue, are hydrophilic because their exterior surface is largely covered with polar or charged groups. Starch, a long-branched polymer of glucose molecules, is also hydrophilic. A hydrophilic colloid particle interacts strongly with water, resulting in a shell of tightly bound water molecules that prevents the particles from aggregating when they collide. Heating such a colloid can cause aggregation because the particles collide with greater energy and disrupt the protective shell of solvent. Moreover, heat causes protein structures to unfold, exposing previously buried hydrophobic groups that can now interact with other hydrophobic groups and cause the particles to aggregate and precipitate from solution. When an egg is boiled, for example, the egg white, which is primarily a colloidal suspension of a protein called albumin, unfolds and exposes its hydrophobic groups, which aggregate and cause the albumin to precipitate as a white solid.

In some cases, a stable colloid can be transformed to an aggregated suspension by a minor chemical modification. Consider, for example, the behavior of hemoglobin, a major component of red blood cells. Hemoglobin molecules normally form a colloidal suspension inside red blood cells, which typically have a “donut” shape and are easily deformed, allowing them to squeeze through the capillaries to deliver oxygen to tissues. In a common inherited disease called sickle-cell anemia, one of the amino acids in hemoglobin that has a hydrophilic carboxylic acid side chain (glutamate) is replaced by another amino acid that has a hydrophobic side chain (valine, Figure 5.16 "The Structures of 10 Amino Acids"). Under some conditions, the abnormal hemoglobin molecules can aggregate to form long, rigid fibers that cause the red blood cells to deform, adopting a characteristic sickle shape that prevents them from passing through the capillaries (Figure 13.23 "Sickle-Cell Anemia"). The reduction in blood flow results in severe cramps, swollen joints, and liver damage. Until recently, many patients with sickle-cell anemia died before the age of 30 from infection, blood clots, or heart or kidney failure, although individuals with the sickle-cell genetic trait are more resistant to malaria than are those with “normal” hemoglobin.

Figure 13.23 Sickle-Cell Anemia

The characteristic shape of sickled red blood cells is the result of fibrous aggregation of hemoglobin molecules inside the cell.

Figure 13.24 Formation of New Land by the Destabilization of a Colloid Suspension

This satellite photograph shows the Mississippi River delta from New Orleans (top) to the Gulf of Mexico (bottom). Where seawater mixes with freshwater from the Mississippi River, colloidal clay particles in the river water precipitate (tan area).

Aggregation and precipitation can also result when the outer, charged layer of a particle is neutralized by ions with the opposite charge. In inland waterways, clay particles, which have a charged surface, form a colloidal suspension. High salt concentrations in seawater neutralize the charge on the particles, causing them to precipitate and form land at the mouths of large rivers, as seen in the satellite view in Figure 13.24 "Formation of New Land by the Destabilization of a Colloid Suspension". Charge neutralization is also an important strategy for precipitating solid particles from gaseous colloids such as smoke, and it is widely used to reduce particulate emissions from power plants that burn fossil fuels.

Emulsions

EmulsionsA dispersion of one liquid phase in another liquid with which it is immiscible. are colloids formed by the dispersion of a hydrophobic liquid in water, thereby bringing two mutually insoluble liquids, such as oil and water, in close contact. Various agents have been developed to stabilize emulsions, the most successful being molecules that combine a relatively long hydrophobic “tail” with a hydrophilic “head”:

Examples of such emulsifying agents include soaps, which are salts of long-chain carboxylic acids, such as sodium stearate [CH3(CH2)16CO2Na+], and detergents, such as sodium dodecyl sulfate [CH3(CH2)11OSO3Na+], whose structures are as follows:

When you wash your laundry, the hydrophobic tails of soaps and detergents interact with hydrophobic particles of dirt or grease through dispersion forces, dissolving in the interior of the hydrophobic particle. The hydrophilic group is then exposed at the surface of the particle, which enables it to interact with water through ion–dipole forces and hydrogen bonding. This causes the particles of dirt or grease to disperse in the wash water and allows them to be removed by rinsing. Similar agents are used in the food industry to stabilize emulsions such as mayonnaise.

A related mechanism allows us to absorb and digest the fats in buttered popcorn and French fries. To solubilize the fats so that they can be absorbed, the gall bladder secretes a fluid called bile into the small intestine. Bile contains a variety of bile salts, detergent-like molecules that emulsify the fats.

Micelles

Detergents and soaps are surprisingly soluble in water in spite of their hydrophobic tails. The reason for their solubility is that they do not, in fact, form simple solutions. Instead, above a certain concentration they spontaneously form micellesA spherical or cylindrical aggregate of detergents or soaps in water that minimizes contact between the hydrophobic tails of the detergents or soaps and water., which are spherical or cylindrical aggregates that minimize contact between the hydrophobic tails and water. In a micelle, only the hydrophilic heads are in direct contact with water, and the hydrophobic tails are in the interior of the aggregate (part (a) in Figure 13.25 "Micelles and a Phospholipid Bilayer").

Figure 13.25 Micelles and a Phospholipid Bilayer

(a) Soaps and detergents, which contain a single hydrophobic tail on each molecule, form spherical micelles with the intertwined tails in the interior and the hydrophilic head groups on the exterior. (b) Phospholipids, which have two hydrophobic tails, tend to form extended double layers in which the hydrophobic tails are sandwiched between the hydrophilic head groups.

A large class of biological molecules called phospholipidsA large class of biological, detergent-like molecules that have a hydrophilic head and two hydrophobic tails and that form bilayers. consists of detergent-like molecules with a hydrophilic head and two hydrophobic tails, as can be seen in the molecule of phosphatidylcholine. The additional tail results in a cylindrical shape that prevents phospholipids from forming a spherical micelle. Consequently, phospholipids form bilayersA two-dimensional sheet consisting of a double layer of phospholipid molecules arranged tail to tail., extended sheets consisting of a double layer of molecules. As shown in part (b) in Figure 13.25 "Micelles and a Phospholipid Bilayer", the hydrophobic tails are in the center of the bilayer, where they are not in contact with water, and the hydrophilic heads are on the two surfaces, in contact with the surrounding aqueous solution.

A cell membraneA mixture of phospholipids that form a phospholipid bilayer around the cell. is essentially a mixture of phospholipids that form a phospholipid bilayer. One definition of a cellA collection of molecules, capable of reproducing itself, that is surrounded by a phospholipid bilayer. is a collection of molecules surrounded by a phospholipid bilayer that is capable of reproducing itself. The simplest cells are bacteria, which consist of only a single compartment surrounded by a single membrane. Animal and plant cells are much more complex, however, and contain many different kinds of compartments, each surrounded by a membrane and able to carry out specialized tasks.

Summary

A suspension is a heterogeneous mixture of particles of one substance distributed throughout a second phase; the dispersed particles separate from the dispersing phase on standing. In contrast, the particles in a colloid are smaller and do not separate on standing. A colloid can be classified as a sol, a dispersion of solid particles in a liquid or solid; a gel, a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an aerosol, a dispersion of solid or liquid particles in a gas; or an emulsion, a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect. Hydrophilic colloids contain an outer shell of groups that interact favorably with water, whereas hydrophobic colloids have an outer surface with little affinity for water. Emulsions are prepared by dispersing a hydrophobic liquid in water. In the absence of a dispersed hydrophobic liquid phase, solutions of detergents in water form organized spherical aggregates called micelles. Phospholipids are a class of detergent-like molecules that have two hydrophobic tails attached to a hydrophilic head. A bilayer is a two-dimensional sheet consisting of a double layer of phospholipid molecules arranged tail to tail with a hydrophobic interior and a hydrophilic exterior. Cells are collections of molecules that are surrounded by a phospholipid bilayer called a cell membrane and are able to reproduce themselves.

Key Takeaway

  • Colloids and suspensions are mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures.

Conceptual Problems

  1. How does a colloid differ from a suspension? Which has a greater effect on solvent properties, such as vapor pressure?

  2. Is homogenized milk a colloid or a suspension? Is human plasma a colloid or a suspension? Justify your answers.

  3. How would you separate the components of an emulsion of fat dispersed in an aqueous solution of sodium chloride?

13.7 End-of-Chapter Material

Application Problems

    Problems marked with a ♦ involve multiple concepts.

  1. ♦ Scuba divers utilize high-pressure gas in their tanks to allow them to breathe under water. At depths as shallow as 100 ft (30 m), the pressure exerted by water is 4.0 atm. At 25°C the values of Henry’s law constants for N2, O2, and He in blood are as follows: N2 = 6.5 × 10−4 mol/(L·atm), O2 = 1.28 × 10−3 mol/(L·atm), and He = 3.7 × 10−4 mol/(L·atm).

    1. What would be the concentration of nitrogen and oxygen in blood at sea level where the air is 21% oxygen and 79% nitrogen?
    2. What would be the concentration of nitrogen and oxygen in blood at a depth of 30 m, assuming that the diver is breathing compressed air?
  2. ♦ Many modern batteries take advantage of lithium ions dissolved in suitable electrolytes. Typical batteries have lithium concentrations of 0.10 M. Which aqueous solution has the higher concentration of ion pairs: 0.08 M LiCl or 1.4 M LiCl? Why? Does an increase in the number of ion pairs correspond to a higher or lower van’t Hoff factor? Batteries rely on a high concentration of unpaired Li+ ions. Why is using a more concentrated solution not an ideal strategy in this case?

  3. Hydrogen sulfide, which is extremely toxic to humans, can be detected at a concentration of 2.0 ppb. At this level, headaches, dizziness, and nausea occur. At higher concentrations, however, the sense of smell is lost, and the lack of warning can result in coma and death can result. What is the concentration of H2S in milligrams per liter at the detection level? The lethal dose of hydrogen sulfide by inhalation for rats is 7.13 × 10−4 g/L. What is this lethal dose in ppm? The density of air is 1.2929 g/L.

  4. One class of antibiotics consists of cyclic polyethers that can bind alkali metal cations in aqueous solution. Given the following antibiotics and cation selectivities, what conclusion can you draw regarding the relative sizes of the cavities?

    Antibiotic Cation Selectivity
    nigericin K+ > Rb+ > Na+ > Cs+ > Li+
    lasalocid Ba2+ >> Cs+ > Rb+, K+ > Na+, Ca2+, Mg2+
  5. Phenylpropanolamine hydrochloride is a common nasal decongestant. An aqueous solution of phenylpropanolamine hydrochloride that is sold commercially as a children’s decongestant has a concentration of 6.67 × 10−3 M. If a common dose is 1.0 mL/12 lb of body weight, how many moles of the decongestant should be given to a 26 lb child?

  6. The “freeze-thaw” method is often used to remove dissolved oxygen from solvents in the laboratory. In this technique, a liquid is placed in a flask that is then sealed to the atmosphere, the liquid is frozen, and the flask is evacuated to remove any gas and solvent vapor in the flask. The connection to the vacuum pump is closed, the liquid is warmed to room temperature and then refrozen, and the process is repeated. Why is this technique effective for degassing a solvent?

  7. Suppose that, on a planet in a galaxy far, far away, a species has evolved whose biological processes require even more oxygen than we do. The partial pressure of oxygen on this planet, however, is much less than that on Earth. The chemical composition of the “blood” of this species is also different. Do you expect their “blood” to have a higher or lower value of the Henry’s law constant for oxygen at standard temperature and pressure? Justify your answer.

  8. A car owner who had never taken general chemistry decided that he needed to put some ethylene glycol antifreeze in his car’s radiator. After reading the directions on the container, however, he decided that “more must be better.” Instead of using the recommended mixture (30% ethylene glycol/70% water), he decided to reverse the amounts and used a 70% ethylene glycol/30% water mixture instead. Serious engine problems developed. Why?

  9. The ancient Greeks produced “Attic ware,” pottery with a characteristic black and red glaze. To separate smaller clay particles from larger ones, the powdered clay was suspended in water and allowed to settle. This process yielded clay fractions with coarse, medium, and fine particles, and one of these fractions was used for painting. Which size of clay particles forms a suspension, which forms a precipitate, and which forms a colloidal dispersion? Would the colloidal dispersion be better characterized as an emulsion? Why or why not? Which fraction of clay particles was used for painting?

  10. The Tyndall effect is often observed in movie theaters, where it makes the beam of light from the projector clearly visible. What conclusions can you draw about the quality of the air in a movie theater where you observe a large Tyndall effect?

  11. Aluminum sulfate is the active ingredient in styptic pencils, which can be used to stop bleeding from small cuts. The Al3+ ions induce aggregation of colloids in the blood, which facilitates formation of a blood clot. How can Al3+ ions induce aggregation of a colloid? What is the probable charge on the colloidal particles in blood?

  12. ♦ The liver secretes bile, which is essential for the digestion of fats. As discussed in , fats are biomolecules with long hydrocarbon chains. The globules of fat released by partial digestion of food particles in the stomach and lower intestine are too large to be absorbed by the intestine unless they are emulsified by bile salts, such as glycocholate. Explain why a molecule like glycocholate is effective at creating an aqueous dispersion of fats in the digestive tract.

Answers

    1. 1 atm: 2.7 × 10−4 M O2 and 5.1 × 10−4 M N2
    2. 4 atm: 1.1 × 10−3 M O2 and 2.1 × 10−3 M N2
  1. 2.6 × 10−6 mg/L, 550 ppm

  2. 1.4 × 10−5 mol

  3. To obtain the same concentration of dissolved oxygen in their “blood” at a lower partial pressure of oxygen, the value of the Henry’s law constant would have to be higher.

  4. The large, coarse particles would precipitate, the medium particles would form a suspension, and the fine ones would form a colloid. A colloid consists of solid particles in a liquid medium, so it is not an emulsion, which consists of small particles of one liquid suspended in another liquid. The finest particles would be used for painting.