This content is copyright Flat World Knowledge and the author(s). This content was captured in November 2012, from http://catalog.flatworldknowledge.com/bookhub/reader/4309. A Creative Commons Attribution-NonCommercial-ShareAlike license was clearly declared and displayed. Creative Commons BY-NC-SA 3.0 Logo
Read License Information
Full Legal Code.

This content is being redistributed in accordance with the permissions of that license.

Chapter 14 Chemical Kinetics

The gases, liquids, solids, and solutions that you learned to describe quantitatively in Chapter 10 "Gases", Chapter 11 "Liquids", Chapter 12 "Solids", and Chapter 13 "Solutions", respectively, are systems whose chemical compositions do not change with time. Now we will present a quantitative description of a far more common situation in which the chemical composition of a system is not constant with time. An example of such a system is the stratosphere, where chemicals rising from the ground level initiate reactions that lead to decreases in the concentration of stratospheric ozone—the so-called ozone hole. (For more information about the ozone hole, see Chapter 3 "Chemical Reactions", Section 3.6 "Chemical Reactions in the Atmosphere".) Another example involves the production of polyethylene, in which the properties of the plastic are determined by the relative speeds of events that occur during the polymerization reaction. (For more information about the polymerization reaction, see Chapter 12 "Solids", Section 12.8 "Polymeric Solids".) The techniques you are about to learn will enable you to describe the speed of many such changes and predict how the composition of each system will change in response to changing conditions.

The Belousov-Zhabotinsky reaction, a chemical reaction that oscillates in time and space. When a very thin layer of an acidic solution containing potassium bromate (KBrO3), cerium ammonium nitrate [(NH4)2Ce(NO3)6], malonic acid (HO2CCH2CO2H), and an indicator is poured into a shallow dish, local fluctuations in the concentration of the reactants and a complex series of reactions cause striking geometric patterns of concentric circles and spirals to propagate across the dish.

We begin Chapter 14 "Chemical Kinetics" with a discussion of chemical kineticsThe study of reaction rates., which is the study of reaction ratesThe changes in concentrations of reactants and products with time., or the changes in the concentrations of reactants and products with time. As you learn about the factors that affect reaction rates, the methods chemists use for reporting and calculating those rates, and the clues that reaction rates provide about events at the molecular level, you will also discover the answers to questions such as the following: How can normally stable substances such as flour and coal cause devastating explosions? How do archaeologists use isotopic composition to estimate the ages of ancient artifacts? How do the catalysts used in catalytic converters, some laundry detergents, and meat tenderizers work?

Summary

Chemical kinetics is the study of reaction rates, the changes in the concentrations of reactants and products with time.

14.1 Factors That Affect Reaction Rates

Learning Objective

  1. To understand the factors that affect reaction rates.

Although a balanced chemical equation for a reaction describes the quantitative relationships between the amounts of reactants present and the amounts of products that can be formed, it gives us no information about whether or how fast a given reaction will occur. This information is obtained by studying the chemical kinetics of a reaction, which depend on various factors: reactant concentrations, temperature, physical states and surface areas of reactants, and solvent and catalyst properties if either are present. By studying the kinetics of a reaction, chemists gain insights into how to control reaction conditions to achieve a desired outcome.

Concentration Effects

Two substances cannot possibly react with each other unless their constituent particles (molecules, atoms, or ions) come into contact. If there is no contact, the reaction rate will be zero. Conversely, the more reactant particles that collide per unit time, the more often a reaction between them can occur. Consequently, the reaction rate usually increases as the concentration of the reactants increases. One example of this effect is the reaction of sucrose (table sugar) with sulfuric acid, which is shown in .

Figure 14.1 The Effect of Concentration on Reaction Rates

Mixing sucrose with dilute sulfuric acid in a beaker (a, right) produces a simple solution. Mixing the same amount of sucrose with concentrated sulfuric acid (a, left) results in a dramatic reaction (b) that eventually produces a column of black porous graphite (c) and an intense smell of burning sugar.

Temperature Effects

You learned in that increasing the temperature of a system increases the average kinetic energy of its constituent particles. As the average kinetic energy increases, the particles move faster, so they collide more frequently per unit time and possess greater energy when they collide. Both of these factors increase the reaction rate. Hence the reaction rate of virtually all reactions increases with increasing temperature. Conversely, the reaction rate of virtually all reactions decreases with decreasing temperature. For example, refrigeration retards the rate of growth of bacteria in foods by decreasing the reaction rates of biochemical reactions that enable bacteria to reproduce. shows how temperature affects the light emitted by two chemiluminescent light sticks.

Figure 14.2 The Effect of Temperature on Reaction Rates

At high temperature, the reaction that produces light in a chemiluminescent light stick occurs more rapidly, producing more photons of light per unit time. Consequently, the light glows brighter in hot water (left) than in ice water (right).

In systems where more than one reaction is possible, the same reactants can produce different products under different reaction conditions. For example, in the presence of dilute sulfuric acid and at temperatures around 100°C, ethanol is converted to diethyl ether:

Equation 14.1

2CH3CH2OHH2SO4CH3CH2OCH2CH3+H2O

At 180°C, however, a completely different reaction occurs, which produces ethylene as the major product:

Equation 14.2

CH3CH2OHH2SO4C2H4+H2O

Phase and Surface Area Effects

When two reactants are in the same fluid phase, their particles collide more frequently than when one or both reactants are solids (or when they are in different fluids that do not mix). If the reactants are uniformly dispersed in a single homogeneous solution, then the number of collisions per unit time depends on concentration and temperature, as we have just seen. If the reaction is heterogeneous, however, the reactants are in two different phases, and collisions between the reactants can occur only at interfaces between phases. The number of collisions between reactants per unit time is substantially reduced relative to the homogeneous case, and, hence, so is the reaction rate. The reaction rate of a heterogeneous reaction depends on the surface area of the more condensed phase.

Automobile engines use surface area effects to increase reaction rates. Gasoline is injected into each cylinder, where it combusts on ignition by a spark from the spark plug. The gasoline is injected in the form of microscopic droplets because in that form it has a much larger surface area and can burn much more rapidly than if it were fed into the cylinder as a stream. Similarly, a pile of finely divided flour burns slowly (or not at all), but spraying finely divided flour into a flame produces a vigorous reaction (). Similar phenomena are partially responsible for dust explosions that occasionally destroy grain elevators or coal mines.

Figure 14.3 The Effect of Surface Area on Reaction Rates

A pile of flour is only scorched by a flame (right), but when the same flour is sprayed into the flame, it burns rapidly (left).

Solvent Effects

The nature of the solvent can also affect the reaction rates of solute particles. For example, a sodium acetate solution reacts with methyl iodide in an exchange reaction to give methyl acetate and sodium iodide.

Equation 14.3

CH3CO2Na(soln) + CH3I(l) → CH3CO2CH3(soln) + NaI(soln)

This reaction occurs 10 million times more rapidly in the organic solvent dimethylformamide [DMF; (CH3)2NCHO] than it does in methanol (CH3OH). Although both are organic solvents with similar dielectric constants (36.7 for DMF versus 32.6 for methanol), methanol is able to hydrogen bond with acetate ions, whereas DMF cannot. Hydrogen bonding reduces the reactivity of the oxygen atoms in the acetate ion.

Solvent viscosity is also important in determining reaction rates. In highly viscous solvents, dissolved particles diffuse much more slowly than in less viscous solvents and can collide less frequently per unit time. Thus the reaction rates of most reactions decrease rapidly with increasing solvent viscosity.

Catalyst Effects

You learned in that a catalyst is a substance that participates in a chemical reaction and increases the reaction rate without undergoing a net chemical change itself. Consider, for example, the decomposition of hydrogen peroxide in the presence and absence of different catalysts (). Because most catalysts are highly selective, they often determine the product of a reaction by accelerating only one of several possible reactions that could occur.

Figure 14.4 The Effect of Catalysts on Reaction Rates

A solution of hydrogen peroxide (H2O2) decomposes in water so slowly that the change is not noticeable (left). Iodide ion acts as a catalyst for the decomposition of H2O2, producing oxygen gas. The solution turns brown because of the reaction of H2O2 with I, which generates small amounts of I3 (center). The enzyme catalase is about 3 billion times more effective than iodide as a catalyst. Even in the presence of very small amounts of enzyme, the decomposition is vigorous (right).

Most of the bulk chemicals produced in industry are formed with catalyzed reactions. Recent estimates indicate that about 30% of the gross national product of the United States and other industrialized nations relies either directly or indirectly on the use of catalysts.

Summary

Factors that influence the reaction rates of chemical reactions include the concentration of reactants, temperature, the physical state of reactants and their dispersion, the solvent, and the presence of a catalyst.

Key Takeaway

  • The reaction rate depends on the concentrations of the reactants, the temperature of the reaction, the phase and surface area of the reactants, the solvent, and the presence or the absence of a catalyst.

Conceptual Problems

  1. What information can you obtain by studying the chemical kinetics of a reaction? Does a balanced chemical equation provide the same information? Why or why not?

  2. If you were tasked with determining whether to proceed with a particular reaction in an industrial facility, why would studying the chemical kinetics of the reaction be important to you?

  3. What is the relationship between each of the following factors and the reaction rate: reactant concentration, temperature of the reaction, physical properties of the reactants, physical and chemical properties of the solvent, and the presence of a catalyst?

  4. A slurry is a mixture of a finely divided solid with a liquid in which it is only sparingly soluble. As you prepare a reaction, you notice that one of your reactants forms a slurry with the solvent, rather than a solution. What effect will this have on the reaction rate? What steps can you take to try to solve the problem?

  5. Why does the reaction rate of virtually all reactions increase with an increase in temperature? If you were to make a glass of sweetened iced tea the old-fashioned way, by adding sugar and ice cubes to a glass of hot tea, which would you add first?

  6. In a typical laboratory setting, a reaction is carried out in a ventilated hood with air circulation provided by outside air. A student noticed that a reaction that gave a high yield of a product in the winter gave a low yield of that same product in the summer, even though his technique did not change and the reagents and concentrations used were identical. What is a plausible explanation for the different yields?

  7. A very active area of chemical research involves the development of solubilized catalysts that are not made inactive during the reaction process. Such catalysts are expected to increase reaction rates significantly relative to the same reaction run in the presence of a heterogeneous catalyst. What is the reason for anticipating that the relative rate will increase?

  8. Water has a dielectric constant more than two times greater than that of methanol (80.1 for H2O and 33.0 for CH3OH). Which would be your solvent of choice for a substitution reaction between an ionic compound and a polar reagent, both of which are soluble in either methanol or water? Why?

Answers

  1. Kinetics gives information on the reaction rate and reaction mechanism; the balanced chemical equation gives only the stoichiometry of the reaction.

  2. Reaction rates generally increase with increasing reactant concentration, increasing temperature, and the addition of a catalyst. Physical properties such as high solubility also increase reaction rates. Solvent polarity can either increase or decrease the reaction rate of a reaction, but increasing solvent viscosity generally decreases reaction rates.

  3. Increasing the temperature increases the average kinetic energy of molecules and ions, causing them to collide more frequently and with greater energy, which increases the reaction rate. First dissolve sugar in the hot tea, and then add the ice.

14.2 Reaction Rates and Rate Laws

Learning Objectives

  1. To determine the reaction rate.
  2. To understand the meaning of a rate law.

The factors discussed in Section 14.1 "Factors That Affect Reaction Rates" affect the reaction rate of a chemical reaction, which may determine whether a desired product is formed. In this section, we will show you how to quantitatively determine the reaction rate.

Reaction Rates

Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time.

The progress of a simple reaction (A → B) is shown in Figure 14.5 "The Progress of a Simple Reaction (A → B)", where the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in Figure 14.5 "The Progress of a Simple Reaction (A → B)". The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time.

Figure 14.5 The Progress of a Simple Reaction (A → B)

The mixture initially contains only A molecules (purple). With increasing time, the number of A molecules decreases and more B molecules (green) are formed (top). The graph shows the change in the number of A and B molecules in the reaction as a function of time over a 1 min period (bottom).

Equation 14.4

rate=Δ[B]Δt=Δ[A]Δt

Just as in Chapter 4 "Reactions in Aqueous Solution" and Chapter 5 "Energy Changes in Chemical Reactions", square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate we would calculate for the reaction A → B using Equation 14.4 would be different for each interval. (This is not true for every reaction, as you will see later.) A much greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, which means that the reaction rate is fastest at first. This is consistent with the concentration effects described in Section 14.1 "Factors That Affect Reaction Rates" because the concentration of A is greatest at the beginning of the reaction.

Note the Pattern

Reaction rates generally decrease with time as reactant concentrations decrease.

Determining the Reaction Rate of Hydrolysis of Aspirin

We can use Equation 14.4 to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world. (More than 25,000,000 kg are produced annually worldwide.) Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid.

Figure 14.6

Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table 14.1 "Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*" and are shown in the graph in Figure 14.7 "The Hydrolysis of Aspirin". These data were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid).

Table 14.1 Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*

Time (h) [Aspirin] (M) [Salicylic Acid] (M)
0 5.55 × 10−3 0
2.0 5.51 × 10−3 0.040 × 10−3
5.0 5.45 × 10−3 0.10 × 10−3
10 5.35 × 10−3 0.20 × 10−3
20 5.15 × 10−3 0.40 × 10−3
30 4.96 × 10−3 0.59 × 10−3
40 4.78 × 10−3 0.77 × 10−3
50 4.61 × 10−3 0.94 × 10−3
100 3.83 × 10−3 1.72 × 10−3
200 2.64 × 10−3 2.91 × 10−3
300 1.82 × 10−3 3.73 × 10−3
*The reaction at pH 7.0 is very slow. It is much faster under acidic conditions, such as those found in the stomach.

Figure 14.7 The Hydrolysis of Aspirin

This graph shows the concentrations of aspirin and salicylic acid as a function of time, based on the hydrolysis data in Table 14.1 "Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*". The time dependence of the concentration of the other product, acetate, is not shown, but based on the stoichiometry of the reaction, it is identical to the data for salicylic acid.

We can calculate the average reaction rateThe reaction rate calculated for a given time interval from the concentrations of either the reactant or one of the products at the beginning of the interval time (t0) and at the end of the interval (t1). for a given time interval from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t0) and at the end of the interval (t1). Using salicylic acid, for example, we find the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial):

rate(t=02.0 h)=[salicyclic acid]2[salicyclic acid]02.0 h0 h=0.040×103 M0 M2.0 h=2.0×105 M/h

We can also calculate the reaction rate from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases:

rate(t=02.0 h)=[aspirin]2[aspirin]02.0 h0 h=(5.51×103 M)(5.55×103 M)2.0 h=2×105 M/h

If we now calculate the reaction rate during the last interval given in Table 14.1 "Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*" (the interval between 200 h and 300 h after the start of the reaction), we find that the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h):

rate(t=200300h)=[salicyclic acid]300[salicyclic acid]200300 h200 h=(3.73×103 M)(2.91×103 M)100 h=8.2×106 M/h

(You should verify from the data in Table 14.1 "Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*" that you get the same rate using the concentrations of aspirin measured at 200 h and 300 h.)

Calculating the Reaction Rate of Fermentation of Sucrose

In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Let us look at a reaction in which the coefficients are not all the same: the fermentation of sucrose to ethanol and carbon dioxide, which we encountered in Chapter 3 "Chemical Reactions".

Equation 14.5

C12H22O11(aq)sucrose+H2O(l)4C2H5OH(aq)+4CO2(g)

The coefficients show us that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, we can find the reaction rate by looking at the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and, as you learned in Chapter 10 "Gases", the volume of CO2 gas formed will depend on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed:

Equation 14.6

Δ[C2H5OH]Δt=4Δ[sucrose]Δt

The concentration of the reactant—in this case sucrose—decreases with increasing time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in Equation 14.6 so that the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration increases with increasing time, so its rate of change is automatically expressed as a positive value.

Often the reaction rate is expressed in terms of the reactant or product that has the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction (Equation 14.5) corresponds to sucrose, so the reaction rate is generally defined as follows:

Equation 14.7

rate=Δ[sucrose]Δt=14(Δ[C2H5OH]Δt)

Example 1

Consider the thermal decomposition of gaseous N2O5 to NO2 and O2 via the following equation:

2N2O5(g)Δ4NO2(g)+O2(g)

Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time.

Given: balanced chemical equation

Asked for: reaction rate expressions

Strategy:

A Choose the species in the equation that has the smallest coefficient. Then write an expression for the rate of change of that species with time.

B For the remaining species in the equation, use molar ratios to obtain equivalent expressions for the reaction rate.

Solution:

A Because O2 has the smallest coefficient in the balanced chemical equation for the reaction, we define the reaction rate as the rate of change in the concentration of O2 and write that expression.

B We know from the balanced chemical equation that 2 mol of N2O5 must decompose for each 1 mol of O2 produced and that 4 mol of NO2 are produced for every 1 mol of O2 produced. The molar ratios of O2 to N2O5 and to NO2 are thus 1:2 and 1:4, respectively. This means that we divide the rate of change of [N2O5] and [NO2] by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO2 is produced at four times the rate of O2, we must divide the rate of production of NO2 by 4. The reaction rate expressions are as follows:

rate=Δ[O2]Δt=Δ[NO2]4Δt=Δ[N2O5]2Δt

Exercise

The key step in the industrial production of sulfuric acid is the reaction of SO2 with O2 to produce SO3.

2SO2(g) + O2(g) → 2SO3(g)

Write expressions for the reaction rate in terms of the rate of change of the concentration of each species.

Answer:

rate=Δ[O2]Δt=Δ[SO2]2Δt=Δ[SO3]2Δt

Example 2

Using the reaction shown in Example 1, calculate the reaction rate from the following data taken at 56°C:

2N2O5(g) → 4NO2(g) + O2(g)
Time (s) [N2O5] (M) [NO2] (M) [O2] (M)
240 0.0388 0.0314 0.00792
600 0.0197 0.0699 0.0175

Given: balanced chemical equation and concentrations at specific times

Asked for: reaction rate

Strategy:

A Using the equations in Example 1, subtract the initial concentration of a species from its final concentration and substitute that value into the equation for that species.

B Substitute the value for the time interval into the equation. Make sure your units are consistent.

Solution:

A We are asked to calculate the reaction rate in the interval between t1 = 240 s and t2 = 600 s. From Example 1, we see that we can evaluate the reaction rate using any of three expressions:

rate=Δ[O2]Δt=Δ[NO2]4Δt=Δ[N2O5]2Δt

Subtracting the initial concentration from the final concentration of N2O5 and inserting the corresponding time interval into the rate expression for N2O5,

rate=Δ[N2O5]2Δt=[N2O5]600[N2O5]2402(600 s240 s)

B Substituting actual values into the expression,

rate=0.0197 M0.0388 M2(360 s)=2.65×105 M/s

Similarly, we can use NO2 to calculate the reaction rate:

rate=Δ[NO2]4Δt=[NO2]600[NO2]2404(600 s240 s)=0.0699 M0.0314 M4(360 s)=2.67×105 M/s

If we allow for experimental error, this is the same rate we obtained using the data for N2O5, as it should be because the reaction rate should be the same no matter which concentration is used. We can also use the data for O2:

rate=Δ[O2]Δt=[O2]600[O2]240600 s 240 s=0.0175 M0.00792 M360 s=2.66×105 M/s

Again, this is the same value we obtained from the N2O5 and NO2 data. Thus the reaction rate does not depend on which reactant or product is used to measure it.

Exercise

Using the data in the following table, calculate the reaction rate of SO2(g) with O2(g) to give SO3(g).

2SO2(g) + O2(g) → 2SO3(g)
Time (s) [SO2] (M) [O2] (M) [SO3] (M)
300 0.0270 0.0500 0.0072
720 0.0194 0.0462 0.0148

Answer: 9.0 × 10−6 M/s

Instantaneous Rates of Reaction

So far, we have determined average reaction rates over particular intervals of time. We can also determine the instantaneous rateThe reaction rate of a chemical reaction at any given point in time. of a reaction, which is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate.If you have studied calculus, you may recognize that the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time.

Think of the distinction between the instantaneous and average rates of a reaction as being similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although you may travel for a long time at 65 mph on an interstate highway during a long trip, there may be times when you travel only 25 mph in construction zones or 0 mph if you stop for meals or gas. Thus your average speed on the trip may be only 50 mph, whereas your instantaneous speed on the interstate at a given moment may be 65 mph. Whether you are able to stop the car in time to avoid an accident depends on your instantaneous speed, not your average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval normally has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time.

In chemical kinetics, we generally focus on one particular instantaneous rate, which is the initial reaction rate, t = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to t = 0.

Rate Laws

In Section 14.1 "Factors That Affect Reaction Rates", you learned that reaction rates generally decrease with time because reactant concentrations decrease as reactants are converted to products. You also learned that reaction rates generally increase when reactant concentrations are increased. We now examine the mathematical expressions called rate lawsMathematical expressions that describe the relationships between reactant rates and reactant concentrations in a chemical reaction., which describe the relationships between reactant rates and reactant concentrations. Rate laws are laws as defined in Chapter 1 "Introduction to Chemistry"; that is, they are mathematical descriptions of experimentally verifiable data.

Rate laws may be written from either of two different but related perspectives. A differential rate lawA rate law that expresses the reaction rate in terms of changes in the concentration of one or more reactants (Δ[R]) over a specific time interval (Δt). expresses the reaction rate in terms of changes in the concentration of one or more reactants (Δ[R]) over a specific time interval (Δt). In contrast, an integrated rate lawA rate law that expresses the reaction rate in terms of the initial concentration ([R0]) and the measured concentration of one or more reactants ([R]) after a given amount of time (t). describes the reaction rate in terms of the initial concentration ([R]0) and the measured concentration of one or more reactants ([R]) after a given amount of time (t); we will discuss integrated rate laws in Section 14.3 "Methods of Determining Reaction Order". The integrated rate law can be found by using calculus to integrate the differential rate law, although the method of doing so is beyond the scope of this text. Whether you use a differential rate law or integrated rate law, always make sure that the rate law gives the proper units for the reaction rate, usually moles per liter per second (M/s).

Reaction Orders

For a reaction with the general equation

Equation 14.8

aA + bB → cC + dD

the experimentally determined rate law usually has the following form:

Equation 14.9

rate = k[A]m[B]n

The proportionality constant (k) is called the rate constantA proportionality constant whose value is characteristic of the reaction and the reaction conditions and whose numerical value does not change as the reaction progresses under a given set of conditions., and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular value of the rate constant under a given set of conditions, such as temperature, pressure, and solvent; varying the temperature or the solvent usually changes the value of the rate constant. The numerical value of k, however, does not change as the reaction progresses under a given set of conditions.

Thus the reaction rate depends on the rate constant for the given set of reaction conditions and the concentration of A and B raised to the powers m and n, respectively. The values of m and n are derived from experimental measurements of the changes in reactant concentrations over time and indicate the reaction orderNumbers that indicate the degree to which the reaction rate depends on the concentration of each reactant., the degree to which the reaction rate depends on the concentration of each reactant; m and n need not be integers. For example, Equation 14.9 tells us that Equation 14.8 is mth order in reactant A and nth order in reactant B. It is important to remember that n and m are not related to the stoichiometric coefficients a and b in the balanced chemical equation and must be determined experimentally. The overall reaction order is the sum of all the exponents in the rate law: m + n.

Note the Pattern

Under a given set of conditions, the value of the rate constant does not change as the reaction progresses.

Although differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, integrated rate laws are used to determine the reaction order and the value of the rate constant from experimental measurements. (We present general forms for integrated rate laws in Section 14.3 "Methods of Determining Reaction Order".) To illustrate how chemists interpret a differential rate law, we turn to the experimentally derived rate law for the hydrolysis of t-butyl bromide in 70% aqueous acetone. This reaction produces t-butanol according to the following equation:

Equation 14.10

(CH3)3CBr(soln) + H2O(soln) → (CH3)3COH(soln) + HBr(soln)

Combining the rate expression in Equation 14.4 and Equation 14.9 gives us a general expression for the differential rate law:

Equation 14.11

rate=Δ[A]Δt=k[A]m[B]n

Inserting the identities of the reactants into Equation 14.11 gives the following expression for the differential rate law for the reaction:

Equation 14.12

rate=Δ[(CH3)3CBr]Δt=k[(CH3)3CBr]m[H2O]n

Experiments done to determine the rate law for the hydrolysis of t-butyl bromide show that the reaction rate is directly proportional to the concentration of (CH3)3CBr but is independent of the concentration of water. Thus m and n in Equation 14.12 are 1 and 0, respectively, and

Equation 14.13

rate = k[(CH3)3CBr]1[H2O]0 = k[(CH3)3CBr]

Because the exponent for the reactant is 1, the reaction is first order in (CH3)3CBr. It is zeroth order in water because the exponent for [H2O] is 0. (Recall that anything raised to the zeroth power equals 1.) Thus the overall reaction order is 1 + 0 = 1. What the reaction orders tell us in practical terms is that doubling the concentration of (CH3)3CBr doubles the reaction rate of the hydrolysis reaction, halving the concentration of (CH3)3CBr halves the reaction rate, and so on. Conversely, increasing or decreasing the concentration of water has no effect on the reaction rate. (Again, when you work with rate laws, there is no simple correlation between the stoichiometry of the reaction and the rate law. The values of k, m, and n in the rate law must be determined experimentally.) Experimental data show that k has the value 5.15 × 10−4 s−1 at 25°C. The rate constant has units of reciprocal seconds (s−1) because the reaction rate is defined in units of concentration per unit time (M/s). The units of a rate constant depend on the rate law for a particular reaction.

Under conditions identical to those for the t-butyl bromide reaction, the experimentally derived differential rate law for the hydrolysis of methyl bromide (CH3Br) is as follows:

Equation 14.14

rate=Δ[CH3Br]Δt=k[CH3Br]

This reaction also has an overall reaction order of 1, but the rate constant in Equation 14.14 is approximately 106 times smaller than that for t-butyl bromide. Thus methyl bromide hydrolyzes about 1 million times more slowly than t-butyl bromide, and this information tells chemists how the reactions differ on a molecular level.

Frequently, changes in reaction conditions also produce changes in a rate law. In fact, chemists often change reaction conditions to obtain clues about what is occurring during a reaction. For example, when t-butyl bromide is hydrolyzed in an aqueous acetone solution containing OH ions rather than in aqueous acetone alone, the differential rate law for the hydrolysis reaction does not change. For methyl bromide, in contrast, the differential rate law becomes rate = k″[CH3Br][OH], with an overall reaction order of 2. Although the two reactions proceed similarly in neutral solution, they proceed very differently in the presence of a base, which again provides clues as to how the reactions differ on a molecular level.

Note the Pattern

Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements.

Example 3

We present three reactions and their experimentally determined differential rate laws. For each reaction, give the units of the rate constant, give the reaction order with respect to each reactant, give the overall reaction order, and predict what happens to the reaction rate when the concentration of the first species in each chemical equation is doubled.

  1. 2HI(g)PtH2(g)+I2(g)rate=12(Δ[HI]Δt)=k[HI]2
  2. 2N2O(g)Δ2N2(g)+O2(g)rate=12(Δ[N2O]Δt)=k
  3. cyclopropane(g) → propane(g)rate=Δ[cyclopropane]Δt=k[cyclopropane]

Given: balanced chemical equations and differential rate laws

Asked for: units of rate constant, reaction orders, and effect of doubling reactant concentration

Strategy:

A Express the reaction rate as moles per liter per second [mol/(L·s), or M/s]. Then determine the units of each chemical species in the rate law. Divide the units for the reaction rate by the units for all species in the rate law to obtain the units for the rate constant.

B Identify the exponent of each species in the rate law to determine the reaction order with respect to that species. Sum all exponents to obtain the overall reaction order.

C Use the mathematical relationships as expressed in the rate law to determine the effect of doubling the concentration of a single species on the reaction rate.

Solution:

  1. A [HI]2 will give units of (moles per liter)2. For the reaction rate to have units of moles per liter per second, the rate constant must have reciprocal units [1/(M·s)]:

    kM2=Msk=M/sM2=1Ms=M1s1

    B The exponent in the rate law is 2, so the reaction is second order in HI. Because HI is the only reactant and the only species that appears in the rate law, the reaction is also second order overall.

    C If the concentration of HI is doubled, the reaction rate will increase from k[HI]02 to k(2[HI])02 = 4k[HI]02. The reaction rate will therefore quadruple.

  2. A Because no concentration term appears in the rate law, the rate constant must have M/s units for the reaction rate to have M/s units.

    B The rate law tells us that the reaction rate is constant and independent of the N2O concentration. That is, the reaction is zeroth order in N2O and zeroth order overall.

    C Because the reaction rate is independent of the N2O concentration, doubling the concentration will have no effect on the reaction rate.

  3. A The rate law contains only one concentration term raised to the first power. Hence the rate constant must have units of reciprocal seconds (s−1) to have units of moles per liter per second for the reaction rate: M·s−1 = M/s.

    B The only concentration in the rate law is that of cyclopropane, and its exponent is 1. This means that the reaction is first order in cyclopropane. Cyclopropane is the only species that appears in the rate law, so the reaction is also first order overall.

    C Doubling the initial cyclopropane concentration will increase the reaction rate from k[cyclopropane]0 to 2k[cyclopropane]0. This doubles the reaction rate.

Exercise

Given the following two reactions and their experimentally determined differential rate laws: determine the units of the rate constant if time is in seconds, determine the reaction order with respect to each reactant, give the overall reaction order, and predict what will happen to the reaction rate when the concentration of the first species in each equation is doubled.

  1. CH3N=NCH3(g) C2H6(g)+N2(g)rate = Δ[CH3N=NCH3]Δt=k[CH3N=NCH3]
  2. 2NO2(g)+F2(g)2NO2F(g)rate = Δ[F2]Δt=12(Δ[NO2]Δt)=k[NO2][F2]

Answer:

  1. s−1; first order in CH3N=NCH3; first order overall; doubling [CH3N=NCH3] will double the reaction rate.
  2. M−1·s−1; first order in NO2, first order in F2; second order overall; doubling [NO2] will double the reaction rate.

Summary

Reaction rates are reported either as the average rate over a period of time or as the instantaneous rate at a single time.

The rate law for a reaction is a mathematical relationship between the reaction rate and the concentrations of species in solution. Rate laws can be expressed either as a differential rate law, describing the change in reactant or product concentrations as a function of time, or as an integrated rate law, describing the actual concentrations of reactants or products as a function of time.

The rate constant (k) of a rate law is a constant of proportionality between the reaction rate and the reactant concentration. The power to which a concentration is raised in a rate law indicates the reaction order, the degree to which the reaction rate depends on the concentration of a particular reactant.

Key Takeaways

  • Reaction rates can be determined over particular time intervals or at a given point in time.
  • A rate law describes the relationship between reactant rates and reactant concentrations.

Key Equations

general definition of rate for A → B

Equation 14.4: rate=Δ[B]Δt=Δ[A]Δt

general form of rate law when A and B are reactants

Equation 14.9: rate=k[A]m[B]n

Conceptual Problems

  1. Explain why the reaction rate is generally fastest at early time intervals. For the second-order A + B → C, what would the plot of the concentration of C versus time look like during the course of the reaction?

  2. Explain the differences between a differential rate law and an integrated rate law. What two components do they have in common? Which form is preferred for obtaining a reaction order and a rate constant? Why?

  3. Diffusion-controlled reactions have rates that are determined only by the reaction rate at which two reactant molecules can diffuse together. These reactions are rapid, with second-order rate constants typically on the order of 1010 L/(mol·s). Would you expect the reactions to be faster or slower in solvents that have a low viscosity? Why? Consider the reactions H3O+ + OH → 2H2O and H3O+ + N(CH3)3 → H2O + HN(CH3)3+ in aqueous solution. Which would have the higher rate constant? Why?

  4. What information can you get from the reaction order? What correlation does the reaction order have with the stoichiometry of the overall equation?

  5. During the hydrolysis reaction A + H2O → B + C, the concentration of A decreases much more rapidly in a polar solvent than in a nonpolar solvent. How do you expect this effect to be reflected in the overall reaction order?

Answers

  1. Reactant concentrations are highest at the beginning of a reaction. The plot of [C] versus t is a curve with a slope that becomes steadily less positive.

  2. Faster in a less viscous solvent because the rate of diffusion is higher; the H3O+/OH reaction is faster due to the decreased relative size of reactants and the higher electrostatic attraction between the reactants.

Numerical Problems

  1. The reaction rate of a particular reaction in which A and B react to make C is as follows:

    rate=Δ[A]Δt=12(Δ[C]Δt)

    Write a reaction equation that is consistent with this rate law. What is the rate expression with respect to time if 2A are converted to 3C?

  2. While commuting to work, a person drove for 12 min at 35 mph, then stopped at an intersection for 2 min, continued the commute at 50 mph for 28 min, drove slowly through traffic at 38 mph for 18 min, and then spent 1 min pulling into a parking space at 3 mph. What was the average rate of the commute? What was the instantaneous rate at 13 min? at 28 min?

  3. Why do most studies of chemical reactions use the initial rates of reaction to generate a rate law? How is this initial rate determined? Given the following data, what is the reaction order? Estimate.

    Time (s) [A] (M)
    120 0.158
    240 0.089
    360 0.062
  4. Predict how the reaction rate will be affected by doubling the concentration of the first species in each equation.

    1. C2H5I → C2H4 + HI: rate = k[C2H5I]
    2. SO + O2 → SO2 + O: rate = k[SO][O2]
    3. 2CH3 → C2H6: rate = k[CH3]2
    4. ClOO → Cl + O2: rate = k
  5. Cleavage of C2H6 to produce two CH3· radicals is a gas-phase reaction that occurs at 700°C. This reaction is first order, with k = 5.46 × 10−4 s−1. How long will it take for the reaction to go to 15% completion? to 50% completion?

  6. Three chemical processes occur at an altitude of approximately 100 km in Earth’s atmosphere.

    N2++O2k1N2+O2+ O2++Ok2O2+O+ O++N2k3NO++N

    Write a rate law for each elementary reaction. If the rate law for the overall reaction were found to be rate = k[N2+][O2], which one of the steps is rate limiting?

  7. The oxidation of aqueous iodide by arsenic acid to give I3 and arsenous acid proceeds via the following reaction:

    H3AsO4(aq)+3I(aq)+2H+(aq)krkfH3AsO3(aq)+I3(aq)+H2O(l)

    Write an expression for the initial rate of decrease of [I3], Δ[I3]/Δt. When the reaction rate of the forward reaction is equal to that of the reverse reaction: kf/kr = [H3AsO3][I3]/[H3AsO4][I]3[H+]2. Based on this information, what can you say about the nature of the rate-determining steps for the reverse and the forward reactions?

Answer

  1. 298 s; 1270 s

14.3 Methods of Determining Reaction Order

Learning Objective

  1. To know how to determine the reaction order from experimental data.

In the examples in this text, the exponents in the rate law are almost always the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth, first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data.

Zeroth-Order Reactions

A zeroth-order reactionA reaction whose rate is independent of concentration. is one whose rate is independent of concentration; its differential rate law is rate = k. We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:

Equation 14.15

rate=Δ[A]Δt=k[reactant]0=k(1)=k

Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of −k. The value of k is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of k, a positive value.

The graph of a zeroth-order reaction. The change in concentration of reactant and product with time produces a straight line.

The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form

Equation 14.16

[A] = [A]0kt

where [A]0 is the initial concentration of reactant A. ( has the form of the algebraic equation for a straight line, y = mx + b, with y = [A], mx = −kt, and b = [A]0.) In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second.

Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N2O on a platinum (Pt) surface to produce N2 and O2, which occurs at temperatures ranging from 200°C to 400°C:

Equation 14.17

2N2O(g)Pt2N2(g)+O2(g)

Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N2O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N2O to react with the entire Pt surface, doubling or quadrupling the N2O concentration will have no effect on the reaction rate.At very low concentrations of N2O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N2O concentration. The reaction rate is as follows:

Equation 14.18

rate=12(Δ[N2O]Δt)=12(Δ[N2]Δt)=Δ[O2]Δt=k[N2O]0=k

Thus the rate at which N2O is consumed and the rates at which N2 and O2 are produced are independent of concentration. As shown in , the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N2O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally.

Figure 14.8 A Zeroth-Order Reaction

This graph shows the concentrations of reactants and products versus time for the zeroth-order catalyzed decomposition of N2O to N2 and O2 on a Pt surface. The change in the concentrations of all species with time is linear.

Note the Pattern

If a plot of reactant concentration versus time is linear, then the reaction is zeroth order in that reactant.

A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the enzymeA catalyst that occurs naturally in living organisms and catalyzes biological reactions. alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is

Figure 14.9

 

where NAD+ (nicotinamide adenine dinucleotide) and NADH (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (part (a) in ). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver.Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in ).

Figure 14.10 The Catalyzed Oxidation of Ethanol

(a) The concentration of ethanol in human blood decreases linearly with time, which is typical of a zeroth-order reaction. (b) The rate at which ethanol is oxidized is constant until the ethanol concentration reaches essentially zero, at which point the reaction rate drops to zero.

These examples illustrate two important points:

  1. In a zeroth-order reaction, the reaction rate does not depend on the reactant concentration.
  2. A linear change in concentration with time is a clear indication of a zeroth-order reaction.

First-Order Reactions

In a first-order reactionA reaction whose rate is directly proportional to the concentration of one reactant., the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:

Equation 14.19

rate=Δ[A]Δt=k[A]

If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s−1).

The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:

Equation 14.20

[A] = [A]0e−kt

where [A]0 is the initial concentration of reactant A at t = 0; k is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places. (Essential Skills 6 in , , discusses natural logarithms.) Recall that an integrated rate law gives the relationship between reactant concentration and time. predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and t:

Equation 14.21

ln[A] = ln[A]0kt

Because has the form of the algebraic equation for a straight line, y = mx + b, with y = ln[A] and b = ln[A]0, a plot of ln[A] versus t for a first-order reaction should give a straight line with a slope of −k and an intercept of ln[A]0. Either the differential rate law () or the integrated rate law () can be used to determine whether a particular reaction is first order.

Graphs of a first-order reaction. The expected shapes of the curves for plots of reactant concentration versus time (top) and the natural logarithm of reactant concentration versus time (bottom) for a first-order reaction.

First-order reactions are very common. In this chapter, we have already encountered two examples of first-order reactions: the hydrolysis of aspirin () and the reaction of t-butyl bromide with water to give t-butanol (). Another reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin.

Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows:

Figure 14.11

Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in have been studied extensively to find ways of maximizing the concentration of the active species.

Note the Pattern

If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order.

The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in . The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin. Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in shows that the reaction rate doubles [(1.8 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = k[cisplatin]1. Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of . For example, substituting the values for Experiment 3 into ,

3.6 × 10−5 M/min = k(0.024 M) 1.5 × 10−3 min−1 = k

Table 14.2 Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C

Experiment [Cisplatin]0 (M) Initial Rate (M/min)
1 0.0060 9.0 × 10−6
2 0.012 1.8 × 10−5
3 0.024 3.6 × 10−5
4 0.030 4.5 × 10−5

Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.

Example 4

At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction:

CH3CH2Cl(g)ΔHCl(g)+C2H4(g)

Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction.

Experiment [CH3CH2Cl]0 (M) Initial Rate (M/s)
1 0.010 1.6 × 10−8
2 0.015 2.4 × 10−8
3 0.030 4.8 × 10−8
4 0.040 6.4 × 10−8

Given: balanced chemical equation, initial concentrations of reactant, and initial rates of reaction

Asked for: reaction order and rate constant

Strategy:

A Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species.

B Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.

C Use measured concentrations and rate data from any of the experiments to find the rate constant.

Solution:

The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.

A Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH3CH2Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH3CH2Cl].

B This behavior is characteristic of a first-order reaction, for which the rate law is rate = k[CH3CH2Cl].

C We can calculate the rate constant (k) using any row in the table. Selecting Experiment 1 gives the following:

1.60 × 10−8 M/s = k(0.010 M) 1.6 × 10−6 s−1 = k

Exercise

Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 by the following reaction:

SO2Cl2(g) → SO2(g) + Cl2(g)

Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.

Experiment [SO2Cl2]0 (M) Initial Rate (M/s)
1 0.0050 1.10 × 10−7
2 0.0075 1.65 × 10−7
3 0.0100 2.20 × 10−7
4 0.0125 2.75 × 10−7

Answer: first order; k = 2.2 × 10−5 s−1

Figure 14.12 The Hydrolysis of Cisplatin, a First-Order Reaction

These plots show hydrolysis of cisplatin at pH 7.0 and 25°C as (a) the experimentally determined concentrations of cisplatin and chloride ions versus time and (b) the natural logarithm of the cisplatin concentration versus time. The straight line in (b) is expected for a first-order reaction.

We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Part (a) in shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C. The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in . The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in for t = 100 min ([cisplatin] = 0.0086 M) and t = 1000 min ([cisplatin] = 0.0022 M),

slope=ln [cisplatin]1000ln [cisplatin]1001000 min100 mink=ln 0.0022ln 0.00861000 min 100 min=6.12(4.76)900 min=1.51×103 min1k=1.5×103 min1

The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min−1 because the times plotted on the horizontal axes in parts (a) and (b) in are in minutes rather than seconds.

The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions.

Example 5

Refer back to Example 4. If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M? (Recall that we calculated the rate constant for this reaction in Example 4.)

Given: initial concentration, rate constant, and time interval

Asked for: concentration at specified time and time required to obtain particular concentration

Strategy:

A Substitute values for the initial concentration ([A]0) and the calculated rate constant for the reaction (k) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time t.

B Given a concentration [A], solve the integrated rate law for time t.

Solution:

The exponential form of the integrated rate law for a first-order reaction () is [A] = [A]0ekt.

A Having been given the initial concentration of ethyl chloride ([A]0) and having calculated the rate constant in Example 4 (k = 1.6 × 10−6 s−1), we can use the rate law to calculate the concentration of the reactant at a given time t. Substituting the known values into the integrated rate law,

[CH3CH2Cl]10 h=[CH3CH2Cl]0ekt=0.0200 M(e(1.6×106 s1)[(10 h)(60 min/h)(60 s/min)])=0.0189 M

We could also have used the logarithmic form of the integrated rate law ():

ln[CH3CH2Cl]10 h=ln [CH3CH2Cl]0kt=ln 0.0200(1.6×106 s1)[(10 h)(60 min/h)(60 s/min)]=3.9120.0576=3.970[CH3CH2Cl]10 h=e3.970 M=0.0189 M

B To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for t. gives the following:

ln[CH3CH2Cl]t=ln[CH3CH2Cl]0ktkt=ln[CH3CH2Cl]0ln[CH3CH2Cl]t=ln[CH3CH2Cl]0[CH3CH2Cl]tt=1k(ln[CH3CH2Cl]0[CH3CH2Cl]t)=11.6×106 s1(ln 0.0200 M0.0050 M)=ln 4.01.6×106 s1=8.7×105 s=240 h=2.4×102 h

Exercise

In the exercise in Example 4, you found that the decomposition of sulfuryl chloride (SO2Cl2) is first order, and you calculated the rate constant at 320°C. Use the form(s) of the integrated rate law to find the amount of SO2Cl2 that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C. How long would it take for 90% of the SO2Cl2 to decompose?

Answer: 0.0252 M; 29 h

Second-Order Reactions

The simplest kind of second-order reactionA reaction whose rate is proportional to the square of the concentration of the reactant (for a reaction with the general form 2A → products) or is proportional to the product of the concentrations of two reactants (for a reaction with the general form A + B → products). is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form 2A → products. A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).

The differential rate law for the simplest second-order reaction in which 2A → products is as follows:

Equation 14.22

rate=Δ[A]2Δt=k[A]2

Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).

For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:

Equation 14.23

1[A]=1[A]0+kt

Because has the form of an algebraic equation for a straight line, y = mx + b, with y = 1/[A] and b = 1/[A]0, a plot of 1/[A] versus t for a simple second-order reaction is a straight line with a slope of k and an intercept of 1/[A]0.

Note the Pattern

Second-order reactions generally have the form 2A → products or A + B → products.

Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO2 to NO and O2 and the decomposition of HI to I2 and H2. Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.

Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:

 

Figure 14.13  

For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively.The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law () or the integrated rate law ().

To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in . From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:

5.0×105 M/min1.8×105 M/min=2.8and3.4×103 M2.0×103 M=1.7

Table 14.3 Rates of Reaction as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M

Time (min) [Monomer] (M) Instantaneous Rate (M/min)
10 0.0044 8.0 × 10−5
26 0.0034 5.0 × 10−5
44 0.0027 3.1 × 10−5
70 0.0020 1.8 × 10−5
120 0.0014 8.0 × 10−6

Because (1.7)2 = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.

rate ∝ [monomer]2

This means that the reaction is second order in the monomer. Using and the data from any row in , we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:

rate=k[A]28.0×105 M/min=k(4.4×103 M)24.1 M1min1=k

We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in part (a) in . The measurements show that the concentration of the monomer (initially 5.4 × 10−3 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus t should be a straight line, as shown in part (b) in . Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, k = 4.1 M−1·min−1, which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.

Figure 14.14 Dimerization of a Monomeric Compound, a Second-Order Reaction

These plots correspond to dimerization of the monomer in as (a) the experimentally determined concentration of monomer versus time and (b) 1/[monomer] versus time. The straight line in (b) is expected for a simple second-order reaction.

For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders. The differential and integrated rate laws for zeroth-, first-, and second-order reactions and their corresponding graphs are shown in in .

Example 6

At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.

2NO2(g)Δ2NO(g)+O2(g)

Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:

Experiment [NO2]0 (M) Initial Rate (M/s)
1 0.015 1.22 × 10−4
2 0.010 5.40 × 10−5
3 0.0080 3.46 × 10−5
4 0.0050 1.35 × 10−5

Determine the reaction order and the rate constant.

Given: balanced chemical equation, initial concentrations, and initial rates

Asked for: reaction order and rate constant

Strategy:

A From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.

B Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant (k).

Solution:

A We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO2 concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10−5) ÷ (1.35 × 10−5) = 4.0], which means that the reaction rate is proportional to [NO2]2. Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO2]2. This behavior is characteristic of a second-order reaction.

B We have rate = k[NO2]2. We can calculate the rate constant (k) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:

rate=k[NO2]25.40×105 M/s=k(0.010 M)20.54 M1s1=k

Exercise

When the highly reactive species HO2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:

2HO2(g) → H2O2(g) + O2(g)

The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:

Experiment [HO2]0 (M) Initial Rate (M/s)
1 1.1 × 10−8 1.7 × 10−7
2 2.5 × 10−8 8.8 × 10−7
3 3.4 × 10−8 1.6 × 10−6
4 5.0 × 10−8 3.5 × 10−6

Determine the reaction order and the rate constant.

Answer: second order in HO2; k = 1.4 × 109 M−1·s−1

Note the Pattern

If a plot of reactant concentration versus time is not linear but a plot of 1/reaction concentration versus time is linear, then the reaction is second order.

Example 7

If a flask that initially contains 0.056 M NO2 is heated at 300°C, what will be the concentration of NO2 after 1.0 h? How long will it take for the concentration of NO2 to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction () and the rate constant calculated in Example 6.

Given: balanced chemical equation, rate constant, time interval, and initial concentration

Asked for: final concentration and time required to reach specified concentration

Strategy:

A Given k, t, and [A]0, use the integrated rate law for a second-order reaction to calculate [A].

B Setting [A] equal to 1/10 of [A]0, use the same equation to solve for t.

Solution:

A We know k and [NO2]0, and we are asked to determine [NO2] at t = 1 h (3600 s). Substituting the appropriate values into ,

1[NO2]3600=1[NO2]0+kt=10.056 M+[(0.54 M1s1)(3600 s)]=2.0×103 M1

Thus [NO2]3600 = 5.1 × 10−4 M.

B In this case, we know k and [NO2]0, and we are asked to calculate at what time [NO2] = 0.1[NO2]0 = 0.1(0.056 M) = 0.0056 M. To do this, we solve for t, using the concentrations given.

t=(1/[NO2])(1/[NO2]0)k=(1/0.0056 M)(1/0.056 M)0.54 M1s1=3.0×102 s=5.0 min

NO2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.

Exercise

In the exercise in Example 6, you calculated the rate constant for the decomposition of HO2 as k = 1.4 × 109 M−1·s−1. This high rate constant means that HO2 decomposes rapidly under the reaction conditions given in the problem. In fact, the HO2 molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO2, calculate the concentration of HO2 that remains after 1.0 h at 25°C. How long will it take for 90% of the HO2 to decompose? Use the integrated rate law for a second-order reaction () and the rate constant calculated in the exercise in Example 6.

Answer: 2.0 × 10−13 M; 6.4 × 10−6 s

In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form A + B → products, in which the reaction is first order in A and first order in B. The differential rate law for this reaction is as follows:

Equation 14.24

rate=Δ[A]Δt=Δ[B]Δt=k[A][B]

Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant. We presented one example at the end of , the reaction of CH3Br with OH to produce CH3OH.

Determining the Rate Law of a Reaction

The number of fundamentally different mechanisms (sets of steps in a reaction) is actually rather small compared to the large number of chemical reactions that can occur. Thus understanding reaction mechanismsThe sequence of events that occur at the molecular level during a reaction. can simplify what might seem to be a confusing variety of chemical reactions. The first step in discovering the reaction mechanism is to determine the reaction’s rate law. This can be done by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction A + B → products, for example, we need to determine k and the exponents m and n in the following equation:

Equation 14.25

rate = k[A]m[B]n

To do this, we might keep the initial concentration of B constant while varying the initial concentration of A and calculating the initial reaction rate. This information would permit us to deduce the reaction order with respect to A. Similarly, we could determine the reaction order with respect to B by studying the initial reaction rate when the initial concentration of A is kept constant while the initial concentration of B is varied. In earlier examples, we determined the reaction order with respect to a given reactant by comparing the different rates obtained when only the concentration of the reactant in question was changed. An alternative way of determining reaction orders is to set up a proportion using the rate laws for two different experiments.

Rate data for a hypothetical reaction of the type A + B → products are given in . The general rate law for the reaction is given in . We can obtain m or n directly by using a proportion of the rate laws for two experiments in which the concentration of one reactant is the same, such as Experiments 1 and 3 in .

Table 14.4 Rate Data for a Hypothetical Reaction of the Form A + B → Products

Experiment [A] (M) [B] (M) Initial Rate (M/min)
1 0.50 0.50 8.5 × 10−3
2 0.75 0.50 19 × 10−3
3 1.00 0.50 34 × 10−3
4 0.50 0.75 8.5 × 10−3
5 0.50 1.00 8.5 × 10−3
rate1rate3=k[A1]m[B1]nk[A3]m[B3]n

Inserting the appropriate values from ,

8.5×103 M/min34×103 M/min=k[0.50 M]m[0.50 M]nk[1.00 M]m[0.50 M]n

Because 1.00 to any power is 1, [1.00 M]m = 1.00 M. We can cancel like terms to give 0.25 = [0.50]m, which can also be written as 1/4 = [1/2]m. Thus we can conclude that m = 2 and that the reaction is second order in A. By selecting two experiments in which the concentration of B is the same, we were able to solve for m.

Conversely, by selecting two experiments in which the concentration of A is the same (e.g., Experiments 5 and 1), we can solve for n.

rate1rate5=k[A1]m[B1]nk[A5]m[B5]n

Substituting the appropriate values from ,

8.5×103 M/min8.5×103 M/min=k[0.50 M]m[0.50M]nk[0.50 M]m[1.00M]n

Canceling leaves 1.0 = [0.50]n, which gives n = 0; that is, the reaction is zeroth order in B. The experimentally determined rate law is therefore

rate = k[A]2[B]0 = k[A]2

We can now calculate the rate constant by inserting the data from any row of into the experimentally determined rate law and solving for k. Using Experiment 2, we obtain

19 × 10−3 M/min = k(0.75 M)2 3.4 × 10−2 M−1·min−1 = k

You should verify that using data from any other row of gives the same rate constant. This must be true as long as the experimental conditions, such as temperature and solvent, are the same.

Example 8

Nitric oxide is produced in the body by several different enzymes and acts as a signal that controls blood pressure, long-term memory, and other critical functions. The major route for removing NO from biological fluids is via reaction with O2 to give NO2, which then reacts rapidly with water to give nitrous acid and nitric acid:

 

These reactions are important in maintaining steady levels of NO. The following table lists kinetics data for the reaction of NO with O2 at 25°C:

2NO(g) + O2(g) → 2NO2(g)

Determine the rate law for the reaction and calculate the rate constant.

Experiment [NO]0 (M) [O2]0 (M) Initial Rate (M/s)
1 0.0235 0.0125 7.98 × 10−3
2 0.0235 0.0250 15.9 × 10−3
3 0.0470 0.0125 32.0 × 10−3
4 0.0470 0.0250 63.5 × 10−3

Given: balanced chemical equation, initial concentrations, and initial rates

Asked for: rate law and rate constant

Strategy:

A Compare the changes in initial concentrations with the corresponding changes in rates of reaction to determine the reaction order for each species. Write the rate law for the reaction.

B Using data from any experiment, substitute appropriate values into the rate law. Solve the rate equation for k.

Solution:

A Comparing Experiments 1 and 2 shows that as [O2] is doubled at a constant value of [NO2], the reaction rate approximately doubles. Thus the reaction rate is proportional to [O2]1, so the reaction is first order in O2. Comparing Experiments 1 and 3 shows that the reaction rate essentially quadruples when [NO] is doubled and [O2] is held constant. That is, the reaction rate is proportional to [NO]2, which indicates that the reaction is second order in NO. Using these relationships, we can write the rate law for the reaction:

rate = k[NO]2[O2]

B The data in any row can be used to calculate the rate constant. Using Experiment 1, for example, gives

k=rate[NO]2[O2]=7.98×103 M/s(0.0235 M)2(0.0125 M)=1.16×103 M2s1

The overall reaction order (m + n) is 3, so this is a third-order reaction, a reaction whose rate is determined by three reactants. The units of the rate constant become more complex as the overall reaction order increases.

Exercise

The peroxydisulfate ion (S2O82−) is a potent oxidizing agent that reacts rapidly with iodide ion in water:

S2O82−(aq) + 3I(aq) → 2SO42−(aq) + I3(aq)

The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant.

Experiment [S2O82−]0 (M) [I]0 (M) Initial Rate (M/s)
1 0.27 0.38 2.05
2 0.40 0.38 3.06
3 0.40 0.22 1.76

Answer: rate = k[S2O82−][I]; k = 20 M−1·s−1

Summary

The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.

Key Takeaway

  • Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data.

Key Equations

zeroth-order reaction

: rate=Δ[A]Δt=k

: [A] = [A]0kt

first-order reaction

: rate=Δ[A]Δt=k[A]

: [A] = [A]0ekt

: ln[A] = ln[A]0kt

second-order reaction

: rate=Δ[A]Δt=k[A]2

: 1[A]=1[A]0+kt

Conceptual Problems

  1. What are the characteristics of a zeroth-order reaction? Experimentally, how would you determine whether a reaction is zeroth order?

  2. Predict whether the following reactions are zeroth order and explain your reasoning.

    1. a substitution reaction of an alcohol with HCl to form an alkyl halide and water
    2. catalytic hydrogenation of an alkene
    3. hydrolysis of an alkyl halide to an alcohol
    4. enzymatic conversion of nitrate to nitrite in a soil bacterium
  3. In a first-order reaction, what is the advantage of using the integrated rate law expressed in natural logarithms over the rate law expressed in exponential form?

  4. If the reaction rate is directly proportional to the concentration of a reactant, what does this tell you about (a) the reaction order with respect to the reactant and (b) the overall reaction order?

  5. The reaction of NO with O2 is found to be second order with respect to NO and first order with respect to O2. What is the overall reaction order? What is the effect of doubling the concentration of each reagent on the reaction rate?

Numerical Problems

  1. Iodide reduces Fe(III) according to the following reaction:

    2Fe3+(soln) + 2I(soln) → 2Fe2+(soln) + I2(soln)

    Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order?

  2. Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows:

    Experiment [Benzoyl Peroxide]0 (M) Initial Rate (M/s)
    1 1.00 2.22 × 10−4
    2 0.70 1.64 × 10−4
    3 0.50 1.12 × 10−4
    4 0.25 0.59 × 10−4

    What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction?

  3. 1-Bromopropane is a colorless liquid that reacts with S2O32− according to the following reaction:

    C3H7Br + S2O32− → C3H7S2O3 + Br

    The reaction is first order in 1-bromopropane and first order in S2O32−, with a rate constant of 8.05 × 10−4 M−1·s−1. If you began a reaction with 40 mmol/100 mL of C3H7Br and an equivalent concentration of S2O32−, what would the initial reaction rate be? If you were to decrease the concentration of each reactant to 20 mmol/100 mL, what would the initial reaction rate be?

  4. The experimental rate law for the reaction 3A + 2B → C + D was found to be Δ[C]/Δt = k[A]2[B] for an overall reaction that is third order. Because graphical analysis is difficult beyond second-order reactions, explain the procedure for determining the rate law experimentally.

Answers

  1. First order in Fe3+; second order in I; third order overall; rate = k[Fe3+][I]2.

  2. 1.29 × 10−4 M/s; 3.22 × 10−5 M/s

14.4 Using Graphs to Determine Rate Laws, Rate Constants, and Reaction Orders

Learning Objective

  1. To use graphs to analyze the kinetics of a reaction.

In , you learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order.

We will illustrate the use of these graphs by considering the thermal decomposition of NO2 gas at elevated temperatures, which occurs according to the following reaction:

Equation 14.26

2NO2(g)Δ2NO(g)+O2(g)

Experimental data for this reaction at 330°C are listed in ; they are provided as [NO2], ln[NO2], and 1/[NO2] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively. The actual concentrations of NO2 are plotted versus time in part (a) in . Because the plot of [NO2] versus t is not a straight line, we know the reaction is not zeroth order in NO2. A plot of ln[NO2] versus t (part (b) in ) shows us that the reaction is not first order in NO2 because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO2] versus t (part (c) in ). This plot is a straight line, indicating that the reaction is second order in NO2.

Table 14.5 Concentration of NO2 as a Function of Time at 330°C

Time (s) [NO2] (M) ln[NO2] 1/[NO2] (M−1)
0 1.00 × 10−2 −4.605 100
60 6.83 × 10−3 −4.986 146
120 5.18 × 10−3 −5.263 193
180 4.18 × 10−3 −5.477 239
240 3.50 × 10−3 −5.655 286
300 3.01 × 10−3 −5.806 332
360 2.64 × 10−3 −5.937 379

Figure 14.15 The Decomposition of NO2

These plots show the decomposition of a sample of NO2 at 330°C as (a) the concentration of NO2 versus t, (b) the natural logarithm of [NO2] versus t, and (c) 1/[NO2] versus t.

We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in , the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method described in required multiple experiments at different NO2 concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions.

Figure 14.16 Properties of Reactions That Obey Zeroth-, First-, and Second-Order Rate Laws

Example 9

Dinitrogen pentoxide (N2O5) decomposes to NO2 and O2 at relatively low temperatures in the following reaction:

2N2O5(soln) → 4NO2(soln) + O2(g)

This reaction is carried out in a CCl4 solution at 45°C. The concentrations of N2O5 as a function of time are listed in the following table, together with the natural logarithms and reciprocal N2O5 concentrations. Plot a graph of the concentration versus t, ln concentration versus t, and 1/concentration versus t and then determine the rate law and calculate the rate constant.

Time (s) [N2O5] (M) ln[N2O5] 1/[N2O5] (M−1)
0 0.0365 −3.310 27.4
600 0.0274 −3.597 36.5
1200 0.0206 −3.882 48.5
1800 0.0157 −4.154 63.7
2400 0.0117 −4.448 85.5
3000 0.00860 −4.756 116
3600 0.00640 −5.051 156

Given: balanced chemical equation, reaction times, and concentrations

Asked for: graph of data, rate law, and rate constant

Strategy:

A Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in to determine the reaction order.

B Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction.

Solution:

A Here are plots of [N2O5] versus t, ln[N2O5] versus t, and 1/[N2O5] versus t:

The plot of ln[N2O5] versus t gives a straight line, whereas the plots of [N2O5] versus t and 1/[N2O5] versus t do not. This means that the decomposition of N2O5 is first order in [N2O5].

B The rate law for the reaction is therefore

rate = k[N2O5]

Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N2O5] versus t. Using the points for t = 0 and 3000 s,

slope=ln[N2O5]3000ln[N2O5]03000 s0 s=(4.756)(3.310)3000 s=4.820×104 s1

Thus k = 4.820 × 10−4 s−1.

Exercise

1,3-Butadiene (CH2=CH—CH=CH2; C4H6) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C4H6 as a function of time at 326°C are listed in the following table along with ln[C4H6] and the reciprocal concentrations. Graph the data as concentration versus t, ln concentration versus t, and 1/concentration versus t. Then determine the reaction order in C4H6, the rate law, and the rate constant for the reaction.

Time (s) [C4H6] (M) ln[C4H6] 1/[C4H6] (M−1)
0 1.72 × 10−2 −4.063 58.1
900 1.43 × 10−2 −4.247 69.9
1800 1.23 × 10−2 −4.398 81.3
3600 9.52 × 10−3 −4.654 105
6000 7.30 × 10−3 −4.920 137

Answer:

second order in C4H6; rate = k[C4H6]2; k = 1.3 × 10−2 M−1·s−1

Summary

For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of −k. For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of −k. For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.

Key Takeaway

  • Plotting the concentration of a reactant as a function of time produces a graph with a characteristic shape that can be used to identify the reaction order in that reactant.

Conceptual Problems

  1. Compare first-order differential and integrated rate laws with respect to the following. Is there any information that can be obtained from the integrated rate law that cannot be obtained from the differential rate law?

    1. the magnitude of the rate constant
    2. the information needed to determine the order
    3. the shape of the graphs
  2. In the single-step, second-order reaction 2A → products, how would a graph of [A] versus time compare to a plot of 1/[A] versus time? Which of these would be the most similar to the same set of graphs for A during the single-step, second-order reaction A + B → products? Explain.

  3. For reactions of the same order, what is the relationship between the magnitude of the rate constant and the reaction rate? If you were comparing reactions with different orders, could the same arguments be made? Why?

Answers

    1. For a given reaction under particular conditions, the magnitude of the first-order rate constant does not depend on whether a differential rate law or an integrated rate law is used.
    2. The differential rate law requires multiple experiments to determine reactant order; the integrated rate law needs only one experiment.
    3. Using the differential rate law, a graph of concentration versus time is a curve with a slope that becomes less negative with time, whereas for the integrated rate law, a graph of ln[reactant] versus time gives a straight line with slope = −k. The integrated rate law allows you to calculate the concentration of a reactant at any time during the reaction; the differential rate law does not.
  1. The reaction rate increases as the rate constant increases. We cannot directly compare reaction rates and rate constants for reactions of different orders because they are not mathematically equivalent.

Numerical Problems

  1. One method of using graphs to determine reaction order is to use relative rate information. Plotting the log of the relative rate versus log of relative concentration provides information about the reaction. Here is an example of data from a zeroth-order reaction:

    Relative [A] (M) Relative Rate (M/s)
    1 1
    2 1
    3 1

    Varying [A] does not alter the reaction rate. Using the relative rates in the table, generate plots of log(rate) versus log(concentration) for zeroth-, first- and second-order reactions. What does the slope of each line represent?

  2. The table below follows the decomposition of N2O5 gas by examining the partial pressure of the gas as a function of time at 45°C. What is the reaction order? What is the rate constant? How long would it take for the pressure to reach 105 mmHg at 45°C?

    Time (s) Pressure (mmHg)
    0 348
    400 276
    1600 156
    3200 69
    4800 33

14.5 Half-Lives and Radioactive Decay Kinetics

Learning Objective

  1. To know how to use half-lives to describe the rates of first-order reactions.

Half-Lives

Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the half-lifeThe period of time it takes for the concentration of a reactant to decrease to one-half its initial value. of the reaction, written as t1/2. Thus the half-life of a reaction is the time required for the reactant concentration to decrease from [A]0 to [A]0/2. If two reactions have the same order, the faster reaction will have a shorter half-life, and the slower reaction will have a longer half-life.

The half-life of a first-order reaction under a given set of reaction conditions is a constant. This is not true for zeroth- and second-order reactions. The half-life of a first-order reaction is independent of the concentration of the reactants. This becomes evident when we rearrange the integrated rate law for a first-order reaction () to produce the following equation:

Equation 14.27

ln[A]0[A]=kt

Substituting [A]0/2 for [A] and t1/2 for t (to indicate a half-life) into gives

ln[A]0[A]0/2=ln 2=kt1/2

The natural logarithm of 2 (to three decimal places) is 0.693. Substituting this value into the equation, we obtain the expression for the half-life of a first-order reaction:

Equation 14.28

t1/2=0.693k

Thus, for a first-order reaction, each successive half-life is the same length of time, as shown in , and is independent of [A].

Figure 14.17 The Half-Life of a First-Order Reaction

This plot shows the concentration of the reactant in a first-order reaction as a function of time and identifies a series of half-lives, intervals in which the reactant concentration decreases by a factor of 2. In a first-order reaction, every half-life is the same length of time.

If we know the rate constant for a first-order reaction, then we can use half-lives to predict how much time is needed for the reaction to reach a certain percent completion.

Number of Half-Lives Percentage of Reactant Remaining
1 100%2=50% 12(100%)=50%
2 50%2=25% 12(12)(100%)=25%
3 25%2=12.5% 12(12)(12)(100%)=12.5%
n 100%2n (12)n(100%)=(12)n%

As you can see from this table, the amount of reactant left after n half-lives of a first-order reaction is (1/2)n times the initial concentration.

Note the Pattern

For a first-order reaction, the concentration of the reactant decreases by a constant with each half-life and is independent of [A].

Example 10

The anticancer drug cisplatin hydrolyzes in water with a rate constant of 1.5 × 10−3 min−1 at pH 7.0 and 25°C. Calculate the half-life for the hydrolysis reaction under these conditions. If a freshly prepared solution of cisplatin has a concentration of 0.053 M, what will be the concentration of cisplatin after 5 half-lives? after 10 half-lives? What is the percent completion of the reaction after 5 half-lives? after 10 half-lives?

Given: rate constant, initial concentration, and number of half-lives

Asked for: half-life, final concentrations, and percent completion

Strategy:

A Use to calculate the half-life of the reaction.

B Multiply the initial concentration by 1/2 to the power corresponding to the number of half-lives to obtain the remaining concentrations after those half-lives.

C Subtract the remaining concentration from the initial concentration. Then divide by the initial concentration, multiplying the fraction by 100 to obtain the percent completion.

Solution:

A We can calculate the half-life of the reaction using :

t1/2=0.693k=0.6931.5×103 min1=4.6×102 min

Thus it takes almost 8 h for half of the cisplatin to hydrolyze.

B After 5 half-lives (about 38 h), the remaining concentration of cisplatin will be as follows:

0.053 M25=0.053 M32=0.0017 M

After 10 half-lives (77 h), the remaining concentration of cisplatin will be as follows:

0.053 M210=0.053 M1024=5.2×105 M

C The percent completion after 5 half-lives will be as follows:

percent completion=(0.053 M0.0017 M)(100)0.053=97%

The percent completion after 10 half-lives will be as follows:

percent completion=(0.053 M5.2×105 M)(100)0.053 M=100%

Thus a first-order chemical reaction is 97% complete after 5 half-lives and 100% complete after 10 half-lives.

Exercise

In Example 4 you found that ethyl chloride decomposes to ethylene and HCl in a first-order reaction that has a rate constant of 1.6 × 10−6 s−1 at 650°C. What is the half-life for the reaction under these conditions? If a flask that originally contains 0.077 M ethyl chloride is heated at 650°C, what is the concentration of ethyl chloride after 4 half-lives?

Answer: 4.3 × 105 s = 120 h = 5.0 days; 4.8 × 10−3 M

Radioactive Decay Rates

As you learned in , radioactivity, or radioactive decay, is the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. The rate of radioactive decay is an intrinsic property of each radioactive isotope that is independent of the chemical and physical form of the radioactive isotope. The rate is also independent of temperature. In this section, we will describe radioactive decay rates and how half-lives can be used to monitor radioactive decay processes.

In any sample of a given radioactive substance, the number of atoms of the radioactive isotope must decrease with time as their nuclei decay to nuclei of a more stable isotope. Using N to represent the number of atoms of the radioactive isotope, we can define the rate of decayThe decrease in the number of a radioisotope’s nuclei per unit time. of the sample, which is also called its activity (A)The decrease in the number of a radioisotope’s nuclei per unit time: A=ΔN/Δt. as the decrease in the number of the radioisotope’s nuclei per unit time:

Equation 14.29

A=ΔNΔt

Activity is usually measured in disintegrations per second (dps) or disintegrations per minute (dpm).

The activity of a sample is directly proportional to the number of atoms of the radioactive isotope in the sample:

Equation 14.30

A = kN

Here, the symbol k is the radioactive decay constant, which has units of inverse time (e.g., s−1, yr−1) and a characteristic value for each radioactive isotope. If we combine and , we obtain the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample:

Equation 14.31

ΔNΔt=kN

is the same as the equation for the reaction rate of a first-order reaction (), except that it uses numbers of atoms instead of concentrations. In fact, radioactive decay is a first-order process and can be described in terms of either the differential rate law () or the integrated rate law:

N = N0ekt

Equation 14.32

ln NN0=kt

Because radioactive decay is a first-order process, the time required for half of the nuclei in any sample of a radioactive isotope to decay is a constant, called the half-life of the isotope. The half-life tells us how radioactive an isotope is (the number of decays per unit time); thus it is the most commonly cited property of any radioisotope. For a given number of atoms, isotopes with shorter half-lives decay more rapidly, undergoing a greater number of radioactive decays per unit time than do isotopes with longer half-lives. The half-lives of several isotopes are listed in , along with some of their applications.

Table 14.6 Half-Lives and Applications of Some Radioactive Isotopes

Radioactive Isotope Half-Life Typical Uses
hydrogen-3 (tritium) 12.32 yr biochemical tracer
carbon-11 20.33 min positron emission tomography (biomedical imaging)
carbon-14 5.70 × 103 yr dating of artifacts
sodium-24 14.951 h cardiovascular system tracer
phosphorus-32 14.26 days biochemical tracer
potassium-40 1.248 × 109 yr dating of rocks
iron-59 44.495 days red blood cell lifetime tracer
cobalt-60 5.2712 yr radiation therapy for cancer
technetium-99m* 6.006 h biomedical imaging
iodine-131 8.0207 days thyroid studies tracer
radium-226 1.600 × 103 yr radiation therapy for cancer
uranium-238 4.468 × 109 yr dating of rocks and Earth’s crust
americium-241 432.2 yr smoke detectors
*The m denotes metastable, where an excited state nucleus decays to the ground state of the same isotope.

Note the Pattern

Radioactive decay is a first-order process.

Radioisotope Dating Techniques

In our earlier discussion, we used the half-life of a first-order reaction to calculate how long the reaction had been occurring. Because nuclear decay reactions follow first-order kinetics and have a rate constant that is independent of temperature and the chemical or physical environment, we can perform similar calculations using the half-lives of isotopes to estimate the ages of geological and archaeological artifacts. The techniques that have been developed for this application are known as radioisotope dating techniques.

The most common method for measuring the age of ancient objects is carbon-14 dating. The carbon-14 isotope, created continuously in the upper regions of Earth’s atmosphere, reacts with atmospheric oxygen or ozone to form 14CO2. As a result, the CO2 that plants use as a carbon source for synthesizing organic compounds always includes a certain proportion of 14CO2 molecules as well as nonradioactive 12CO2 and 13CO2. Any animal that eats a plant ingests a mixture of organic compounds that contains approximately the same proportions of carbon isotopes as those in the atmosphere. When the animal or plant dies, the carbon-14 nuclei in its tissues decay to nitrogen-14 nuclei by a radioactive process known as beta decay, which releases low-energy electrons (β particles) that can be detected and measured:

Equation 14.33

14C → 14N + β

The half-life for this reaction is 5700 ± 30 yr.

The 14C/12C ratio in living organisms is 1.3 × 10−12, with a decay rate of 15 dpm/g of carbon (). Comparing the disintegrations per minute per gram of carbon from an archaeological sample with those from a recently living sample enables scientists to estimate the age of the artifact, as illustrated in Example 11.Using this method implicitly assumes that the 14CO2/12CO2 ratio in the atmosphere is constant, which is not strictly correct. Other methods, such as tree-ring dating, have been used to calibrate the dates obtained by radiocarbon dating, and all radiocarbon dates reported are now corrected for minor changes in the 14CO2/12CO2 ratio over time.

Figure 14.18 Radiocarbon Dating

A plot of the specific activity of 14C versus age for a number of archaeological samples shows an inverse linear relationship between 14C content (a log scale) and age (a linear scale).

Example 11

In 1990, the remains of an apparently prehistoric man were found in a melting glacier in the Italian Alps. Analysis of the 14C content of samples of wood from his tools gave a decay rate of 8.0 dpm/g carbon. How long ago did the man die?

Given: isotope and final activity

Asked for: elapsed time

Strategy:

A Use to calculate N0/N. Then substitute the value for the half-life of 14C into to find the rate constant for the reaction.

B Using the values obtained for N0/N and the rate constant, solve to obtain the elapsed time.

Solution:

We know the initial activity from the isotope’s identity (15 dpm/g), the final activity (8.0 dpm/g), and the half-life, so we can use the integrated rate law for a first-order nuclear reaction () to calculate the elapsed time (the amount of time elapsed since the wood for the tools was cut and began to decay).

ln NN0=ktln(N/N0)k=t

A From , we know that A = kN. We can therefore use the initial and final activities (A0 = 15 dpm and A = 8.0 dpm) to calculate N0/N:

A0A=kN0kN=N0N=158.0

Now we need only calculate the rate constant for the reaction from its half-life (5730 yr) using :

t1/2=0.693k

This equation can be rearranged as follows:

k=0.693t1/2=0.6935730 yr=1.22×104 yr1

B Substituting into the equation for t,

t=ln(N0/N)k=ln(15/8.0)1.22×104 yr1=5.2×103 yr

From our calculations, the man died 5200 yr ago.

Exercise

It is believed that humans first arrived in the Western Hemisphere during the last Ice Age, presumably by traveling over an exposed land bridge between Siberia and Alaska. Archaeologists have estimated that this occurred about 11,000 yr ago, but some argue that recent discoveries in several sites in North and South America suggest a much earlier arrival. Analysis of a sample of charcoal from a fire in one such site gave a 14C decay rate of 0.4 dpm/g of carbon. What is the approximate age of the sample?

Answer: 30,000 yr

Summary

The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k.

Radioactive decay reactions are first-order reactions. The rate of decay, or activity, of a sample of a radioactive substance is the decrease in the number of radioactive nuclei per unit time.

Key Takeaways

  • The half-life of a first-order reaction is independent of the concentration of the reactants.
  • The half-lives of radioactive isotopes can be used to date objects.

Key Equations

half-life of first-order reaction

: t1/2=0.693k

radioactive decay

: A = kN

Conceptual Problems

  1. What do chemists mean by the half-life of a reaction?

  2. If a sample of one isotope undergoes more disintegrations per second than the same number of atoms of another isotope, how do their half-lives compare?

Numerical Problems

  1. Half-lives for the reaction A + B → C were calculated at three values of [A]0, and [B] was the same in all cases. The data are listed in the following table:

    [A]0 (M) t½ (s)
    0.50 420
    0.75 280
    1.0 210

    Does this reaction follow first-order kinetics? On what do you base your answer?

  2. Ethyl-2-nitrobenzoate (NO2C6H4CO2C2H5) hydrolyzes under basic conditions. A plot of [NO2C6H4CO2C2H5] versus t was used to calculate t½, with the following results:

    [NO2C6H4CO2C2H5] (M/cm3) t½ (s)
    0.050 240
    0.040 300
    0.030 400

    Is this a first-order reaction? Explain your reasoning.

  3. Azomethane (CH3N2CH3) decomposes at 600 K to C2H6 and N2. The decomposition is first order in azomethane. Calculate t½ from the data in the following table:

    Time (s) PCH3N2CH3 (atm)
    0 8.2 × 10−2
    2000 3.99 × 10−2
    4000 1.94 × 10−2

    How long will it take for the decomposition to be 99.9% complete?

  4. The first-order decomposition of hydrogen peroxide has a half-life of 10.7 h at 20°C. What is the rate constant (expressed in s−1) for this reaction? If you started with a solution that was 7.5 × 10−3 M H2O2, what would be the initial rate of decomposition (M/s)? What would be the concentration of H2O2 after 3.3 h?

Answers

  1. No; the reaction is second order in A because the half-life decreases with increasing reactant concentration according to t1/2 = 1/k[A0].

  2. t1/2 = 1.92 × 103 s or 1920 s; 19100 s or 5.32 hrs.

14.6 Reaction Rates—A Microscopic View

Learning Objective

  1. To determine the individual steps of a simple reaction.

One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism.

In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water:

Equation 14.34

2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)

For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an elementary reactionEach of the complex series of reactions that take place in a stepwise fashion to convert reactants to products., involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction.

Molecularity and the Rate-Determining Step

To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide.

Equation 14.35

NO2(g) + CO(g) → NO(g) + CO2(g)

From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of NO2 with a molecule of CO that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows:

Equation 14.36

rate = k[NO2]2

The fact that the reaction is second order in [NO2] and independent of [CO] tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be rate = k[NO2][CO].

The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2:

step 1NO2+NO2slowNO3+NOelementary reactionstep 2NO3+CONO2+CO2_elementary reactionsumNO2+CONO+CO2overall reaction

According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The NO3 molecule is an intermediateA species in a reaction mechanism that does not appear in the balanced chemical equation for the overall reaction. in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step.

Note the Pattern

The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction.

Using Molecularity to Describe a Rate Law

The molecularityThe number of molecules that collide during any step in a reaction mechanism. of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as unimolecular; if there are two reactant molecules, it is bimolecular; and if there are three reactant molecules (a relatively rare situation), it is termolecular. Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!)

Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity (). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is rate = k[A]. For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in . For a bimolecular elementary reaction of the form A + B → products, the general rate law is rate = k[A][B].

Table 14.7 Common Types of Elementary Reactions and Their Rate Laws

Elementary Reaction Molecularity Rate Law Reaction Order
A → products unimolecular rate = k[A] first
2A → products bimolecular rate = k[A]2 second
A + B → products bimolecular rate = k[A][B] second
2A + B → products termolecular rate = k[A]2[B] third
A + B + C → products termolecular rate = k[A][B][C] third

Figure 14.19 The Basis for Writing Rate Laws of Elementary Reactions

This diagram illustrates how the number of possible collisions per unit time between two reactant species, A and B, depends on the number of A and B particles present. The number of collisions between A and B particles increases as the product of the number of particles, not as the sum. This is why the rate law for an elementary reaction depends on the product of the concentrations of the species that collide in that step.

Identifying the Rate-Determining Step

Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the rate-determining stepThe slowest step in a reaction mechanism., that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions.

Rate-determining step. The phenomenon of a rate-determining step can be compared to a succession of funnels. The smallest-diameter funnel controls the rate at which the bottle is filled, whether it is the first or the last in the series. Pouring liquid into the first funnel faster than it can drain through the smallest results in an overflow.

Look at the rate laws for each elementary reaction in our example as well as for the overall reaction.

step 1NO2+NO2k1NO3+NOrate=k1[NO2]2 (predicted)step 2NO3+COk2NO2+CO2_rate=k2[NO3][CO] (predicted)sumNO2+COkNO+CO2rate=k[NO2]2 (observed)

The experimentally determined rate law for the reaction of NO2 with CO is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so k for the overall reaction must equal k1. That is, NO3 is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2.

Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect.

Example 12

In an alternative mechanism for the reaction of NO2 with CO, N2O4 appears as an intermediate.

step 1NO2+NO2k1N2O4step 2N2O4+COk2NO+NO2+CO2_sumNO2+CONO+CO2

Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = k[NO2]2)?

Given: elementary reactions

Asked for: rate law for each elementary reaction and overall rate law

Strategy:

A Determine the rate law for each elementary reaction in the reaction.

B Determine which rate law corresponds to the experimentally determined rate law for the reaction. This rate law is the one for the rate-determining step.

Solution:

A The rate law for step 1 is rate = k1[NO2]2; for step 2, it is rate = k2[N2O4][CO].

B If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = k1[NO2]2. This is the same as the experimentally determined rate law. Hence this mechanism, with N2O4 as an intermediate, and the one described previously, with NO3 as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO3 and N2O4, directly.

Exercise A

Iodine monochloride (ICl) reacts with H2 as follows:

2ICl(l) + H2(g) → 2HCl(g) + I2(s)

The experimentally determined rate law is rate = k[ICl][H2]. Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: HI is an intermediate.)

Answer:

step 1ICl+H2k1HCl+HIrate=k1[ICl][H2] (slow)step 2HI+IClk2HCl+I2_rate=k2[HI][ICl] (fast)sum2ICl+H22HCl+I2

This mechanism is consistent with the experimental rate law if the first step is the rate-determining step.

Exercise B

The reaction between NO and H2 occurs via a three-step process:

step 1NO+NOk1N2O2(fast)step 2N2O2+H2k2N2O+H2O(slow)step 3N2O+H2k3N2+H2O(fast)

Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction: rate = k[NO]2[H2]2?

Answer:

  • rate = k1[NO]2;
  • rate = k2[N2O2][H2];
  • rate = k3[N2O][H2];
  • 2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
  • step 2
  • Yes, because the rate of formation of [N2O2] = k1[NO]2. Substituting k1[NO]2 for [N2O2] in the rate law for step 2 gives the experimentally derived rate law for the overall chemical reaction, where k = k1k2.

Chain Reactions

Many reaction mechanisms, like those discussed so far, consist of only two or three elementary reactions. Many others consist of long series of elementary reactions. The most common mechanisms are chain reactionsA reaction mechanism in which one or more elementary reactions that contain a highly reactive species repeat again and again during the reaction process., in which one or more elementary reactions that contain a highly reactive species repeat again and again during the reaction process. Chain reactions occur in fuel combustion, explosions, the formation of many polymers, and the tissue changes associated with aging. They are also important in the chemistry of the atmosphere.

Chain reactions are described as having three stages. The first is initiation, a step that produces one or more reactive intermediates. Often these intermediates are radicalsSpecies that have one or more unpaired valence electrons., species that have an unpaired valence electron. In the second stage, propagation, reactive intermediates are continuously consumed and regenerated while products are formed. Intermediates are also consumed but not regenerated in the final stage of a chain reaction, termination, usually by forming stable products.

Let us look at the reaction of methane with chlorine at elevated temperatures (400°C–450°C), a chain reaction used in industry to manufacture methyl chloride (CH3Cl), dichloromethane (CH2Cl2), chloroform (CHCl3), and carbon tetrachloride (CCl4):

CH4 + Cl2 → CH3Cl + HCl CH3Cl + Cl2 → CH2Cl2 + HCl CH2Cl2 + Cl2 → CHCl3 + HCl CHCl3 + Cl2 → CCl4 + HCl

Direct chlorination generally produces a mixture of all four carbon-containing products, which must then be separated by distillation. In our discussion, we will examine only the chain reactions that lead to the preparation of CH3Cl.

In the initiation stage of this reaction, the relatively weak Cl–Cl bond cleaves at temperatures of about 400°C to produce chlorine atoms (Cl·):

Cl2 → 2Cl·

During propagation, a chlorine atom removes a hydrogen atom from a methane molecule to give HCl and CH3·, the methyl radical:

Cl· + CH4 → CH3· + HCl

The methyl radical then reacts with a chlorine molecule to form methyl chloride and another chlorine atom, Cl·:

CH3· + Cl2 → CH3Cl + Cl·

The sum of the propagation reactions is the same as the overall balanced chemical equation for the reaction:

Cl·+CH4CH3·+HClCH3·+Cl2CH3Cl+Cl·_Cl2+CH4CH3Cl+HCl

Without a chain-terminating reaction, propagation reactions would continue until either the methane or the chlorine was consumed. Because radical species react rapidly with almost anything, however, including each other, they eventually form neutral compounds, thus terminating the chain reaction in any of three ways:

CH3· + Cl· → CH3Cl CH3· + CH3· → H3CCH3 Cl· + Cl· → Cl2

Here is the overall chain reaction, with the desired product (CH3Cl) in bold:

Initiation: Cl2 → 2Cl·
Propagation: Cl· + CH4 → CH3· + HCl
CH3· + Cl2CH3Cl + Cl·
Termination: CH3· + Cl· → CH3Cl
CH3· + CH3· → H3CCH3
Cl· + Cl· → Cl2

The chain reactions responsible for explosions generally have an additional feature: the existence of one or more chain branching steps, in which one radical reacts to produce two or more radicals, each of which can then go on to start a new chain reaction. Repetition of the branching step has a cascade effect such that a single initiation step generates large numbers of chain reactions. The result is a very rapid reaction or an explosion.

The reaction of H2 and O2, used to propel rockets, is an example of a chain branching reaction:

Initiation: H2 + O2 → HO2· + H·
Propagation: HO2· + H2 → H2O + OH·
OH· + H2 → H2O + H·
Termination: H· + O2 → OH· + ·O·
·O· + H2 → OH· + H·

Termination reactions occur when the extraordinarily reactive H· or OH· radicals react with a third species. The complexity of a chain reaction makes it unfeasible to write a rate law for the overall reaction.

Summary

A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its molecularity, the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step. Chain reactions consist of three kinds of reactions: initiation, propagation, and termination. Intermediates in chain reactions are often radicals, species that have an unpaired valence electron.

Key Takeaway

  • A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law.

Conceptual Problems

  1. How does the term molecularity relate to elementary reactions? How does it relate to the overall balanced chemical equation?

  2. What is the relationship between the reaction order and the molecularity of a reaction? What is the relationship between the reaction order and the balanced chemical equation?

  3. When you determine the rate law for a given reaction, why is it valid to assume that the concentration of an intermediate does not change with time during the course of the reaction?

  4. If you know the rate law for an overall reaction, how would you determine which elementary reaction is rate determining? If an intermediate is contained in the rate-determining step, how can the experimentally determined rate law for the reaction be derived from this step?

  5. Give the rate-determining step for each case.

    1. Traffic is backed up on a highway because two lanes merge into one.
    2. Gas flows from a pressurized cylinder fitted with a gas regulator and then is bubbled through a solution.
    3. A document containing text and graphics is downloaded from the Internet.
  6. Before being sent on an assignment, an aging James Bond was sent off to a health farm where part of the program’s focus was to purge his body of radicals. Why was this goal considered important to his health?

Numerical Problems

  1. Cyclopropane, a mild anesthetic, rearranges to propylene via a collision that produces and destroys an energized species. The important steps in this rearrangement are as follows:

    where M is any molecule, including cyclopropane. Only those cyclopropane molecules with sufficient energy (denoted with an asterisk) can rearrange to propylene. Which step determines the rate constant of the overall reaction?

  2. Above approximately 500 K, the reaction between NO2 and CO to produce CO2 and NO follows the second-order rate law Δ[CO2]/Δt = k[NO2][CO]. At lower temperatures, however, the rate law is Δ[CO2]/Δt = k′[NO2]2, for which it is known that NO3 is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest?

  3. Nitramide (O2NNH2) decomposes in aqueous solution to N2O and H2O. What is the experimental rate law (Δ[N2O]/Δt) for the decomposition of nitramide if the mechanism for the decomposition is as follows?

    O2NNH2k1k1O2NNH+H+(fast)O2NNHk2N2O+OH(slow)H++OHk3H2O(fast)

    Assume that the rates of the forward and reverse reactions in the first equation are equal.

  4. The following reactions are given:

    A+Bk1k1C+DD+Ek2F

    What is the relationship between the relative magnitudes of k−1 and k2 if these reactions have the rate law Δ[F]/Δt = k[A][B][E]/[C]? How does the magnitude of k1 compare to that of k2? Under what conditions would you expect the rate law to be Δ[F]/Δt = k′[A][B]? Assume that the rates of the forward and reverse reactions in the first equation are equal.

Answers

  1. The k2 step is likely to be rate limiting; the rate cannot proceed any faster than the second step.

  2.  

    rate=k2k1[O2NNH2]k1[H+]=k[O2NNH2][H+]

14.7 The Collision Model of Chemical Kinetics

Learning Objective

  1. To understand why and how chemical reactions occur.

In Section 14.6 "Reaction Rates—A Microscopic View", you saw that it is possible to use kinetics studies of a chemical system, such as the effect of changes in reactant concentrations, to deduce events that occur on a microscopic scale, such as collisions between individual particles. Such studies have led to the collision model of chemical kinetics, which is a useful tool for understanding the behavior of reacting chemical species. According to the collision model, a chemical reaction can occur only when the reactant molecules, atoms, or ions collide with more than a certain amount of kinetic energy and in the proper orientation. The collision model explains why, for example, most collisions between molecules do not result in a chemical reaction. Nitrogen and oxygen molecules in a single liter of air at room temperature and 1 atm of pressure collide about 1030 times per second. If every collision produced two molecules of NO, the atmosphere would have been converted to NO and then NO2 a long time ago. Instead, in most collisions, the molecules simply bounce off one another without reacting, much as marbles bounce off each other when they collide. The collision model also explains why such chemical reactions occur more rapidly at higher temperatures. For example, the reaction rates of many reactions that occur at room temperature approximately double with a temperature increase of only 10°C. In this section, we will use the collision model to analyze this relationship between temperature and reaction rates.

Activation Energy

In Chapter 10 "Gases", we discussed the kinetic molecular theory of gases, which showed that the average kinetic energy of the particles of a gas increases with increasing temperature. Because the speed of a particle is proportional to the square root of its kinetic energy, increasing the temperature will also increase the number of collisions between molecules per unit time. What the kinetic molecular theory of gases does not explain is why the reaction rate of most reactions approximately doubles with a 10°C temperature increase. This result is surprisingly large considering that a 10°C increase in the temperature of a gas from 300 K to 310 K increases the kinetic energy of the particles by only about 4%, leading to an increase in molecular speed of only about 2% and a correspondingly small increase in the number of bimolecular collisions per unit time.

The collision model of chemical kinetics explains this behavior by introducing the concept of activation energy (Ea)The energy barrier or threshold that corresponds to the minimum amount of energy the particles in a reaction must have to react when they colllide.. We will define this concept using the reaction of NO with ozone, which plays an important role in the depletion of ozone in the ozone layer:

Equation 14.37

NO(g) + O3(g) → NO2(g) + O2(g)

Increasing the temperature from 200 K to 350 K causes the rate constant for this particular reaction to increase by a factor of more than 10, whereas the increase in the frequency of bimolecular collisions over this temperature range is only 30%. Thus something other than an increase in the collision rate must be affecting the reaction rate.

The reaction rate, not the rate constant, will vary with concentration. The rate constant, however, does vary with temperature. Figure 14.20 "Rate Constant versus Temperature for the Reaction of NO with O" shows a plot of the rate constant of the reaction of NO with O3 at various temperatures. The relationship is not linear but instead resembles the relationships seen in graphs of vapor pressure versus temperature (Chapter 11 "Liquids") and of conductivity versus temperature (Chapter 12 "Solids"). In all three cases, the shape of the plots results from a distribution of kinetic energy over a population of particles (electrons in the case of conductivity; molecules in the case of vapor pressure; and molecules, atoms, or ions in the case of reaction rates). Only a fraction of the particles have sufficient energy to overcome an energy barrier.

Figure 14.20 Rate Constant versus Temperature for the Reaction of NO with O3

The nonlinear shape of the curve is caused by a distribution of kinetic energy over a population of molecules. Only a fraction of the particles have enough energy to overcome an energy barrier, but as the temperature is increased, the size of that fraction increases.

In the case of vapor pressure, particles must overcome an energy barrier to escape from the liquid phase to the gas phase. This barrier corresponds to the energy of the intermolecular forces that hold the molecules together in the liquid. In conductivity, the barrier is the energy gap between the filled and empty bands. In chemical reactions, the energy barrier corresponds to the amount of energy the particles must have to react when they collide. This energy threshold, called the activation energy, was first postulated in 1888 by the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry 1903). It is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. Molecules that collide with less than the threshold energy bounce off one another chemically unchanged, with only their direction of travel and their speed altered by the collision. Molecules that are able to overcome the energy barrier are able to react and form an arrangement of atoms called the activated complexAlso called the transition state, the arrangement of atoms that first forms when molecules are able to overcome the activation energy and react. or the transition stateAlso called the activated complex, the arrangement of atoms that first forms when molecules are able to overcome the activation energy and react. of the reaction. The activated complex is not a reaction intermediate; it does not last long enough to be detected readily.

Note the Pattern

Any phenomenon that depends on the distribution of thermal energy in a population of particles has a nonlinear temperature dependence.

Graphing Energy Changes during a Reaction

We can graph the energy of a reaction by plotting the potential energy of the system as the reaction progresses. Figure 14.21 "Energy of the Activated Complex for the NO–O" shows a plot for the NO–O3 system, in which the vertical axis is potential energy and the horizontal axis is the reaction coordinate, which indicates the progress of the reaction with time. The activated complex is shown in brackets with an asterisk. The overall change in potential energy for the reaction (ΔE) is negative, which means that the reaction releases energy. (In this case, ΔE is −200.8 kJ/mol.) To react, however, the molecules must overcome the energy barrier to reaction (Ea is 9.6 kJ/mol). That is, 9.6 kJ/mol must be put into the system as the activation energy. Below this threshold, the particles do not have enough energy for the reaction to occur.

Figure 14.21 Energy of the Activated Complex for the NO–O3 System

The diagram shows how the energy of this system varies as the reaction proceeds from reactants to products. Note the initial increase in energy required to form the activated complex.

Part (a) in Figure 14.22 "Differentiating between " illustrates the general situation in which the products have a lower potential energy than the reactants. In contrast, part (b) in Figure 14.22 "Differentiating between " illustrates the case in which the products have a higher potential energy than the reactants, so the overall reaction requires an input of energy; that is, it is energetically uphill, and ΔE > 0. Although the energy changes that result from a reaction can be positive, negative, or even zero, in all cases an energy barrier must be overcome before a reaction can occur. This means that the activation energy is always positive.

Figure 14.22 Differentiating between Ea and ΔE

The potential energy diagrams for a reaction with (a) ΔE < 0 and (b) ΔE > 0 illustrate the change in the potential energy of the system as reactants are converted to products. Ea is always positive. For a reaction such as the one shown in (b), Ea must be greater than ΔE.

Note the Pattern

For similar reactions under comparable conditions, the one with the smallest Ea will occur most rapidly.

Whereas ΔE is related to the tendency of a reaction to occur spontaneously, Ea gives us information about the reaction rate and how rapidly the reaction rate changes with temperature. (For more information on spontaneous reactions, see Chapter 18 "Chemical Thermodynamics".) For two similar reactions under comparable conditions, the reaction with the smallest Eawill occur more rapidly.

Even when the energy of collisions between two reactant species is greater than Ea, however, most collisions do not produce a reaction. The probability of a reaction occurring depends not only on the collision energy but also on the spatial orientation of the molecules when they collide. For NO and O3 to produce NO2 and O2, a terminal oxygen atom of O3 must collide with the nitrogen atom of NO at an angle that allows O3 to transfer an oxygen atom to NO to produce NO2 (Figure 14.23 "The Effect of Molecular Orientation on the Reaction of NO and O"). All other collisions produce no reaction. Because fewer than 1% of all possible orientations of NO and O3 result in a reaction at kinetic energies greater than Ea, most collisions of NO and O3 are unproductive. The fraction of orientations that result in a reaction is called the steric factor (p)The fraction of orientations of particles that result in a chemical reaction., and, in general, its value can range from 0 (no orientations of molecules result in reaction) to 1 (all orientations result in reaction).

Figure 14.23 The Effect of Molecular Orientation on the Reaction of NO and O3

Most collisions of NO and O3 molecules occur with an incorrect orientation for a reaction to occur. Only those collisions in which the N atom of NO collides with one of the terminal O atoms of O3 are likely to produce NO2 and O2, even if the molecules collide with E > Ea.

The Arrhenius Equation

Figure 14.24 "Surmounting the Energy Barrier to a Reaction" shows both the kinetic energy distributions and a potential energy diagram for a reaction. The shaded areas show that at the lower temperature (300 K), only a small fraction of molecules collide with kinetic energy greater than Ea; however, at the higher temperature (500 K) a much larger fraction of molecules collide with kinetic energy greater than Ea. Consequently, the reaction rate is much slower at the lower temperature because only a relatively few molecules collide with enough energy to overcome the potential energy barrier.

Figure 14.24 Surmounting the Energy Barrier to a Reaction

This chart juxtaposes the energy distributions of lower-temperature (300 K) and higher-temperature (500 K) samples of a gas against the potential energy diagram for a reaction. Only those molecules in the shaded region of the energy distribution curve have E > Ea and are therefore able to cross the energy barrier separating reactants and products. The fraction of molecules with E > Ea is much greater at 500 K than at 300 K, so the reaction will occur much more rapidly at 500 K.

For an A + B elementary reaction, all the factors that affect the reaction rate can be summarized in a single series of relationships:

rate = (collision frequency)(steric factor)(fraction of collisions with E > Ea)

where

Equation 14.38

rate = k[A][B]

Arrhenius used these relationships to arrive at an equation that relates the magnitude of the rate constant for a reaction to the temperature, the activation energy, and the constant, A, called the frequency factorA constant in the Arrhenius equation, it converts concentrations to collisions per second.:

Equation 14.39

k=AeEa/RT

The frequency factor is used to convert concentrations to collisions per second.Because the frequency of collisions depends on the temperature, A is actually not constant. Instead, A increases slightly with temperature as the increased kinetic energy of molecules at higher temperatures causes them to move slightly faster and thus undergo more collisions per unit time. Equation 14.39 is known as the Arrhenius equationAn expression that summarizes the collision model of chemical kinetics: k=AeEa/RT. and summarizes the collision model of chemical kinetics, where T is the absolute temperature (in K) and R is the ideal gas constant [8.314 J/(K·mol)]. Ea indicates the sensitivity of the reaction to changes in temperature. The reaction rate with a large Ea increases rapidly with increasing temperature, whereas the reaction rate with a smaller Ea increases much more slowly with increasing temperature.

If we know the reaction rate at various temperatures, we can use the Arrhenius equation to calculate the activation energy. Taking the natural logarithm of both sides of Equation 14.39,

Equation 14.40

ln k=ln A+(EaRT)=ln A+[(EaR)(1T)]

Equation 14.40 is the equation of a straight line, y = mx + b, where y = ln k and x = 1/T. This means that a plot of ln k versus 1/T is a straight line with a slope of −Ea/R and an intercept of ln A. In fact, we need to measure the reaction rate at only two temperatures to estimate Ea.

Knowing the Ea at one temperature allows us to predict the reaction rate at other temperatures. This is important in cooking and food preservation, for example, as well as in controlling industrial reactions to prevent potential disasters. The procedure for determining Ea from reaction rates measured at several temperatures is illustrated in Example 13.

Example 13

Many people believe that the rate of a tree cricket’s chirping is related to temperature. To see whether this is true, biologists have carried out accurate measurements of the rate of tree cricket chirping (f) as a function of temperature (T). Use the data in the following table, along with the graph of ln[chirping rate] versus 1/T in Figure 14.25 "Graphical Determination of ", to calculate Ea for the biochemical reaction that controls cricket chirping. Then predict the chirping rate on a very hot evening, when the temperature is 308 K (35°C, or 95°F).

Frequency (f; chirps/min) ln f T (K) 1/T (K)
200 5.30 299 3.34 × 10−3
179 5.19 298 3.36 × 10−3
158 5.06 296 3.38 × 10−3
141 4.95 294 3.40 × 10−3
126 4.84 293 3.41 × 10−3
112 4.72 292 3.42 × 10−3
100 4.61 290 3.45 × 10−3
89 4.49 289 3.46 × 10−3
79 4.37 287 3.48 × 10−3

Given: chirping rate at various temperatures

Asked for: activation energy and chirping rate at specified temperature

Strategy:

A From the plot of ln f versus 1/T in Figure 14.25 "Graphical Determination of ", calculate the slope of the line (−Ea/R) and then solve for the activation energy.

B Express Equation 14.40 in terms of k1 and T1 and then in terms of k2 and T2.

C Subtract the two equations; rearrange the result to describe k2/k1 in terms of T2 and T1.

D Using measured data from the table, solve the equation to obtain the ratio k2/k1. Using the value listed in the table for k1, solve for k2.

Solution:

A If cricket chirping is controlled by a reaction that obeys the Arrhenius equation, then a plot of ln f versus 1/T should give a straight line (Figure 14.25 "Graphical Determination of "). Also, the slope of the plot of ln f versus 1/T should be equal to −Ea/R. We can use the two endpoints in Figure 14.25 "Graphical Determination of " to estimate the slope:

slope=Δln fΔ(1/T)=5.304.373.34×103 K13.48×103 K1=0.93.014×103 K1=6.6×103 K

A computer best-fit line through all the points has a slope of −6.67 × 103 K, so our estimate is very close. We now use it to solve for the activation energy:

Ea=(slope)(R)=(6.6×103 K)(8.314 JKmol)(1 KJ1000 J)=55 kJmol

B If the activation energy of a reaction and the rate constant at one temperature are known, then we can calculate the reaction rate at any other temperature. We can use Equation 14.40 to express the known rate constant (k1) at the first temperature (T1) as follows:

ln k1=ln AEaRT1

Similarly, we can express the unknown rate constant (k2) at the second temperature (T2) as follows:

ln k2=ln AEaRT2

C These two equations contain four known quantities (Ea, T1, T2, and k1) and two unknowns (A and k2). We can eliminate A by subtracting the first equation from the second:

ln k2ln k1=(ln AEaRT2)(ln AEaRT1)=EaRT2+EaRT1

Then

ln k2k1=EaR(1T11T2)

D To obtain the best prediction of chirping rate at 308 K (T2), we try to choose for T1 and k1 the measured rate constant and corresponding temperature in the data table that is closest to the best-fit line in the graph. Choosing data for T1 = 296 K, where f = 158, and using the Ea calculated previously,

ln kT2kT1=EaR(1T11T2)=55 kJ/mol8.314 J/(Kmol)(1000 J1 kJ)(1296 K1308 K)=0.87

Thus k308/k296 = 2.4 and k308 = (2.4)(158) = 380, and the chirping rate on a night when the temperature is 308 K is predicted to be 380 chirps per minute.

Exercise

The equation for the decomposition of NO2 to NO and O2 is second order in NO2:

2NO2(g) → 2NO(g) + O2(g)

Data for the reaction rate as a function of temperature are listed in the following table. Calculate Ea for the reaction and the rate constant at 700 K.

T (K) k (M−1·s−1)
592 522
603 755
627 1700
652 4020
656 5030

Answer: Ea = 114 kJ/mol; k700 = 18,600 M−1·s−1 = 1.86 × 104 M−1·s−1.

What Ea results in a doubling of the reaction rate with a 10°C increase in temperature from 20° to 30°C?

Answer: about 51 kJ/mol

Figure 14.25 Graphical Determination of Ea for Tree Cricket Chirping

When the natural logarithm of the rate of tree cricket chirping is plotted versus 1/T, a straight line results. The slope of the line suggests that the chirping rate is controlled by a single reaction with an Ea of 55 kJ/mol.

Summary

A minimum energy (activation energy, Ea) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the activated complex, or transition state, of the reaction. At a given temperature, the higher the Ea, the slower the reaction. The fraction of orientations that result in a reaction is the steric factor. The frequency factor, steric factor, and activation energy are related to the rate constant in the Arrhenius equation: k=AeEa/RT. A plot of the natural logarithm of k versus 1/T is a straight line with a slope of −Ea/R.

Key Takeaway

  • For a chemical reaction to occur, an energy threshold must be overcome, and the reacting species must also have the correct spatial orientation.

Key Equation

Arrhenius equation

Equation 14.39: k=AeEa/RT

Conceptual Problems

  1. Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy?

  2. For any given reaction, what is the relationship between the activation energy and each of the following?

    1. electrostatic repulsions
    2. bond formation in the activated complex
    3. the nature of the activated complex
  3. If you are concerned with whether a reaction will occur rapidly, why would you be more interested in knowing the magnitude of the activation energy than the change in potential energy for the reaction?

  4. The product C in the reaction A + B → C + D can be separated easily from the reaction mixture. You have been given pure A and pure B and are told to determine the activation energy for this reaction to determine whether the reaction is suitable for the industrial synthesis of C. How would you do this? Why do you need to know the magnitude of the activation energy to make a decision about feasibility?

  5. Above Ea, molecules collide with enough energy to overcome the energy barrier for a reaction. Is it possible for a reaction to occur at a temperature less than that needed to reach Ea? Explain your answer.

  6. What is the relationship between A, Ea, and T? How does an increase in A affect the reaction rate?

  7. Of two highly exothermic reactions with different values of Ea, which would need to be monitored more carefully: the one with the smaller value or the one with the higher value? Why?

Numerical Problems

  1. What happens to the approximate rate of a reaction when the temperature of the reaction is increased from 20°C to 30°C? What happens to the reaction rate when the temperature is raised to 70°C? For a given reaction at room temperature (20°C), what is the shape of a plot of reaction rate versus temperature as the temperature is increased to 70°C?

  2. Acetaldehyde, used in silvering mirrors and some perfumes, undergoes a second-order decomposition between 700 and 840 K. From the data in the following table, would you say that acetaldehyde follows the general rule that each 10 K increase in temperature doubles the reaction rate?

    T (K) k (M−1·s−1)
    720 0.024
    740 0.051
    760 0.105
    800 0.519
  3. Bromoethane reacts with hydroxide ion in water to produce ethanol. The activation energy for this reaction is 90 kJ/mol. If the reaction rate is 3.6 × 10−5 M/s at 25°C, what would the reaction rate be at the following temperatures?

    1. 15°C
    2. 30°C
    3. 45°C
  4. An enzyme-catalyzed reaction has an activation energy of 15 kcal/mol. How would the value of the rate constant differ between 20°C and 30°C? If the enzyme reduced the Ea from 25 kcal/mol to 15 kcal/mol, by what factor has the enzyme increased the reaction rate at each temperature?

  5. The data in the following table are the rate constants as a function of temperature for the dimerization of 1,3-butadiene. What is the activation energy for this reaction?

    T (K) k (M−1·min−1)
    529 1.4
    560 3.7
    600 25
    645 82
  6. The reaction rate at 25°C is 1.0 × 10−4 M/s. Increasing the temperature to 75°C causes the reaction rate to increase to 7.0 × 10−2 M/s. Estimate Ea for this process. If Ea were 25 kJ/mol and the reaction rate at 25°C is 1.0 × 10−4 M/s, what would be the reaction rate at 75°C?

Answers

  1. The reaction rate will approximately double: 20°C to 30°C, the reaction rate increases by about 21 = 2; 20°C to 70°C, the reaction rate increases by about 25 = 32-fold. A plot of reaction rate versus temperature will give an exponential increase: rate ∝ 2ΔT/10.

    1. 1.0 × 10−5 M/s
    2. 6.6 × 10−5 M/s
    3. 3.5 × 10−4 M/s
  2. 100 kJ/mol

14.8 Catalysis

Learning Objective

  1. To understand how catalysts increase the reaction rate and the selectivity of chemical reactions.

described catalystsA substance that participates in a reaction and causes it to occur more rapidly but that can be recovered unchanged at the end of the reaction and reused. Catalysts may also control which products are formed in a reaction. as substances that increase the reaction rate of a chemical reaction without being consumed in the process. A catalyst, therefore, does not appear in the overall stoichiometry of the reaction it catalyzes, but it must appear in at least one of the elementary reactions in the mechanism for the catalyzed reaction. The catalyzed pathway has a lower Ea, but the net change in energy that results from the reaction (the difference between the energy of the reactants and the energy of the products) is not affected by the presence of a catalyst (). Nevertheless, because of its lower Ea, the reaction rate of a catalyzed reaction is faster than the reaction rate of the uncatalyzed reaction at the same temperature. Because a catalyst decreases the height of the energy barrier, its presence increases the reaction rates of both the forward and the reverse reactions by the same amount. In this section, we will examine the three major classes of catalysts: heterogeneous catalysts, homogeneous catalysts, and enzymes.

Note the Pattern

A catalyst affects Ea, not ΔE.

Figure 14.26 Lowering the Activation Energy of a Reaction by a Catalyst

This graph compares potential energy diagrams for a single-step reaction in the presence and absence of a catalyst. The only effect of the catalyst is to lower the activation energy of the reaction. The catalyst does not affect the energy of the reactants or products (and thus does not affect ΔE).

Heterogeneous Catalysis

In heterogeneous catalysisA catalytic reaction in which the catalyst is in a different phase from the reactants., the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called adsorption in such a way that a chemical bond in the reactant becomes weak and then breaks. Poisons are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency.

An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in , the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called desorption. Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H2 is substantially lower on the catalyst surface.

Figure 14.27 Hydrogenation of Ethylene on a Heterogeneous Catalyst

When a molecule of hydrogen adsorbs to the catalyst surface, the H–H bond breaks, and new M–H bonds are formed. The individual H atoms are more reactive than gaseous H2. When a molecule of ethylene interacts with the catalyst surface, it reacts with the H atoms in a stepwise process to eventually produce ethane, which is released.

shows a process called hydrogenation, in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter.

Several important examples of industrial heterogeneous catalytic reactions are in . Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface.

Table 14.8 Some Commercially Important Reactions that Employ Heterogeneous Catalysts

Commercial Process Catalyst Initial Reaction Final Commercial Product
contact process V2O5 or Pt 2SO2 + O2 → 2SO3 H2SO4
Haber process Fe, K2O, Al2O3 N2 + 3H2 → 2NH3 NH3
Ostwald process Pt and Rh 4NH3 + 5O2 → 4NO + 6H2O HNO3
water–gas shift reaction Fe, Cr2O3, or Cu CO + H2O → CO2 + H2 H2 for NH3, CH3OH, and other fuels
steam reforming Ni CH4 + H2O → CO + 3H2 H2
methanol synthesis ZnO and Cr2O3 CO + 2H2 → CH3OH CH3OH
Sohio process bismuth phosphomolybdate CH2=CHCH3+NH3+32O2CH2=CHCN+3H2O CH2=CHCNacrylonitrile
catalytic hydrogenation Ni, Pd, or Pt RCH=CHR′ + H2 → RCH2—CH2R′ partially hydrogenated oils for margarine, and so forth

Homogeneous Catalysis

In homogeneous catalysisA catalytic reaction in which the catalyst is uniformly dispersed throughout the reactant mixture to form a solution., the catalyst is in the same phase as the reactant(s). The number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture. Many homogeneous catalysts in industry are transition metal compounds (), but recovering these expensive catalysts from solution has been a major challenge. As an added barrier to their widespread commercial use, many homogeneous catalysts can be used only at relatively low temperatures, and even then they tend to decompose slowly in solution. Despite these problems, a number of commercially viable processes have been developed in recent years. High-density polyethylene and polypropylene are produced by homogeneous catalysis.

Table 14.9 Some Commercially Important Reactions that Employ Homogeneous Catalysts

Commercial Process Catalyst Reactants Final Product
Union Carbide [Rh(CO)2I2] CO + CH3OH CH3CO2H
hydroperoxide process Mo(VI) complexes CH3CH=CH2 + R–O–O–H

 

hydroformylation Rh/PR3 complexes RCH=CH2 + CO + H2 RCH2CH2CHO
adiponitrile process Ni/PR3 complexes 2HCN + CH2=CHCH=CH2 NCCH2CH2CH2CH2CN used to synthesize nylon
olefin polymerization (RC5H5)2ZrCl2 CH2=CH2 –(CH2CH2–)n: high-density polyethylene

Enzymes

Enzymes, catalysts that occur naturally in living organisms, are almost all protein molecules with typical molecular masses of 20,000–100,000 amu. Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism. Others are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings. The reactant in an enzyme-catalyzed reaction is called a substrateThe reactant in an enzyme-catalyzed reaction..

Because enzymes can increase reaction rates by enormous factors (up to 1017 times the uncatalyzed rate) and tend to be very specific, typically producing only a single product in quantitative yield, they are the focus of active research. At the same time, enzymes are usually expensive to obtain, they often cease functioning at temperatures greater than 37°C, have limited stability in solution, and have such high specificity that they are confined to turning one particular set of reactants into one particular product. This means that separate processes using different enzymes must be developed for chemically similar reactions, which is time-consuming and expensive. Thus far, enzymes have found only limited industrial applications, although they are used as ingredients in laundry detergents, contact lens cleaners, and meat tenderizers. The enzymes in these applications tend to be proteases, which are able to cleave the amide bonds that hold amino acids together in proteins. Meat tenderizers, for example, contain a protease called papain, which is isolated from papaya juice. It cleaves some of the long, fibrous protein molecules that make inexpensive cuts of beef tough, producing a piece of meat that is more tender. Some insects, like the bombadier beetle, carry an enzyme capable of catalyzing the decomposition of hydrogen peroxide to water ().

Enzyme inhibitorsSubstances that decrease the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of the enzyme, thus slowing or preventing a reaction from occurring. cause a decrease in the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring. Irreversible inhibitors are therefore the equivalent of poisons in heterogeneous catalysis. One of the oldest and most widely used commercial enzyme inhibitors is aspirin, which selectively inhibits one of the enzymes involved in the synthesis of molecules that trigger inflammation. The design and synthesis of related molecules that are more effective, more selective, and less toxic than aspirin are important objectives of biomedical research.

Figure 14.28 A Catalytic Defense Mechanism

The scalding, foul-smelling spray emitted by this bombardier beetle is produced by the catalytic decomposition of H2O2.

Summary

Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts that produce large increases in reaction rates and tend to be specific for certain reactants and products. The reactant in an enzyme-catalyzed reaction is called a substrate. Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction.

Key Takeaway

  • Catalysts allow a reaction to proceed via a pathway that has a lower activation energy.

Conceptual Problems

  1. What effect does a catalyst have on the activation energy of a reaction? What effect does it have on the frequency factor (A)? What effect does it have on the change in potential energy for the reaction?

  2. How is it possible to affect the product distribution of a reaction by using a catalyst?

  3. A heterogeneous catalyst works by interacting with a reactant in a process called adsorption. What occurs during this process? Explain how this can lower the activation energy.

  4. What effect does increasing the surface area of a heterogeneous catalyst have on a reaction? Does increasing the surface area affect the activation energy? Explain your answer.

  5. Identify the differences between a heterogeneous catalyst and a homogeneous catalyst in terms of the following.

    1. ease of recovery
    2. collision frequency
    3. temperature sensitivity
    4. cost
  6. An area of intensive chemical research involves the development of homogeneous catalysts, even though homogeneous catalysts generally have a number of operational difficulties. Propose one or two reasons why a homogenous catalyst may be preferred.

  7. Consider the following reaction between cerium(IV) and thallium(I) ions:

    2Ce4+ + Tl+ → 2Ce3+ + Tl3+

    This reaction is slow, but Mn2+ catalyzes it, as shown in the following mechanism:

    Ce4+ + Mn2+ → Ce3+ + Mn3+ Ce4+ + Mn3+ → Ce3+ + Mn4+ Mn4+ + Tl+ → Tl3+ + Mn2+

    In what way does Mn2+ increase the reaction rate?

  8. The text identifies several factors that limit the industrial applications of enzymes. Still, there is keen interest in understanding how enzymes work for designing catalysts for industrial applications. Why?

  9. Most enzymes have an optimal pH range; however, care must be taken when determining pH effects on enzyme activity. A decrease in activity could be due to the effects of changes in pH on groups at the catalytic center or to the effects on groups located elsewhere in the enzyme. Both examples are observed in chymotrypsin, a digestive enzyme that is a protease that hydrolyzes polypeptide chains. Explain how a change in pH could affect the catalytic activity due to (a) effects at the catalytic center and (b) effects elsewhere in the enzyme. (Hint: remember that enzymes are composed of functional amino acids.)

Answers

  1. A catalyst lowers the activation energy of a reaction. Some catalysts can also orient the reactants and thereby increase the frequency factor. Catalysts have no effect on the change in potential energy for a reaction.

  2. In adsorption, a reactant binds tightly to a surface. Because intermolecular interactions between the surface and the reactant weaken or break bonds in the reactant, its reactivity is increased, and the activation energy for a reaction is often decreased.

    1. Heterogeneous catalysts are easier to recover.
    2. Collision frequency is greater for homogeneous catalysts.
    3. Homogeneous catalysts are often more sensitive to temperature.
    4. Homogeneous catalysts are often more expensive.
  3. The Mn2+ ion donates two electrons to Ce4+, one at a time, and then accepts two electrons from Tl+. Because Mn can exist in three oxidation states separated by one electron, it is able to couple one-electron and two-electron transfer reactions.

Numerical Problems

  1. At some point during an enzymatic reaction, the concentration of the activated complex, called an enzyme–substrate complex (ES), and other intermediates involved in the reaction is nearly constant. When a single substrate is involved, the reaction can be represented by the following sequence of equations:

    enzyme (E) + substrate (S)enzyme-substrate complex (ES)enzyme (E) + product (P)

    This can also be shown as follows:

    E + Sk1k1ESk2k2E+P

    Using molar concentrations and rate constants, write an expression for the rate of disappearance of the enzyme–substrate complex. Typically, enzyme concentrations are small, and substrate concentrations are high. If you were determining the rate law by varying the substrate concentrations under these conditions, what would be your apparent reaction order?

  2. A particular reaction was found to proceed via the following mechanism:

    A + B → C + D 2C → E E + A → 3B + F

    What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates.

  3. A particular reaction has two accessible pathways (A and B), each of which favors conversion of X to a different product (Y and Z, respectively). Under uncatalyzed conditions pathway A is favored, but in the presence of a catalyst pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst.

  4. The kinetics of an enzyme-catalyzed reaction can be analyzed by plotting the reaction rate versus the substrate concentration. This type of analysis is referred to as a Michaelis–Menten treatment. At low substrate concentrations, the plot shows behavior characteristic of first-order kinetics, but at very high substrate concentrations, the behavior shows zeroth-order kinetics. Explain this phenomenon.

Answers

  1. Δ[ES]Δt=(k2+k1)[ES]+k1[E][S]+k2[E][P]0; zeroth order in substrate.

  2. In both cases, the product of pathway A is favored. All of the Z produced in the catalyzed reversible pathway B will eventually be converted to X as X is converted irreversibly to Y by pathway A.

    ZBXAY

14.9 End-of-Chapter Material

Application Problems

    Problems marked with a ♦ involve multiple concepts.

  1. Atmospheric chemistry in the region below the clouds of Venus appears to be dominated by reactions of sulfur and carbon-containing compounds. Included in representative elementary reactions are the following:

    SO2 + CO → SO + CO2 SO + CO → S + CO2 SO + SO2 → S + SO3

    For each elementary reaction, write an expression for the net rate of reaction in terms of the concentrations of reactants and products.

  2. In acid, nitriles hydrolyze to produce a carboxylic acid and ammonium ion. For example, acetonitrile, a substance used to extract fatty acids from fish liver oils, is hydrolyzed to acetic acid via the following reaction:

    Express the reaction rate in terms of changes in the concentrations of each reactant and each product with time.

  3. ♦ Ozone production occurs at lower altitudes according to the elementary reaction O + O2 → O3, with an estimated rate of ozone production of 4.86 × 1031 molecules·s−1 worldwide. What is the overall reaction order? If the reaction rate of loss of O3 due to absorption of UV light () is 0.89 × 1031 molecules·s−1, and 0.06 × 1031 molecules·s−1 of ozone is transported to other atmospheric regions, is ozone being produced faster than it is being destroyed? Measurements show that ozone concentrations are not increasing rapidly. What conclusion can you draw from these data?

  4. ♦ The water in a fishery became polluted when toxic waste was dumped into its pond, causing the fish population to substantially decline. The percentage of fish that survived is recorded in the following table.

    Day 1 2 3 4 5
    % survival 79 55 38 31 19

    What is the reaction order of live fish → dead fish? What is the rate constant? If the fish continue to die at this rate, how many fish will be alive after 10 days?

  5. Until 200 yr ago, manufactured iron contained charcoal produced from freshly cut wood that was added during the smelting process. As a result of this practice, older samples of iron can be dated accurately using the carbon-14 method. An archaeologist found a cast iron specimen that she believed dated to the period between 480 and 221 BC in Hunan, China. Radiocarbon dating of the sample indicated a 24% reduction in carbon-14 content. Was the archaeologist correct?

  6. ♦ Because of its short half-life, 32P-labeled compounds must be shipped as quickly as possible so that they can be used as radioactive tags in biological studies. A 50 g sample that contained 0.60% 32P by mass was shipped at 11 a.m. on Monday morning. The package was delivered to a chemist via an overnight delivery service such that it arrived the next day.

    1. What would be the mass of 32P remaining in the sample if he received the package on Tuesday afternoon but was unable to use it until 9 a.m. on Wednesday?
    2. What would be the mass of 32P present in the sample if the shipper had not delivered the sample until Friday afternoon and then it sat on a loading dock until 9 a.m. on Monday morning?
    3. The late shipment was used immediately on Monday morning, but the biological samples were not analyzed until Thursday at 5 p.m. What percentage of the sample still consists of 32P?
  7. ♦ Tritium (3H) is a radioactive isotope that is commonly used to follow biochemical reactions.

    1. Using the data in , calculate the radioactive decay constant (k) for tritium.
    2. Use the value of k to determine the mass of tritium that is still present in a 5.00 g sample of NaB3H4 that is 17.57 yr old.
  8. ♦ L-Aspartic acid is an amino acid found in fossil bone. It can convert to a geometrically different form (D-aspartic acid) at 20°C, with a half-life corresponding to the conversion of L → D of 14,000–20,000 yr. If the temperature of an archaeological site is constant, then the extent of the conversion can be used to date fossils. In one such case, archaeologists dated the arrival of humans on the North American continent to be 20,000 yr ago, but the conversion of L-aspartic acid to D-aspartic acid in human fossils indicated that Paleo-Indians were living in California at least 48,000 yr ago. What would be the relative concentrations of the L- and D-forms that produced this result? Carbon-14 has a half-life of approximately 5730 yr. What percentage of the carbon-14 originally present would have been found in the bones?

    The technique described is frequently used in conjunction with radiocarbon dating. In cases where the results from the two techniques are in gross disagreement, what information can you get by comparing the two results?

  9. ♦ Peroxides are able to initiate the radical polymerization of alkenes. Polyethylene, for example, is a high-molecular-weight polymer used as a film in packaging, as kitchenware, and as tubing. It is produced by heating ethylene at high pressure in the presence of oxygen or peroxide. It is formed by the following radical process:

    RO:ORΔ2RO RO+CH2=CH2ROCH2CH2 RO―CH2―CH2· + CH2=CH2 → RO―CH2―CH2―CH2―CH2·
    1. Label the steps that correspond to initiation and propagation.
    2. Show all available chain-terminating steps.
    3. The polymerization of styrene (C6H5CH=CH2) occurs by a similar process to produce polystyrene, which is used as a packaging material. Draw the structure of the polymer that results from five propagation cycles.
  10. Lucite and Plexiglas are transparent polymers used as a glass substitute when a plastic material is preferred for safety. The compound used to synthesize Lucite and Plexiglas is methyl methacrylate, which is shown here. During the polymerization reaction, light produces a radical initiator from hydrogen peroxide (H2O2 → 2HO·). Show the mechanism for the polymerization, being sure to include the initiation and propagation steps.

  11. ♦ At higher altitudes ozone is converted to O2 by the reaction O + O3 → 2O2, with a rate constant at 220 K of 6.8 × 10−16 cm3·molecule−1·s−1.

    1. What is the overall reaction order?
    2. What is Ea for this reaction if A = 8 × 10−12 cm3·molecule−1·s−1?

      If Cl is present, the rate constant at 220 K becomes 3.7 × 10−11 cm3·molecule−1 · s−1, with A = 4.7 × 10−11 cm3·molecule−1·s−1.

    3. Calculate Ea for the depletion of ozone in the presence of Cl.
    4. Show an energy-level diagram for these two processes, clearly labeling reactants, products, and activation energies.
    5. If you were an environmental scientist using these data to explain the effects of Cl on ozone concentration, what would be your conclusions?
  12. ♦ Nitric acid is produced commercially by the catalytic oxidation of ammonia by air over platinum gauze at approximately 900°C. The following reactions occur:

    NH3(g)+54O2(g)NO(g)+32H2O(g)ΔH°=226.3 kJ/molNO(g) + 12O2(g)NO2(g)ΔH°=57.1 kJ/mol3NO2(g)+H2O(g)2HNO3(l)+NO(g)ΔH°=71.7 kJ/mol

    Why is platinum gauze rather than platinum wire used for the initial reaction? The reaction 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g) has ΔH° = −316.6 kJ/mol. What would occur if the catalyst were not present? If the gas leaving the catalyst is not free of NH3, the following reaction takes place: 6NO(g) + 4NH3(g) → 5N2(g) + 6H2O(g). If this occurs, what will be the overall reaction?

  13. illustrates the mechanism for the reduction of ethylene on a platinum surface to produce ethane. Industrially important silanes are synthesized using a related mechanism and are used to increase adhesion between layers of glass fiber and between layers of silicone rubber. Predict the products of the following reactions:

  14. ♦ In catalysis, if a molecule forms strong bonds to the catalyst, then the catalyst may become poisoned. Experiments on various catalysts showed the following results:

    1. Fe, Ru, and Os form weak bonds with N2; however, O2, alkynes, alkenes, CO, H2, and CO2 interact more strongly.
    2. CO2 and H2 form weak bonds with a Co or Ni surface.
    3. Rh, Pd, Ir, and Pt form weak bonds with H2 but do not bond with CO2.
    4. Cu, Ag, and Au form weak bonds with CO and ethylene.
    1. Explain why Fe was chosen as a catalyst to convert nitrogen and hydrogen to ammonia. Why is Fe more suitable than Ru or Os?
    2. Because alkenes generally interact more strongly with metal surfaces than does H2, what catalyst would you choose for hydrogenation of an alkene such as ethylene?
    3. Although platinum is used in catalytic converters for automobile exhaust, it was not found to be a particularly effective catalyst for the reaction of H2 with a mixture of carbon monoxide and carbon dioxide to produce methane. Why?
    4. If you were interested in developing a catalyst to reversibly bind ethylene, which of the catalysts listed here would you choose?
  15. Nonstoichiometric metal oxides can be effective catalysts for oxidation–reduction reactions. One such catalyst is Ni1−xO, found to be effective for converting CO to CO2 when oxygen is present. Why is it so effective?

  16. The chemical reactions in an organism can be controlled by regulating the activity of certain enzymes. Efficient regulation results in an enzyme being active only when it is needed. For example, if a cell needed histidine, the nine enzymes needed to synthesize histidine would all be active. If the cell had adequate histidine, however, those enzymes would be inactive. The following diagram illustrates a situation in which three amino acids (D, F, H) are all synthesized from a common species, A. The numbers above the arrows refer to the enzymes that catalyze each step. Which enzymes would need to be regulated to produce D? F? H?

  17. ♦ Because phosphorus-32 is incorporated into deoxyribonucleic acid (DNA), it can be used to detect DNA fragments. Consequently, it is used extensively in biological research, including the Human Genome Project, whose goal was to determine the complete sequence of human DNA. If you were to start with a 20 g sample of phosphorus that contained 10% 32P by mass, converted it into DNA via several chemical steps that had an overall yield of 75% and took 25 days, and then incorporated it into bacteria and allowed them to grow for 5 more days, what mass of 32P would be available for analysis at the end of this time?

  18. The enzyme urease contains two atoms of nickel and catalyzes the hydrolysis of urea by the following reaction:

    H2NC(O)NH2 + H2O → 2NH3 + CO2

    Urease is one of the most powerful catalysts known. It lowers the activation energy for the hydrolysis of urea from 137 kJ/mol to only 37 kJ/mol. Calculate the ratio of the reaction rate of the catalyzed reaction to the reaction rate of the uncatalyzed reaction at 37°C. Assume that the frequency factor is the same for both reactions.

  19. As noted in , the reaction rate for the hydrogenation of ethylene to give ethane can be increased by heterogeneous catalysts such as Pt or Ni:

    H2(g)+H2C=CH2(g)Pt,NiH3C–CH3(g)

    The activation energy for the uncatalyzed reaction is large (188 kJ/mol), so the reaction is very slow at room temperature. In the presence of finely divided metallic Ni, the activation energy is only 84 kJ/mol. Calculate the ratio of the reaction rate of the catalyzed reaction to the reaction rate of the uncatalyzed reaction at 75°C.

Answers

  1. rate = kf[SO2][CO] − kr[SO][CO2]; rate = kf[SO][CO] − kr[S][CO2]; rate = kf[SO][SO2] − kr[S][SO3]

  2. The reaction is second order: first order in O and first order in O3. Ozone is being produced faster than it is being destroyed. If ozone concentrations are not increasing, then either some other reaction must be consuming some of the ozone produced in this reaction or the ozone-producing reaction does not operate at this rate continuously.

  3. Yes; the object is about 2300 yr old.

    1. k = 0.05626 yr−1
    2. 0.487 g of 3H
    1. second order, first order in O and first order in O3;
    2. 17 kJ/mol;
    3. 0.44 kJ/mol;
    4.  

    5. Cl is a potent catalyst for ozone destruction because there is a large decrease in Ea when Cl is present.
  4. Ni1−xO is a nonstoichiometric oxide that contains a fraction of Ni(I) sites. These can react with oxygen to form a Ni(III)-oxide site, which is reduced by CO to give Ni(I) and CO2.

  5. 0.35 g of 32P

  6. 4.1 × 1015

Chapter 15 Chemical Equilibrium

In , we discussed the principles of chemical kinetics, which deal with the rate of change, or how quickly a given chemical reaction occurs. We now turn our attention to the extent to which a reaction occurs and how reaction conditions affect the final concentrations of reactants and products. For most of the reactions that we have discussed so far, you may have assumed that once reactants are converted to products, they are likely to remain that way. In fact, however, virtually all chemical reactions are reversible to some extent. That is, an opposing reaction occurs in which the products react, to a greater or lesser degree, to re-form the reactants. Eventually, the forward and reverse reaction rates become the same, and the system reaches chemical equilibriumThe point at which the forward and reverse reaction rates become the same so that the net composition of the system no longer changes with time., the point at which the composition of the system no longer changes with time.

A smoggy sunset in Shenzhen, China. The reaction of O2 with N2 at high temperature in an internal combustion engine produces small amounts of NO, which reacts with atmospheric O2 to form NO2, an important component of smog. The reddish-brown color of NO2 is responsible for the characteristic color of smog, as shown in this true-color photo.

We introduced the concept of equilibrium in , where you learned that a liquid and a vapor are in equilibrium when the number of molecules evaporating from the surface of the liquid per unit time is the same as the number of molecules condensing from the vapor phase. Vapor pressure is an example of a physical equilibrium because only the physical form of the substance changes. Similarly, in , we discussed saturated solutions, another example of a physical equilibrium, in which the rate of dissolution of a solute is the same as the rate at which it crystallizes from solution.

In this chapter, we describe the methods chemists use to quantitatively describe the composition of chemical systems at equilibrium, and we discuss how factors such as temperature and pressure influence the equilibrium composition. As you study these concepts, you will also learn how urban smog forms and how reaction conditions can be altered to produce H2 rather than the combustion products CO2 and H2O from the methane in natural gas. You will discover how to control the composition of the gases emitted in automobile exhaust and how synthetic polymers such as the polyacrylonitrile used in sweaters and carpets are produced on an industrial scale.

15.1 The Concept of Chemical Equilibrium

Learning Objective

  1. To understand what is meant by chemical equilibrium.

Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide (N2O4) to nitrogen dioxide (NO2). You may recall from Chapter 14 "Chemical Kinetics" that NO2 is responsible for the brown color we associate with smog. When a sealed tube containing solid N2O4 (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of NO2 appears (Figure 15.1 "The "). The reaction can be followed visually because the product (NO2) is colored, whereas the reactant (N2O4) is colorless:

Equation 15.1

N2O(g)colorless2NO2(g)red-brown

The double arrow indicates that both the forward and reverse reactions are occurring simultaneously; it is read “is in equilibrium with.”

Figure 15.1 The N2O(g)2NO2(g) System at Different Temperatures

(left) At dry ice temperature (−78.4°C), the system contains essentially pure solid N2O4, which is colorless. (center) As the system is warmed above the melting point of N2O4 (−9.3°C), the N2O4 melts and then evaporates, and some of the vapor dissociates to red-brown NO2. (right) Eventually the sample reaches room temperature, and a mixture of gaseous N2O4 and NO2 is present. The composition of the mixture and hence the color do not change further with time: the system has reached equilibrium at the new temperature.

Figure 15.2 "The Composition of N" shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of NO2 were zero, then it increases as the concentration of N2O4 decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no N2O4 but an initial NO2 concentration twice the initial concentration of N2O4 in part (a) in Figure 15.2 "The Composition of N", in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition, as shown in part (b) in Figure 15.2 "The Composition of N". Thus equilibrium can be approached from either direction in a chemical reaction.

Figure 15.2 The Composition of N2O4/NO2 Mixtures as a Function of Time at Room Temperature

(a) Initially, this idealized system contains 0.0500 M gaseous N2O4 and no gaseous NO2. The concentration of N2O4 decreases with time as the concentration of NO2 increases. (b) Initially, this system contains 0.1000 M NO2 and no N2O4. The concentration of NO2 decreases with time as the concentration of N2O4 increases. In both cases, the final concentrations of the substances are the same: [N2O4] = 0.0422 M and [NO2] = 0.0156 M at equilibrium.

Figure 15.3 "The Forward and Reverse Reaction Rates as a Function of Time for the " shows the forward and reverse reaction rates for a sample that initially contains pure NO2. Because the initial concentration of N2O4 is zero, the forward reaction rate (dissociation of N2O4) is initially zero as well. In contrast, the reverse reaction rate (dimerization of NO2) is initially very high (2.0 × 106 M/s), but it decreases rapidly as the concentration of NO2 decreases. (Recall from Chapter 14 "Chemical Kinetics" that the reaction rate of the dimerization reaction is expected to decrease rapidly because the reaction is second order in NO2: rate = kr[NO2]2, where kr is the rate constant for the reverse reaction shown in Equation 15.1.) As the concentration of N2O4 increases, the rate of dissociation of N2O4 increases—but more slowly than the dimerization of NO2—because the reaction is only first order in N2O4 (rate = kf[N2O4], where kf is the rate constant for the forward reaction in Equation 15.1). Eventually, the forward and reverse reaction rates become identical, kF = kr, and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium.

Figure 15.3 The Forward and Reverse Reaction Rates as a Function of Time for the N2O4(g)2NO2(g) System Shown in Part (b) in Figure 15.2 "The Composition of N"

The rate of dimerization of NO2 (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of N2O4 is zero, the rate of the dissociation reaction (forward reaction) at t = 0 is also zero. As the dimerization reaction proceeds, the N2O4 concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of N2O4 and NO2 no longer change.

Note the Pattern

At equilibrium, the forward reaction rate is equal to the reverse reaction rate.

Example 1

The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation 2AB, where the blue circles are A and the purple ovals are B. Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium?

Given: three reaction systems

Asked for: relative time to reach chemical equilibrium

Strategy:

Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium.

Solution:

In systems 1 and 3, the concentration of A decreases from t0 through t2 but is the same at both t2 and t3. Thus systems 1 and 3 are at equilibrium by t3. In system 2, the concentrations of A and B are still changing between t2 and t3, so system 2 may not yet have reached equilibrium by t3. Thus system 2 took the longest to reach chemical equilibrium.

Exercise

In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals; the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium?

Answer: system 2

Summary

Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached.

Key Takeaway

  • At equilibrium, the forward and reverse reactions of a system proceed at equal rates.

Conceptual Problems

  1. What is meant when a reaction is described as “having reached equilibrium”? What does this statement mean regarding the forward and reverse reaction rates? What does this statement mean regarding the concentrations or amounts of the reactants and the products?

  2. Is it correct to say that the reaction has “stopped” when it has reached equilibrium? Explain your answer and support it with a specific example.

  3. Why is chemical equilibrium described as a dynamic process? Describe this process in the context of a saturated solution of NaCl in water. What is occurring on a microscopic level? What is happening on a macroscopic level?

  4. Which of these systems exists in a state of chemical equilibrium?

    1. oxygen and hemoglobin in the human circulatory system
    2. iodine crystals in an open beaker
    3. the combustion of wood
    4. the amount of 14C in a decomposing organism

Answer

  1. Both forward and reverse reactions occur but at the same rate. Na+ and Cl ions continuously leave the surface of an NaCl crystal to enter solution, while at the same time Na+ and Cl ions in solution precipitate on the surface of the crystal.

15.2 The Equilibrium Constant

Learning Objectives

  1. To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions.
  2. To write an equilibrium constant expression for any reaction.

Because an equilibrium state is achieved when the forward reaction rate equals the reverse reaction rate, under a given set of conditions there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction (represented by rate constants). We can show this relationship using the system described in Equation 15.1, the decomposition of N2O4 to NO2. Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows:

Equation 15.2

forward rate = kf[N2O4]

Equation 15.3

reverse rate = kr[NO2]2

At equilibrium, the forward rate equals the reverse rate:

Equation 15.4

kf[N2O4] = kr[NO2]2

so

Equation 15.5

kfkr=[NO2]2[N2O4]

The ratio of the rate constants gives us a new constant, the equilibrium constant (K)The ratio of the rate constants for the forward reaction and the reverse reaction; that is, K=kf/kr. It is also the equilibrium constant calculated from solution concentrations: K=[C]c[D]d/[A]a[B]b for the general reaction aA+bBcC+dD, in which each component is in solution., which is defined as follows:

Equation 15.6

K=kfkr

Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: under a given set of conditions, the composition of the equilibrium mixture is determined by the magnitudes of the rate constants for the forward and the reverse reactions.

Note the Pattern

The equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction.

Table 15.1 "Initial and Equilibrium Concentrations for " lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation 15.1. At equilibrium the magnitude of the quantity [NO2]2/[N2O4] is essentially the same for all five experiments. In fact, no matter what the initial concentrations of NO2 and N2O4 are, at equilibrium the quantity [NO2]2/[N2O4] will always be 6.53 ± 0.03 × 10−3 at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions. That is, at a given temperature, the equilibrium constant for a reaction always has the same value, even though the specific concentrations of the reactants and products vary depending on their initial concentrations.

Table 15.1 Initial and Equilibrium Concentrations for NO2/N204 Mixtures at 25°C

Initial Concentrations Concentrations at Equilibrium
Experiment [N2O4] (M) [NO2] (M) [N2O4] (M) [NO2] (M) K = [NO2]2/[N2O4]
1 0.0500 0.0000 0.0417 0.0165 6.54 × 10−3
2 0.0000 0.1000 0.0417 0.0165 6.54 × 10−3
3 0.0750 0.0000 0.0647 0.0206 6.56 × 10−3
4 0.0000 0.0750 0.0304 0.0141 6.54 × 10−3
5 0.0250 0.0750 0.0532 0.0186 6.50 × 10−3

Developing an Equilibrium Constant Expression

In 1864, the Norwegian chemists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for any reversible reaction of the general form

Equation 15.7

aA+bBcC+dD

where A and B are reactants, C and D are products, and a, b, c, and d are the stoichiometric coefficients in the balanced chemical equation for the reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced chemical equation) is always a constant under a given set of conditions. This relationship is known as the law of mass actionFor the general balanced chemical equation aA+bBcC+dD, the equilibrium constant expression is K=[C]c[D]d/[A]a[B]b. and can be stated as follows:

Equation 15.8

K=[C]c[D]d[A]a[B]b

where K is the equilibrium constant for the reaction. Equation 15.7 is called the equilibrium equationFor the general balanced chemical equation aA+bBcC+dD, the equilibrium constant expression is K=[C]c[D]d/[A]a[B]b., and the right side of Equation 15.8 is called the equilibrium constant expressionFor a balanced chemical equation, the ratio is [C]c[D]d/[A]a[B]b for the general reaction aA+bBcC+dD.. The relationship shown in Equation 15.8 is true for any pair of opposing reactions regardless of the mechanism of the reaction or the number of steps in the mechanism.

The equilibrium constant can vary over a wide range of values. The values of K shown in Table 15.2 "Equilibrium Constants for Selected Reactions*", for example, vary by 60 orders of magnitude. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of K greater than 103 indicate a strong tendency for reactants to form products. In this case, chemists say that equilibrium lies to the right as written, favoring the formation of products. An example is the reaction between H2 and Cl2 to produce HCl, which has an equilibrium constant of 1.6 × 1033 at 300 K. Because H2 is a good reductant and Cl2 is a good oxidant, the reaction proceeds essentially to completion. In contrast, values of K less than 10−3 indicate that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants.

Table 15.2 Equilibrium Constants for Selected Reactions*

Reaction Temperature (K) Equilibrium Constant (K)
S(s)+O2(g)SO2(g) 300 4.4 × 1053
2H2(g)+O2(g)2H2O(g) 500 2.4 × 1047
H2(g)+Cl2(g)2HCl(g) 300 1.6 × 1033
H2(g)+Br2(g)2HBr(g) 300 4.1 × 1018
2NO(g)+O2(g)2NO2(g) 300 4.2 × 1013
3H2(g)+N2(g)2NH3(g) 300 2.7 × 108
H2(g)+D2(g)2HD(g) 100 1.92
H2(g)+I2(g)2HI(g) 300 2.9 × 10−1
I2(g)2I(g) 800 4.6 × 10−7
Br2(g)2Br(g) 1000 4.0 × 10−7
Cl2(g)2Cl(g) 1000 1.8 × 10−9
F2(g)2F(g) 500 7.4 × 10−13
*Equilibrium constants vary with temperature. The K values shown are for systems at the indicated temperatures.

You will also notice in Table 15.2 "Equilibrium Constants for Selected Reactions*" that equilibrium constants have no units, even though Equation 15.8 suggests that the units of concentration might not always cancel because the exponents may vary. In fact, equilibrium constants are calculated using “effective concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M. As shown in Equation 15.9, the units of concentration cancel, which makes K unitless as well:

Equation 15.9

[A]measured[A]standard state=MM=mol/Lmol/L

Many reactions have equilibrium constants between 1000 and 0.001 (103K ≥ 10−3), neither very large nor very small. At equilibrium, these systems tend to contain significant amounts of both products and reactants, indicating that there is not a strong tendency to form either products from reactants or reactants from products. An example of this type of system is the reaction of gaseous hydrogen and deuterium, a component of high-stability fiber-optic light sources used in ocean studies, to form HD:

Equation 15.10

H2(g)+D2(g)2HD(g)

The equilibrium constant expression for this reaction is [HD]2/[H2][D2], and K is between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of H2, D2, and HD contains significant concentrations of both product and reactants.

Figure 15.4 "The Relationship between the Composition of the Mixture at Equilibrium and the Magnitude of the Equilibrium Constant" summarizes the relationship between the magnitude of K and the relative concentrations of reactants and products at equilibrium for a general reaction, written as reactantsproducts. Because there is a direct relationship between the kinetics of a reaction and the equilibrium concentrations of products and reactants (Equation 15.9 and Equation 15.8), when kf >> kr, K is a large number, and the concentration of products at equilibrium predominate. This corresponds to an essentially irreversible reaction. Conversely, when kf << kr, K is a very small number, and the reaction produces almost no products as written. Systems for which kfkr have significant concentrations of both reactants and products at equilibrium.

Figure 15.4 The Relationship between the Composition of the Mixture at Equilibrium and the Magnitude of the Equilibrium Constant

The larger the K, the farther the reaction proceeds to the right before equilibrium is reached, and the greater the ratio of products to reactants at equilibrium.

Note the Pattern

A large value of the equilibrium constant K means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium.

Example 2

Write the equilibrium constant expression for each reaction.

  1. N2(g)+3H2(g)2NH3(g)
  2. CO(g)+12O2(g)CO2(g)
  3. 2CO2(g)2CO(g)+O2(g)

Given: balanced chemical equations

Asked for: equilibrium constant expressions

Strategy:

Refer to Equation 15.8. Place the arithmetic product of the concentrations of the products (raised to their stoichiometric coefficients) in the numerator and the product of the concentrations of the reactants (raised to their stoichiometric coefficients) in the denominator.

Solution:

  1. The only product is ammonia, which has a coefficient of 2. For the reactants, N2 has a coefficient of 1 and H2 has a coefficient of 3. The equilibrium constant expression is as follows:

    [NH3]2[N2][H2]3
  2. The only product is carbon dioxide, which has a coefficient of 1. The reactants are CO, with a coefficient of 1, and O2, with a coefficient of 12. Thus the equilibrium constant expression is as follows:

    [CO2][CO][O2]1/2
  3. This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for O2. The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2:

    [CO]2[O2][CO2]2

Exercise

Write the equilibrium constant expression for each reaction.

  1. N2O(g)N2(g)+12O2(g)
  2. 2C8H18(g)+25O2(g)16CO2(g)+18H2O(g)
  3. H2(g)+I2(g)2HI(g)

Answer:

  1. [N2][O2]1/2[N2O]
  2. [CO2]16[H2O]18[C8H18]2[O2]25
  3. [HI]2[H2][I2]

Example 3

Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants.

  1. H2(g)+I2(g)2HI(g)K(700K)=54
  2. 2CO2(g)2CO(g)+O2(g)K(1200K)=3.1×1018
  3. PCl5(g)PCl3(g)+Cl2(g)K(613K)=97
  4. 2O3(g)3O2(g)K(298 K)=5.9×1055

Given: systems and values of K

Asked for: composition of systems at equilibrium

Strategy:

Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both.

Solution:

  1. Only system 4 has K >> 103, so at equilibrium it will consist of essentially only products.
  2. System 2 has K << 10−3, so the reactants have little tendency to form products under the conditions specified; thus, at equilibrium the system will contain essentially only reactants.
  3. Both systems 1 and 3 have equilibrium constants in the range 103 ≥ K ≥ 10−3, indicating that the equilibrium mixtures will contain appreciable amounts of both products and reactants.

Exercise

Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation:

3H2(g)+N2(g)2NH3(g)

Values of the equilibrium constant at various temperatures were reported as K25°C = 3.3 × 108, K177°C = 2.6 × 103, and K327°C = 4.1.

  1. At which temperature would you expect to find the highest proportion of H2 and N2 in the equilibrium mixture?
  2. Assuming that the reaction rates are fast enough so that equilibrium is reached quickly, at what temperature would you design a commercial reactor to operate to maximize the yield of ammonia?

Answer:

  1. 327°C, where K is smallest
  2. 25°C

Variations in the Form of the Equilibrium Constant Expression

Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. For example, if we write the reaction described in Equation 15.7 in reverse, we obtain the following:

Equation 15.11

cC+dDaA+bB

The corresponding equilibrium constant K′ is as follows:

Equation 15.12

K=[A]a[B]b[C]c[D]d

This expression is the inverse of the expression for the original equilibrium constant, so K′ = 1/K. That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction N2O42NO2 is as follows:

Equation 15.13

K=[NO2]2[N2O4]

but for the opposite reaction, 2NO2N2O4, the equilibrium constant K′ is given by the inverse expression:

Equation 15.14

K=[N2O4][NO2]2

Consider another example, the formation of water: 2H2(g)+O2(g)2H2O(g). Because H2 is a good reductant and O2 is a good oxidant, this reaction has a very large equilibrium constant (K = 2.4 × 1047 at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form O2 and H2, is very small: K′ = 1/K = 1/(2.4 × 1047) = 4.2 × 10−48. As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into H2 and O2.

Note the Pattern

The equilibrium constant for a reaction written in reverse is the inverse of the equilibrium constant for the reaction as written originally.

Writing an equation in different but chemically equivalent forms also causes both the equilibrium constant expression and the magnitude of the equilibrium constant to be different. For example, we could write the equation for the reaction 2NO2N2O4 as NO212N2O4, for which the equilibrium constant K″ is as follows:

Equation 15.15

K=[N2O4]1/2[NO2]

The values for K′ (Equation 15.14) and K″ are related as follows:

Equation 15.16

K=(K)1/2=K

In general, if all the coefficients in a balanced chemical equation are subsequently multiplied by n, then the new equilibrium constant is the original equilibrium constant raised to the nth power.

Example 4

At 745 K, K is 0.118 for the following reaction:

N2(g)+3H2(g)2NH3(g)

What is the equilibrium constant for each related reaction at 745 K?

  1. 2NH3(g)N2(g)+3H2(g)
  2. 12N2(g)+32H2(g)NH3(g)

Given: balanced equilibrium equation, K at a given temperature, and equations of related reactions

Asked for: values of K for related reactions

Strategy:

Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate K for each reaction.

Solution:

The equilibrium constant expression for the given reaction of N2(g) with H2(g) to produce NH3(g) at 745 K is as follows:

K=[NH3]2[N2][H2]3=0.118
  1. This reaction is the reverse of the one given, so its equilibrium constant expression is as follows:

    K=1K=[N2][H2]3[NH3]2=10.118=8.47
  2. In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows:

    K=[NH3][N2]1/2[H2]3/2=K1/2=K=0.118=0.344

Exercise

At 527°C, the equilibrium constant for the reaction

2SO2(g)+O2(g)2SO3(g)

is 7.9 × 104. Calculate the equilibrium constant for the following reaction at the same temperature:

SO3(g)SO2(g)+12O2(g)

Answer: 3.6 × 10−3

Equilibrium Constant Expressions for Systems that Contain Gases

For reactions that involve species in solution, the concentrations used in equilibrium calculations are usually expressed in moles/liter. For gases, however, the concentrations are usually expressed in terms of partial pressures rather than molarity, where the standard state is 1 atm of pressure. The symbol KpAn equilibrium constant expressed as the ratio of the partial pressures of the products and reactants, each raised to its coefficient in the chemical equation. is used to denote equilibrium constants calculated from partial pressures. For the general reaction aA+bBcC+dD, in which all the components are gases, we can write the equilibrium constant expression as the ratio of the partial pressures of the products and reactants (each raised to its coefficient in the chemical equation):

Equation 15.17

Kp=(PC)c(PD)d(PA)a(PB)b

Thus Kp for the decomposition of N2O4 (Equation 15.1) is as follows:

Equation 15.18

Kp=(PNO2)2PN2O4

Like K, Kp is a unitless quantity because the quantity that is actually used to calculate it is an “effective pressure,” the ratio of the measured pressure to a standard state of 1 bar (approximately 1 atm), which produces a unitless quantity.The “effective pressure” is called the fugacity, just as activity is the effective concentration.

Because partial pressures are usually expressed in atmospheres or mmHg, the molar concentration of a gas and its partial pressure do not have the same numerical value. Consequently, the numerical values of K and Kp are usually different. They are, however, related by the ideal gas constant (R) and the temperature (T):

Equation 15.19

Kp = K(RT)Δn

where K is the equilibrium constant expressed in units of concentration and Δn is the difference between the numbers of moles of gaseous products and gaseous reactants (npnr). The temperature is expressed as the absolute temperature in kelvins. According to Equation 15.19, Kp = K only if the moles of gaseous products and gaseous reactants are the same (i.e., Δn = 0). For the decomposition of N2O4, there are 2 mol of gaseous product and 1 mol of gaseous reactant, so Δn = 1. Thus, for this reaction, Kp = K(RT)1 = KRT.

Example 5

The equilibrium constant for the reaction of nitrogen and hydrogen to give ammonia is 0.118 at 745 K. The balanced equilibrium equation is as follows:

N2(g)+3H2(g)2NH3(g)

What is Kp for this reaction at the same temperature?

Given: equilibrium equation, equilibrium constant, and temperature

Asked for: K p

Strategy:

Use the coefficients in the balanced chemical equation to calculate Δn. Then use Equation 15.19 to calculate K from Kp.

Solution:

This reaction has 2 mol of gaseous product and 4 mol of gaseous reactants, so Δn = (2 − 4) = −2. We know K, and T = 745 K. Thus, from Equation 15.16, we have the following:

Kp=K(RT)2=K(RT)2=0.118{[0.08206(Latm)/(molK)][745 K]}2=3.16×105

Because Kp is a unitless quantity, the answer is Kp = 3.16 × 10−5.

Exercise

Calculate Kp for the reaction 2SO2(g)+O2(g)2SO3(g) at 527°C, if K = 7.9 × 104 at this temperature.

Answer: Kp = 1.2 × 103

Homogeneous and Heterogeneous Equilibriums

When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibriumAn equilibrium in which the reactants and products of an equilibrium reaction form a single phase, whether gas or liquid.. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibriumAn equilibrium in which the reactants of an equilibrium reaction, the products, or both are in more than one phase., such as the reaction of a gas with a solid or liquid.

Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, their concentrations are treated as constants, which allows us to simplify equilibrium constant expressions that involve pure solids or liquids.The reference states for pure solids and liquids are those forms stable at 1 bar (approximately 1 atm), which are assigned an activity of 1. (Recall from Chapter 11 "Liquids", for example, that the density of water, and thus its volume, changes by only a few percentage points between 0°C and 100°C.)

Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes:

Equation 15.20

CO2(g)+C(s)2CO(g)

The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows:

Equation 15.21

K=[CO]2[CO2][C]

Because graphite is a solid, however, its molar concentration, determined from its density and molar mass, is essentially constant and has the following value:

Equation 15.22

[C]=2.26 g/cm312.01 g/mol×1000 cm3/L=188 mol/L=188 M

We can rearrange Equation 15.18 so that the constant terms are on one side:

Equation 15.23

K[C]=K(188)=[CO]2[CO2]

Incorporating the constant value of [C] into the equilibrium equation for the reaction in Equation 15.17,

Equation 15.24

K=[CO]2[CO2]

The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases:

Equation 15.25

Kp=(PCO)2PCO2

Incorporating all the constant values into K′ or Kp allows us to focus on the substances whose concentrations change during the reaction.

Although the concentrations of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of CO and CO2, the system described in Equation 15.17 will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in Figure 15.5 "Effect of the Amount of Solid Present on Equilibrium in a Heterogeneous Solid–Gas System", it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium.

Figure 15.5 Effect of the Amount of Solid Present on Equilibrium in a Heterogeneous Solid–Gas System

In the system, the equilibrium composition of the gas phase at a given temperature, 1000 K in this case, is the same whether a small amount of solid carbon (left) or a large amount (right) is present.

Example 6

Write each expression for K, incorporating all constants, and Kp for the following equilibrium reactions.

  1. PCl3(l)+Cl2(g)PCl5(s)
  2. Fe3O4(s)+4H2(g)3Fe(s)+4H2O(g)

Given: balanced equilibrium equations

Asked for: expressions for K and Kp

Strategy:

Find K by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation. Then express Kp as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation.

Solution:

  1. This reaction contains a pure solid (PCl5) and a pure liquid (PCl3). Their concentrations do not appear in the equilibrium constant expression because they do not change significantly. So

    K=1[Cl2] and Kp=1PCl2
  2. This reaction contains two pure solids (Fe3O4 and Fe), which do not appear in the equilibrium constant expressions. The two gases do, however, appear in the expressions:

    K=[H2O]4[H2]4 and Kp=(PH2O)4(PH2)4

Exercise

Write the expressions for K and Kp for the following reactions.

  1. CaCO3(s)CaO(s)+CO2(g)
  2. C6H12O6(s)glucose+6O2(g)6CO2(g)+6H2O(g)

Answer:

  1. K = [CO2]; Kp=PCO2
  2. K=[CO2]6[H2O]6[O2]6Kp=(PCO2)6(PH2O)6(PO2)6

For reactions carried out in solution, the concentration of the solvent is omitted from the equilibrium constant expression even when the solvent appears in the balanced chemical equation for the reaction. The concentration of the solvent is also typically much greater than the concentration of the reactants or products (recall that pure water is about 55.5 M, and pure ethanol is about 17 M). Consequently, the solvent concentration is essentially constant during chemical reactions, and the solvent is therefore treated as a pure liquid. The equilibrium constant expression for a reaction contains only those species whose concentrations could change significantly during the reaction.

Note the Pattern

The concentrations of pure solids, pure liquids, and solvents are omitted from equilibrium constant expressions because they do not change significantly during reactions when enough is present to reach equilibrium.

Equilibrium Constant Expressions for the Sums of Reactions

Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.

To illustrate this procedure, let’s consider the reaction of N2 with O2 to give NO2. As we stated in Section 15.1 "The Concept of Chemical Equilibrium", this reaction is an important source of the NO2 that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (1), N2 reacts with O2 at the high temperatures inside an internal combustion engine to give NO. The released NO then reacts with additional O2 to give NO2 (2). The equilibrium constant for each reaction at 100°C is also given.

  1. N2(g)+O2(g)2NO(g)K1=2.0 × 1025
  2. 2NO(g)+O2(g)2NO2(g)K2=6.4×109

    Summing reactions (1) and (2) gives the overall reaction of N2 with O2:

  3. N2(g)+2O2(g)2NO2(g)K3=?

The equilibrium constant expressions for the reactions are as follows:

K1=[NO]2[N2][O2]K2=[NO2]2[NO]2[O2]K3=[NO2]2[N2][O2]2

What is the relationship between K1, K2, and K3, all at 100°C? The expression for K1 has [NO]2 in the numerator, the expression for K2 has [NO]2 in the denominator, and [NO]2 does not appear in the expression for K3. Multiplying K1 by K2 and canceling the [NO]2 terms,

K1K2=[NO]2[N2][O2]×[NO2]2[NO]2[O2]=[NO2]2[N2][O2]2=K3

Thus the product of the equilibrium constant expressions for K1 and K2 is the same as the equilibrium constant expression for K3:

K3 = K1K2 = (2.0 × 10−25)(6.4 × 109) = 1.3 × 10−15

The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to Hess’s Law, ΔH for the sum of two or more reactions is the sum of the ΔH values for the individual reactions.

Note the Pattern

To determine K for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.

Example 7

The following reactions occur at 1200°C:

  1. CO(g)+3H2(g)CH4(g)+H2O(g)K1=9.17×102
  2. CH4(g)+2H2S(g)CS2(g)+4H2(g)K2=3.3×104

    Calculate the equilibrium constant for the following reaction at the same temperature.

  3. CO(g)+2H2S(g)CS2(g)+H2O(g)+H2(g)K3=?

Given: two balanced equilibrium equations, values of K, and an equilibrium equation for the overall reaction

Asked for: equilibrium constant for the overall reaction

Strategy:

Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of K for that equation. Calculate K for the overall equation by multiplying the equilibrium constants for the individual equations.

Solution:

The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2:

CO(g)+3H2(g)CH4(g)+H2O(g)CH4(g)+2H2S(g)CS2(g)+3H2(g)+H2(g)CO(g)+2H2S(g)CS2(g)+H2O(g)+H2(g)

The values for K1 and K2 are given, so it is straightforward to calculate K3:

K3 = K1K2 = (9.17 × 10−2)(3.3 × 104) = 3.03 × 103

Exercise

In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature.

  1. 18S8(s)+O2(g)SO2(g)K1=4.4×1053
  2. SO2(g)+12O2(g)SO3(g)K2=2.6×1012
  3. 18S8(s)+32O2(g)SO3(g)K3=?

Answer: K3 = 1.1 × 1066

Summary

The ratio of the rate constants for the forward and reverse reactions at equilibrium is the equilibrium constant (K), a unitless quantity. The composition of the equilibrium mixture is therefore determined by the magnitudes of the forward and reverse rate constants at equilibrium. Under a given set of conditions, a reaction will always have the same K. For a system at equilibrium, the law of mass action relates K to the ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the equilibrium equation. The ratio is called the equilibrium constant expression. When a reaction is written in the reverse direction, K and the equilibrium constant expression are inverted. For gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to a power matching its coefficient in the chemical equation. An equilibrium constant calculated from partial pressures (Kp) is related to K by the ideal gas constant (R), the temperature (T), and the change in the number of moles of gas during the reaction. An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions.

Key Takeaways

  • The law of mass action describes a system at equilibrium in terms of the concentrations of the products and the reactants.
  • For a system involving one or more gases, either the molar concentrations of the gases or their partial pressures can be used.

Key Equations

Definition of equilibrium constant in terms of forward and reverse rate constants

Equation 15.6: K=kfkr

Equilibrium constant expression (law of mass action)

Equation 15.8: K=[C]c[D]d[A]a[B]b

Equilibrium constant expression for reactions involving gases using partial pressures

Equation 15.17: Kp=(PC)c(PD)d(PA)a(PB)b

Relationship between K p and K

Equation 15.19: Kp= K(RT)Δn

Conceptual Problems

  1. For an equilibrium reaction, what effect does reversing the reactants and products have on the value of the equilibrium constant?

  2. Which of the following equilibriums are homogeneous and which are heterogeneous?

    1. 2HF(g)H2(g)+F2(g)
    2. C(s) + 2H2(g)CH4(g)
    3. H2C=CH2(g)+H2(g)C2H6(g)
    4. 2Hg(l) + O2(g)2HgO(s)
  3. Classify each equilibrium system as either homogeneous or heterogeneous.

    1. NH4CO2NH2(s)2NH3(g)+CO2(g)
    2. C(s) + O2(g)CO2(g)
    3. 2Mg(s) + O2(g)2MgO(s)
    4. AgCl(s)Ag+(aq)+Cl(aq)
  4. If an equilibrium reaction is endothermic, what happens to the equilibrium constant if the temperature of the reaction is increased? if the temperature is decreased?

  5. Industrial production of NO by the reaction N2(g)+O2(g)2NO(g) is carried out at elevated temperatures to drive the reaction toward the formation of product. After sufficient product has formed, the reaction mixture is quickly cooled. Why?

  6. How would you differentiate between a system that has reached chemical equilibrium and one that is reacting so slowly that changes in concentration are difficult to observe?

  7. What is the relationship between the equilibrium constant, the concentration of each component of the system, and the rate constants for the forward and reverse reactions?

  8. Write the equilibrium constant expressions for K and Kp for each reaction.

    1. CO(g) + H2O(g)CO2(g)+H2(g)
    2. PCl3(g)+Cl2(g)PCl5(g)
    3. 2O3(g)3O2(g)
  9. Write the equilibrium constant expressions for K and Kp as appropriate for each reaction.

    1. 2NO(g)+O2(g)2NO2(g)
    2. 12H2(g)+12I2(g)HI(g)
    3. cis-stilbene(soln)trans-stilbene(soln)
  10. Why is it incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression?

  11. Write the equilibrium constant expressions for K and Kp for each equilibrium reaction.

    1. 2S(s)+3O2(g)2SO3(g)
    2. C(s) + CO2(g)2CO(g)
    3. 2ZnS(s)+3O2(g)2ZnO(s)+2SO2(g)
  12. Write the equilibrium constant expressions for K and Kp for each equilibrium reaction.

    1. 2HgO(s)2Hg(l)+O2(g)
    2. H2(g)+I2(s)2HI(g)
    3. NH4CO2NH2(s)2NH3(g)+CO2(g)
  13. At room temperature, the equilibrium constant for the reaction 2A(g)B(g) is 1. What does this indicate about the concentrations of A and B at equilibrium? Would you expect K and Kp to vary significantly from each other? If so, how would their difference be affected by temperature?

  14. For a certain series of reactions, if [OH][HCO3]/[CO32−] = K1 and [OH][H2CO3]/[HCO3] = K2, what is the equilibrium constant expression for the overall reaction? Write the overall equilibrium equation.

  15. In the equation for an enzymatic reaction, ES represents the complex formed between the substrate S and the enzyme protein E. In the final step of the following oxidation reaction, the product P dissociates from the ESO2 complex, which regenerates the active enzyme:

    E + SESK1 ES + O2ESO2K2 ESO2E+PK3

    Give the overall reaction equation and show that K = K1 × K2 × K3.

Answers

  1. The equilibrium constant for the reaction written in reverse: K′ = 1/K.

  2. Each system is heterogeneous.

  3. Rapid cooling “quenches” the reaction mixture and prevents the system from reverting to the low-temperature equilibrium composition that favors the reactants.

  4.  

    K=kf/krK=[C]c[D]d[A]a[B]b
  5.  

    1. K=[NO2]2[NO]2[O2]Kp=(PN2O)2(PNO)2(PO2)
    2. K=[HI][H2]1/2[I2]1/2Kp=PHI(PH2)1/2(PI2)
    3. K=[trans-stilbene][cis-stilbene]
  6.  

    1. K=[SO3]2[O2]3Kp=(PSO3)2(PO2)3
    2. K=[CO]2[CO2]Kp=(PCO)2PCO2
    3. K=[SO2]2[O2]3Kp=(PSO2)2(PO2)3
  7. At equilibrium,

    [A]=BΔn=1, so Kp=K(RT)Δn=KRT;

    the difference increases as T increases.

Numerical Problems

  1. Explain what each of the following values for K tells you about the relative concentrations of the reactants versus the products in a given equilibrium reaction: K = 0.892; K = 3.25 × 108; K = 5.26 × 10−11. Are products or reactants favored at equilibrium?

  2. Write the equilibrium constant expression for each reaction. Are these equilibrium constant expressions equivalent? Explain.

    1. N2O4(g)2NO2(g)
    2. 12N2O4(g)NO2(g)
  3. Write the equilibrium constant expression for each reaction.

    1. 12N2(g)+32H2(g)NH3(g)
    2. 13N2(g)+H2(g)23NH3(g)

    How are these two expressions mathematically related to the equilibrium constant expression for

    N2(g)+3H2(g)2NH3(g)?
  4. Write an equilibrium constant expression for each reaction.

    1. C(s) + 2H2O(g)CO2(g)+2H2(g)
    2. SbCl3(g)+Cl2(g)SbCl5(g)
    3. 2O3(g)3O2(g)
  5. Give an equilibrium constant expression for each reaction.

    1. 2NO(g) + O2(g) ⇌ 2NO2(g)
    2. 12H2(g)+12I2(g)HI(g)
    3. CaCO3(s) + 2HOCl(aq) ⇌ Ca2+(aq) + 2OCl(aq) + H2O(l) + CO2(g)
  6. Calculate K and Kp for each reaction.

    1. 2NOBr(g)2NO(g)+Br(g): at 727°C, the equilibrium concentration of NO is 1.29 M, Br2 is 10.52 M, and NOBr is 0.423 M.
    2. C(s) + CO2(g)2CO(g): at 1200 K, a 2.00 L vessel at equilibrium has partial pressures of 93.5 atm CO2 and 76.8 atm CO, and the vessel contains 3.55 g of carbon.
  7. Calculate K and Kp for each reaction.

    1. N2O4(g)2NO2(g): at the equilibrium temperature of −40°C, a 0.150 M sample of N2O4 undergoes a decomposition of 0.456%.
    2. CO(g)+2H2(g)CH3OH(g): an equilibrium is reached at 227°C in a 15.5 L reaction vessel with a total pressure of 6.71 × 102 atm. It is found to contain 37.8 g of hydrogen gas, 457.7 g of carbon monoxide, and 7193 g of methanol.
  8. Determine K and Kp (where applicable) for each reaction.

    1. 2H2S(g)2H2(g)+S2(g): at 1065°C, an equilibrium mixture consists of 1.00 × 10−3 M H2, 1.20 × 10−3 M S2, and 3.32 × 10−3 M H2S.
    2. Ba(OH)2(s)2OH(aq)+Ba2+(aq): at 25°C, a 250 mL beaker contains 0.330 mol of barium hydroxide in equilibrium with 0.0267 mol of barium ions and 0.0534 mol of hydroxide ions.
  9. Determine K and Kp for each reaction.

    1. 2NOCl(g)2NO(g)+Cl2(g): at 500 K, a 24.3 mM sample of NOCl has decomposed, leaving an equilibrium mixture that contains 72.7% of the original amount of NOCl.
    2. Cl2(g)+PCl3(g)PCl5(g): at 250°C, a 500 mL reaction vessel contains 16.9 g of Cl2 gas, 0.500 g of PCl3, and 10.2 g of PCl5 at equilibrium.
  10. The equilibrium constant expression for a reaction is [CO2]2/[SO2]2[O2]. What is the balanced chemical equation for the overall reaction if one of the reactants is Na2CO3(s)?

  11. The equilibrium constant expression for a reaction is [NO][H2O]3/2/[NH3][O2]5/4. What is the balanced chemical equation for the overall reaction?

  12. Given K = kf/kr, what happens to the magnitude of the equilibrium constant if the reaction rate of the forward reaction is doubled? What happens if the reaction rate of the reverse reaction for the overall reaction is decreased by a factor of 3?

  13. The value of the equilibrium constant for

    2H2(g)+S2(g)2H2S(g)

    is 1.08 × 107 at 700°C. What is the value of the equilibrium constant for the following related reactions?

    1. H2(g)+12S2(g)H2S(g)
    2. 4H2(g)+2S2(g)4H2S(g)
    3. H2S(g)H2(g)+12S2(g)

Answers

  1. K = 0.892: the concentrations of the products and the reactants are approximately equal at equilibrium so neither is favored; K = 3.25 × 108: the ratio of the concentration of the products to the reactants at equilibrium is very large so the formation of products is favored; K = 5.26 × 10−11: the ratio of the concentration of the products to the reactants at equilibrium is very small so the formation of products is not favored.

    1. K=[NH3][N2]1/2[H2]3/2
    2. K=[NH3]2/3[N2]1/3[H2]K=[NH3]2[N2][H2]3; K′ = K1/2, and K″ = K1/3
    1. K=[NO2]2[NO]2[O2]
    2. K=[HI][H2]1/2[I2]1/2
    3. K=[Ca2+][OCl]2[PCO2][HOCl]2
    1. K = 1.25 × 10−5; Kp = 2.39 × 10−4
    2. K = 9.43; Kp = 5.60 × 10−3
    1. K=[Cl2][NO]2[NOCl]2=4.59×104Kp=1.88×102
    2. K=[PCl5][PCl3][Cl2]=28.3Kp=0.658
  2. NH3 + 54O2NO +32H2O, which can also be written as follows: 4NH3(g) + 5O2(g)4NO(g) + 6H2O(g)

    1. 3.29 × 103
    2. 1.17 × 1014
    3. 3.04 × 10−4

15.3 Solving Equilibrium Problems

Learning Objective

  1. To solve quantitative problems involving chemical equilibriums.

There are two fundamental kinds of equilibrium problems: (1) those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and (2) those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems.

Calculating an Equilibrium Constant from Equilibrium Concentrations

We saw in the exercise in Example 6 in Section 15.2 "The Equilibrium Constant" that the equilibrium constant for the decomposition of CaCO3(s) to CaO(s) and CO2(g) is K = [CO2]. At 800°C, the concentration of CO2 in equilibrium with solid CaCO3 and CaO is 2.5 × 10−3 M. Thus K at 800°C is 2.5 × 10−3. (Remember that equilibrium constants are unitless.)

A more complex example of this type of problem is the conversion of n-butane, an additive used to increase the volatility of gasoline, to isobutane (2-methylpropane). This reaction can be written as follows:

Equation 15.26

n-butane(g)isobutane(g)

and the equilibrium constant K = [isobutane]/[n-butane]. At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. Substituting these concentrations into the equilibrium constant expression,

Equation 15.27

K=[isobutane][n-butane]=0.041 M0.016 M=2.6

Thus the equilibrium constant for the reaction as written is 2.6.

Example 8

The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:

2SO2(g)+O2(g)2SO3(g)

A mixture of SO2 and O2 was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained 5.0 × 10−2 M SO3, 3.5 × 10−3 M O2, and 3.0 × 10−3 M SO2. Calculate K and Kp at this temperature.

Given: balanced equilibrium equation and composition of equilibrium mixture

Asked for: equilibrium constant

Strategy:

Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain K.

Solution:

Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,

K=[SO3]2[SO2]2[O2]=(5.0×102)2(3.0×103)2(3.5×103)=7.9×104

To solve for Kp, we use Equation 16.18, where Δn = 2 − 3 = −1:

Kp=K(RT)Δn=7.9×104[(0.08206 Latm/molK)(800 K)]1=1.2×103

Exercise

Hydrogen gas and iodine react to form hydrogen iodide via the reaction

H2(g)+I2(g)2HI(g)

A mixture of H2 and I2 was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained 1.37 × 10−2 M HI, 6.47 × 10−3 M H2, and 5.94 × 10−4 M I2. Calculate K and Kp for this reaction.

Answer: K = 48.8; Kp = 48.8

Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example 9 shows one way to do this.

Example 9

A 1.00 mol sample of NOCl was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of Cl2. Calculate K at this temperature. The equation for the decomposition of NOCl to NO and Cl2 is as follows:

2NOCl(g)2NO(g)+Cl2(g)

Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium

Asked for: K

Strategy:

A Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations).

B Calculate all possible initial concentrations from the data given and insert them in the table.

C Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table.

D Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction.

Solution:

A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:

K=[NO]2[Cl2][NOCl]2

To obtain the concentrations of NOCl, NO, and Cl2 at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations.

2NOCl(g)2NO(g)+Cl2(g)
[NOCl] [NO] [Cl2]
initial
change
final

B Initially, the system contains 1.00 mol of NOCl in a 2.00 L container. Thus [NOCl]i = 1.00 mol/2.00 L = 0.500 M. The initial concentrations of NO and Cl2 are 0 M because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of Cl2 in a 2.00 L container, so [Cl2]f = 0.056 mol/2.00 L = 0.028 M. We insert these values into the following table:

2NOCl(g)2NO(g)+Cl2(g)
[NOCl] [NO] [Cl2]
initial 0.500 0 0
change
final 0.028

C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of Cl2, the substance for which initial and final concentrations are known:

Δ[Cl2] = [0.028 M (final) − 0.00 M (initial)] = +0.028 M

According to the coefficients in the balanced chemical equation, 2 mol of NO are produced for every 1 mol of Cl2, so the change in the NO concentration is as follows:

Δ[NO]=(0.028 molCl2L)(2 mol NO1 molCl2)=0.056 M

Similarly, 2 mol of NOCl are consumed for every 1 mol of Cl2 produced, so the change in the NOCl concentration is as follows:

Δ[NOCl]=(0.028 mol Cl2L)(2 mol NOCl1 mol Cl2)=0.056 M

We insert these values into our table:

2NOCl(g)2NO(g)+Cl2(g)
[NOCl] [NO] [Cl2]
initial 0.500 0 0
change −0.056 +0.056 +0.028
final 0.028

D We sum the numbers in the [NOCl] and [NO] columns to obtain the final concentrations of NO and NOCl:

[NO]f = 0.000 M + 0.056 M = 0.056 M [NOCl]f = 0.500 M + (−0.056 M) = 0.444 M

We can now complete the table:

2NOCl(g)2NO(g) +Cl2(g)
[NOCl] [NO] [Cl2]
initial 0.500 0 0
change −0.056 +0.056 +0.028
final 0.444 0.056 0.028

We can now calculate the equilibrium constant for the reaction:

K=[NO]2[Cl2][NOCl]2=(0.056)2(0.028)(0.444)2=4.5×104

Exercise

The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (NH3) by reacting 0.1248 M H2 and 0.0416 M N2 at about 500°C. At equilibrium, the mixture contained 0.00272 M NH3. What is K for the reaction N2+3H22NH3 at this temperature? What is Kp?

Answer: K = 0.105; Kp = 2.61 × 10−5

Calculating Equilibrium Concentrations from the Equilibrium Constant

To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n-butane to isobutane (Equation 15.26), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n-butane, we can determine the concentration of n-butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example 9.

n-butane(g)isobutane(g)
[n-Butane] [Isobutane]
initial
change
final

The original laboratory apparatus designed by Fritz Haber and Robert Le Rossignol in 1908 for synthesizing ammonia from its elements. A metal catalyst bed, where ammonia was produced, is in the large cylinder at the left. The Haber-Bosch process used for the industrial production of ammonia uses essentially the same process and components but on a much larger scale. Unfortunately, Haber’s process enabled Germany to prolong World War I when German supplies of nitrogen compounds, which were used for explosives, had been exhausted in 1914.

The initial concentrations of the reactant and product are both known: [n-butane]i = 1.00 M and [isobutane]i = 0 M. We need to calculate the equilibrium concentrations of both n-butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as +x, then the change in the concentration of n-butane is Δ[n-butane] = −x. This is because the balanced chemical equation for the reaction tells us that 1 mol of n-butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone.

n-butane(g)isobutane(g)
[n-Butane] [Isobutane]
initial 1.00 0
change x +x
final (1.00 − x) (0 + x) = x

Substituting the expressions for the final concentrations of n-butane and isobutane from the table into the equilibrium equation,

K=[isobutane][n-butane]=x1.00x=2.6

Rearranging and solving for x,

x=2.6(1.00x)=2.62.6xx+2.6x=2.6x=0.72

We obtain the final concentrations by substituting this x value into the expressions for the final concentrations of n-butane and isobutane listed in the table:

[n-butane]f = (1.00 − x) M = (1.00 − 0.72) M = 0.28 M [isobutane]f = (0.00 + x) M = (0.00 + 0.72) M = 0.72 M

We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same K that we used in the calculation:

K=[isobutane][n-butane]=0.72 M0.28 M=2.6

This is the same K we were given, so we can be confident of our results.

Example 10 illustrates a common type of equilibrium problem that you are likely to encounter.

Example 10

The water–gas shift reaction is important in several chemical processes, such as the production of H2 for fuel cells. This reaction can be written as follows:

H2(g)+CO2(g)H2O(g)+CO(g)

K = 0.106 at 700 K. If a mixture of gases that initially contains 0.0150 M H2 and 0.0150 M CO2 is allowed to equilibrate at 700 K, what are the final concentrations of all substances present?

Given: balanced equilibrium equation, K, and initial concentrations

Asked for: final concentrations

Strategy:

A Construct a table showing what is known and what needs to be calculated. Define x as the change in the concentration of one substance. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of x. From the values in the table, calculate the final concentrations.

B Write the equilibrium equation for the reaction. Substitute appropriate values from the table to obtain x.

C Calculate the final concentrations of all species present. Check your answers by substituting these values into the equilibrium constant expression to obtain K.

Solution:

A The initial concentrations of the reactants are [H2]i = [CO2]i = 0.0150 M. Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of H2O as x, then Δ[H2O] = +x. We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of x. For example, 1 mol of CO is produced for every 1 mol of H2O, so the change in the CO concentration can be expressed as Δ[CO] = +x. Similarly, for every 1 mol of H2O produced, 1 mol each of H2 and CO2 are consumed, so the change in the concentration of the reactants is Δ[H2] = Δ[CO2] = −x. We enter the values in the following table and calculate the final concentrations.

H2(g)+CO2(g)H2O(g)+CO(g)
[H2] [CO2] [H2O] [CO]
initial 0.0150 0.0150 0 0
change x x +x +x
final (0.0150 − x) (0.0150 − x) x x

B We can now use the equilibrium equation and the given K to solve for x:

K=[H2O][CO][H2][CO2]=(x)(x)(0.0150x)(0.0150x)=x2(0.0150x)2=0.106

We could solve this equation with the quadratic formula, but it is far easier to solve for x by recognizing that the left side of the equation is a perfect square; that is,

x2(0.0150x)2=(x0.0150x)2=0.106

(The quadratic formula is presented in Essential Skills 7 in Section 15.7 "Essential Skills".) Taking the square root of the middle and right terms,

x(0.0150x)=(0.106)1/2=0.326x=(0.326)(0.0150)0.326x1.326x=0.00489x=0.00369=3.69×103

C The final concentrations of all species in the reaction mixture are as follows:

[H2]f=[H2]i+Δ[H2]=(0.01500.00369) M=0.0113 M[CO2]f=[CO2]i+Δ[CO2]=(0.01500.00369) M=0.0113 M[H2O]f=[H2O]i+Δ[H2O]=(0+0.00369) M=0.00369 M[CO]f=[CO]i+Δ[CO]=(0+0.00369) M=0.00369 M

We can check our work by inserting the calculated values back into the equilibrium constant expression:

K=[H2O][CO][H2][CO2]=(0.00369)2(0.0113)2=0.107

To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed.

Exercise

Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation:

H2(g)+I2(g)2HI(g)

K = 54 at 425°C. If 0.172 M H2 and I2 are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture?

Answer: [HI]f = 0.270 M; [H2]f = [I2]f = 0.037 M

In Example 10, the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example 11.

Example 11

In the water–gas shift reaction shown in Example 10, a sample containing 0.632 M CO2 and 0.570 M H2 is allowed to equilibrate at 700 K. At this temperature, K = 0.106. What is the composition of the reaction mixture at equilibrium?

Given: balanced equilibrium equation, concentrations of reactants, and K

Asked for: composition of reaction mixture at equilibrium

Strategy:

A Write the equilibrium equation. Construct a table showing the initial concentrations of all substances in the mixture. Complete the table showing the changes in the concentrations (x) and the final concentrations.

B Write the equilibrium constant expression for the reaction. Substitute the known K value and the final concentrations to solve for x.

C Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain K.

Solution:

A [CO2]i = 0.632 M and [H2]i = 0.570 M. Again, x is defined as the change in the concentration of H2O: Δ[H2O] = +x. Because 1 mol of CO is produced for every 1 mol of H2O, the change in the concentration of CO is the same as the change in the concentration of H2O, so Δ[CO] = +x. Similarly, because 1 mol each of H2 and CO2 are consumed for every 1 mol of H2O produced, Δ[H2] = Δ[CO2] = −x. The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium.

H2(g)+CO2(g)H2O(g)+CO(g)
[H2] [CO2] [H2O] [CO]
initial 0.570 0.632 0 0
change x x +x +x
final (0.570 − x) (0.632 − x) x x

B We can now use the equilibrium equation and the known K value to solve for x:

K=[H2O][CO][H2][CO2]=x2(0.570x)(0.632x)=0.106

In contrast to Example 10, however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator:

x2 = 0.106(0.360 − 1.20x + x2)

Collecting terms on one side of the equation,

0.894x2 + 0.127x − 0.0382 = 0

This equation can be solved using the quadratic formula:

x=b±b24ac2a=0.127±(0.127)24(0.894)(0.0382)2(0.894)=0.1480.290

Only the answer with the positive value has any physical significance, so Δ[H2O] = Δ[CO] = +0.148 M, and Δ[H2] = Δ[CO2] = −0.148 M.

C The final concentrations of all species in the reaction mixture are as follows:

[H2]f=[H2]i+Δ[H2]=0.570 M0.148 M=0.422 M[CO2]f=[CO2]i+Δ[CO2]=0.632 M0.148 M=0.484 M[H2O]f=[H2O]i+Δ[H2O]=0 M+0.148 M=0.148 M[CO]f=[CO]i+Δ[CO]=0 M+0.148 M=0.148 M

We can check our work by substituting these values into the equilibrium constant expression:

K=[H2O][CO][H2][CO2]=(0.148)2(0.422)(0.484)=0.107

Because K is essentially the same as the value given in the problem, our calculations are confirmed.

Exercise

The exercise in Example 8 showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which K = 54 at 425°C. If a sample containing 0.200 M H2 and 0.0450 M I2 is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture?

Answer: [HI]f = 0.0882 M; [H2]f = 0.156 M; [I2]f = 9.2 × 10−4 M

In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small (K ≤ 10−3) or very large (K ≥ 103), which means that the change in the concentration (defined as x) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example 12.

Example 12

Atmospheric nitrogen and oxygen react to form nitric oxide:

N2(g)+O2(g)2NO(g)

Kp = 2.0 × 10−31 at 25°C. What is the partial pressure of NO in equilibrium with N2 and O2 in the atmosphere (at 1 atm, PN2 = 0.78 atm and PO2 = 0.21 atm)?

Given: balanced equilibrium equation and values of Kp, PO2, and PN2

Asked for: partial pressure of NO

Strategy:

A Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances.

B Write the equilibrium equation for the reaction. Then substitute values from the table to solve for the change in concentration (x).

C Calculate the partial pressure of NO. Check your answer by substituting values into the equilibrium equation and solving for K.

Solution:

A Because we are given Kp and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of O2 is 0.21 atm and that of N2 is 0.78 atm. If we define the change in the partial pressure of NO as 2x, then the change in the partial pressure of O2 and of N2 is −x because 1 mol each of N2 and of O2 is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.

N2(g)+O2(g)2NO(g)
PN2 (atm) PO2 (atm) PNO (atm)
initial P 0.78 0.21 0
change in P x x +2x
final P (0.78 − x) (0.21 − x) 2x

B Substituting these values into the equation for the equilibrium constant,

Kp=(PNO)2(PN2)(PO2)=(2x)2(0.78x)(0.21x)=2.0×1031

In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the x value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, (0.78 − x) = 0.78 and (0.21 − x) = 0.21. Substituting these expressions into our original equation,

(2x)2(0.78)(0.21)=2.0×10314x20.16=2.0×1031x2=0.33×10314x=9.1×1017

C Substituting this value of x into our expressions for the final partial pressures of the substances,

PNO=2x atm=1.8×1016 atmPN2=(0.78x) atm=0.78 atmPO2=(0.21x) atm=0.21 atm

From these calculations, we see that our initial assumption regarding x was correct: given two significant figures, 2.0 × 10−16 is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if x is less than about 5% of the total, or 10−3 > K > 103, then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic NO, an ingredient of smog, does not form from atmospheric concentrations of N2 and O2 to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation:

Kp=(PNO)2(PN2)(PO2)=(1.8×1016)2(0.78)(0.21)=2.0×1031

The final Kp agrees with the value given at the beginning of this example.

Exercise

Under certain conditions, oxygen will react to form ozone, as shown in the following equation:

3O2(g)2O3(g)

Kp = 2.5 × 10−59 at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere (PO2=0.21 atm)?

Answer: 4.8 × 10−31 atm

Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large (K ≥ 103). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example 13.

Example 13

The chemical equation for the reaction of hydrogen with ethylene (C2H4) to give ethane (C2H6) is as follows:

H2(g)+C2H4(g)NiC2H6(g)

K = 9.6 × 1018 at 25°C. If a mixture of 0.200 M H2 and 0.155 M C2H4 is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture?

Given: balanced chemical equation, K, and initial concentrations of reactants

Asked for: equilibrium concentrations

Strategy:

A Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations.

B Write the equilibrium constant expression for the reaction. Then substitute values from the table into the expression to solve for x (the change in concentration).

C Calculate the equilibrium concentrations. Check your answers by substituting these values into the equilibrium equation.

Solution:

A From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example 12. If we define −x as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is +x. The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions.

H2(g)+C2H4(g)NiC2H6(g)
[H2] [C2H4] [C2H6]
initial 0.200 0.155 0
assuming 100% reaction 0.045 0 0.155
change +x +x x
final (0.045 + x) (0 + x) (0.155 − x)

B Substituting values into the equilibrium constant expression,

K=[C2H6][H2][C2H4]=0.155x(0.045+x)x=9.6×1018

Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus x is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified [(0.045 + x) = 0.045 and (0.155 − x) = 0.155] as follows:

K=0.1550.045x=9.6×1018x=3.6×1019

C The small x value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table:

[C2H6]f = (0.155 − x) M = 0.155 [C2H4]f = x M = 3.6 × 10−19 M [H2]f = (0.045 + x) M = 0.045 M

We can verify our calculations by substituting the final concentrations into the equilibrium constant expression:

K=[C2H6][H2][C2H4]=0.155(0.045)(3.6×1019)=9.6×1018

This K value agrees with our initial value at the beginning of the example.

Exercise

Hydrogen reacts with chlorine gas to form hydrogen chloride:

H2(g)+Cl2(g)2HCl(g)

Kp = 4.0 × 1031 at 47°C. If a mixture of 0.257 M H2 and 0.392 M Cl2 is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture?

Answer: [H2]f = 4.8 × 10−32 M; [Cl2]f = 0.135 M; [HCl]f = 0.514 M

Summary

When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium.

Key Takeaway

  • Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture.

Conceptual Problems

  1. Describe how to determine the magnitude of the equilibrium constant for a reaction when not all concentrations of the substances are known.

  2. Calculations involving systems with very small or very large equilibrium constants can be dramatically simplified by making certain assumptions about the concentrations of products and reactants. What are these assumptions when K is (a) very large and (b) very small? Illustrate this technique using the system A+2BC for which you are to calculate the concentration of the product at equilibrium starting with only A and B. Under what circumstances should simplifying assumptions not be used?

Numerical Problems

    Please be sure you are familiar with the topics discussed in Essential Skills 7 (Section 15.7 "Essential Skills") before proceeding to the Numerical Problems.

  1. In the equilibrium reaction A+BC, what happens to K if the concentrations of the reactants are doubled? tripled? Can the same be said about the equilibrium reaction 2AB+C?

  2. The following table shows the reported values of the equilibrium PO2 at three temperatures for the reaction Ag2O(s)2Ag(s)+12O2(g), for which ΔH° = 31 kJ/mol. Are these data consistent with what you would expect to occur? Why or why not?

    T (°C) PO2 (mmHg)
    150 182
    184 143
    191 126
  3. Given the equilibrium system N2O4(g)2NO2(g), what happens to Kp if the initial pressure of N2O4 is doubled? If Kp is 1.7 × 10−1 at 2300°C, and the system initially contains 100% N2O4 at a pressure of 2.6 × 102 atm, what is the equilibrium pressure of each component?

  4. At 430°C, 4.20 mol of HI in a 9.60 L reaction vessel reaches equilibrium according to the following equation: H2(g)+I2(g)2HI(g). At equilibrium, [H2] = 0.047 M and [HI] = 0.345 M. What are K and Kp for this reaction?

  5. Methanol, a liquid used as an automobile fuel additive, is commercially produced from carbon monoxide and hydrogen at 300°C according to the following reaction: CO(g)+2H2(g)CH3OH(g), and Kp = 1.3 × 10−4. If 56.0 g of CO is mixed with excess hydrogen in a 250 mL flask at this temperature, and the hydrogen pressure is continuously maintained at 100 atm, what would be the maximum percent yield of methanol? What pressure of hydrogen would be required to obtain a minimum yield of methanol of 95% under these conditions?

  6. Starting with pure A, if the total equilibrium pressure is 0.969 atm for the reaction A(s)2B(g)+C(g), what is Kp?

  7. The decomposition of ammonium carbamate to NH3 and CO2 at 40°C is written as NH4CO2NH2(s)2NH3(g)+CO2(g). If the partial pressure of NH3 at equilibrium is 0.242 atm, what is the equilibrium partial pressure of CO2? What is the total gas pressure of the system? What is Kp?

  8. At 375 K, Kp for the reaction SO2Cl2(g)SO2(g)+Cl2(g) is 2.4, with pressures expressed in atmospheres. At 303 K, Kp is 2.9 × 10−2.

    1. What is K for the reaction at each temperature?
    2. If a sample at 375 K has 0.100 M Cl2 and 0.200 M SO2 at equilibrium, what is the concentration of SO2Cl2?
    3. If the sample given in part b is cooled to 303 K, what is the pressure inside the bulb?
  9. For the gas-phase reaction aAbB, show that Kp = K(RT)Δn assuming ideal gas behavior.

  10. For the gas-phase reaction I22I, show that the total pressure is related to the equilibrium pressure by the following equation:

    PT=KpPI2+PI2
  11. Experimental data on the system Br2(l)Br2(aq) are given in the following table. Graph [Br2] versus moles of Br2(l) present; then write the equilibrium constant expression and determine K.

    Grams Br2 in 100 mL Water [Br2] (M)
    1.0 0.0626
    2.5 0.156
    3.0 0.188
    4.0 0.219
    4.5 0.219
  12. Data accumulated for the reaction n-butane(g)isobutane(g) at equilibrium are shown in the following table. What is the equilibrium constant for this conversion? If 1 mol of n-butane is allowed to equilibrate under the same reaction conditions, what is the final number of moles of n-butane and isobutane?

    Moles n-butane Moles Isobutane
    0.5 1.25
    1.0 2.5
    1.50 3.75
  13. Solid ammonium carbamate (NH4CO2NH2) dissociates completely to ammonia and carbon dioxide when it vaporizes:

    NH4CO2NH2(s)2NH3(g)+CO2(g)

    At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas? What is Kp? If the concentration of CO2 is doubled and then equilibrates to its initial equilibrium partial pressure +x atm, what change in the NH3 concentration is necessary for the system to restore equilibrium?

  14. The equilibrium constant for the reaction COCl2(g)CO(g)+Cl2(g) is Kp = 2.2 × 10−10 at 100°C. If the initial concentration of COCl2 is 3.05 × 10−3 M, what is the partial pressure of each gas at equilibrium at 100°C? What assumption can be made to simplify your calculations?

  15. Aqueous dilution of IO4 results in the following reaction: IO4(aq)+2H2O(l)H4IO6(aq), and K = 3.5 × 10−2. If you begin with 50 mL of a 0.896 M solution of IO4 that is diluted to 250 mL with water, how many moles of H4IO6 are formed at equilibrium?

  16. Iodine and bromine react to form IBr, which then sublimes. At 184.4°C, the overall reaction proceeds according to the following equation:

    I2(g)+Br2(g)2IBr(g)

    Kp = 1.2 × 102. If you begin the reaction with 7.4 g of I2 vapor and 6.3 g of Br2 vapor in a 1.00 L container, what is the concentration of IBr(g) at equilibrium? What is the partial pressure of each gas at equilibrium? What is the total pressure of the system?

  17. For the reaction C(s) + 12N2(g)+52H2(g)CH3NH2(g), K = 1.8 × 10−6. If you begin the reaction with 1.0 mol of N2, 2.0 mol of H2, and sufficient C(s) in a 2.00 L container, what are the concentrations of N2 and CH3NH2 at equilibrium? What happens to K if the concentration of H2 is doubled?

15.4 Nonequilibrium Conditions

Learning Objective

  1. To predict in which direction a reaction will proceed.

In , we saw that knowing the magnitude of the equilibrium constant under a given set of conditions allows chemists to predict the extent of a reaction. Often, however, chemists must decide whether a system has reached equilibrium or if the composition of the mixture will continue to change with time. In this section, we describe how to quantitatively analyze the composition of a reaction mixture to make this determination.

The Reaction Quotient

To determine whether a system has reached equilibrium, chemists use a quantity called the reaction quotient (Q)A quantity derived from a set of values measured at any time during the reaction of any mixture of reactants and products, regardless of whether the system is at equilibrium: Q=[C]c[D]d/[A]a[B]b for the general balanced chemical equation aA+bBcC+dD.. The expression for the reaction quotient has precisely the same form as the equilibrium constant expression, except that Q may be derived from a set of values measured at any time during the reaction of any mixture of the reactants and the products, regardless of whether the system is at equilibrium. Therefore, for the following general reaction:

aA+bBcC+dD

the reaction quotient is defined as follows:

Equation 15.28

Q=[C]c[D]d[A]a[B]b

The reaction quotient (Qp)A quantity derived from a set of values measured at any time during the reaction of any mixture of reactants and products in the gas phase, regardless of whether the system is at equilibrium: Qp=(PC)c(PD)d/(PA)a(PB)b for the general balanced chemical equation aA+bBcC+dD., which is analogous to Kp, can be written for any reaction that involves gases by using the partial pressures of the components.

To understand how information is obtained using a reaction quotient, consider the dissociation of dinitrogen tetroxide to nitrogen dioxide, N2O4(g)2NO2(g), for which K = 4.65 × 10−3 at 298 K. We can write Q for this reaction as follows:

Equation 15.29

Q=[NO2]2[N2O4]

The following table lists data from three experiments in which samples of the reaction mixture were obtained and analyzed at equivalent time intervals, and the corresponding values of Q were calculated for each. Each experiment begins with different proportions of product and reactant:

Experiment [NO2] (M) [N2O4] (M) Q = [NO2]2/[N2O4]
1 0 0.0400 020.0400=0
2 0.0600 0 (0.0600)20=undefined
3 0.0200 0.0600 (0.0200)20.0600=6.67×103

As these calculations demonstrate, Q can have any numerical value between 0 and infinity (undefined); that is, Q can be greater than, less than, or equal to K.

Comparing the magnitudes of Q and K enables us to determine whether a reaction mixture is already at equilibrium and, if it is not, predict how its composition will change with time to reach equilibrium (i.e., whether the reaction will proceed to the right or to the left as written). All you need to remember is that the composition of a system not at equilibrium will change in a way that makes Q approach K. If Q = K, for example, then the system is already at equilibrium, and no further change in the composition of the system will occur unless the conditions are changed. If Q < K, then the ratio of the concentrations of products to the concentrations of reactants is less than the ratio at equilibrium. Therefore, the reaction will proceed to the right as written, forming products at the expense of reactants. Conversely, if Q > K, then the ratio of the concentrations of products to the concentrations of reactants is greater than at equilibrium, so the reaction will proceed to the left as written, forming reactants at the expense of products. These points are illustrated graphically in .

Figure 15.6 Two Different Ways of Illustrating How the Composition of a System Will Change Depending on the Relative Values of Q and K

(a) Both Q and K are plotted as points along a number line: the system will always react in the way that causes Q to approach K. (b) The change in the composition of a system with time is illustrated for systems with initial values of Q > K, Q < K, and Q = K.

Note the Pattern

If Q < K, the reaction will proceed to the right as written. If Q > K, the reaction will proceed to the left as written. If Q = K, then the system is at equilibrium.

Example 14

At elevated temperatures, methane (CH4) reacts with water to produce hydrogen and carbon monoxide in what is known as a steam-reforming reaction:

CH4(g)+H2O(g)CO(g)+3H2(g)

K = 2.4 × 10−4 at 900 K. Huge amounts of hydrogen are produced from natural gas in this way and are then used for the industrial synthesis of ammonia. If 1.2 × 10−2 mol of CH4, 8.0 × 10−3 mol of H2O, 1.6 × 10−2 mol of CO, and 6.0 × 10−3 mol of H2 are placed in a 2.0 L steel reactor and heated to 900 K, will the reaction be at equilibrium or will it proceed to the right to produce CO and H2 or to the left to form CH4 and H2O?

Given: balanced chemical equation, K, amounts of reactants and products, and volume

Asked for: direction of reaction

Strategy:

A Calculate the molar concentrations of the reactants and the products.

B Use to determine Q. Compare Q and K to determine in which direction the reaction will proceed.

Solution:

A We must first find the initial concentrations of the substances present. For example, we have 1.2 × 10−2 mol of CH4 in a 2.0 L container, so

[CH4]=1.2×102 mol2.0 L=6.0×103 M

We can calculate the other concentrations in a similar way: [H2O] = 4.0 × 10−3 M, [CO] = 8.0 × 10−3 M, and [H2] = 3.0 × 10−3 M.

B We now compute Q and compare it with K:

Q=[CO][H2]3[CH4][H2O]=(8.0×103)(3.0×103)3(6.0×103)(4.0×103)=9.0×106

Because K = 2.4 × 10−4, we see that Q < K. Thus the ratio of the concentrations of products to the concentrations of reactants is less than the ratio for an equilibrium mixture. The reaction will therefore proceed to the right as written, forming H2 and CO at the expense of H2O and CH4.

Exercise

In the water–gas shift reaction introduced in Example 10, carbon monoxide produced by steam-reforming reaction of methane reacts with steam at elevated temperatures to produce more hydrogen:

CO(g)+H2O(g)CO2(g)+H2(g)

K = 0.64 at 900 K. If 0.010 mol of both CO and H2O, 0.0080 mol of CO2, and 0.012 mol of H2 are injected into a 4.0 L reactor and heated to 900 K, will the reaction proceed to the left or to the right as written?

Answer: Q = 0.96 (Q > K), so the reaction will proceed to the left, and CO and H2O will form.

Predicting the Direction of a Reaction with a Graph

By graphing a few equilibrium concentrations for a system at a given temperature and pressure, we can readily see the range of reactant and product concentrations that correspond to equilibrium conditions, for which Q = K. Such a graph allows us to predict what will happen to a reaction when conditions change so that Q no longer equals K, such as when a reactant concentration or a product concentration is increased or decreased.

Lead carbonate decomposes to lead oxide and carbon dioxide according to the following equation:

Equation 15.30

PbCO3(s)PbO(s)+CO2(g)

Because PbCO3 and PbO are solids, the equilibrium constant is simply K = [CO2]. At a given temperature, therefore, any system that contains solid PbCO3 and solid PbO will have exactly the same concentration of CO2 at equilibrium, regardless of the ratio or the amounts of the solids present. This situation is represented in , which shows a plot of [CO2] versus the amount of PbCO3 added. Initially, the added PbCO3 decomposes completely to CO2 because the amount of PbCO3 is not sufficient to give a CO2 concentration equal to K. Thus the left portion of the graph represents a system that is not at equilibrium because it contains only CO2(g) and PbO(s). In contrast, when just enough PbCO3 has been added to give [CO2] = K, the system has reached equilibrium, and adding more PbCO3 has no effect on the CO2 concentration: the graph is a horizontal line. Thus any CO2 concentration that is not on the horizontal line represents a nonequilibrium state, and the system will adjust its composition to achieve equilibrium, provided enough PbCO3 and PbO are present. For example, the point labeled A in lies above the horizontal line, so it corresponds to a [CO2] that is greater than the equilibrium concentration of CO2 (Q > K). To reach equilibrium, the system must decrease [CO2], which it can do only by reacting CO2 with solid PbO to form solid PbCO3. Thus the reaction in will proceed to the left as written, until [CO2] = K. Conversely, the point labeled B in lies below the horizontal line, so it corresponds to a [CO2] that is less than the equilibrium concentration of CO2 (Q < K). To reach equilibrium, the system must increase [CO2], which it can do only by decomposing solid PbCO3 to form CO2 and solid PbO. The reaction in will therefore proceed to the right as written, until [CO2] = K.

Figure 15.7 The Concentration of Gaseous CO2 in a Closed System at Equilibrium as a Function of the Amount of Solid PbCO3 Added

Initially the concentration of CO2(g) increases linearly with the amount of solid PbCO3 added, as PbCO3 decomposes to CO2(g) and solid PbO. Once the CO2 concentration reaches the value that corresponds to the equilibrium concentration, however, adding more solid PbCO3 has no effect on [CO2], as long as the temperature remains constant.

In contrast, the reduction of cadmium oxide by hydrogen gives metallic cadmium and water vapor:

Equation 15.31

CdO(s)+H2(g)Cd(s)+H2O(g)

and the equilibrium constant K is [H2O]/[H2]. If [H2O] is doubled at equilibrium, then [H2] must also be doubled for the system to remain at equilibrium. A plot of [H2O] versus [H2] at equilibrium is a straight line with a slope of K (). Again, only those pairs of concentrations of H2O and H2 that lie on the line correspond to equilibrium states. Any point representing a pair of concentrations that does not lie on the line corresponds to a nonequilibrium state. In such cases, the reaction in will proceed in whichever direction causes the composition of the system to move toward the equilibrium line. For example, point A in lies below the line, indicating that the [H2O]/[H2] ratio is less than the ratio of an equilibrium mixture (Q < K). Thus the reaction in will proceed to the right as written, consuming H2 and producing H2O, which causes the concentration ratio to move up and to the left toward the equilibrium line. Conversely, point B in lies above the line, indicating that the [H2O]/[H2] ratio is greater than the ratio of an equilibrium mixture (Q > K). Thus the reaction in will proceed to the left as written, consuming H2O and producing H2, which causes the concentration ratio to move down and to the right toward the equilibrium line.

Figure 15.8 The Concentration of Water Vapor versus the Concentration of Hydrogen for the CdO(s)+H2(g)Cd(s)+H2O(g) System at Equilibrium

For any equilibrium concentration of H2O(g), there is only one equilibrium concentration of H2(g). Because the magnitudes of the two concentrations are directly proportional, a large [H2O] at equilibrium requires a large [H2] and vice versa. In this case, the slope of the line is equal to K.

In another example, solid ammonium iodide dissociates to gaseous ammonia and hydrogen iodide at elevated temperatures:

Equation 15.32

NH4I(s)NH3(g)+HI(g)

For this system, K is equal to the product of the concentrations of the two products: [NH3][HI]. If we double the concentration of NH3, the concentration of HI must decrease by approximately a factor of 2 to maintain equilibrium, as shown in . As a result, for a given concentration of either HI or NH3, only a single equilibrium composition that contains equal concentrations of both NH3 and HI is possible, for which [NH3] = [HI] = K1/2. Any point that lies below and to the left of the equilibrium curve (such as point A in ) corresponds to Q < K, and the reaction in will therefore proceed to the right as written, causing the composition of the system to move toward the equilibrium line. Conversely, any point that lies above and to the right of the equilibrium curve (such as point B in ) corresponds to Q > K, and the reaction in will therefore proceed to the left as written, again causing the composition of the system to move toward the equilibrium line. By graphing equilibrium concentrations for a given system at a given temperature and pressure, we can predict the direction of reaction of that mixture when the system is not at equilibrium.

Figure 15.9 The Concentration of NH3(g) versus the Concentration of HI(g) for the NH4I(s)NH3(g)+HI(g) System at Equilibrium

Only one equilibrium concentration of NH3(g) is possible for any given equilibrium concentration of HI(g). In this case, the two are inversely proportional. Thus a large [HI] at equilibrium requires a small [NH3] at equilibrium and vice versa.

Le Châtelier’s Principle

When a system at equilibrium is perturbed in some way, the effects of the perturbation can be predicted qualitatively using Le Châtelier’s principleIf a stress is applied to a system at equilibrium, the composition of the system will change to relieve the applied stress. (named after the French chemist Henri Louis Le Châtelier, 1850–1936).The name is pronounced “Luh SHOT-lee-ay.” This principle can be stated as follows: if a stress is applied to a system at equilibrium, the composition of the system will change to counteract the applied stress. Stress occurs when any change in a system affects the magnitude of Q or K. In , for example, increasing [NH3] produces a stress on the system that requires a decrease in [HI] for the system to return to equilibrium. As a further example, consider esters, which are one of the products of an equilibrium reaction between a carboxylic acid and an alcohol. (For more information on this type of reaction, see , .) Esters are responsible for the scents we associate with fruits (such as oranges and bananas), and they are also used as scents in perfumes. Applying a stress to the reaction of a carboxylic acid and an alcohol will change the composition of the system, leading to an increase or a decrease in the amount of ester produced. In and , we explore how chemists control reactions conditions to affect equilibrium concentrations.

Note the Pattern

In all reactions, if a stress is applied to a system at equilibrium, the composition of the system will change to counteract the applied stress (Le Châtelier’s principle).

Example 15

Write an equilibrium constant expression for each reaction and use this expression to predict what will happen to the concentration of the substance in bold when the indicated change is made if the system is to maintain equilibrium.

  1. 2HgO(s) ⇌ 2Hg(l) + O2(g): the amount of HgO is doubled.
  2. NH4HS(s) ⇌ NH3(g) + H2S(g): the concentration of H2S is tripled.
  3. n-butane(g) ⇌ isobutene(g): the concentration of isobutane is halved.

Given: equilibrium systems and changes

Asked for: equilibrium constant expressions and effects of changes

Strategy:

Write the equilibrium constant expression, remembering that pure liquids and solids do not appear in the expression. From this expression, predict the change that must occur to maintain equilibrium when the indicated changes are made.

Solution:

  1. Because HgO(s) and Hg(l) are pure substances, they do not appear in the equilibrium constant expression. Thus, for this reaction, K = [O2]. The equilibrium concentration of O2 is a constant and does not depend on the amount of HgO present. Hence adding more HgO will not affect the equilibrium concentration of O2, so no compensatory change is necessary.
  2. NH4HS does not appear in the equilibrium constant expression because it is a solid. Thus K = [NH3][H2S], which means that the concentrations of the products are inversely proportional. If adding H2S triples the H2S concentration, for example, then the NH3 concentration must decrease by about a factor of 3 for the system to remain at equilibrium so that the product of the concentrations equals K.
  3. For this reaction, K = [isobutane]/[n-butane], so halving the concentration of isobutane means that the n-butane concentration must also decrease by about half if the system is to maintain equilibrium.

Exercise

Write an equilibrium constant expression for each reaction. What must happen to the concentration of the substance in bold when the indicated change occurs if the system is to maintain equilibrium?

  1. HBr(g) + NaH(s) ⇌ NaBr(s) + H2(g): the concentration of HBr is decreased by a factor of 3.
  2. 6Li(s) + N2(g) ⇌ 2Li3N(s): the amount of Li is tripled.
  3. SO2(g) + Cl2(g) ⇌ SO2Cl2(l): the concentration of Cl2 is doubled.

Answer:

  1. K = [H2]/[HBr]; [H2] must decrease by about a factor of 3.
  2. K = 1/[N2]; solid lithium does not appear in the equilibrium constant expression, so no compensatory change is necessary.
  3. K = 1/[SO2][Cl2]; [SO2] must decrease by about half.

Summary

The reaction quotient (Q or Qp) has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, Q = K. Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Points that do not lie on the line or curve represent nonequilibrium states, and the system will adjust, if it can, to achieve equilibrium. Le Châtelier’s principle states that if a stress is applied to a system at equilibrium, the composition of the system will adjust to counteract the stress.

Key Takeaway

  • The reaction quotient (Q) is used to determine whether a system is at equilibrium and if it is not, to predict the direction of reaction.

Key Equation

Reaction quotient

: Q=[C]c[D]d[A]a[B]b

Conceptual Problems

  1. During a set of experiments, graphs were drawn of [reactants] versus [products] at equilibrium. Using and as your guides, sketch the shape of each graph using appropriate labels.

    1. H2O(l)H2O(g)
    2. 2MgO(s)2Mg(s)+O2(g)
    3. 2O3(g)3O2(g)
    4. 2PbS(s)+3O2(g)2PbO(s)+2SO2(g)
  2. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?

    1. 2NaHCO3(s) ⇌ Na2CO3(s) + CO2(g) + H2O(g): [CO2] is doubled.
    2. N2F4(g) ⇌ 2NF2(g): [NF] is decreased by a factor of 2.
    3. H2(g) + I2(g) ⇌ 2HI(g): [I2] is doubled.
  3. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?

    1. CS2(g) + 4H2(g) ⇌ CH4(g) + 2H2S(g): [CS2] is doubled.
    2. PCl5(g) ⇌ PCl3(g) + Cl2(g): [Cl2] is decreased by a factor of 2.
    3. 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g): [NO] is doubled.

Answer

    1. K=[CH4][H2S]2[CS2][H2]4; doubling [CS2] would require decreasing [H2] by a factor of 24 1.189.
    2. K=[PCl3][Cl2][PCl5]; if [Cl2] is halved, [PCl5] must also be halved.
    3. K=[NO]4[H2O]6[NH3][O2]5; if [NO] is doubled, [H2O] is multiplied by 22/31.587.

Numerical Problems

  1. The data in the following table were collected at 450°C for the reaction N2(g)+3H2(g)2NH3(g):

    Equilibrium Partial Pressure (atm)
    P (atm) NH3 N2 H2
    30 (equilibrium) 1.740 6.588 21.58
    100 15.20 19.17 65.13
    600 321.6 56.74 220.8

    The reaction equilibrates at a pressure of 30 atm. The pressure on the system is first increased to 100 atm and then to 600 atm. Is the system at equilibrium at each of these higher pressures? If not, in which direction will the reaction proceed to reach equilibrium?

  2. For the reaction 2AB+3C, K at 200°C is 2.0. A 6.00 L flask was used to carry out the reaction at this temperature. Given the experimental data in the following table, all at 200°C, when the data for each experiment were collected, was the reaction at equilibrium? If it was not at equilibrium, in which direction will the reaction proceed?

    Experiment A B C
    1 2.50 M 2.50 M 2.50 M
    2 1.30 atm 1.75 atm 14.15 atm
    3 12.61 mol 18.72 mol 6.51 mol
  3. The following two reactions are carried out at 823 K:

    CoO(s)+H2(g)Co(s)+H2O(g)K=67 CoO(s)+CO(g)Co(s)+CO2(g)K=490
    1. Write the equilibrium expression for each reaction.
    2. Calculate the partial pressure of both gaseous components at equilibrium in each reaction if a 1.00 L reaction vessel initially contains 0.316 mol of H2 or CO plus 0.500 mol CoO.
    3. Using the information provided, calculate Kp for the following reaction:

      H2(g)+CO2(g)CO(g)+H2O(g)
    4. Describe the shape of the graphs of [reactants] versus [products] as the amount of CoO changes.
  4. Hydrogen iodide (HI) is synthesized via H2(g)+I2(g)2HI(g), for which Kp = 54.5 at 425°C. Given a 2.0 L vessel containing 1.12 × 10−2 mol of H2 and 1.8 × 10−3 mol of I2 at equilibrium, what is the concentration of HI? Excess hydrogen is added to the vessel so that the vessel now contains 3.64 × 10−1 mol of H2. Calculate Q and then predict the direction in which the reaction will proceed. What are the new equilibrium concentrations?

Answers

  1. Not at equilibrium; in both cases, the sum of the equilibrium partial pressures is less than the total pressure, so the reaction will proceed to the right to decrease the pressure.

    1. K=[H2O][H2]K=[CO2][CO]
    2. PH2O = 21.0 atm; PH2 = 0.27 atm; PCO2 = 21.3 atm; PCO = 0.07 atm
    3. Kp = 0.14
    4. The amount of CoO has no effect on the shape of a graph of products versus reactants as long as some solid CoO is present.

15.5 Factors That Affect Equilibrium

Learning Objective

  1. To predict the effects of stresses on a system at equilibrium.

Chemists use various strategies to increase the yield of the desired products of reactions. When synthesizing an ester, for example, how can a chemist control the reaction conditions to obtain the maximum amount of the desired product? Only three types of stresses can change the composition of an equilibrium mixture: (1) a change in the concentrations (or partial pressures) of the components by adding or removing reactants or products, (2) a change in the total pressure or volume, and (3) a change in the temperature of the system. In this section, we explore how changes in reaction conditions can affect the equilibrium composition of a system. We will explore each of these possibilities in turn.

Changes in Concentration

If we add a small volume of carbon tetrachloride (CCl4) solvent to a flask containing crystals of iodine, we obtain a saturated solution of I2 in CCl4, along with undissolved crystals:

Equation 15.33

I2(s)solventI2(soln)

The system reaches equilibrium, with K = [I2]. If we add more CCl4, thereby diluting the solution, Q is now less than K. Le Châtelier’s principle tells us that the system will react to relieve the stress—but how? Adding solvent stressed the system by decreasing the concentration of dissolved I2. Hence more crystals will dissolve, thereby increasing the concentration of dissolved I2 until the system again reaches equilibrium if enough solid I2 is available (). By adding solvent, we drove the reaction shown in to the right as written.

Figure 15.10 The Concentration of Dissolved I2 as a Function of Time Following the Addition of More Solvent to a Saturated Solution in Contact with Excess Solid I2

The concentration of I2 decreases initially due to dilution but returns to its original value as long as solid I2 is present.

We encounter a more complex system in the reaction of hydrogen and nitrogen to form ammonia:

Equation 15.34

N2(g)+3H2(g)2NH3(g)

The Kp for this reaction is 2.14 × 10−2 at about 540 K. Under one set of equilibrium conditions, the partial pressure of ammonia is PNH3 = 0.454 atm, that of hydrogen is PH2 = 2.319 atm, and that of nitrogen is PN2 = 0.773 atm. If an additional 1 atm of hydrogen is added to the reactor to give PH2 = 3.319 atm, how will the system respond? Because the stress is an increase in PH2, the system must respond in some way that decreases the partial pressure of hydrogen to counteract the stress. The reaction will therefore proceed to the right as written, consuming H2 and N2 and forming additional NH3. Initially, the partial pressures of H2 and N2 will decrease, and the partial pressure of NH3 will increase until the system eventually reaches a new equilibrium composition, which will have a net increase in PH2.

We can confirm that this is indeed what will happen by evaluating Qp under the new conditions and comparing its value with Kp. The equations used to evaluate Kp and Qp have the same form: substituting the values after adding hydrogen into the expression for Qp results in the following:

Qp=(PNH3)2(PN2)(PH2)3=(0.454)2(0.773)(2.319+1.00)3=7.29×103

Thus Qp < Kp, which tells us that the ratio of products to reactants is less than at equilibrium. To reach equilibrium, the reaction must proceed to the right as written: the partial pressures of the products will increase, and the partial pressures of the reactants will decrease. Qp will thereby increase until it equals Kp, and the system will once again be at equilibrium. Changes in the partial pressures of the various substances in the reaction mixture () as a function of time are shown in .

Figure 15.11 The Partial Pressures of H2, N2, and NH3 as a Function of Time Following the Addition of More H2 to an Equilibrium Mixture

Some of the added hydrogen is consumed by reacting with nitrogen to produce more ammonia, allowing the system to reach a new equilibrium composition.

We can force a reaction to go essentially to completion, regardless of the magnitude of K, by continually removing one of the products from the reaction mixture. Consider, for example, the methanation reaction, in which hydrogen reacts with carbon monoxide to form methane and water:

Equation 15.35

CO(g)+3H2(g)CH4(g)+H2O(g)

This reaction is used for the industrial production of methane, whereas the reverse reaction is used for the production of H2 (Example 14). The expression for Q has the following form:

Equation 15.36

Q=[CH4][H2O][CO][H2]3

Regardless of the magnitude of K, if either H2O or CH4 can be removed from the reaction mixture so that [H2O] or [CH4] is approximately zero, then Q ≈ 0. In other words, when product is removed, the system is stressed (Q << K), and more product will form to counter the stress. Because water (bp = 100°C) is much less volatile than methane, hydrogen, or carbon monoxide (all of which have boiling points below −100°C), passing the gaseous reaction mixture through a cold coil will cause the water vapor to condense to a liquid that can be drawn off. Continuing to remove water from the system forces the reaction to the right as the system attempts to equilibrate, thus enriching the reaction mixture in methane. This technique, referred to as driving a reaction to completion, can be used to force a reaction to completion even if K is relatively small. For example, esters are usually synthesized by removing water. The products of the condensation reaction are shown here. In , we will describe the thermodynamic basis for the change in the equilibrium position caused by changes in the concentrations of reaction components.

Example 16

For each equilibrium system, predict the effect of the indicated stress on the specified quantity.

  1. 2SO2(g)+O2(g)2SO3(g): (1) the effect of removing O2 on PSO2; (2) the effect of removing O2 on PSO3
  2. CaCO3(s)CaO(s)+CO2(g): (1) the effect of removing CO2 on the amount of CaCO3; (2) the effect of adding CaCO3 on PCO2

Given: balanced chemical equations and changes

Asked for: effects of indicated stresses

Strategy:

Use Q and K to predict the effect of the stress on each reaction.

Solution:

  1. (1) Removing O2 will decrease PO2, thereby decreasing the denominator in the reaction quotient and making Qp > Kp. The reaction will proceed to the left as written, increasing the partial pressures of SO2 and O2 until Qp once again equals Kp. (2) Removing O2 will decrease PO2 and thus increase Qp, so the reaction will proceed to the left. The partial pressure of SO3 will decrease.
  2. Kp and Qp are both equal to PCO2. (1) Removing CO2 from the system causes more CaCO3 to react to produce CO2, which increases PCO2 to the partial pressure required by Kp. (2) Adding (or removing) solid CaCO3 has no effect on PCO2 because it does not appear in the expression for Kp (or Qp).

Exercise

For each equilibrium system, predict the effect that the indicated stress will have on the specified quantity.

  1. H2(g)+CO2(g)H2O(g)+CO(g): (1) the effect of adding CO on [H2]; (2) the effect of adding CO2 on [H2]
  2. CuO(s)+CO(g)Cu(s)+CO2(g): (1) the effect of adding CO on the amount of Cu; (2) the effect of adding CO2 on [CO]

Answer:

  1. (1) [H2] increases; (2) [H2] decreases.
  2. (1) the amount of Cu increases; (2) [CO] increases.

Changes in Total Pressure or Volume

Because liquids are relatively incompressible, changing the pressure above a liquid solution has little effect on the concentrations of dissolved substances. Consequently, changes in external pressure have very little effect on equilibrium systems that contain only solids or liquids. In contrast, because gases are highly compressible, their concentrations vary dramatically with pressure. From the ideal gas law, PV = nRT, described in , the concentration (C) of a gas is related to its pressure as follows:

Equation 15.37

C=nV=PRT

Hence the concentration of any gaseous reactant or product is directly proportional to the applied pressure (P) and inversely proportional to the total volume (V). Consequently, the equilibrium compositions of systems that contain gaseous substances are quite sensitive to changes in pressure, volume, and temperature.

These principles can be illustrated using the reversible dissociation of gaseous N2O4 to gaseous NO2 (). The syringe shown in initially contains an equilibrium mixture of colorless N2O4 and red-brown NO2. Decreasing the volume by 50% causes the mixture to become darker because all concentrations have doubled. Decreasing the volume also constitutes a stress, however, as we can see by examining the effect of a change in volume on Q. At equilibrium, Q = K = [NO2]2/[N2O4] (). If the volume is decreased by half, the concentrations of the substances in the mixture are doubled, so the new reaction quotient is as follows:

Equation 15.38

Q=(2[NO2]i)22[N2O4]i=4([NO2]i)22[N2O4]i=2K

Because Q is now greater than K, the system is no longer at equilibrium. The stress can be relieved if the reaction proceeds to the left, consuming 2 mol of NO2 for every 1 mol of N2O4 produced. This will decrease the concentration of NO2 and increase the concentration of N2O4, causing Q to decrease until it once again equals K. Thus, as shown in part (c) in , the intensity of the brown color due to NO2 decreases with time following the change in volume.

Figure 15.12 The Effect of Changing the Volume (and Thus the Pressure) of an Equilibrium Mixture of N2O4 and NO2 at Constant Temperature

(a) The syringe with a total volume of 15 mL contains an equilibrium mixture of N2O4 and NO2; the red-brown color is proportional to the NO2 concentration. (b) If the volume is rapidly decreased by a factor of 2 to 7.5 mL, the initial effect is to double the concentrations of all species present, including NO2. Hence the color becomes more intense. (c) With time, the system adjusts its composition in response to the stress as predicted by Le Châtelier’s principle, forming colorless N2O4 at the expense of red-brown NO2, which decreases the intensity of the color of the mixture.

Note the Pattern

Increasing the pressure of a system (or decreasing the volume) favors the side of the reaction that has fewer gaseous molecules and vice versa.

In general, if a balanced chemical equation contains different numbers of gaseous reactant and product molecules, the equilibrium will be sensitive to changes in volume or pressure. Increasing the pressure on a system (or decreasing the volume) will favor the side of the reaction that has fewer gaseous molecules and vice versa.

Example 17

For each equilibrium system, write the reaction quotient for the system if the pressure is decreased by a factor of 2 (i.e., if the volume is doubled) at constant temperature and then predict the direction of the reaction.

  1. N2(g)+3H2(g)2NH3(g)
  2. C2H2(g)+C2H6(g)2C2H4(g)
  3. 2NO2(g)2NO(g)+O2(g)

Given: balanced chemical equations

Asked for: direction of reaction if pressure is halved

Strategy:

Use Le Châtelier’s principle to predict the effect of the stress.

Solution:

  1. Two moles of gaseous products are formed from 4 mol of gaseous reactants. Decreasing the pressure will cause the reaction to shift to the left because that side contains the larger number of moles of gas. Thus the pressure increases, counteracting the stress. K for this reaction is [NH3]2/[N2][H2]3. When the pressure is decreased by a factor of 2, the concentrations are halved, which means that the new reaction quotient is as follows:

    Q=[1/2NH3]2[1/2N2][1/2H2]3=1/4NH321/16[N2][H2]3=4K

  2. Two moles of gaseous products form from 2 mol of gaseous reactants. Decreasing the pressure will have no effect on the equilibrium composition because both sides of the balanced chemical equation have the same number of moles of gas. Here K = [C2H4]2/[C2H2][C2H6]. The new reaction quotient is as follows:

    Q=[C2H4]2[C2H2][C2H6]=[1/2C2H4]2[1/2C2H2][1/2C2H6]=1/4[C2H4]21/4[C2H2][C2H6]=K

  3. Three moles of gaseous products are formed from 2 mol of gaseous reactants. Decreasing the pressure will favor the side that contains more moles of gas, so the reaction will shift toward the products to increase the pressure. For this reaction K = [NO]2[O2]/[NO2]2. Under the new reaction conditions the reaction quotient is as follows:

    Q=[1/2NO]2[1/2O2][1/2NO2]2=1/8[NO]2[O2]1/4[NO2]2=1/2K

Exercise

For each equilibrium system, write a new reaction quotient for the system if the pressure is increased by a factor of 2 (i.e., if the volume is halved) at constant temperature and then predict the direction in which the reaction will shift.

  1. H2O(g)+CO(g)H2(g)+CO2(g)
  2. H2(g)+C2H4(g)C2H6(g)
  3. 2SO2(g)+O2(g)2SO3(g)

Answer:

  1. Q = K; no effect
  2. Q = 1/2 K; to the right
  3. Q = 1/2 K; to the right

Changes in Temperature

In all the cases we have considered so far, the magnitude of the equilibrium constant, K or Kp, was constant. Changes in temperature can, however, change the value of the equilibrium constant without immediately affecting the reaction quotient (QK). In this case, the system is no longer at equilibrium; the composition of the system will change until Q equals K at the new temperature.

To predict how an equilibrium system will respond to a change in temperature, we must know something about the enthalpy change of the reaction (ΔHrxn). As you learned in , heat is released to the surroundings in an exothermic reaction (ΔHrxn < 0), and heat is absorbed from the surroundings in an endothermic reaction (ΔHrxn > 0). We can express these changes in the following way:

Equation 15.39

Exothermic: reactantsproducts+heat(ΔH<0)

Equation 15.40

Endothermic: reactants + heatproducts(ΔH>0)

Thus heat can be thought of as a product in an exothermic reaction and as a reactant in an endothermic reaction. Increasing the temperature of a system corresponds to adding heat. Le Châtelier’s principle predicts that an exothermic reaction will shift to the left (toward the reactants) if the temperature of the system is increased (heat is added). Conversely, an endothermic reaction will shift to the right (toward the products) if the temperature of the system is increased. If a reaction is thermochemically neutral (ΔHrxn = 0), then a change in temperature will not affect the equilibrium composition.

We can examine the effects of temperature on the dissociation of N2O4 to NO2, for which ΔH = +58 kJ/mol. This reaction can be written as follows:

Equation 15.41

58 kJ+N2O4(g)2NO2(g)

Increasing the temperature (adding heat to the system) is a stress that will drive the reaction to the right, as illustrated in . Thus increasing the temperature increases the ratio of NO2 to N2O4 at equilibrium, which increases K.

Figure 15.13 The Effect of Temperature on the Equilibrium between Gaseous N2O4 and NO2

(center) A tube containing a mixture of N2O4 and NO2 in the same proportion at room temperature is red-brown due to the NO2 present. (left) Immersing the tube in ice water causes the mixture to become lighter in color due to a shift in the equilibrium composition toward colorless N2O4. (right) In contrast, immersing the same tube in boiling water causes the mixture to become darker due to a shift in the equilibrium composition toward the highly colored NO2.

The effect of increasing the temperature on a system at equilibrium can be summarized as follows: increasing the temperature increases the magnitude of the equilibrium constant for an endothermic reaction, decreases the equilibrium constant for an exothermic reaction, and has no effect on the equilibrium constant for a thermally neutral reaction. shows the temperature dependence of the equilibrium constants for the synthesis of ammonia from hydrogen and nitrogen, which is an exothermic reaction with ΔH° = −91.8 kJ/mol. The values of both K and Kp decrease dramatically with increasing temperature, as predicted for an exothermic reaction.

Table 15.3 Temperature Dependence of K and Kp for N2(g)+3H2(g)2NH3(g)

Temperature (K) K K p
298 3.3 × 108 5.6 × 105
400 3.9 × 104 3.6 × 101
450 2.6 × 103 1.9
500 1.7 × 102 1.0 × 10−1
550 2.6 × 101 1.3 × 10−2
600 4.1 1.7 × 10−3

Note the Pattern

Increasing the temperature causes endothermic reactions to favor products and exothermic reactions to favor reactants.

Example 18

For each equilibrium reaction, predict the effect of decreasing the temperature:

  1. N2(g)+3H2(g)2NH3(g)    ΔHrxn=91.8kJ/mol
  2. CaCO3(s)CaO(s)+CO2(g)    ΔHrxn=178kJ/mol

Given: balanced chemical equations and values of ΔHrxn

Asked for: effects of decreasing temperature

Strategy:

Use Le Châtelier’s principle to predict the effect of decreasing the temperature on each reaction.

Solution:

  1. The formation of NH3 is exothermic, so we can view heat as one of the products:

    N2(g)+3H2(g)2NH3(g)+91.8 kJ

    If the temperature of the mixture is decreased, heat (one of the products) is being removed from the system, which causes the equilibrium to shift to the right. Hence the formation of ammonia is favored at lower temperatures.

  2. The decomposition of calcium carbonate is endothermic, so heat can be viewed as one of the reactants:

    CaCO3(s)+178 kJCaO(s)+CO2(g)

    If the temperature of the mixture is decreased, heat (one of the reactants) is being removed from the system, which causes the equilibrium to shift to the left. Hence the thermal decomposition of calcium carbonate is less favored at lower temperatures.

Exercise

For each equilibrium system, predict the effect of increasing the temperature on the reaction mixture:

  1. 2SO2(g)+O2(g)2SO3(g)    ΔHrxn=198 kJ/mol
  2. N2(g)+O2(g)2NO(g)    ΔHrxn=+181 kJ/mol

Answer:

  1. Reaction shifts to the left.
  2. Reaction shifts to the right.

Summary

Three types of stresses can alter the composition of an equilibrium system: adding or removing reactants or products, changing the total pressure or volume, and changing the temperature of the system. A reaction with an unfavorable equilibrium constant can be driven to completion by continually removing one of the products of the reaction. Equilibriums that contain different numbers of gaseous reactant and product molecules are sensitive to changes in volume or pressure; higher pressures favor the side with fewer gaseous molecules. Removing heat from an exothermic reaction favors the formation of products, whereas removing heat from an endothermic reaction favors the formation of reactants.

Key Takeaway

  • Equilibriums are affected by changes in concentration, total pressure or volume, and temperature.

Conceptual Problems

  1. If an equilibrium reaction is endothermic in the forward direction, what is the expected change in the concentration of each component of the system if the temperature of the reaction is increased? If the temperature is decreased?

  2. Write the equilibrium equation for the following system:

    4NH3(g)+5O2(g)4NO(g)+6H2O(g)

    Would you expect the equilibrium to shift toward the products or reactants with an increase in pressure? Why?

  3. The reaction rate approximately doubles for every 10°C rise in temperature. What happens to K?

  4. The formation of A2B2(g) via the equilibrium reaction 2AB(g)A2B2(g) is exothermic. What happens to the ratio kf/kr if the temperature is increased? If both temperature and pressure are increased?

  5. In each system, predict the effect that the indicated change will have on the specified quantity at equilibrium:

    1. H2(g)+I2(g)2HI(g)

      H2 is removed; what is the effect on PI2?

    2. 2NOBr(g)2NO(g)+Br2(g)

      Br2 is removed; what is the effect on PNOBr?

    3. 2NaHCO3(s)Na2CO3(g)+CO2(g)+H2O(g)

      CO2 is removed; what is the effect on PNaHCO3?

  6. What effect will the indicated change have on the specified quantity at equilibrium?

    1. NH4Cl(s)NH3(g)+HCl(g)

      NH4Cl is increased; what is the effect on PHCl?

    2. 2H2O(g)2H2(g)+O2(g)

      O2 is added; what is the effect on PH2?

    3. PCl3(g)+Cl2(g)PCl5(g)

      Cl2 is removed; what is the effect on PPCl5?

Numerical Problems

  1. For each equilibrium reaction, describe how Q and K change when the pressure is increased, the temperature is increased, the volume of the system is increased, and the concentration(s) of the reactant(s) is increased.

    1. A(g)B(g)ΔH=20.6 kJ/mol
    2. 2A(g)B(g)ΔH=0.3 kJ/mol
    3. A(g)+B(g)2C(g)ΔH=46 kJ/mol
  2. For each equilibrium reaction, describe how Q and K change when the pressure is decreased, the temperature is increased, the volume of the system is decreased, and the concentration(s) of the reactant(s) is increased.

    1. 2A(g)B(g)ΔH=80 kJ/mol
    2. A(g)2B(g)ΔH=0.3 kJ/mol
    3. 2A(g)2B(g)+C(g)ΔH=46 kJ/mol
  3. Le Châtelier’s principle states that a system will change its composition to counteract stress. For the system CO(g) + Cl2(g)COCl2(g), write the equilibrium constant expression Kp. What changes in the values of Q and K would you anticipate when (a) the volume is doubled, (b) the pressure is increased by a factor of 2, and (c) COCl2 is removed from the system?

  4. For the equilibrium system 3O2(g)2O3(g), ΔH° = 284 kJ, write the equilibrium constant expression Kp. What happens to the values of Q and K if the reaction temperature is increased? What happens to these values if both the temperature and pressure are increased?

  5. Carbon and oxygen react to form CO2 gas via C(s) + O2(g)CO2(g), for which K = 1.2 × 1069. Would you expect K to increase or decrease if the volume of the system were tripled? Why?

  6. The reaction COCl2(g)CO(g)+Cl2(g) has K = 2.2 × 10−10 at 100°C. Starting with an initial PCOCl2 of 1.0 atm, you determine the following values of PCO at three successive time intervals: 6.32 × 10−6 atm, 1.78 × 10−6 atm, and 1.02 × 10−5 atm. Based on these data, in which direction will the reaction proceed after each measurement? If chlorine gas is added to the system, what will be the effect on Q?

  7. The following table lists experimentally determined partial pressures at three temperatures for the reaction Br2(g)2Br(g).

    T (K) 1123 1173 1273
    PBr2 (atm) 3.000 0.3333 6.755 × 10−2
    PBr (atm) 3.477 × 10−2 2.159 × 10−2 2.191 × 10−2

    Is this an endothermic or an exothermic reaction? Explain your reasoning.

  8. The dissociation of water vapor proceeds according to the following reaction: H2O(g)12O2(g)+H2(g). At 1300 K, there is 0.0027% dissociation, whereas at 2155 K, the dissociation is 1.18%. Calculate K and Kp. Is this an endothermic reaction or an exothermic reaction? How do the magnitudes of the two equilibriums compare? Would increasing the pressure improve the yield of H2 gas at either temperature? (Hint: assume that the system initially contains 1.00 mol of H2O in a 1.00 L container.)

  9. When 1.33 mol of CO2 and 1.33 mol of H2 are mixed in a 0.750 L container and heated to 395°C, they react according to the following equation: CO2(g)+H2(g)CO(g)+H2O(g). If K = 0.802, what are the equilibrium concentrations of each component of the equilibrium mixture? What happens to K if H2O is removed during the course of the reaction?

  10. The equilibrium reaction H2(g)+Br2(g)2HBr(g) has Kp = 2.2 × 109 at 298 K. If you begin with 2.0 mol of Br2 and 2.0 mol of H2 in a 5.0 L container, what is the partial pressure of HBr at equilibrium? What is the partial pressure of H2 at equilibrium? If H2 is removed from the system, what is the effect on the partial pressure of Br2?

  11. Iron(II) oxide reacts with carbon monoxide according to the following equation: FeO(s) + CO(g)Fe(s)+CO2(g). At 800°C, K = 0.34; at 1000°C, K = 0.40.

    1. A 20.0 L container is charged with 800.0 g of CO2, 1436 g of FeO, and 1120 g of iron. What are the equilibrium concentrations of all components of the mixture at each temperature?
    2. What are the partial pressures of the gases at each temperature?
    3. If CO were removed, what would be the effect on PCO2 at each temperature?
  12. The equilibrium constant K for the reaction C(s) + CO2(g)2CO(g) is 1.9 at 1000 K and 0.133 at 298 K.

    1. If excess C is allowed to react with 25.0 g of CO2 in a 3.00 L flask, how many grams of CO are produced at each temperature?
    2. What are the partial pressures of each gas at 298 K? at 1000 K?
    3. Would you expect K to increase or decrease if the pressure were increased at constant temperature and volume?
  13. Data for the oxidation of methane, CH4(g)+2O2(g)CO2(g)+2H2O(g), in a closed 5.0 L vessel are listed in the following table. Fill in the blanks and determine the missing values of Q and K (indicated by ?) as the reaction is driven to completion.

    CH4 O2 CO2 H2O Q K
    initial (moles) 0.45 0.90 0 0 ?
    at equilibrium 1.29
    add 0.50 mol of methane 0.95 ?
    new equilibrium ?
    remove water 0 ?
    new equilibrium 1.29

Answers

  1.  

    Kp=PCOCl2PCO·PCl2

    None of the changes would affect K; (a) Q doubles; (b) Q is halved; Q decreases.

  2. K would not change; it does not depend on volume.

  3. [CO] = [H2O] = 0.839 M, [CO2] = [H2] = 0.930 M; no effect on K

    1. At 800°C, [CO] = 0.678 M, [CO2] = 0.231 M; at 1000°C, [CO] = 0.645 M, [CO2] = 0.264 M
    2. At 800°C, PCO = 59.7 atm, PCO2 = 20.3 atm; at 1000°C, PCO = 67.4 atm, PCO2 = 27.6 atm.
    3. Removing CO would cause the reaction to shift to the right, causing PCO2 to decrease.
  4. CH4 O2 CO2 H2O Q K
    initial (moles) 0.45 0.90 0 0 0 1.29
    at equilibrium 0.215 0.43 0.235 0.47 K 1.29
    add 0.50 mol of methane 0.715 0.43 0.235 0.47 0.39 1.29
    new equilibrium 0.665 0.33 0.285 0.57 K 1.29
    remove water 0.665 0.33 0.285 0 0 1.29
    new equilibrium 0.57 0.14 0.38 0.19 K 1.29

15.6 Controlling the Products of Reactions

Learning Objective

  1. To understand different ways to control the products of a reaction.

Whether in the synthetic laboratory or in industrial settings, one of the primary goals of modern chemistry is to control the identity and quantity of the products of chemical reactions. For example, a process aimed at synthesizing ammonia is designed to maximize the amount of ammonia produced using a given amount of energy. Alternatively, other processes may be designed to minimize the creation of undesired products, such as pollutants emitted from an internal combustion engine. To achieve these goals, chemists must consider the competing effects of the reaction conditions that they can control.

One way to obtain a high yield of a desired compound is to make the reaction rate of the desired reaction much faster than the reaction rates of any other possible reactions that might occur in the system. Altering reaction conditions to control reaction rates, thereby obtaining a single product or set of products, is called kinetic controlThe altering of reaction conditions to control reaction rates, thereby obtaining a single desired product or set of products.. A second approach, called thermodynamic controlThe altering of reaction conditions so that a single desired product or set of products is present in significant quantities at equilibrium., consists of adjusting conditions so that at equilibrium only the desired products are present in significant quantities.

An example of thermodynamic control is the Haber-Bosch processKarl Bosch (1874–1940) was a German chemical engineer who was responsible for designing the process that took advantage of Fritz Haber’s discoveries regarding the N2 + H2/NH3 equilibrium to make ammonia synthesis via this route cost-effective. He received the Nobel Prize in Chemistry in 1931 for his work. The industrial process is called either the Haber process or the Haber-Bosch process. used to synthesize ammonia via the following reaction:

Equation 15.42

N2(g)+3H2(g)2NH3(g)ΔHrxn=91.8 kJ/mol

Because the reaction converts 4 mol of gaseous reactants to only 2 mol of gaseous product, Le Châtelier’s principle predicts that the formation of NH3 will be favored when the pressure is increased. The reaction is exothermic, however (ΔHrxn = −91.8 kJ/mol), so the equilibrium constant decreases with increasing temperature, which causes an equilibrium mixture to contain only relatively small amounts of ammonia at high temperatures (). Taken together, these considerations suggest that the maximum yield of NH3 will be obtained if the reaction is carried out at as low a temperature and as high a pressure as possible. Unfortunately, at temperatures less than approximately 300°C, where the equilibrium yield of ammonia would be relatively high, the reaction is too slow to be of any commercial use. The industrial process therefore uses a mixed oxide (Fe2O3/K2O) catalyst that enables the reaction to proceed at a significant rate at temperatures of 400°C–530°C, where the formation of ammonia is less unfavorable than at higher temperatures.

Figure 15.14 Effect of Temperature and Pressure on the Equilibrium Composition of Two Systems that Originally Contained a 3:1 Mixture of Hydrogen and Nitrogen

At all temperatures, the total pressure in the systems was initially either 4 atm (purple curves) or 200 atm (green curves). Note the dramatic decrease in the proportion of NH3 at equilibrium at higher temperatures in both cases, as well as the large increase in the proportion of NH3 at equilibrium at any temperature for the system at higher pressure (green) versus lower pressure (purple). Commercial plants that use the Haber-Bosch process to synthesize ammonia on an industrial scale operate at temperatures of 400°C–530°C (indicated by the darker gray band) and total pressures of 130–330 atm.

Because of the low value of the equilibrium constant at high temperatures (e.g., K = 0.039 at 800 K), there is no way to produce an equilibrium mixture that contains large proportions of ammonia at high temperatures. We can, however, control the temperature and the pressure while using a catalyst to convert a fraction of the N2 and H2 in the reaction mixture to NH3, as is done in the Haber-Bosch process. This process also makes use of the fact that the product—ammonia—is less volatile than the reactants. Because NH3 is a liquid at room temperature at pressures greater than 10 atm, cooling the reaction mixture causes NH3 to condense from the vapor as liquid ammonia, which is easily separated from unreacted N2 and H2. The unreacted gases are recycled until complete conversion of hydrogen and nitrogen to ammonia is eventually achieved. is a simplified layout of a Haber-Bosch process plant.

Figure 15.15 A Schematic Diagram of an Industrial Plant for the Production of Ammonia via the Haber-Bosch Process

A 3:1 mixture of gaseous H2 and N2 is compressed to 130–330 atm, heated to 400°C–530°C, and passed over an Fe2O3/K2O catalyst, which results in partial conversion to gaseous NH3. The resulting mixture of gaseous NH3, H2, and N2 is passed through a heat exchanger, which uses the hot gases to prewarm recycled N2 and H2, and a condenser to cool the NH3, giving a liquid that is readily separated from unreacted N2 and H2. (Although the normal boiling point of NH3 is −33°C, the boiling point increases rapidly with increasing pressure, to 20°C at 8.5 atm and 126°C at 100 atm.) The unreacted N2 and H2 are recycled to form more NH3.

The Sohio acrylonitrile process, in which propene and ammonia react with oxygen to form acrylonitrile, is an example of a kinetically controlled reaction:

Equation 15.43

CH2=CHCH3(g)+NH3(g)+32O2(g)CH2=CHCN(g)+3H2O(g)

Like most oxidation reactions of organic compounds, this reaction is highly exothermic (ΔH° = −519 kJ/mol) and has a very large equilibrium constant (K = 1.2 × 1094). Nonetheless, the reaction shown in is not the reaction a chemist would expect to occur when propene or ammonia is heated in the presence of oxygen. Competing combustion reactions that produce CO2 and N2 from the reactants, such as those shown in and , are even more exothermic and have even larger equilibrium constants, thereby reducing the yield of the desired product, acrylonitrile:

Equation 15.44

CH2=CHCH3(g)+92O2(g)3CO2(g)+3H2O(g)ΔH°=-1926.1kJ/mol, K=4.5×10338

Equation 15.45

2NH3(g)+3O2(g)N2(g)+6H2O(g)ΔH°=1359.2 kJ/mol, K=4.4×10234

In fact, the formation of acrylonitrile in is accompanied by the release of approximately 760 kJ/mol of heat due to partial combustion of propene during the reaction.

The Sohio process uses a catalyst that selectively accelerates the rate of formation of acrylonitrile without significantly affecting the reaction rates of competing combustion reactions. Consequently, acrylonitrile is formed more rapidly than CO2 and N2 under the optimized reaction conditions (approximately 1.5 atm and 450°C). The reaction mixture is rapidly cooled to prevent further oxidation or combustion of acrylonitrile, which is then washed out of the vapor with a liquid water spray. Thus controlling the kinetics of the reaction causes the desired product to be formed under conditions where equilibrium is not established. In industry, this reaction is carried out on an enormous scale. Acrylonitrile is the building block of the polymer called polyacrylonitrile, found in all the products referred to collectively as acrylics, whose wide range of uses includes the synthesis of fibers woven into clothing and carpets.

Note the Pattern

Controlling the amount of product formed requires that both thermodynamic and kinetic factors be considered.

Example 19

Recall that methanation is the reaction of hydrogen with carbon monoxide to form methane and water:

CO(g) +3H2(g)CH4(g)+H2O(g)

This reaction is the reverse of the steam reforming of methane described in Example 14. The reaction is exothermic (ΔH° = −206 kJ/mol), with an equilibrium constant at room temperature of Kp = 7.6 × 1024. Unfortunately, however, CO and H2 do not react at an appreciable rate at room temperature. What conditions would you select to maximize the amount of methane formed per unit time by this reaction?

Given: balanced chemical equation and values of ΔH° and K

Asked for: conditions to maximize yield of product

Strategy:

Consider the effect of changes in temperature and pressure and the addition of an effective catalyst on the reaction rate and equilibrium of the reaction. Determine which combination of reaction conditions will result in the maximum production of methane.

Solution:

The products are highly favored at equilibrium, but the rate at which equilibrium is reached is too slow to be useful. You learned in that the reaction rate can often be increased dramatically by increasing the temperature of the reactants. Unfortunately, however, because the reaction is quite exothermic, an increase in temperature will shift the equilibrium to the left, causing more reactants to form and relieving the stress on the system by absorbing the added heat. If we increase the temperature too much, the equilibrium will no longer favor methane formation. (In fact, the equilibrium constant for this reaction is very temperature sensitive, decreasing to only 1.9 × 10−3 at 1000°C.) To increase the reaction rate, we can try to find a catalyst that will operate at lower temperatures where equilibrium favors the formation of products. Higher pressures will also favor the formation of products because 4 mol of gaseous reactant are converted to only 2 mol of gaseous product. Very high pressures should not be needed, however, because the equilibrium constant favors the formation of products. Thus optimal conditions for the reaction include carrying it out at temperatures greater than room temperature (but not too high), adding a catalyst, and using pressures greater than atmospheric pressure.

Industrially, catalytic methanation is typically carried out at pressures of 1–100 atm and temperatures of 250°C–450°C in the presence of a nickel catalyst. (At 425°C°C, Kp is 3.7 × 103, so the formation of products is still favored.) The synthesis of methane can also be favored by removing either H2O or CH4 from the reaction mixture by condensation as they form.

Exercise

As you learned in Example 10, the water–gas shift reaction is as follows:

H2(g)+CO2(g)H2O(g)+CO(g)

Kp = 0.106 and ΔH = 41.2 kJ/mol at 700 K. What reaction conditions would you use to maximize the yield of carbon monoxide?

Answer: high temperatures to increase the reaction rate and favor product formation, a catalyst to increase the reaction rate, and atmospheric pressure because the equilibrium will not be greatly affected by pressure

Summary

Changing conditions to affect the reaction rates to obtain a single product is called kinetic control of the system. In contrast, thermodynamic control is adjusting the conditions to ensure that only the desired product or products are present in significant concentrations at equilibrium.

Key Takeaway

  • Both kinetic and thermodynamic factors can be used to control reaction products.

Conceptual Problems

  1. A reaction mixture will produce either product A or B depending on the reaction pathway. In the absence of a catalyst, product A is formed; in the presence of a catalyst, product B is formed. What conclusions can you draw about the forward and reverse rates of the reaction that produces A versus the reaction that produces B in (a) the absence of a catalyst and (b) the presence of a catalyst?

  2. Describe how you would design an experiment to determine the equilibrium constant for the synthesis of ammonia:

    N2(g)+3H2(g)2NH3(g)

    The forward reaction is exothermic (ΔH° = −91.8 kJ). What effect would an increase in temperature have on the equilibrium constant?

  3. What effect does a catalyst have on each of the following?

    1. the equilibrium position of a reaction
    2. the rate at which equilibrium is reached
    3. the equilibrium constant?
  4. How can the ratio Q/K be used to determine in which direction a reaction will proceed to reach equilibrium?

  5. Industrial reactions are frequently run under conditions in which competing reactions can occur. Explain how a catalyst can be used to achieve reaction selectivity. Does the ratio Q/K for the selected reaction change in the presence of a catalyst?

Numerical Problems

  1. The oxidation of acetylene via 2C2H2(g)+5O2(g)4CO2(g)+2H2O(l) has ΔH° = −2600 kJ. What strategy would you use with regard to temperature, volume, and pressure to maximize the yield of product?

  2. The oxidation of carbon monoxide via CO(g)  + 12O2(g)CO2(g) has ΔH° = −283 kJ. If you were interested in maximizing the yield of CO2, what general conditions would you select with regard to temperature, pressure, and volume?

  3. You are interested in maximizing the product yield of the system

    2SO2(g)+O2(g)2SO3(g)

    K = 280 and ΔH° = −158 kJ. What general conditions would you select with regard to temperature, pressure, and volume? If SO2 has an initial concentration of 0.200 M and the amount of O2 is stoichiometric, what amount of SO3 is produced at equilibrium?

Answer

  1. Use low temperature and high pressure (small volume).

15.7 Essential Skills

Topic

  • The quadratic formula

Previous Essential Skills sections introduced many of the mathematical operations you need to solve chemical problems. We now introduce the quadratic formula, a mathematical relationship involving sums of powers in a single variable that you will need to apply to solve some of the problems in this chapter.

The Quadratic Formula

Mathematical expressions that involve a sum of powers in one or more variables (e.g., x) multiplied by coefficients (such as a) are called polynomials. Polynomials of a single variable have the general form

anxn + ··· + a2x2 + a1x + a0

The highest power to which the variable in a polynomial is raised is called its order. Thus the polynomial shown here is of the nth order. For example, if n were 3, the polynomial would be third order.

A quadratic equation is a second-order polynomial equation in a single variable x:

ax2 + bx + c = 0

According to the fundamental theorem of algebra, a second-order polynomial equation has two solutions—called roots—that can be found using a method called completing the square. In this method, we solve for x by first adding −c to both sides of the quadratic equation and then divide both sides by a:

x2+bax=ca

We can convert the left side of this equation to a perfect square by adding b2/4a2, which is equal to (b/2a)2:

Left side: x2+bax+b24a2=(x+b2a)2

Having added a value to the left side, we must now add that same value, b2 ⁄ 4a2, to the right side:

(x+b2a)2=ca+b24a2

The common denominator on the right side is 4a2. Rearranging the right side, we obtain the following:

(x+b2a)2=b24ac4a2

Taking the square root of both sides and solving for x,

x+b2a=±b24ac2a x=b±b24ac2a

This equation, known as the quadratic formula, has two roots:

x=b+b24ac2a and x=bb24ac2a

Thus we can obtain the solutions to a quadratic equation by substituting the values of the coefficients (a, b, c) into the quadratic formula.

When you apply the quadratic formula to obtain solutions to a quadratic equation, it is important to remember that one of the two solutions may not make sense or neither may make sense. There may be times, for example, when a negative solution is not reasonable or when both solutions require that a square root be taken of a negative number. In such cases, we simply discard any solution that is unreasonable and only report a solution that is reasonable. Skill Builder ES1 gives you practice using the quadratic formula.

Skill Builder ES1

Use the quadratic formula to solve for x in each equation. Report your answers to three significant figures.

  1. x2 + 8x − 5 = 0
  2. 2x2 − 6x + 3 = 0
  3. 3x2 − 5x − 4 = 6
  4. 2x(−x + 2) + 1 = 0
  5. 3x(2x + 1) − 4 = 5

Solution:

  1. x=8+824(1)(5)2(1)=0.583 and x=8824(1)(5)2(1)=8.58
  2. x=(6)+(62)4(2)(3)2(2)=2.37 and x=(6)(62)4(2)(3)2(2)=0.634
  3. x=(5)+(52)4(3)(10)2(3)=2.84 and x=(5)(52)4(3)(10)2(3)=1.17
  4. x=4+424(2)(1)2(2)=0.225 and x=4424(2)(1)2((2))=2.22
  5. x=1+124(2)(3)2(2)=1.00 and x=1124(2)(3)2(2)=1.50

15.8 End-of-Chapter Material

Application Problems

    Problems marked with a ♦ involve multiple concepts.

  1. ♦ The total concentrations of dissolved Al in a soil sample represent the sum of “free” Al3+ and bound forms of Al that are stable enough to be considered definite chemical species. The distribution of aluminum among its possible chemical forms can be described using equilibrium constants such as the following:

    K1 = [AlOH2+]/[Al3+][OH] = 1.0 × 109 K2 = [AlSO4+]/[Al3+][SO42−] = 1.0 × 103 K3 = [AlF2+]/[Al3+][F] = 1.0 × 107
    1. Write an equilibrium equation for each expression.
    2. Which anion has the highest affinity for Al3+: OH, SO42−, or F? Explain your reasoning.
    3. A 1.0 M solution of Al3+ is mixed with a 1.0 M solution of each of the anions. Which mixture has the lowest Al3+ concentration?
  2. Many hydroxy acids form lactones (cyclic esters) that contain a 5- or 6-membered ring. Common hydroxy acids found in nature are glycolic acid, a constituent of cane sugar juice; lactic acid, which has the characteristic odor and taste of sour milk; and citric acid, found in fruit juices. The general reaction for lactone formation can be written as follows:

    hydroxy acidlactone+H2O

    Use the information in the following table to calculate the equilibrium constant for lactone formation for each hydroxy acid given and determine which ring size is most stable.

    At Equilibrium
    Hydroxy Acid Formula Size of Lactone Ring (atoms) Hydroxy Acid (M) Lactone (M)
    HOCH2CH2COOH 4 4.99 × 10−3 5.00 × 10−5
    HOCH2CH2CH2COOH 5 8.10 × 10−5 2.19 × 10−4
    HOCH2CH2CH2CH2COOH 6 5.46 × 10−2 5.40 × 10−9
    HOCH2CH2CH2CH2CH2COOH 7 9.90 × 10−3 1.00 × 10−4
  3. ♦ Phosphorus pentachloride, an important reagent in organic chemistry for converting alcohols to alkyl chlorides (ROH → RCl), is hydrolyzed in water to form phosphoric acid and hydrogen chloride. In the gaseous state, however, PCl5 can decompose at 250°C according to PCl5(g)PCl3(g)+Cl2(g), for which K = 0.0420.

    1. Are products or reactants favored in the decomposition of PCl5(g)?
    2. If a 2.00 L flask containing 104.1 g PCl5 is heated to 250°C, what is the equilibrium concentration of each species in this reaction?
    3. What effect would an increase in pressure have on the equilibrium position? Why?
    4. If a 1.00 × 103 L vessel containing 2.00 × 103 kg of PCl3 with a constant chlorine pressure of 2.00 atm is allowed to reach equilibrium, how many kilograms of PCl5 are produced? What is the percent yield of PCl5?
  4. ♦ Carbon disulfide (CS2) is used in the manufacture of rayon and in electronic vacuum tubes. However, extreme caution must be used when handling CS2 in its gaseous state because it is extremely toxic and can cause fatal convulsions. Chronic toxicity is marked by psychic disturbances and tremors. CS2 is used to synthesize H2S at elevated T via the following reaction:

    CS2(g)+4H2(g)CH4(g)+2H2S(g)K=3.3×104
    1. If the equilibrium concentration of methane in this reaction is 2.5 × 10−2 M and the initial concentration of each reactant is 0.1635 M, what is the concentration of H2S at equilibrium?
    2. Exposure to CS2 concentrations greater than 300 ppm for several hours can start to produce adverse effects. After working for several hours in a laboratory that contains large quantities of CS2, you notice that the fume hoods were off and there was not enough ventilation to remove any CS2 vapor. Given the equilibrium CS2(l)CS2(g), where T = 20°C and Kp = 0.391, determine whether you are in any danger.
  5. ♦ Chloral hydrate, a sedative commonly referred to as “knockout drops,” is in equilibrium with trichloroacetaldehyde in highly concentrated aqueous solutions:

    The equilibrium constant for this reaction as written is 3 × 104. Are the products or the reactants favored? Write an equilibrium expression for this reaction. How could you drive this reaction to completion?

  6. Hydrogen cyanide is commercially produced in the United States by the following reaction: CH4(g)+NH3(g)+32O2(g)HCN(g)+3H2O(g), where HCN is continuously removed from the system. This reaction is carried out at approximately 1100°C in the presence of a catalyst; however, the high temperature causes other reactions to occur. Why is it necessary to run this reaction at such an elevated temperature? Does the presence of the catalyst affect the equilibrium position?

  7. ♦ Hemoglobin transports oxygen from the lungs to the capillaries and consists of four subunits, each capable of binding a single molecule of O2. In the lungs, PO2 is relatively high (100 mmHg), so hemoglobin becomes nearly saturated with O2. In the tissues, however, PO2 is relatively low (40 mmHg), so hemoglobin releases about half of its bound oxygen. Myoglobin, a protein in muscle, resembles a single subunit of hemoglobin. The plots show the percent O2 saturation versus PO2 for hemoglobin and myoglobin. Based on these plots, which molecule has the higher affinity for oxygen? What advantage does hemoglobin have over myoglobin as the oxygen transporter? Why is it advantageous to have myoglobin in muscle tissue? Use equilibrium to explain why it is more difficult to exercise at high altitudes where the partial pressure of oxygen is lower.

  8. ♦ Sodium sulfate is widely used in the recycling industry as well as in the detergent and glass industries. This compound combines with H2SO4 via Na2SO4(s)+H2SO4(g)2NaHSO4(s). Sodium hydrogen sulfate is used as a cleaning agent because it is water soluble and acidic.

    1. Write an expression for K for this reaction.
    2. Relate this equilibrium constant to the equilibrium constant for the related reaction: 2Na2SO4(s)+2H2SO4(g)4NaHSO4(s).
    3. The dissolution of Na2SO4 in water produces the equilibrium reaction SO42(aq)+H2O(l)HSO4(aq)+OH(aq) with K = 8.33 × 10−13. What is the concentration of OH in a solution formed from the dissolution of 1.00 g of sodium sulfate to make 150.0 mL of aqueous solution? Neglect the autoionization of water in your answer.
  9. ♦ One of the Venera orbiter satellites measured S2 concentrations at the surface of Venus. The resulting thermochemical data suggest that S2 formation at the planet’s surface occurs via the following equilibrium reaction: 4CO(g) +2SO2(g)4CO2(g)+S2(g). Write an expression for K for this reaction and then relate this expression to those for the following reactions:

    1. 2CO(g) + SO2(g)2CO2(g)+12S2(g)
    2. CO(g) +12SO2(g)CO2(g)+14S2(g)
    3. At 450°C, the equilibrium pressure of CO2 is 85.0 atm, SO2 is 1.0 atm, CO is 1.0 atm, and S2 is 3.0 × 10−8 atm. What are K and Kp at this temperature? What is the concentration of S2?
  10. ♦ Until the early part of the 20th century, commercial production of sulfuric acid was carried out by the “lead-chamber” process, in which SO2 was oxidized to H2SO4 in a lead-lined room. This process may be summarized by the following sequence of reactions:

    1. NO(g) + NO2(g)+2H2SO4(l)2NOHSO4(s)+H2O(l)K1
    2. 2NOHSO4(s)+SO2(g)+2H2O(l)3H2SO4(l)+2NO(g)K2
    1. Write the equilibrium constant expressions for reactions 1 and 2 and the sum of the reactions (reaction 3).
    2. Show that K3 = K1 × K2.
    3. If insufficient water is added in reaction 2 such that the reaction becomes NOHSO4(s)+12SO2(g)+H2O(l)32H2SO4(l)+NO(g), does K3 still equal K1 × K2?
    4. Based on part c, write the equilibrium constant expression for K2.
  11. Phosgene (carbonic dichloride, COCl2) is a colorless, highly toxic gas with an odor similar to that of moldy hay. Used as a lethal gas in war, phosgene can be immediately fatal; inhalation can cause either pneumonia or pulmonary edema. For the equilibrium reaction COCl2(g)CO(g)+Cl2(g), Kp is 0.680 at −10°C. If the initial pressure of COCl2 is 0.681 atm, what is the partial pressure of each component of this equilibrium system? Is the formation of products or reactant favored in this reaction?

  12. ♦ British bituminous coal has a high sulfur content and produces much smoke when burned. In 1952, burning of this coal in London led to elevated levels of smog containing high concentrations of sulfur dioxide, a lung irritant, and more than 4000 people died. Sulfur dioxide emissions can be converted to SO3 and ultimately to H2SO4, which is the cause of acid rain. The initial reaction is 2SO2(g)+O2(g)2SO3(g), for which Kp = 44.

    1. Given this Kp, are product or reactants favored in this reaction?
    2. What is the partial pressure of each species under equilibrium conditions if the initial pressure of each reactant is 0.50 atm?
    3. Would an increase in pressure favor the formation of product or reactants? Why?
  13. Oxyhemoglobin is the oxygenated form of hemoglobin, the oxygen-carrying pigment of red blood cells. Hemoglobin is built from α and β protein chains. Assembly of the oxygenated (oxy) and deoxygenated (deoxy) β-chains has been studied with the following results:

    4β(oxy)β4(oxy)K=9.07×1015 4β(deoxy)β4(deoxy)K=9.20×1013

    Is it more likely that hemoglobin β chains assemble in an oxygenated or deoxygenated state? Explain your answer.

  14. ♦ Inorganic weathering reactions can turn silicate rocks, such as diopside (CaMgSi2O6), to carbonate via the following reaction:

    CaMgSi2O6+CO2(g)MgSiO3(s)+CaCO3(s)+SiO2(s)

    Write an expression for the equilibrium constant. Although this reaction occurs on both Earth and Venus, the high surface temperature of Venus causes the reaction to be driven in one direction on that planet. Predict whether high temperatures will drive the reaction to the right or the left and then justify your answer. The estimated partial pressure of carbon dioxide on Venus is 85 atm due to the dense Venusian atmosphere. How does this pressure influence the reaction?

  15. Silicon and its inorganic compounds are widely used to manufacture textile glass fibers, cement, ceramic products, and synthetic fillers. Two of the most important industrially utilized silicon halides are SiCl4 and SiHCl3, formed by reaction of elemental silicon with HCl at temperatures greater than 300°C:

    Si(s) +4HCl(g)SiCl4(g)+2H2(g) Si(g) +3HCl(g)SiHCl3(g)+H2(g)

    Which of these two reactions is favored by increasing [HCl]? by decreasing the volume of the system?

  16. ♦ The first step in the utilization of glucose in humans is the conversion of glucose to glucose-6-phosphate via the transfer of a phosphate group from ATP (adenosine triphosphate), which produces glucose-6-phosphate and ADP (adenosine diphosphate):

    glucose +ATPglucose-6-phosphate+ADPK=680 at 25°C
    1. Is the formation of products or reactants favored in this reaction?
    2. Would K increase, decrease, or remain the same if the glucose concentration were doubled?
    3. If −RT ln K = −RT′ ln K', what would K be if the temperature were decreased to 0°C?
    4. Is the formation of products favored by an increase or a decrease in the temperature of the system?
  17. In the presence of O2, the carbon atoms of glucose can be fully oxidized to CO2 with a release of free energy almost 20 times greater than that possible under conditions in which O2 is not present. In many animal cells, the TCA cycle (tricarboxylic acid cycle) is the second stage in the complete oxidation of glucose. One reaction in the TCA cycle is the conversion of citrate to isocitrate, for which K = 0.08 in the forward direction. Speculate why the cycle continues despite this unfavorable value of K. What happens if the citrate concentration increases?

  18. ♦ Soil is an open system, subject to natural inputs and outputs that may change its chemical composition. Aqueous-phase, adsorbed, and solid-phase forms of Al(III) are of critical importance in controlling the acidity of soils, although industrial effluents, such as sulfur and nitrogen oxide gases, and fertilizers containing nitrogen can also have a large effect. Dissolution of the mineral gibbsite, which contains Al3+ in the form Al(OH)3(s), occurs in soil according to the following reaction:

    Al(OH)3(s)+3H+(aq)Al3+(aq)+3H2O(l)

    When gibbsite is in a highly crystalline state, K = 9.35 for this reaction at 298 K. In the microcrystalline state, K = 8.11. Is this change consistent with the increased surface area of the microcrystalline state?

Answers

    1. reactant
    2. [Cl2] = [PCl3] = 0.0836 M; [PCl5] = 0.166 M
    3. increasing pressure favors reactant (PCl5)
    4. 1.59 × 103 kg; 52.5%
  1. Products are favored; K=[chloral hydrate][Cl3CHO][H2O]; high concentrations of water will favor chloral hydrate formation.

  2. K=[CO2]4[S2][CO]4[SO2]2

    1. K=[CO2]2[S2]1/2[CO]2[SO2]=K1/2
    2. K=[CO2][S2]1/4[CO][SO2]1/2=K1/4
    3. Kp = 1.6; K = 93; [S2] = 5.1 × 10−10 M
  3. PCO=PCl2 = 0.421 atm; PCOCl2 = 0.260 atm; reactants are slightly favored.

  4. Both reactions are favored by increasing [HCl] and decreasing volume.

Chapter 16 Aqueous Acid–Base Equilibriums

Many vital chemical and physical processes take place exclusively in aqueous solution, including the complex biochemical reactions that occur in living organisms and the reactions that rust and corrode steel objects, such as bridges, ships, and automobiles. Among the most important reactions in aqueous solution are those that can be categorized as acid–base, precipitation, and complexation reactions. So far, our discussions of these reactions have been largely qualitative. In this chapter and , however, we take a more quantitative approach to understanding such reactions, using the concept of chemical equilibrium that we developed in for simple gas-phase reactions. We will begin by revisiting acid–base reactions in a qualitative fashion and then develop quantitative methods to describe acid–base equilibriums. In , we will use the same approach to describe the equilibriums involved in the dissolution of sparingly soluble solids and the formation of metal complexes.

Indicators are used to monitor changes in pH. The pH of a solution can be monitored using an acid–base indicator, a substance that undergoes a color change within a specific pH range that is characteristic of that indicator. The color changes for seven commonly used indicators over a pH range of 1–10 are shown here.

In , we described how acid rain can adversely affect the survival of marine life and plant growth. Many significant phenomena, such as acid rain, can be understood only in terms of the acid–base behavior of chemical species. As we expand our discussion of acid–base behavior in this chapter, you will learn why lemon slices are served with fish, why the strengths of acids and bases can vary over many orders of magnitude, and why rhubarb leaves are toxic to humans. You will also understand how the pH of your blood is kept constant, even though you produce large amounts of acid when you exercise.

16.1 The Autoionization of Water

Learning Objectives

  1. To understand the autoionization reaction of liquid water.
  2. To know the relationship among pH, pOH, and pKw.

As you learned in and , acids and bases can be defined in several different ways (). Recall that the Arrhenius definition of an acid is a substance that dissociates in water to produce H+ ions (protons), and an Arrhenius base is a substance that dissociates in water to produce OH (hydroxide) ions. According to this view, an acid–base reaction involves the reaction of a proton with a hydroxide ion to form water. Although Brønsted and Lowry defined an acid similarly to Arrhenius by describing an acid as any substance that can donate a proton, the Brønsted–Lowry definition of a base is much more general than the Arrhenius definition. In Brønsted–Lowry terms, a base is any substance that can accept a proton, so a base is not limited to just a hydroxide ion. This means that for every Brønsted–Lowry acid, there exists a corresponding conjugate base with one fewer proton, as we demonstrated in . Consequently, all Brønsted–Lowry acid–base reactions actually involve two conjugate acid–base pairs and the transfer of a proton from one substance (the acid) to another (the base). In contrast, the Lewis definition of acids and bases, discussed in , focuses on accepting or donating pairs of electrons rather than protons. A Lewis base is an electron-pair donor, and a Lewis acid is an electron-pair acceptor.

Table 16.1 Definitions of Acids and Bases

Acids Bases
Arrhenius H+ donor OH donor
Brønsted–Lowry H+ donor H+ acceptor
Lewis electron-pair acceptor electron-pair donor

Because this chapter deals with acid–base equilibriums in aqueous solution, our discussion will use primarily the Brønsted–Lowry definitions and nomenclature. Remember, however, that all three definitions are just different ways of looking at the same kind of reaction: a proton is an acid, and the hydroxide ion is a base—no matter which definition you use. In practice, chemists tend to use whichever definition is most helpful to make a particular point or understand a given system. If, for example, we refer to a base as having one or more lone pairs of electrons that can accept a proton, we are simply combining the Lewis and Brønsted–Lowry definitions to emphasize the characteristic properties of a base.

In , we also introduced the acid–base properties of water, its autoionization reaction, and the definition of pH. The purpose of this section is to review those concepts and describe them using the concepts of chemical equilibrium developed in .

Acid–Base Properties of Water

The structure of the water molecule, with its polar O–H bonds and two lone pairs of electrons on the oxygen atom, was described in and , and the structure of liquid water was discussed in . Recall that because of its highly polar structure, liquid water can act as either an acid (by donating a proton to a base) or a base (by using a lone pair of electrons to accept a proton). For example, when a strong acid such as HCl dissolves in water, it dissociates into chloride ions (Cl) and protons (H+). As you learned in , the proton, in turn, reacts with a water molecule to form the hydronium ion (H3O+):

Equation 16.1

HCl(aq) acid+H2O(l)baseH3O+(aq) acid+Cl(aq)base

In this reaction, HCl is the acid, and water acts as a base by accepting an H+ ion. The reaction in is often written in a simpler form by removing H2O from each side:

Equation 16.2

HCl(aq) → H+(aq) + Cl(aq)

In , the hydronium ion is represented by H+, although free H+ ions do not exist in liquid water.

Water can also act as an acid, as shown in . In this equilibrium reaction, H2O donates a proton to NH3, which acts as a base:

Equation 16.3

H2O(l) acid + NH3(aq)baseNH4+(aq) acid+OH(aq)base

Thus water is amphiproticSubstances that can behave as either an acid or a base in a chemical reaction, depending on the nature of the other reactant(s)., meaning that it can behave as either an acid or a base, depending on the nature of the other reactant. Notice that is an equilibrium reaction as indicated by the double arrow.

The Ion-Product Constant of Liquid Water

Because water is amphiprotic, one water molecule can react with another to form an OH ion and an H3O+ ion in an autoionization process:

Equation 16.4

2H2O(l)H3O+(aq)+OH(aq)

The equilibrium constant K for this reaction can be written as follows:

Equation 16.5

K=[H3O+][OH][H2O]2

When pure liquid water is in equilibrium with hydronium and hydroxide ions at 25°C, the concentrations of the hydronium ion and the hydroxide ion are equal: [H3O+] = [OH] = 1.003 × 10−7 M. Thus the number of dissociated water molecules is very small indeed, approximately 2 ppb. We can calculate [H2O] at 25°C from the density of water at this temperature (0.997 g/mL):

Equation 16.6

[H2O]=molL=(0.997 gmL)(1 mol18.02 g)(1000 mLL)=55.3 M

With so few water molecules dissociated, the equilibrium of the autoionization reaction () lies far to the left. Consequently, [H2O] is essentially unchanged by the autoionization reaction and can be treated as a constant. Incorporating this constant into the equilibrium expression allows us to rearrange to define a new equilibrium constant, the ion-product constant of liquid water (Kw)An equilibrium constant for the autoionization of water, 2H2O(l)H3O+(aq) + OH(aq), in which the concentration of water is treated as a constant: Kw = [H3O+][OH] = 1.006×1014.:

Equation 16.7

K[H2O]2=[H3O+][OH]Kw=[H3O+][OH]

Substituting the values for [H3O+] and [OH] at 25°C into this expression,

Equation 16.8

Kw=(1.003×107)(1.003×107)=1.006×1014

Thus, to three significant figures, Kw = 1.01 × 10−14 M. Like any other equilibrium constant, Kw varies with temperature, ranging from 1.15 × 10−15 at 0°C to 4.99 × 10−13 at 100°C.

In pure water, the concentrations of the hydronium ion and the hydroxide ion are equal, and the solution is therefore neutral. If [H3O+] > [OH], however, the solution is acidic, whereas if [H3O+] < [OH], the solution is basic. For an aqueous solution, the H3O+ concentration is a quantitative measure of acidity: the higher the H3O+ concentration, the more acidic the solution. Conversely, the higher the OH concentration, the more basic the solution. In most situations that you will encounter, the H3O+ and OH concentrations from the dissociation of water are so small (1.003 × 10−7 M) that they can be ignored in calculating the H3O+ or OH concentrations of solutions of acids and bases, but this is not always the case.

The Relationship among pH, pOH, and pKw

The pH scale is a concise way of describing the H3O+ concentration and hence the acidity or basicity of a solution. Recall from that pH and the H+ (H3O+) concentration are related as follows:

Equation 16.9

pH=log10[H+]

Equation 16.10

[H+]=10pH

Because the scale is logarithmic, a pH difference of 1 between two solutions corresponds to a difference of a factor of 10 in their hydronium ion concentrations. (Refer to Essential Skills 3 in , , if you need to refresh your memory about how to use logarithms.) Recall also that the pH of a neutral solution is 7.00 ([H3O+] = 1.0 × 10−7 M), whereas acidic solutions have pH < 7.00 (corresponding to [H3O+] > 1.0 × 10−7) and basic solutions have pH > 7.00 (corresponding to [H3O+] < 1.0 × 10−7).

Similar notation systems are used to describe many other chemical quantities that contain a large negative exponent. For example, chemists use an analogous pOH scale to describe the hydroxide ion concentration of a solution. The pOH and [OH] are related as follows:

Equation 16.11

pOH=log10[OH]

Equation 16.12

[OH]=10pOH

The constant Kw can also be expressed using this notation, where pKw = −log Kw.

Because a neutral solution has [OH] = 1.0 × 10−7, the pOH of a neutral solution is 7.00. Consequently, the sum of the pH and the pOH for a neutral solution at 25°C is 7.00 + 7.00 = 14.00. We can show that the sum of pH and pOH is equal to 14.00 for any aqueous solution at 25°C by taking the negative logarithm of both sides of :

Equation 16.13

logKw=log([H3O+][OH])pKw=(log[H3O+])+(log[OH])pKw=pH+pOH

Thus at any temperature, pH + pOH = pKw, so at 25°C, where Kw = 1.0 × 10−14, pH + pOH = 14.00. More generally, the pH of any neutral solution is half of the pKw at that temperature. The relationship among pH, pOH, and the acidity or basicity of a solution is summarized graphically in over the common pH range of 0 to 14. Notice the inverse relationship between the pH and pOH scales.

Note the Pattern

For any neutral solution, pH + pOH = 14.00 (at 25°C) and pH=12pKw.

Figure 16.1 The Inverse Relationship between the pH and pOH Scales

As pH decreases, [H+] and the acidity increase. As pOH increases, [OH] and the basicity decrease. Common substances have pH values that range from extremely acidic to extremely basic.

Example 1

The Kw for water at 100°C is 4.99 × 10−13. Calculate pKw for water at this temperature and the pH and the pOH for a neutral aqueous solution at 100°C. Report pH and pOH values to two decimal places.

Given: K w

Asked for: pKw, pH, and pOH

Strategy:

A Calculate pKw by taking the negative logarithm of Kw.

B For a neutral aqueous solution, [H3O+] = [OH]. Use this relationship and to calculate [H3O+] and [OH]. Then determine the pH and the pOH for the solution.

Solution:

A Because pKw is the negative logarithm of Kw, we can write

pKw = −log Kw = −log(4.99 × 10−13) = 12.302

The answer is reasonable: Kw is between 10−13 and 10−12, so pKw must be between 12 and 13.

B shows that Kw = [H3O+][OH]. Because [H3O+] = [OH] in a neutral solution, we can let x = [H3O+] = [OH]:

Kw=[H3O+][OH]=(x)(x)=x2x=Kw=4.99×1013=7.06×107 M

Because x is equal to both [H3O+] and [OH],

pH = pOH = −log(7.06 × 10−7) = 6.15 (to two decimal places)

We could obtain the same answer more easily (without using logarithms) by using the pKw. In this case, we know that pKw = 12.302, and from , we know that pKw = pH + pOH. Because pH = pOH in a neutral solution, we can use directly, setting pH = pOH = y. Solving to two decimal places we obtain the following:

pKw=pH+pOH=y+y=2yy=pKw2=12.3022=6.15=pH=pOH

Exercise

Humans maintain an internal temperature of about 37°C. At this temperature, Kw = 3.55 × 10−14. Calculate pKw and the pH and the pOH of a neutral solution at 37°C. Report pH and pOH values to two decimal places.

Answer: pKw = 13.45 pH = pOH = 6.73

Summary

Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion (H3O+). The autoionization of liquid water produces OH and H3O+ ions. The equilibrium constant for this reaction is called the ion-product constant of liquid water (Kw) and is defined as Kw = [H3O+][OH]. At 25°C, Kw is 1.01 × 10−14; hence pH + pOH = pKw = 14.00.

Key Takeaway

  • For any neutral solution, pH + pOH = 14.00 (at 25°C) and pH = 1/2 pKw.

Key Equations

Ion-product constant of liquid water

: Kw = [H3O+][OH]

Definition of pH

: pH = −log10[H+]

: [H+] = 10−pH

Definition of pOH

: pOH = −log10[OH+]

: [OH] = 10−pOH

Relationship among pH, pOH, and p K w

: pKw= pH + pOH

Conceptual Problems

  1. What is the relationship between the value of the equilibrium constant for the autoionization of liquid water and the tabulated value of the ion-product constant of liquid water (Kw)?

  2. The density of liquid water decreases as the temperature increases from 25°C to 50°C. Will this effect cause Kw to increase or decrease? Why?

  3. Show that water is amphiprotic by writing balanced chemical equations for the reactions of water with HNO3 and NH3. In which reaction does water act as the acid? In which does it act as the base?

  4. Write a chemical equation for each of the following.

    1. Nitric acid is added to water.
    2. Potassium hydroxide is added to water.
    3. Calcium hydroxide is added to water.
    4. Sulfuric acid is added to water.
  5. Show that K for the sum of the following reactions is equal to Kw.

    1. HMnO4(aq)H+(aq)+MnO4(aq)
    2. MnO4(aq)+H2O(l)HMnO4(aq)+OH(aq)

Answers

  1.  

    Kauto = [H3O+][OH]/[H2O]2 Kw = [H3O+][OH] = Kauto[H2O]2
  2. H2O(l) + HNO3(g) → H3O+(aq) + NO3(aq); water is the base H2O(l) + NH3(g) → OH(aq) + NH4(aq); water is the acid

Numerical Problems

  1. The autoionization of sulfuric acid can be described by the following chemical equation:

    H2SO4(l)+H2SO4(l)H3SO4+(soln)+HSO4(soln)

    At 25°C, K = 3 × 10−4. Write an equilibrium constant expression for KH2SO4 that is analogous to Kw. The density of H2SO4 is 1.8 g/cm3 at 25°C. What is the concentration of H3SO4+? What fraction of H2SO4 is ionized?

  2. An aqueous solution of a substance is found to have [H3O]+ = 2.48 × 10−8 M. Is the solution acidic, neutral, or basic?

  3. The pH of a solution is 5.63. What is its pOH? What is the [OH]? Is the solution acidic or basic?

  4. State whether each solution is acidic, neutral, or basic.

    1. [H3O+] = 8.6 × 10−3 M
    2. [H3O+] = 3.7 × 10−9 M
    3. [H3O+] = 2.1 × 10−7 M
    4. [H3O+] = 1.4 × 10−6 M
  5. Calculate the pH and the pOH of each solution.

    1. 0.15 M HBr
    2. 0.03 M KOH
    3. 2.3 × 10−3 M HNO3
    4. 9.78 × 10−2 M NaOH
    5. 0.00017 M HCl
    6. 5.78 M HI
  6. Calculate the pH and the pOH of each solution.

    1. 25.0 mL of 2.3 × 10−2 M HCl, diluted to 100 mL
    2. 5.0 mL of 1.87 M NaOH, diluted to 125 mL
    3. 5.0 mL of 5.98 M HCl added to 100 mL of water
    4. 25.0 mL of 3.7 M HNO3 added to 250 mL of water
    5. 35.0 mL of 0.046 M HI added to 500 mL of water
    6. 15.0 mL of 0.0087 M KOH added to 250 mL of water.
  7. The pH of stomach acid is approximately 1.5. What is the [H+]?

  8. Given the pH values in parentheses, what is the [H+] of each solution?

    1. household bleach (11.4)
    2. milk (6.5)
    3. orange juice (3.5)
    4. seawater (8.5)
    5. tomato juice (4.2)
  9. A reaction requires the addition of 250.0 mL of a solution with a pH of 3.50. What mass of HCl (in milligrams) must be dissolved in 250 mL of water to produce a solution with this pH?

  10. If you require 333 mL of a pH 12.50 solution, how would you prepare it using a 0.500 M sodium hydroxide stock solution?

Answers

  1. KH2SO4=[H3SO4+][HSO4]=K[H2SO4]2;

    [H3SO4+] = 0.3 M; the fraction ionized is 0.02.

  2. pOH = 8.37; [OH] = 4.3 × 10−9 M; acidic

    1. pH = 0.82; pOH = 13.18
    2. pH = 12.5; pOH = 1.5
    3. pH = 2.64; pOH = 11.36
    4. pH = 12.990; pOH = 1.010
    5. pH = 3.77; pOH = 10.23
    6. pH = −0.762; pOH = 14.762
  3. 2.9 mg HCl

16.2 A Qualitative Description of Acid–Base Equilibriums

Learning Objectives

  1. To understand the concept of conjugate acid–base pairs.
  2. To know the relationship between acid or base strength and the magnitude of Ka, Kb, pKa, and pKb.
  3. To understand the leveling effect.

We now turn our attention to acid–base reactions to see how the concepts of chemical equilibrium and equilibrium constants can deepen our understanding of this kind of chemical behavior. We begin with a qualitative description of acid–base equilibriums in terms of the Brønsted–Lowry model and then proceed to a quantitative description in Section 16.4 "Quantitative Aspects of Acid–Base Equilibriums".

Conjugate Acid–Base Pairs

We discussed the concept of conjugate acid–base pairs in Chapter 4 "Reactions in Aqueous Solution", using the reaction of ammonia, the base, with water, the acid, as an example. In aqueous solutions, acids and bases can be defined in terms of the transfer of a proton from an acid to a base. Thus for every acidic species in an aqueous solution, there exists a species derived from the acid by the loss of a proton. These two species that differ by only a proton constitute a conjugate acid–base pairAn acid and a base that differ by only one hydrogen ion.. For example, in the reaction of HCl with water (Equation 16.1), HCl, the parent acid, donates a proton to a water molecule, the parent base, thereby forming Cl. Thus HCl and Cl constitute a conjugate acid–base pair. By convention, we always write a conjugate acid–base pair as the acid followed by its conjugate base. In the reverse reaction, the Cl ion in solution acts as a base to accept a proton from H3O+, forming H2O and HCl. Thus H3O+ and H2O constitute a second conjugate acid–base pair. In general, any acid–base reaction must contain two conjugate acid–base pairs, which in this case are HCl/Cl and H3O+/H2O.

Note the Pattern

All acid–base reactions contain two conjugate acid–base pairs.

Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, H3O+ is the acid that donates a proton to the acetate ion, which acts as the base. Once again, we have two conjugate acid–base pairs: the parent acid and its conjugate base (CH3CO2H/CH3CO2) and the parent base and its conjugate acid (H3O+/H2O).

In the reaction of ammonia with water to give ammonium ions and hydroxide ions (Equation 16.3), ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are NH4+/NH3 and H2O/OH. Some common conjugate acid–base pairs are shown in Figure 16.2 "The Relative Strengths of Some Common Conjugate Acid–Base Pairs".

Figure 16.2 The Relative Strengths of Some Common Conjugate Acid–Base Pairs

The strongest acids are at the bottom left, and the strongest bases are at the top right. The conjugate base of a strong acid is a very weak base, and, conversely, the conjugate acid of a strong base is a very weak acid.

Acid–Base Equilibrium Constants: Ka, Kb, pKa, and pKb

The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A is its conjugate base, is as follows:

Equation 16.14

HA(aq)+H2O(l)H3O+(aq)+A(aq)

The equilibrium constant for this dissociation is as follows:

Equation 16.15

K=[H3O+][A][H2O][HA]

As we noted earlier, the concentration of water is essentially constant for all reactions in aqueous solution, so [H2O] in Equation 16.15 can be incorporated into a new quantity, the acid ionization constant (Ka)An equilibrium constant for the ionization (dissociation) of a weak acid (HA) with water, HA(aq) + H2O(l)H3O(aq) + A(aq), in which the concentration of water is treated as a constant: Ka = [H3O+][A]/[HA]., also called the acid dissociation constant:

Equation 16.16

Ka=K[H2O]=[H3O+][A][HA]

Thus the numerical values of K and Ka differ by the concentration of water (55.3 M). Again, for simplicity, H3O+ can be written as H+ in Equation 16.16. Keep in mind, though, that free H+ does not exist in aqueous solutions and that a proton is transferred to H2O in all acid ionization reactions to form H3O+. The larger the Ka, the stronger the acid and the higher the H+ concentration at equilibrium.Like all equilibrium constants, acid–base ionization constants are actually measured in terms of the activities of H+ or OH, thus making them unitless. The values of Ka for a number of common acids are given in Table 16.2 "Values of ".

Table 16.2 Values of Ka, pKa, Kb, and pKb for Selected Acids (HA) and Their Conjugate Bases (A)

Acid HA K a pKa A K b pKb
hydroiodic acid HI 2 × 109 −9.3 I 5.5 × 10−24 23.26
sulfuric acid (1)* H2SO4 1 × 102 −2.0 HSO4 1 × 10−16 16.0
nitric acid HNO3 2.3 × 101 −1.37 NO3 4.3 × 10−16 15.37
hydronium ion H3O+ 1.0 0.00 H2O 1.0 × 10−14 14.00
sulfuric acid (2)* HSO4 1.0 × 10−2 1.99 SO42− 9.8 × 10−13 12.01
hydrofluoric acid HF 6.3 × 10−4 3.20 F 1.6 × 10−11 10.80
nitrous acid HNO2 5.6 × 10−4 3.25 NO2 1.8 × 10−11 10.75
formic acid HCO2H 1.78 × 10−4 3.750 HCO2 5.6 × 10−11 10.25
benzoic acid C6H5CO2H 6.3 × 10−5 4.20 C6H5CO2 1.6 × 10−10 9.80
acetic acid CH3CO2H 1.7 × 10−5 4.76 CH3CO2 5.8 × 10−10 9.24
pyridinium ion C5H5NH+ 5.9 × 10−6 5.23 C5H5N 1.7 × 10−9 8.77
hypochlorous acid HOCl 4.0 × 10−8 7.40 OCl 2.5 × 10−7 6.60
hydrocyanic acid HCN 6.2 × 10−10 9.21 CN 1.6 × 10−5 4.79
ammonium ion NH4+ 5.6 × 10−10 9.25 NH3 1.8 × 10−5 4.75
water H2O 1.0 × 10−14 14.00 OH 1.00 0.00
acetylene C2H2 1 × 10−26 26.0 HC2 1 × 1012 −12.0
ammonia NH3 1 × 10−35 35.0 NH2 1 × 1021 −21.0
*The number in parentheses indicates the ionization step referred to for a polyprotic acid.

Weak bases react with water to produce the hydroxide ion, as shown in the following general equation, where B is the parent base and BH+ is its conjugate acid:

Equation 16.17

B(aq)+H2O(l)BH+(aq)+OH(aq)

The equilibrium constant for this reaction is the base ionization constant (Kb)An equilibrium constant for the reaction of a weak base (B) with water, B(aq) + H2O(l)BH+(aq) + OH(aq), in which the concentration of water is treated as a constant: Kb = [BH+][OH]/[B]., also called the base dissociation constant:

Equation 16.18

Kb=K[H2O]=[BH+][OH][B]

Once again, the concentration of water is constant, so it does not appear in the equilibrium constant expression; instead, it is included in the Kb. The larger the Kb, the stronger the base and the higher the OH concentration at equilibrium. The values of Kb for a number of common weak bases are given in Table 16.3 "Values of ".

Table 16.3 Values of Kb, pKb, Ka, and pKa for Selected Weak Bases (B) and Their Conjugate Acids (BH+)

Base B K b pKb BH+ K a pKa
hydroxide ion OH 1.0 0.00* H2O 1.0 × 10−14 14.00
phosphate ion PO43− 2.1 × 10−2 1.68 HPO42− 4.8 × 10−13 12.32
dimethylamine (CH3)2NH 5.4 × 10−4 3.27 (CH3)2NH2+ 1.9 × 10−11 10.73
methylamine CH3NH2 4.6 × 10−4 3.34 CH3NH3+ 2.2 × 10−11 10.66
trimethylamine (CH3)3N 6.3 × 10−5 4.20 (CH3)3NH+ 1.6 × 10−10 9.80
ammonia NH3 1.8 × 10−5 4.75 NH4+ 5.6 × 10−10 9.25
pyridine C5H5N 1.7 × 10−9 8.77 C5H5NH+ 5.9 × 10−6 5.23
aniline C6H5NH2 7.4 × 10−10 9.13 C6H5NH3+ 1.3 × 10−5 4.87
water H2O 1.0 × 10−14 14.00 H3O+ 1.0* 0.00
*As in Table 16.2 "Values of ".

There is a simple relationship between the magnitude of Ka for an acid and Kb for its conjugate base. Consider, for example, the ionization of hydrocyanic acid (HCN) in water to produce an acidic solution, and the reaction of CN with water to produce a basic solution:

Equation 16.19

HCN(aq)H+(aq)+CN(aq)

Equation 16.20

CN(aq)+H2O(l)OH(aq)+HCN(aq)

The equilibrium constant expression for the ionization of HCN is as follows:

Equation 16.21

Ka=[H+][CN][HCN]

The corresponding expression for the reaction of cyanide with water is as follows:

Equation 16.22

Kb=[OH][HCN][CN]

If we add Equation 16.19 and Equation 16.20, we obtain the following (recall from Chapter 15 "Chemical Equilibrium" that the equilibrium constant for the sum of two reactions is the product of the equilibrium constants for the individual reactions):

HCN(aq)H+(aq)+CN(aq)Ka=[H+][CN]/[HCN]CN(aq)+H2O(l)OH(aq)+HCN(aq)Kb=[OH][HCN]/[CN]H2O(l)H+(aq)+OH(aq)K=Ka×Kb=[H+][OH]

In this case, the sum of the reactions described by Ka and Kb is the equation for the autoionization of water, and the product of the two equilibrium constants is Kw:

Equation 16.23

KaKb = Kw

Thus if we know either Ka for an acid or Kb for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair.

Just as with pH, pOH, and pKw, we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, by defining pKa as follows:

Equation 16.24

pKa = −log10Ka

Equation 16.25

Ka=10pKa

and pKb as

Equation 16.26

pKb = −log10Kb

Equation 16.27

Kb=10pKb

Similarly, Equation 16.23, which expresses the relationship between Ka and Kb, can be written in logarithmic form as follows:

Equation 16.28

pKa + pKb = pKw

At 25°C, this becomes

Equation 16.29

pKa + pKb = 14.00

The values of pKa and pKb are given for several common acids and bases in Table 16.2 "Values of " and Table 16.3 "Values of ", respectively, and a more extensive set of data is provided in Chapter 27 "Appendix C: Dissociation Constants and p" and Chapter 28 "Appendix D: Dissociation Constants and p". Because of the use of negative logarithms, smaller values of pKa correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid (HNO2), with a pKa of 3.25, is about a 1000 times stronger acid than hydrocyanic acid (HCN), with a pKa of 9.21. Conversely, smaller values of pKb correspond to larger base ionization constants and hence stronger bases.

The relative strengths of some common acids and their conjugate bases are shown graphically in Figure 16.2 "The Relative Strengths of Some Common Conjugate Acid–Base Pairs". The conjugate acid–base pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of pKa. This order corresponds to decreasing strength of the conjugate base or increasing values of pKb. At the bottom left of Figure 16.2 "The Relative Strengths of Some Common Conjugate Acid–Base Pairs" are the common strong acids; at the top right are the most common strong bases. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate base. Thus the conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base.

Note the Pattern

The conjugate base of a strong acid is a weak base and vice versa.

We can use the relative strengths of acids and bases to predict the direction of an acid–base reaction by following a single rule: an acid–base equilibrium always favors the side with the weaker acid and base, as indicated by these arrows:

In an acid–base reaction, the proton always reacts with the stronger base.

For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute aqueous solution to produce H3O+ and Cl; only negligible amounts of HCl molecules remain undissociated. Hence the ionization equilibrium lies virtually all the way to the right, as represented by a single arrow:

Equation 16.30

HCl(aq) + H2O(l)H3O+(aq)+Cl(aq)

In contrast, acetic acid is a weak acid, and water is a weak base. Consequently, aqueous solutions of acetic acid contain mostly acetic acid molecules in equilibrium with a small concentration of H3O+ and acetate ions, and the ionization equilibrium lies far to the left, as represented by these arrows:

Figure 16.3

 

Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong base, and ammonia is a weak base, whereas the ammonium ion is a stronger acid than water. Hence this equilibrium also lies to the left:

Figure 16.4

 

Note the Pattern

All acid–base equilibriums favor the side with the weaker acid and base. Thus the proton is bound to the stronger base.

Example 2

  1. Calculate Kb and pKb of the butyrate ion (CH3CH2CH2CO2). The pKa of butyric acid at 25°C is 4.83. Butyric acid is responsible for the foul smell of rancid butter.
  2. Calculate Ka and pKa of the dimethylammonium ion [(CH3)2NH2+]. The base ionization constant Kb of dimethylamine [(CH3)2NH] is 5.4 × 10−4 at 25°C.

Given: pKa and Kb

Asked for: corresponding Kb and pKb, Ka and pKa

Strategy:

The constants Ka and Kb are related as shown in Equation 16.23. The pKa and pKb for an acid and its conjugate base are related as shown in Equation 16.28 and Equation 16.29. Use the relationships pK = −log K and K = 10−pK (Equation 16.24 and Equation 16.26) to convert between Ka and pKa or Kb and pKb.

Solution:

  1. We are given the pKa for butyric acid and asked to calculate the Kb and the pKb for its conjugate base, the butyrate ion. Because the pKa value cited is for a temperature of 25°C, we can use Equation 16.29: pKa + pKb = pKw = 14.00. Substituting the pKa and solving for the pKb,

    4.83+pKb=14.00pKb=14.004.83=9.17

    Because pKb = −log Kb, Kb is 10−9.17 = 6.8 × 10−10.

  2. In this case, we are given Kb for a base (dimethylamine) and asked to calculate Ka and pKa for its conjugate acid, the dimethylammonium ion. Because the initial quantity given is Kb rather than pKb, we can use Equation 16.23: KaKb = Kw. Substituting the values of Kb and Kw at 25°C and solving for Ka,

    Ka(5.4×104)=1.01×1014Ka=1.9×1011

    Because pKa = −log Ka, we have pKa = −log(1.9 × 10−11) = 10.72. We could also have converted Kb to pKb to obtain the same answer:

    pKb=log(5.4×104)=3.27pKa+pKb=14.00pKa=10.73Ka=10pKa=1010.73=1.9×1011

    If we are given any one of these four quantities for an acid or a base (Ka, pKa, Kb, or pKb), we can calculate the other three.

Exercise

Lactic acid [CH3CH(OH)CO2H] is responsible for the pungent taste and smell of sour milk; it is also thought to produce soreness in fatigued muscles. Its pKa is 3.86 at 25°C. Calculate Ka for lactic acid and pKb and Kb for the lactate ion.

Answer: Ka = 1.4 × 10−4 for lactic acid; pKb = 10.14 and Kb = 7.2 × 10−11 for the lactate ion

Solutions of Strong Acids and Bases: The Leveling Effect

You will notice in Table 16.2 "Values of " that acids like H2SO4 and HNO3 lie above the hydronium ion, meaning that they have pKa values less than zero and are stronger acids than the H3O+ ion.Recall from Chapter 4 "Reactions in Aqueous Solution" that the acidic proton in virtually all oxoacids is bonded to one of the oxygen atoms of the oxoanion. Thus nitric acid should properly be written as HONO2. Unfortunately, however, the formulas of oxoacids are almost always written with hydrogen on the left and oxygen on the right, giving HNO3 instead. In fact, all six of the common strong acids that we first encountered in Chapter 4 "Reactions in Aqueous Solution" have pKa values less than zero, which means that they have a greater tendency to lose a proton than does the H3O+ ion. Conversely, the conjugate bases of these strong acids are weaker bases than water. Consequently, the proton-transfer equilibriums for these strong acids lie far to the right, and adding any of the common strong acids to water results in an essentially stoichiometric reaction of the acid with water to form a solution of the H3O+ ion and the conjugate base of the acid.

Although Ka for HI is about 108 greater than Ka for HNO3, the reaction of either HI or HNO3 with water gives an essentially stoichiometric solution of H3O+ and I or NO3. In fact, a 0.1 M aqueous solution of any strong acid actually contains 0.1 M H3O+, regardless of the identity of the strong acid. This phenomenon is called the leveling effectThe phenomenon that makes H3O+ the strongest acid that can exist in water. Any species that is a stronger acid than H3O+ is leveled to the strength of H3O+ in aqueous solution.: any species that is a stronger acid than the conjugate acid of water (H3O+) is leveled to the strength of H3O+ in aqueous solution because H3O+ is the strongest acid that can exist in equilibrium with water. Consequently, it is impossible to distinguish between the strengths of acids such as HI and HNO3 in aqueous solution, and an alternative approach must be used to determine their relative acid strengths.

One method is to use a solvent such as anhydrous acetic acid. Because acetic acid is a stronger acid than water, it must also be a weaker base, with a lesser tendency to accept a proton than H2O. Measurements of the conductivity of 0.1 M solutions of both HI and HNO3 in acetic acid show that HI is completely dissociated, but HNO3 is only partially dissociated and behaves like a weak acid in this solvent. This result clearly tells us that HI is a stronger acid than HNO3. The relative order of acid strengths and approximate Ka and pKa values for the strong acids at the top of Table 16.2 "Values of " were determined using measurements like this and different nonaqueous solvents.

Note the Pattern

In aqueous solutions, [H3O+] is the strongest acid and OH is the strongest base that can exist in equilibrium with H2O.

The leveling effect applies to solutions of strong bases as well: In aqueous solution, any base stronger than OH is leveled to the strength of OH because OH is the strongest base that can exist in equilibrium with water. Salts such as K2O, NaOCH3 (sodium methoxide), and NaNH2 (sodamide, or sodium amide), whose anions are the conjugate bases of species that would lie below water in Table 16.3 "Values of ", are all strong bases that react essentially completely (and often violently) with water, accepting a proton to give a solution of OH and the corresponding cation:

Equation 16.31

K2O(s)+H2O(l)2OH(aq)+2K+(aq)

Equation 16.32

NaOCH3(s)+H2O(l)OH(aq)+Na+(aq)+CH3OH(aq)

Equation 16.33

NaNH2(s)+H2O(l)OH(aq)+Na+(aq)+NH3(aq)

Other examples that you may encounter are potassium hydride (KH) and organometallic compounds such as methyl lithium (CH3Li).

Polyprotic Acids and Bases

As you learned in Chapter 4 "Reactions in Aqueous Solution", polyprotic acids such as H2SO4, H3PO4, and H2CO3 contain more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. Thus acid strength decreases with the loss of subsequent protons, and, correspondingly, the pKa increases. Consider H2SO4, for example:

Figure 16.5

 

Equation 16.34

HSO4(aq)SO42(aq)+H+(aq)pKa=1.99

The equilibrium in the first reaction lies far to the right, consistent with H2SO4 being a strong acid. In contrast, in the second reaction, appreciable quantities of both HSO4 and SO42− are present at equilibrium.

Note the Pattern

For a polyprotic acid, acid strength decreases and the pKa increases with the sequential loss of each proton.

The hydrogen sulfate ion (HSO4) is both the conjugate base of H2SO4 and the conjugate acid of SO42−. Just like water, HSO4 can therefore act as either an acid or a base, depending on whether the other reactant is a stronger acid or a stronger base. Conversely, the sulfate ion (SO42−) is a polyprotic base that is capable of accepting two protons in a stepwise manner:

Figure 16.6

 

Figure 16.7

 

Like any other conjugate acid–base pair, the strengths of the conjugate acids and bases are related by pKa + pKb = pKw. Consider, for example, the HSO4/ SO42− conjugate acid–base pair. From Table 16.2 "Values of ", we see that the pKa of HSO4 is 1.99. Hence the pKb of SO42− is 14.00 − 1.99 = 12.01. Thus sulfate is a rather weak base, whereas OH is a strong base, so the equilibrium shown in Figure 16.6 lies to the left. The HSO4 ion is also a very weak base [pKa of H2SO4 = 2.0, pKb of HSO4 = 14 − (−2.0) = 16], which is consistent with what we expect for the conjugate base of a strong acid. Thus the equilibrium shown in Figure 16.7 also lies almost completely to the left. Once again, equilibrium favors the formation of the weaker acid–base pair.

Example 3

Predict whether the equilibrium for each reaction lies to the left or the right as written.

  1. NH4+(aq)+PO43(aq)NH3(aq)+HPO42(aq)
  2. CH3CH2CO2H(aq)+CN(aq)CH3CH2CO2(aq)+HCN(aq)

Given: balanced chemical equation

Asked for: equilibrium position

Strategy:

Identify the conjugate acid–base pairs in each reaction. Then refer to Table 16.2 "Values of ", Table 16.3 "Values of ", and Figure 16.2 "The Relative Strengths of Some Common Conjugate Acid–Base Pairs" to determine which is the stronger acid and base. Equilibrium always favors the formation of the weaker acid–base pair.

Solution:

  1. The conjugate acid–base pairs are NH4+/NH3 and HPO42−/PO43−. According to Table 16.2 "Values of " and Table 16.3 "Values of ", NH4+ is a stronger acid (pKa = 9.25) than HPO42− (pKa = 12.32), and PO43− is a stronger base (pKb = 1.68) than NH3 (pKb = 4.75). The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:

     

  2. The conjugate acid–base pairs are CH3CH2CO2H/CH3CH2CO2 and HCN/CN. According to Table 16.2 "Values of ", HCN is a weak acid (pKa = 9.21) and CN is a moderately weak base (pKb = 4.79). Propionic acid (CH3CH2CO2H) is not listed in Table 16.2 "Values of ", however. In a situation like this, the best approach is to look for a similar compound whose acid–base properties are listed. For example, propionic acid and acetic acid are identical except for the groups attached to the carbon atom of the carboxylic acid (−CH2CH3 versus −CH3), so we might expect the two compounds to have similar acid–base properties. In particular, we would expect the pKa of propionic acid to be similar in magnitude to the pKa of acetic acid. (In fact, the pKa of propionic acid is 4.87, compared to 4.76 for acetic acid, which makes propionic acid a slightly weaker acid than acetic acid.) Thus propionic acid should be a significantly stronger acid than HCN. Because the stronger acid forms the weaker conjugate base, we predict that cyanide will be a stronger base than propionate. The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:

Exercise

Predict whether the equilibrium for each reaction lies to the left or the right as written.

  1. H2O(l)+HS(aq)OH(aq)+H2S(aq)
  2. HCO2(aq)+HSO4(aq)HCO2H(aq)+SO42(aq)

Answer:

  1. left
  2. left

Acid–Base Properties of Solutions of Salts

We can also use the relative strengths of conjugate acid–base pairs to understand the acid–base properties of solutions of salts. In Chapter 4 "Reactions in Aqueous Solution", you learned that a neutralization reaction can be defined as the reaction of an acid and a base to produce a salt and water. That is, another cation, such as Na+, replaces the proton on the acid. An example is the reaction of CH3CO2H, a weak acid, with NaOH, a strong base:

Equation 16.35

CH3CO2H(l) acid+NaOH(s)baseH2OCH3CO2Na(aq) salt+H2O(l)water

Depending on the acid–base properties of its component ions, however, a salt can dissolve in water to produce a neutral solution, a basic solution, or an acidic solution.

When a salt such as NaCl dissolves in water, it produces Na+(aq) and Cl(aq) ions. Using a Lewis approach, the Na+ ion can be viewed as an acid because it is an electron pair acceptor, although its low charge and relatively large radius make it a very weak acid. The Cl ion is the conjugate base of the strong acid HCl, so it has essentially no basic character. Consequently, dissolving NaCl in water has no effect on the pH of a solution, and the solution remains neutral.

Now let's compare this behavior to the behavior of aqueous solutions of potassium cyanide and sodium acetate. Again, the cations (K+ and Na+) have essentially no acidic character, but the anions (CN and CH3CO2) are weak bases that can react with water because they are the conjugate bases of the weak acids HCN and acetic acid, respectively.

Figure 16.8

 

Figure 16.9

 

Neither reaction proceeds very far to the right as written because the formation of the weaker acid–base pair is favored. Both HCN and acetic acid are stronger acids than water, and hydroxide is a stronger base than either acetate or cyanide, so in both cases, the equilibrium lies to the left. Nonetheless, each of these reactions generates enough hydroxide ions to produce a basic solution. For example, the pH of a 0.1 M solution of sodium acetate or potassium cyanide at 25°C is 8.8 or 11.1, respectively. From Table 16.2 "Values of " and Figure 16.2 "The Relative Strengths of Some Common Conjugate Acid–Base Pairs", we can see that CN is a stronger base (pKb = 4.79) than acetate (pKb = 9.24), which is consistent with KCN producing a more basic solution than sodium acetate at the same concentration.

In contrast, the conjugate acid of a weak base should be a weak acid (Figure 16.2 "The Relative Strengths of Some Common Conjugate Acid–Base Pairs"). For example, ammonium chloride and pyridinium chloride are salts produced by reacting ammonia and pyridine, respectively, with HCl. As you already know, the chloride ion is such a weak base that it does not react with water. In contrast, the cations of the two salts are weak acids that react with water as follows:

Figure 16.10

 

Figure 16.11

 

Figure 16.2 "The Relative Strengths of Some Common Conjugate Acid–Base Pairs" shows that H3O+ is a stronger acid than either NH4+ or C5H5NH+, and conversely, ammonia and pyridine are both stronger bases than water. The equilibrium will therefore lie far to the left in both cases, favoring the weaker acid–base pair. The H3O+ concentration produced by the reactions is great enough, however, to decrease the pH of the solution significantly: the pH of a 0.10 M solution of ammonium chloride or pyridinium chloride at 25°C is 5.13 or 3.12, respectively. This is consistent with the information shown in Figure 16.2 "The Relative Strengths of Some Common Conjugate Acid–Base Pairs", indicating that the pyridinium ion is more acidic than the ammonium ion.

What happens with aqueous solutions of a salt such as ammonium acetate, where both the cation and the anion can react separately with water to produce an acid and a base, respectively? According to Figure 16.10, the ammonium ion will lower the pH, while according to Figure 16.9, the acetate ion will raise the pH. This particular case is unusual, in that the cation is as strong an acid as the anion is a base (pKa ≈ pKb). Consequently, the two effects cancel, and the solution remains neutral. With salts in which the cation is a stronger acid than the anion is a base, the final solution has a pH < 7.00. Conversely, if the cation is a weaker acid than the anion is a base, the final solution has a pH > 7.00.

Solutions of simple salts of metal ions can also be acidic, even though a metal ion cannot donate a proton directly to water to produce H3O+. Instead, a metal ion can act as a Lewis acid and interact with water, a Lewis base, by coordinating to a lone pair of electrons on the oxygen atom to form a hydrated metal ion (part (a) in Figure 16.12 "Effect of a Metal Ion on the Acidity of Water"), as discussed in Chapter 4 "Reactions in Aqueous Solution". A water molecule coordinated to a metal ion is more acidic than a free water molecule for two reasons. First, repulsive electrostatic interactions between the positively charged metal ion and the partially positively charged hydrogen atoms of the coordinated water molecule make it easier for the coordinated water to lose a proton.

Second, the positive charge on the Al3+ ion attracts electron density from the oxygen atoms of the water molecules, which decreases the electron density in the O–H bonds, as shown in part (b) in Figure 16.12 "Effect of a Metal Ion on the Acidity of Water". With less electron density between the O atoms and the H atoms, the O–H bonds are weaker than in a free H2O molecule, making it easier to lose a H+ ion.

Figure 16.12 Effect of a Metal Ion on the Acidity of Water

(a) Reaction of the metal ion Al3+ with water to form the hydrated metal ion is an example of a Lewis acid–base reaction. (b) The positive charge on the aluminum ion attracts electron density from the oxygen atoms, which shifts electron density away from the O–H bonds. The decrease in electron density weakens the O–H bonds in the water molecules and makes it easier for them to lose a proton.

The magnitude of this effect depends on the following two factors (Figure 16.13 "The Effect of the Charge and Radius of a Metal Ion on the Acidity of a Coordinated Water Molecule"):

  1. The charge on the metal ion. A divalent ion (M2+) has approximately twice as strong an effect on the electron density in a coordinated water molecule as a monovalent ion (M+) of the same radius.
  2. The radius of the metal ion. For metal ions with the same charge, the smaller the ion, the shorter the internuclear distance to the oxygen atom of the water molecule and the greater the effect of the metal on the electron density distribution in the water molecule.

Figure 16.13 The Effect of the Charge and Radius of a Metal Ion on the Acidity of a Coordinated Water Molecule

The contours show the electron density on the O atoms and the H atoms in both a free water molecule (left) and water molecules coordinated to Na+, Mg2+, and Al3+ ions. These contour maps demonstrate that the smallest, most highly charged metal ion (Al3+) causes the greatest decrease in electron density of the O–H bonds of the water molecule. Due to this effect, the acidity of hydrated metal ions increases as the charge on the metal ion increases and its radius decreases.

Thus aqueous solutions of small, highly charged metal ions, such as Al3+ and Fe3+, are acidic:

Equation 16.36

[Al(H2O)6]3+(aq)[Al(H2O)5(OH)]2+(aq)+H+(aq)

The [Al(H2O)6]3+ ion has a pKa of 5.0, making it almost as strong an acid as acetic acid. Because of the two factors described previously, the most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. A number of pairs of metal ions that lie on a diagonal line in the periodic table, such as Li+ and Mg2+ or Ca2+ and Y3+, have different sizes and charges but similar charge-to-radius ratios. As a result, these pairs of metal ions have similar effects on the acidity of coordinated water molecules, and they often exhibit other significant similarities in chemistry as well.

Note the Pattern

Solutions of small, highly charged metal ions in water are acidic.

Reactions such as those discussed in this section, in which a salt reacts with water to give an acidic or basic solution, are often called hydrolysis reactionsA chemical reaction in which a salt reacts with water to yield an acidic or a basic solution.. Using a separate name for this type of reaction is unfortunate because it suggests that they are somehow different. In fact, hydrolysis reactions are just acid–base reactions in which the acid is a cation or the base is an anion; they obey the same principles and rules as all other acid–base reactions.

Note the Pattern

A hydrolysis reaction is an acid–base reaction.

Example 4

Predict whether aqueous solutions of these compounds are acidic, basic, or neutral.

  1. KNO3
  2. CrBr3·6H2O
  3. Na2SO4

Given: compound

Asked for: acidity or basicity of aqueous solution

Strategy:

A Assess the acid–base properties of the cation and the anion. If the cation is a weak Lewis acid, it will not affect the pH of the solution. If the cation is the conjugate acid of a weak base or a relatively highly charged metal cation, however, it will react with water to produce an acidic solution.

B If the anion is the conjugate base of a strong acid, it will not affect the pH of the solution. If, however, the anion is the conjugate base of a weak acid, the solution will be basic.

Solution:

  1. A The K+ cation has a small positive charge (+1) and a relatively large radius (because it is in the fourth row of the periodic table), so it is a very weak Lewis acid.

    B The NO3 anion is the conjugate base of a strong acid, so it has essentially no basic character (Table 16.1 "Definitions of Acids and Bases"). Hence neither the cation nor the anion will react with water to produce H+ or OH, and the solution will be neutral.

  2. A The Cr3+ ion is a relatively highly charged metal cation that should behave similarly to the Al3+ ion and form the [Cr(H2O)6]3+ complex, which will behave as a weak acid:

     

    B The Br anion is a very weak base (it is the conjugate base of the strong acid HBr), so it does not affect the pH of the solution. Hence the solution will be acidic.

  3. A The Na+ ion, like the K+, is a very weak acid, so it should not affect the acidity of the solution.

    B In contrast, SO42− is the conjugate base of HSO4, which is a weak acid. Hence the SO42− ion will react with water as shown in Figure 16.6 to give a slightly basic solution.

Exercise

Predict whether aqueous solutions of the following are acidic, basic, or neutral.

  1. KI
  2. Mg(ClO4)2
  3. NaHS

Answer:

  1. neutral
  2. acidic
  3. basic (due to the reaction of HS with water to form H2S and OH)

Summary

Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant (Ka). Similarly, the equilibrium constant for the reaction of a weak base with water is the base ionization constant (Kb). For any conjugate acid–base pair, KaKb = Kw. Smaller values of pKa correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of pKb correspond to larger base ionization constants and hence stronger bases. At 25°C, pKa + pKb = 14.00. Acid–base reactions always proceed in the direction that produces the weaker acid–base pair. No acid stronger than H3O+ and no base stronger than OH can exist in aqueous solution, leading to the phenomenon known as the leveling effect. Polyprotic acids (and bases) lose (and gain) protons in a stepwise manner, with the fully protonated species being the strongest acid and the fully deprotonated species the strongest base. A salt can dissolve in water to produce a neutral, a basic, or an acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion (A), the conjugate acid of a weak base as the cation (BH+), or both. Salts that contain small, highly charged metal ions produce acidic solutions in water. The reaction of a salt with water to produce an acidic or a basic solution is called a hydrolysis reaction.

Key Takeaways

  • Acid–base reactions always contain two conjugate acid–base pairs.
  • Each acid and each base has an associated ionization constant that corresponds to its acid or base strength.

Key Equations

Acid ionization constant

Equation 16.16: Ka=[H3O+][A][HA]

Base ionization constant

Equation 16.18: Kb=[BH+][OH][B]

Relationship between K a and K b of a conjugate acid–base pair

Equation 16.23: KaKb = Kw

Definition of p K a

Equation 16.24: pKa = −log10Ka

Equation 16.25: Ka=10pKa

Definition of p K b

Equation 16.26: pKa = −log10Ka

Equation 16.27: Kb=10pKb

Relationship between p K a and p K b of a conjugate acid–base pair

Equation 16.28: pKa + pKb = pKw

Equation 16.29: pKa + pKb = 14.00 (at 25°C)

Conceptual Problems

  1. Identify the conjugate acid–base pairs in each equilibrium.

    1. HSO4(aq)+H2O(l)SO42(aq)+H3O+(aq)
    2. C3H7NO2(aq)+H3O+(aq)C3H8NO2+(aq)+H2O(l)
    3. CH3CO2H(aq)+NH3(aq)CH3CO2(aq)+NH4+(aq)
    4. SbF5(aq)+2HF(aq)H2F+(aq)+SbF6(aq)
  2. Identify the conjugate acid–base pairs in each equilibrium.

    1. HF(aq)+H2O(l)H3O+(aq)+F(aq)
    2. CH3CH2NH2(aq)+H2O(l)CH3CH2NH3+(aq)+OH(aq)
    3. C3H7NO2(aq)+OH(aq)C3H6NO2(aq)+H2O(l)
    4. CH3CO2H(aq)+2HF(aq)CH3C(OH)2+(aq)+HF2(aq)
  3. Salts such as NaH contain the hydride ion (H). When sodium hydride is added to water, it produces hydrogen gas in a highly vigorous reaction. Write a balanced chemical equation for this reaction and identify the conjugate acid–base pairs.

  4. Write the expression for Ka for each reaction.

    1. HCO3(aq)+H2O(l)CO32(aq)+H3O+(aq)
    2. formic acid(aq)+H2O(l)formate(aq)+H3O+(aq)
    3. H3PO4(aq)+H2O(l)H2PO4(aq)+H3O+(aq)
  5. Write an expression for the ionization constant Kb for each reaction.

    1. OCH3(aq)+H2O(l)HOCH3(aq)+OH(aq)
    2. NH2(aq)+H2O(l)NH3(aq)+OH(aq)
    3. S2(aq)+H2O(l)HS(aq)+OH(aq)
  6. Predict whether each equilibrium lies primarily to the left or to the right.

    1. HBr(aq)+H2O(l)H3O+(aq)+Br(aq)
    2. NaH(soln)+NH3(l)H2(soln)+NaNH2(soln)
    3. OCH3(aq)+NH3(aq)CH3OH(aq)+NH2(aq)
    4. NH3(aq)+HCl(aq)NH4+(aq)+Cl(aq)
  7. Species that are strong bases in water, such as CH3, NH2, and S2−, are leveled to the strength of OH, the conjugate base of H2O. Because their relative base strengths are indistinguishable in water, suggest a method for identifying which is the strongest base. How would you distinguish between the strength of the acids HIO3, H2SO4, and HClO4?

  8. Is it accurate to say that a 2.0 M solution of H2SO4, which contains two acidic protons per molecule, is 4.0 M in H+? Explain your answer.

  9. The alkalinity of soil is defined by the following equation: alkalinity = [HCO3] + 2[CO32−] + [OH] − [H+]. The source of both HCO3 and CO32− is H2CO3. Explain why the basicity of soil is defined in this way.

  10. Why are aqueous solutions of salts such as CaCl2 neutral? Why is an aqueous solution of NaNH2 basic?

  11. Predict whether aqueous solutions of the following are acidic, basic, or neutral.

    1. Li3N
    2. NaH
    3. KBr
    4. C2H5NH3+Cl
  12. When each compound is added to water, would you expect the pH of the solution to increase, decrease, or remain the same?

    1. LiCH3
    2. MgCl2
    3. K2O
    4. (CH3)2NH2+Br
  13. Which complex ion would you expect to be more acidic—Pb(H2O)42+ or Sn(H2O)42+? Why?

  14. Would you expect Sn(H2O)42+ or Sn(H2O)64+ to be more acidic? Why?

  15. Is it possible to arrange the hydrides LiH, RbH, KH, CsH, and NaH in order of increasing base strength in aqueous solution? Why or why not?

Answer

    1. HSO4(aq) acid 1+H2O(l)base 2SO42(aq)base 1 + H3O+(aq)acid 2
    2. C3H7NO2(aq)base 2 + H3O+(aq)acid 1C3H8NO2+(aq)acid 2 + H2O(l)base 1
    3. HOAc(aq)acid 1 + NH3(aq)base 2CH3CO2(aq)base 1 + NH4+(aq)acid 2
    4. SbF5(aq)acid 1 + 2HF(aq)base 2H2F+(aq)acid 2 + SbF6(aq)base 1

Numerical Problems

  1. Arrange these acids in order of increasing strength.

    • acid A: pKa = 1.52
    • acid B: pKa = 6.93
    • acid C: pKa = 3.86

    Given solutions with the same initial concentration of each acid, which would have the highest percent ionization?

  2. Arrange these bases in order of increasing strength:

    • base A: pKb = 13.10
    • base B: pKb = 8.74
    • base C: pKb = 11.87

    Given solutions with the same initial concentration of each base, which would have the highest percent ionization?

  3. Calculate the Ka and the pKa of the conjugate acid of a base with each pKb value.

    1. 3.80
    2. 7.90
    3. 13.70
    4. 1.40
    5. −2.50
  4. Benzoic acid is a food preservative with a pKa of 4.20. Determine the Kb and the pKb for the benzoate ion.

  5. Determine Ka and pKa of boric acid [B(OH)3], solutions of which are occasionally used as an eyewash; the pKb of its conjugate base is 4.80.

Answers

  1. acid B < acid C < acid A (strongest)

    1. Ka = 6.3 × 10−11; pKa = 10.20
    2. Ka = 7.9 × 10−7; pKa = 6.10
    3. Ka = 0.50; pKa = 0.30
    4. Ka = 2.5 × 10−13; pKa = 12.60
    5. Ka = 3.2 × 10−17; pKa = 16.50
  2. Ka = 6.3 × 10−10 pKa = 9.20

16.3 Molecular Structure and Acid–Base Strength

Learning Objective

  1. To understand how molecular structure affects the strength of an acid or base.

We have seen that the strengths of acids and bases vary over many orders of magnitude. In this section, we explore some of the structural and electronic factors that control the acidity or basicity of a molecule.

Bond Strengths

In general, the stronger the A–H or B–H+ bond, the less likely the bond is to break to form H+ ions and thus the less acidic the substance. This effect can be illustrated using the hydrogen halides:

Relative Acid Strength HF < HCl < HBr < HI
H–X Bond Energy (kJ/mol) 570 432 366 298
pKa 3.20 −6.1 −8.9 −9.3

The trend in bond energies is due to a steady decrease in overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, acid strengths of binary hydrides increase as we go down a column of the periodic table. For example, the order of acidity for the binary hydrides of Group 16 is as follows, with pKa values in parentheses: H2O (14.00 = pKw) < H2S (7.05) < H2Se (3.89) < H2Te (2.6).

Stability of the Conjugate Base

Whether we write an acid–base reaction as AHA+H+ or as BH+B+H+, the conjugate base (A or B) contains one more lone pair of electrons than the parent acid (AH or BH+). Any factor that stabilizes the lone pair on the conjugate base favors dissociation of H+ and makes the parent acid a stronger acid. Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the periodic table. The observed order of increasing acidity is the following, with pKa values in parentheses: CH4 (~50) << NH3 (~36) < H2O (14.00) < HF (3.20). Consider, for example, the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of CH4 is CH3, and the conjugate base of HF is F. Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the F ion than carbon can stabilize the negative charge in the CH3 ion. Consequently, HF has a greater tendency to dissociate to form H+ and F than does methane to form H+ and CH3, making HF a much stronger acid than CH4.

The same trend is predicted by analyzing the properties of the conjugate acids. For a series of compounds of the general formula HE, as the electronegativity of E increases, the E–H bond becomes more polar, favoring dissociation to form E and H+. Due to both the increasing stability of the conjugate base and the increasing polarization of the E–H bond in the conjugate acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table.

Note the Pattern

Acid strengths of binary hydrides increase as we go down a column or from left to right across a row of the periodic table.

Inductive Effects

Atoms or groups of atoms in a molecule other than those to which H is bonded can induce a change in the distribution of electrons within the molecule. This is called an inductive effect, and, much like the coordination of water to a metal ion, it can have a major effect on the acidity or basicity of the molecule. For example, the hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In aqueous solution, they all produce the following equilibrium:

Equation 16.37

HOX(aq)H+(aq)+OX(aq)

The acidities of these acids vary by about three orders of magnitude, however, due to the difference in electronegativity of the halogen atoms:

HOX Electronegativity of X pKa
HOCl 3.0 7.40
HOBr 2.8 8.55
HOI 2.5 10.5

As the electronegativity of X increases, the distribution of electron density within the molecule changes: the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as H+.

The acidity of oxoacids, with the general formula HOXOn (n = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom X. As shown in , the Ka values of the oxoacids of chlorine increase by a factor of about 104 to 106 with each oxygen as successive oxygen atoms are added. The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base.

Note the Pattern

Any inductive effect that withdraws electron density from an O–H bond increases the acidity of the compound.

Because oxygen is the second most electronegative element, adding terminal oxygen atoms causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. The colors in show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen atoms increases. In and , blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from HClO to HClO4 (also written as HOClO3), while the H atom becomes steadily bluer, indicating that the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. The decrease in electron density in the O–H bond weakens it, making it easier to lose hydrogen as H+ ions, thereby increasing the strength of the acid.

Figure 16.14 The Relationship between the Acid Strengths of the Oxoacids of Chlorine and the Electron Density on the O–H Unit

These electrostatic potential maps show how the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. Blue corresponds to low electron densities, whereas red corresponds to high electron densities.

At least as important, however, is the effect of delocalization of the negative charge in the conjugate base. As shown in , the number of resonance structures that can be written for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, allowing the single negative charge to be delocalized over successively more oxygen atoms. The electrostatic potential plots in demonstrate that the electron density on the terminal oxygen atoms decreases steadily as their number increases. The oxygen atom in ClO is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in ClO4, indicating that the oxygen atoms are becoming steadily less electron rich through the series. For example, in the perchlorate ion (ClO4), the single negative charge is delocalized over all four oxygen atoms, whereas in the hypochlorite ion (OCl), the negative charge is largely localized on a single oxygen atom (). As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known.

Note the Pattern

Electron delocalization in the conjugate base increases acid strength.

Figure 16.15 The Relationship between Delocalization of the Negative Charge in the Oxoanions of Chlorine and the Number of Terminal Oxygen Atoms

As the number of terminal oxygen atoms increases, the number of resonance structures that can be written for the oxoanions of chlorine also increases, and the single negative charge is delocalized over more oxygen atoms. As these electrostatic potential plots demonstrate, the electron density on the terminal oxygen atoms decreases steadily as their number increases. As the electron density on the oxygen atoms decreases, so does their affinity for a proton, making the anion less basic. As a result, the parent oxoacid is more acidic.

Similar inductive effects are also responsible for the trend in the acidities of oxoacids that have the same number of oxygen atoms as we go across a row of the periodic table from left to right. For example, H3PO4 is a weak acid, H2SO4 is a strong acid, and HClO4 is one of the strongest acids known. The number of terminal oxygen atoms increases steadily across the row, consistent with the observed increase in acidity. In addition, the electronegativity of the central atom increases steadily from P to S to Cl, which causes electrons to be drawn from oxygen to the central atom, weakening the O–H bond and increasing the strength of the oxoacid.

Careful inspection of the data in shows two apparent anomalies: carbonic acid and phosphorous acid. If carbonic acid (H2CO3) were a discrete molecule with the structure (HO)2C=O, it would have a single terminal oxygen atom and should be comparable in acid strength to phosphoric acid (H3PO4), for which pKa1 = 2.16. Instead, the tabulated value of pKa1 for carbonic acid is 6.35, making it about 10,000 times weaker than expected. As we shall see in , however, H2CO3 is only a minor component of the aqueous solutions of CO2 that are referred to as carbonic acid. Similarly, if phosphorous acid (H3PO3) actually had the structure (HO)3P, it would have no terminal oxygen atoms attached to phosphorous. It would therefore be expected to be about as strong an acid as HOCl (pKa = 7.40). In fact, the pKa1 for phosphorous acid is 1.30, and the structure of phosphorous acid is (HO)2P(=O)H with one H atom directly bonded to P and one P=O bond. Thus the pKa1 for phosphorous acid is similar to that of other oxoacids with one terminal oxygen atom, such as H3PO4. Fortunately, phosphorous acid is the only common oxoacid in which a hydrogen atom is bonded to the central atom rather than oxygen.

Table 16.4 Values of pKa for Selected Polyprotic Acids and Bases

Polyprotic Acids Formula p K a1 p K a2 p K a3
carbonic acid* “H2CO3 6.35 10.33
citric acid HO2CCH2C(OH)(CO2H)CH2CO2H 3.13 4.76 6.40
malonic acid HO2CCH2CO2H 2.85 5.70
oxalic acid HO2CCO2H 1.25 3.81
phosphoric acid H3PO4 2.16 7.21 12.32
phosphorous acid H3PO3 1.3 6.70
succinic acid HO2CCH2CH2CO2H 4.21 5.64
sulfuric acid H2SO4 −2.0 1.99
sulfurous acid* “H2SO3 1.85 7.21
  
Polyprotic Bases Formula p K b1 p K b2
ethylenediamine H2N(CH2)2NH2 4.08 7.14
piperazine HN(CH2CH2)2NH 4.27 8.67
propylenediamine H2N(CH2)3NH2 3.45 5.12
*H2CO3 and H2SO3 are at best minor components of aqueous solutions of CO2(g) and SO2(g), respectively, but such solutions are commonly referred to as containing carbonic acid and sulfurous acid, respectively.

Inductive effects are also observed in organic molecules that contain electronegative substituents. The magnitude of the electron-withdrawing effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives:

 pKaCH3CO2H4.76<CH2ClCO2H2.87<CHCl2CO2H1.35<CCl3CO2H0.66<CF3CO2H0.52

As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite large. For instance, replacing the –CH3 group of acetic acid by a –CF3 group results in about a 10,000-fold increase in acidity!

Example 5

Arrange the compounds of each series in order of increasing acid or base strength.

  1. sulfuric acid [H2SO4, or (HO)2SO2], fluorosulfonic acid (FSO3H, or FSO2OH), and sulfurous acid [H2SO3, or (HO)2SO]
  2. ammonia (NH3), trifluoramine (NF3), and hydroxylamine (NH2OH)

The structures are shown here.

Given: series of compounds

Asked for: relative acid or base strengths

Strategy:

Use relative bond strengths, the stability of the conjugate base, and inductive effects to arrange the compounds in order of increasing tendency to ionize in aqueous solution.

Solution:

  1. Although both sulfuric acid and sulfurous acid have two –OH groups, the sulfur atom in sulfuric acid is bonded to two terminal oxygen atoms versus one in sulfurous acid. Because oxygen is highly electronegative, sulfuric acid is the stronger acid because the negative charge on the anion is stabilized by the additional oxygen atom. In comparing sulfuric acid and fluorosulfonic acid, we note that fluorine is more electronegative than oxygen. Thus replacing an –OH by –F will remove more electron density from the central S atom, which will, in turn, remove electron density from the S–OH bond and the O–H bond. Because its O–H bond is weaker, FSO3H is a stronger acid than sulfuric acid. The predicted order of acid strengths given here is confirmed by the measured pKa values for these acids:

    pKaH2SO31.85<H2SO42<FSO3H10
  2. The structures of both trifluoramine and hydroxylamine are similar to that of ammonia. In trifluoramine, all of the hydrogen atoms in NH3 are replaced by fluorine atoms, whereas in hydroxylamine, one hydrogen atom is replaced by OH. Replacing the three hydrogen atoms by fluorine will withdraw electron density from N, making the lone electron pair on N less available to bond to an H+ ion. Thus NF3 is predicted to be a much weaker base than NH3. Similarly, because oxygen is more electronegative than hydrogen, replacing one hydrogen atom in NH3 by OH will make the amine less basic. Because oxygen is less electronegative than fluorine and only one hydrogen atom is replaced, however, the effect will be smaller. The predicted order of increasing base strength shown here is confirmed by the measured pKb values:

     pKbNF3<<NH2OH8.06<NH34.75

    Trifluoramine is such a weak base that it does not react with aqueous solutions of strong acids. Hence its base ionization constant has not been measured.

Exercise

Arrange the compounds of each series in order of

  1. decreasing acid strength: H3PO4, CH3PO3H2, and HClO3.
  2. increasing base strength: CH3S, OH, and CF3S.

Answer:

  1. HClO3 > CH3PO3H2 > H3PO4
  2. CF3S < CH3S < OH

Summary

The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an H+ ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of H+, making the conjugate acid a stronger acid. Atoms or groups of atoms elsewhere in a molecule can also be important in determining acid or base strength through an inductive effect, which can weaken an O–H bond and allow hydrogen to be more easily lost as H+ ions.

Key Takeaway

  • Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound.

Conceptual Problems

  1. presented several factors that affect the relative strengths of acids and bases. For each pair, identify the most important factor in determining which is the stronger acid or base in aqueous solution.

    1. CH3CCl2CH2CO2H versus CH3CH2CH2CO2H
    2. CH3CO2H versus CH3CH2OH
    3. HClO versus HBrO
    4. CH3C(=O)NH2 versus CH3CH2NH2
    5. H3AsO4 versus H3AsO3
  2. The stability of the conjugate base is an important factor in determining the strength of an acid. Which would you expect to be the stronger acid in aqueous solution—C6H5NH3+ or NH4+? Justify your reasoning.

  3. Explain why H2Se is a weaker acid than HBr.

  4. Arrange the following in order of decreasing acid strength in aqueous solution: H3PO4, CH3PO3H2, and HClO3.

  5. Arrange the following in order of increasing base strength in aqueous solution: CH3S, OH, and CF3S.

  6. Arrange the following in order of increasing acid strength in aqueous solution: HClO2, HNO2, and HNO3.

  7. Do you expect H2SO3 or H2SeO3 to be the stronger acid? Why?

  8. Give a plausible explanation for why CF3OH is a stronger acid than CH3OH in aqueous solution. Do you expect CHCl2CH2OH to be a stronger or a weaker acid than CH3OH? Why?

  9. Do you expect Cl2NH or NH3 to be the stronger base in aqueous solution? Why?

Answers

  1. CF3S < CH3S < OH (strongest base)

  2. NH3; Cl atoms withdraw electron density from N in Cl2NH.

16.4 Quantitative Aspects of Acid–Base Equilibriums

Learning Objective

  1. To use Ka and Kb values to calculate the percent ionization and the pH of a solution of an acid or a base.

This section presents a quantitative approach to analyzing acid–base equilibriums. You will learn how to determine the values of Ka and Kb, how to use Ka or Kb to calculate the percent ionization and the pH of an aqueous solution of an acid or a base, and how to calculate the equilibrium constant for the reaction of an acid with a base from the Ka and Kb of the reactants.

Determining Ka and Kb

The ionization constants Ka and Kb are equilibrium constants that are calculated from experimentally measured concentrations, just like the equilibrium constants discussed in . Before proceeding further, it is important to understand exactly what is meant when we describe the concentration of an aqueous solution of a weak acid or a weak base. Suppose, for example, we have a bottle labeled 1.0 M acetic acid or 1.0 M ammonia. As you learned in , such a solution is usually prepared by dissolving 1.0 mol of acetic acid or ammonia in water and adding enough water to give a final volume of exactly 1.0 L. If, however, we were to list the actual concentrations of all the species present in either solution, we would find that none of the values is exactly 1.0 M because a weak acid such as acetic acid or a weak base such as ammonia always reacts with water to some extent. The extent of the reaction depends on the Ka or the Kb, the concentration of the acid or the base, and the temperature. Consequently, only the total concentration of both the ionized and unionized species is equal to 1.0 M.

The analytical concentration (C) is defined as the total concentration of all forms of an acid or a base that are present in solution, regardless of their state of protonation. Thus a “1.0 M” solution of acetic acid has an analytical concentration of 1.0 M, which is the sum of the actual concentrations of unionized acetic acid (CH3CO2H) and the ionized form (CH3CO2):

Equation 16.38

CCH3CO2H=[CH3CO2H]+[CH3CO2]

As we shall see shortly, if we know the analytical concentration and the Ka, we can calculate the actual values of [CH3CO2H] and [CH3CO2].

The equilibrium equations for the reaction of acetic acid and ammonia with water are as follows:

Equation 16.39

Ka=[H+][CH3CO2][CH3CO2H]

Equation 16.40

Kb=[NH4+][OH][NH3]

where Ka and Kb are the ionization constants for acetic acid and ammonia, respectively. In addition to the analytical concentration of the acid (or the base), we must have a way to measure the concentration of at least one of the species in the equilibrium constant expression to determine the Ka (or the Kb). There are two common ways to obtain the concentrations: (1) measure the electrical conductivity of the solution, which is related to the total concentration of ions present, and (2) measure the pH of the solution, which gives [H+] or [OH].

Example 6 and Example 7 illustrate the procedure for determining Ka for a weak acid and Kb for a weak base. In both cases, we will follow the procedure developed in : the analytical concentration of the acid or the base is the initial concentration, and the stoichiometry of the reaction with water determines the change in concentrations. The final concentrations of all species are calculated from the initial concentrations and the changes in the concentrations. Inserting the final concentrations into the equilibrium constant expression enables us to calculate the Ka or the Kb.

Example 6

Electrical conductivity measurements indicate that 0.42% of the acetic acid molecules in a 1.00 M solution are ionized at 25°C. Calculate Ka and pKa for acetic acid at this temperature.

Given: analytical concentration and percent ionization

Asked for: Ka and pKa

Strategy:

A Write the balanced equilibrium equation for the reaction and derive the equilibrium constant expression.

B Use the data given and the stoichiometry of the reaction to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations for all species in the equilibrium constant expression.

C Substitute the final concentrations into the equilibrium constant expression and calculate the Ka. Take the negative logarithm of Ka to obtain the pKa.

Solution:

A The balanced equilibrium equation for the dissociation of acetic acid is as follows:

CH3CO2H(aq)H+(aq)+CH3CO2(aq)

and the equilibrium constant expression is as follows:

Ka=[H+][CH3CO2][CH3CO2H]

B To calculate the Ka, we need to know the equilibrium concentrations of CH3CO2H, CH3CO2, and H+. The most direct way to do this is to construct a table that lists the initial concentrations and the changes in concentrations that occur during the reaction to give the final concentrations, using the procedure introduced in . The initial concentration of unionized acetic acid ([CH3CO2H]i) is the analytical concentration, 1.00 M, and the initial acetate concentration ([CH3CO2]i) is zero. The initial concentration of H+ is not zero, however; [H+]i is 1.00 × 10−7 M due to the autoionization of water. The measured percent ionization tells us that 0.42% of the acetic acid molecules are ionized at equilibrium. Consequently, the change in the concentration of acetic acid is Δ[CH3CO2H] = −(4.2 × 10−3)(1.00 M) = −0.0042 M. Conversely, the change in the acetate concentration is Δ[CH3CO2] = +0.0042 M because every 1 mol of acetic acid that ionizes gives 1 mol of acetate. Because one proton is produced for each acetate ion formed, Δ[H+] = +0.0042 M as well. These results are summarized in the following table.

CH3CO2H(aq)H+(aq)+CH3CO2(aq)
[CH3CO2H] [H+] [CH3CO2]
initial 1.00 1.00 × 10−7 0
change −0.0042 +0.0042 +0.0042
final

The final concentrations of all species are therefore as follows:

[CH3CO2H]f=[CH3CO2H]i+Δ[CH3CO2H]=1.00 M+(0.0042 M)=1.00 M[CH3CO2]f=[CH3CO2]i+Δ[CH3CO2]=0 M+(+0.0042 M)=0.0042 M[H+]f=[H+]i+Δ[H+]=1.00×107 M+(+0.0042 M)=0.0042 M

C We can now calculate Ka by inserting the final concentrations into the equilibrium constant expression:

Ka=[H+][CH3CO2][CH3CO2H]=(0.0042)(0.0042)1.00=1.8×105

The pKa is the negative logarithm of Ka: pKa = −log Ka = −log(1.8 × 10−5) = 4.74.

Exercise

Picric acid is the common name for 2,4,6-trinitrophenol, a derivative of phenol (C6H5OH) in which three H atoms are replaced by nitro (–NO2) groups. The presence of the nitro groups removes electron density from the phenyl ring, making picric acid a much stronger acid than phenol (pKa = 9.99). The nitro groups also make picric acid potentially explosive, as you might expect based on its chemical similarity to 2,4,6-trinitrotoluene, better known as TNT. A 0.20 M solution of picric acid is 73% ionized at 25°C. Calculate Ka and pKa for picric acid.

Answer: Ka = 0.39; pKa = 0.41

Example 7

A 1.0 M aqueous solution of ammonia has a pH of 11.63 at 25°C. Calculate Kb and pKb for ammonia.

Given: analytical concentration and pH

Asked for: Kb and pKb

Strategy:

A Write the balanced equilibrium equation for the reaction and derive the equilibrium constant expression.

B Use the data given and the stoichiometry of the reaction to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations for all species in the equilibrium constant expression.

C Substitute the final concentrations into the equilibrium constant expression and calculate the Kb. Take the negative logarithm of Kb to obtain the pKb.

Solution:

A The balanced equilibrium equation for the reaction of ammonia with water is as follows:

NH3(aq)+H2O(l)NH4+(aq)+OH(aq)

and the equilibrium constant expression is as follows:

Kb=[NH4+][OH][NH3]

Remember that water does not appear in the equilibrium constant expression for Kb.

B To calculate Kb, we need to know the equilibrium concentrations of NH3, NH4+, and OH. The initial concentration of NH3 is the analytical concentration, 1.0 M, and the initial concentrations of NH4+ and OH are 0 M and 1.00 × 10−7 M, respectively. In this case, we are given the pH of the solution, which allows us to calculate the final concentration of one species (OH) directly, rather than the change in concentration. Recall that pKw = pH + pOH = 14.00 at 25°C. Thus pOH = 14.00 − pH = 14.00 − 11.63 = 2.37, and [OH]f = 10−2.37 = 4.3 × 10−3 M. Our data thus far are listed in the following table.

NH3(aq)NH4+(aq)+OH(aq)
[NH3] [NH4+] [OH]
initial 1.0 0 1.00 × 10−7
change
final 4.3 × 10−3

The final [OH] is much greater than the initial [H+], so the change in [OH] is as follows:

Δ[OH] = (4.3 × 10−3 M) − (1.00 × 10−7 M) ≈ 4.3 × 10−3 M

The stoichiometry of the reaction tells us that 1 mol of NH3 is converted to NH4+ for each 1 mol of OH formed, so

Δ[NH4+] = +4.3 × 10−3 M and Δ[NH3] = −4.3 ×10−3 M

We can now insert these values for the changes in concentrations into the table, which enables us to complete the table.

H2O(l)+NH3(aq)NH4+(aq)+OH(aq)
[NH3] [NH4+] [OH]
initial 1.0 0 1.00 × 10−7
change −4.3 × 10−3 +4.3 × 10−3 +4.3 × 10−3
final 1.0 4.3 × 10−3 4.3 × 10−3

C Inserting the final concentrations into the equilibrium constant expression gives Kb:

Kb=[NH4+][OH][NH3]=(4.3×103)21.0=1.8×105

and pKb = −log Kb = 4.74.

The Kb and the pKb for ammonia are almost exactly the same as the Ka and the pKa for acetic acid at 25°C. In other words, ammonia is almost exactly as strong a base as acetic acid is an acid. Consequently, the extent of the ionization reaction in an aqueous solution of ammonia at a given concentration is the same as in an aqueous solution of acetic acid at the same concentration.

Exercise

The pH of a 0.050 M solution of pyridine (C6H5N) is 8.96 at 25°C. Calculate Kb and pKb for pyridine.

 

Answer: Kb = 1.7 × 10−9; pKb = 8.77

Calculating Percent Ionization from Ka or Kb

When carrying out a laboratory analysis, chemists frequently need to know the concentrations of all species in solution. Because the reactivity of a weak acid or a weak base is usually very different from the reactivity of its conjugate base or acid, we often need to know the percent ionization of a solution of an acid or a base to understand a chemical reaction. The percent ionization is defined as follows:

Equation 16.41

percent ionization οf acid=[H+]CHA×100

Equation 16.42

percent ionization οf base=[OH]CB×100

One way to determine the concentrations of species in solutions of weak acids and bases is a variation of the tabular method we used previously to determine Ka and Kb values. As a demonstration, we will calculate the concentrations of all species and the percent ionization in a 0.150 M solution of formic acid at 25°C. The data in show that formic acid (Ka = 1.8 × 10−4 at 25°C) is a slightly stronger acid than acetic acid. The equilibrium equation for the ionization of formic acid in water is as follows:

Equation 16.43

HCO2H(aq)H+(aq)+HCO2(aq)

and the equilibrium constant expression for this reaction is as follows:

Equation 16.44

Ka=[H+][HCO2][HCO2H]

We set the initial concentration of HCO2H equal to 0.150 M, and that of HCO2 is 0 M. The initial concentration of H+ is 1.00 × 10−7 M due to the autoionization of water. Because the equilibrium constant for the ionization reaction is small, the equilibrium will lie to the left, favoring the unionized form of the acid. Hence we can define x as the amount of formic acid that dissociates.

If the change in [HCO2H] is −x, then the change in [H+] and [HCO2] is +x. The final concentration of each species is the sum of its initial concentration and the change in concentration, as summarized in the following table.

HCO2H(aq)H+(aq)+HCO2(aq)
[HCO2H] [H+] [HCO2]
initial 0.150 1.00 × 10−7 0
change x +x +x
final (0.150 − x) (1.00 × 10−7 + x) x

We can calculate x by substituting the final concentrations from the table into the equilibrium constant expression:

Ka=[H+][HCO2][HCO2H]=(1.00×107+x)x0.150x

Because the ionization constant Ka is small, x is likely to be small compared with the initial concentration of formic acid: (0.150 − x) M ≈ 0.150 M. Moreover, [H+] due to the autoionization of water (1.00 × 10−7 M) is likely to be negligible compared with [H+] due to the dissociation of formic acid: (1.00 × 10−7 + x) M ≈ x M. Inserting these values into the equilibrium constant expression and solving for x,

Ka=x20.150=1.8×104x=5.2×103

We can now calculate the concentrations of the species present in a 0.150 M formic acid solution by inserting this value of x into the expressions in the last line of the table:

[HCO2H]=(0.150x) M=0.145 M[HCO2]=x=5.2×103 M[H+]=(1.00×107+x) M=5.2×103 M

Thus the pH of the solution is –log(5.2 × 10−3) = 2.28. We can also use these concentrations to calculate the fraction of the original acid that is ionized. In this case, the percent ionization is the ratio of [H+] (or [HCO2]) to the analytical concentration, multiplied by 100 to give a percentage:

percent ionization=[H+]CHA×100=5.2×103 M0.150×100=3.5%

Always check to make sure that any simplifying assumption was valid. As a general rule of thumb, approximations such as those used here are valid only if the quantity being neglected is no more than about 5% of the quantity to which it is being added or from which it is being subtracted. If the quantity that was neglected is much greater than about 5%, then the approximation is probably not valid, and you should go back and solve the problem using the quadratic formula. In the previous demonstration, both simplifying assumptions were justified: the percent ionization is only 3.5%, which is well below the approximately 5% limit, and the 1.00 × 10−7 M [H+] due to the autoionization of water is much, much less than the 5.2 × 10−3 M [H+] due to the ionization of formic acid.

As a general rule, the [H+] contribution due to the autoionization of water can be ignored as long as the product of the acid or the base ionization constant and the analytical concentration of the acid or the base is at least 10 times greater than the [H+] or [OH] from the autoionization of water—that is, if

Equation 16.45

KaCHA ≥ 10(1.00 × 10−7) = 1.0 × 10−6

or

Equation 16.46

KbCB ≥ 10(1.00 × 10−7) = 1.0 × 10−6

By substituting the appropriate values for the formic acid solution into , we see that the simplifying assumption is valid in this case:

Equation 16.47

KaCHA = (1.8 × 10−4)(0.150) = 2.7 × 10−5 > 1.0 × 10−6

Doing this simple calculation before solving this type of problem saves time and allows you to write simplified expressions for the final concentrations of the species present. In practice, it is necessary to include the [H+] contribution due to the autoionization of water only for extremely dilute solutions of very weak acids or bases.

Example 8 illustrates how the procedure outlined previously can be used to calculate the pH of a solution of a weak base.

Example 8

Calculate the pH and percent ionization of a 0.225 M solution of ethylamine (CH3CH2NH2), which is used in the synthesis of some dyes and medicines. The pKb of ethylamine is 3.19 at 20°C.

Given: concentration and pKb

Asked for: pH and percent ionization

Strategy:

A Write the balanced equilibrium equation for the reaction and the equilibrium constant expression. Calculate Kb from pKb.

B Use to see whether you can ignore [H+] due to the autoionization of water. Then use a tabular format to write expressions for the final concentrations of all species in solution. Substitute these values into the equilibrium equation and solve for [OH]. Use to calculate the percent ionization.

C Use the relationship Kw = [OH][H+] to obtain [H+]. Then calculate the pH of the solution.

Solution:

A We begin by writing the balanced equilibrium equation for the reaction:

CH3CH2NH2(aq)+H2O(l)CH3CH2NH3+(aq)+OH(aq)

The corresponding equilibrium constant expression is as follows:

Kb=[CH3CH2NH3+][OH][CH3CH2NH2]

From the pKb, we have Kb = 10−3.19 = 6.5 × 10−4.

B To calculate the pH, we need to determine the H+ concentration. Unfortunately, H+ does not appear in either the chemical equation or the equilibrium constant expression. However, [H+] and [OH] in an aqueous solution are related by Kw = [H+][OH]. Hence if we can determine [OH], we can calculate [H+] and then the pH. The initial concentration of CH3CH2NH2 is 0.225 M, and the initial [OH] is 1.00 × 10−7 M. Because ethylamine is a weak base, the extent of the reaction will be small, and it makes sense to let x equal the amount of CH3CH2NH2 that reacts with water. The change in [CH3CH2NH2] is therefore −x, and the change in both [CH3CH2NH3+] and [OH] is +x. To see whether the autoionization of water can safely be ignored, we substitute Kb and CB into :

KbCB = (6.5 × 10−4)(0.225) = 1.5 × 10−4 > 1.0 × 10−6

Thus the simplifying assumption is valid, and we will not include [OH] due to the autoionization of water in our calculations.

H2O(1)+CH3CH2NH2(aq)CH3CH2NH3+(aq)+OH(aq)
[CH3CH2NH2] [CH3CH2NH3+] [OH]
initial 0.225 0 1.00 × 10−7
change x +x +x
final (0.225 − x) x x

Substituting the quantities from the last line of the table into the equilibrium constant expression,

Kb=[CH3CH2NH3+][OH][CH3CH2NH2]=(x)(x)0.225x=6.5×104

As before, we assume the amount of CH3CH2NH2 that ionizes is small compared with the initial concentration, so [CH3CH2NH2]f = 0.225 − x ≈ 0.225. With this assumption, we can simplify the equilibrium equation and solve for x:

Kb=x20.225=6.5×104x=0.012=[CH3CH2NH3+]f=[OH]f

The percent ionization is therefore

percent ionization=[OH]CB×100=0.012 M0.225 M×100=5.4%

which is at the upper limit of the approximately 5% range that can be ignored. The final hydroxide concentration is thus 0.012 M.

C We can now determine the [H+] using the expression for Kw:

Kw=[OH][H+]1.01×1014=(0.012 M)[H+]8.4×1013 M=[H+]

The pH of the solution is −log(8.4 × 10−13) = 12.08. Alternatively, we could have calculated pOH as −log(0.012) = 1.92 and determined the pH as follows:

pH + pOH=pKw=14.00pH=14.001.92=12.08

The two methods are equivalent.

Exercise

Aromatic amines, in which the nitrogen atom is bonded directly to a phenyl ring (−C6H5) tend to be much weaker bases than simple alkylamines. For example, aniline (C6H5NH2) has a pKb of 9.13 at 25°C. What is the pH of a 0.050 M solution of aniline?

Answer: 8.78

The previous examples illustrate a key difference between solutions of strong acids and bases and solutions of weak acids and bases. Because strong acids and bases ionize essentially completely in water, the percent ionization is always approximately 100%, regardless of the concentration. In contrast, the percent ionization in solutions of weak acids and bases is small and depends on the analytical concentration of the weak acid or base. As illustrated for benzoic acid in , the percent ionization of a weak acid or a weak base actually increases as its analytical concentration decreases. The percent ionization also increases as the magnitude of Ka and Kb increases.

Figure 16.16 The Relationship between the Analytical Concentration of a Weak Acid and Percent Ionization

As shown here for benzoic acid (C6H5CO2H), the percent ionization decreases as the analytical concentration of a weak acid increases.

Unlike the Ka or the Kb, the percent ionization is not a constant for weak acids and bases but depends on both the Ka or the Kband the analytical concentration. Consequently, the procedure in Example 8 must be used to calculate the percent ionization and pH for solutions of weak acids and bases. Example 9 and its corresponding exercise demonstrate that the combination of a dilute solution and a relatively large Ka or Kb can give a percent ionization much greater than 5%, making it necessary to use the quadratic equation to determine the concentrations of species in solution.

Note the Pattern

The percent ionization in a solution of a weak acid or a weak base increases as the analytical concentration decreases and as the Ka or the Kbincreases.

Example 9

Benzoic acid (C6H5CO2H) is used in the food industry as a preservative and medically as an antifungal agent. Its pKa at 25°C is 4.20, making it a somewhat stronger acid than acetic acid. Calculate the percentage of benzoic acid molecules that are ionized in each solution.

  1. a 0.0500 M solution
  2. a 0.00500 M solution

Given: concentrations and pKa

Asked for: percent ionization

Strategy:

A Write both the balanced equilibrium equation for the ionization reaction and the equilibrium equation (). Use to calculate the Ka from the pKa.

B For both the concentrated solution and the dilute solution, use a tabular format to write expressions for the final concentrations of all species in solution. Substitute these values into the equilibrium equation and solve for [C6H5CO2]f for each solution.

C Use the values of [C6H5CO2]f and to calculate the percent ionization.

Solution:

A If we abbreviate benzoic acid as PhCO2H where Ph = –C6H5, the balanced equilibrium equation for the ionization reaction and the equilibrium equation can be written as follows:

PhCO2H(aq)H+(aq)+PhCO2(aq)Ka=[H+][PhCO2][PhCO2H]

From the pKa, we have Ka = 10−4.20 = 6.3 × 10−5.

  1. B For the more concentrated solution, we set up our table of initial concentrations, changes in concentrations, and final concentrations:

    PhCO2H(aq)H+(aq)+PhCO2(aq)
    [PhCO2H] [H+] [PhCO2]
    initial 0.0500 1.00 × 10−7 0
    change x +x +x
    final (0.0500 − x) (1.00 × 10−7 + x) x

    Inserting the expressions for the final concentrations into the equilibrium equation and making our usual assumptions, that [PhCO2] and [H+] are negligible due to the autoionization of water,

    Ka=[H+][PhCO2][PhCO2H]=(x)(x)0.0500x=x20.0500=6.3×1051.8×103=x

    This value is less than 5% of 0.0500, so our simplifying assumption is justified, and [PhCO2] at equilibrium is 1.8 × 10−3 M. We reach the same conclusion using CHA: KaCHA = (6.3 × 10−5)(0.0500) = 3.2 × 10−6 > 1.0 × 10−6.

    C The percent ionized is the ratio of the concentration of PhCO2 to the analytical concentration, multiplied by 100:

    percent ionized=[PhCO2]CPhCO2H×100=1.8×1030.0500×100=3.6%

    Because only 3.6% of the benzoic acid molecules are ionized in a 0.0500 M solution, our simplifying assumptions are confirmed.

  2. B For the more dilute solution, we proceed in exactly the same manner. Our table of concentrations is therefore as follows:

    PhCO2H(aq)H+(aq)+PhCO2(aq)
    [PhCO2H] [H+] [PhCO2]
    initial 0.00500 1.00 × 10−7 0
    change x +x +x
    final (0.00500 − x) (1.00 × 10−7 + x) x

    Inserting the expressions for the final concentrations into the equilibrium equation and making our usual simplifying assumptions,

    Ka=[H+][PhCO2][PhCO2H]=(x)(x)0.00500x=x20.00500=6.3×1055.6×104=x

    Unfortunately, this number is greater than 10% of 0.00500, so our assumption that the fraction of benzoic acid that is ionized in this solution could be neglected and that (0.00500 − x) ≈ x is not valid. Furthermore, we see that KaCHA = (6.3 × 10−5)(0.00500) = 3.2 × 10−7 < 1.0 × 10−6. Thus the relevant equation is as follows:

    x20.00500x=6.3×105

    which must be solved using the quadratic formula. Multiplying out the quantities,

    x2 = (6.3 × 10−5)(0.00500 − x) = (3.2 × 10−7) − (6.3 × 10−5)x

    Rearranging the equation to fit the standard quadratic equation format,

    x2 + (6.3 × 10−5)x − (3.2 × 10−7) = 0

    This equation can be solved by using the quadratic formula:

    x=b±b24ac2a=(6.3×105)±(6.3×105)24(1)(3.2×107)2(1)=(6.3×105)±(1.1×103)2=5.3×104 or5.9×104

    Because a negative x value corresponds to a negative [PhCO2], which is not physically meaningful, we use the positive solution: x = 5.3 × 10−4. Thus [PhCO2] = 5.3 × 10−4 M.

    C The percent ionized is therefore

    percent ionized=[PhCO2]CPhCO2H×100=5.3×1040.00500×100=11%

    In the more dilute solution (C = 0.00500 M), 11% of the benzoic acid molecules are ionized versus only 3.6% in the more concentrated solution (C = 0.0500 M). Decreasing the analytical concentration by a factor of 10 results in an approximately threefold increase in the percentage of benzoic acid molecules that are ionized.

Exercise

Lactic acid (CH3CH(OH)CO2H) is a weak acid with a pKa of 3.86 at 25°C. What percentage of the lactic acid is ionized in each solution?

  1. a 0.10 M solution
  2. a 0.0020 M solution

Answer:

  1. 3.7%
  2. 23%

Determining Keq from Ka and Kb

In , you learned how to use Ka and Kb values to qualitatively predict whether reactants or products are favored in an acid–base reaction. Tabulated values of Ka (or pKa) and Kb (or pKb), plus the Kw, enable us to quantitatively determine the direction and extent of reaction for a weak acid and a weak base by calculating K for the reaction. To illustrate how to do this, we begin by writing the dissociation equilibriums for a weak acid and a weak base and then summing them:

Equation 16.48

acidHAH++AKabase+ H2OHB++OHKbsumHA + B+ H2OH++A+HB++OHKsum=KaKb

The overall reaction has H2O on the left and H+ and OH on the right, which means it involves the autoionization of water (H2OH++OH) in addition to the acid–base equilibrium in which we are interested. We can obtain an equation that includes only the acid–base equilibrium by simply adding the equation for the reverse of the autoionization of water (H++OHH2O), for which K = 1/Kw, to the overall equilibrium in and canceling:

Equation 16.49

HA + B+ H2OH++A+HB++OHKsum=KaKbH++OHH2O1/KwHA + BA+HB+K=(KaKb)/Kw

Thus the equilibrium constant for the reaction of a weak acid with a weak base is the product of the ionization constants of the acid and the base divided by Kw. Example 10 illustrates how to calculate the equilibrium constant for the reaction of a weak acid with a weak base.

Example 10

Fish tend to spoil rapidly, even when refrigerated. The cause of the resulting “fishy” odor is a mixture of amines, particularly methylamine (CH3NH2), a volatile weak base (pKb = 3.34). Fish is often served with a wedge of lemon because lemon juice contains citric acid, a triprotic acid with pKa values of 3.13, 4.76, and 6.40 that can neutralize amines. Calculate the equilibrium constant for the reaction of excess citric acid with methylamine, assuming that only the first dissociation constant of citric acid is important.

Given: pKb for base and pKa for acid

Asked for: K

Strategy:

A Write the balanced equilibrium equation and the equilibrium constant expression for the reaction.

B Convert pKa and pKb to Ka and Kb and then use to calculate K.

Solution:

A If we abbreviate citric acid as H3citrate, the equilibrium equation for its reaction with methylamine is as follows:

CH3NH2(aq)+H3citrate(aq)CH3NH3+(aq)+H2citrate(aq)

The equilibrium constant expression for this reaction is as follows:

K=[CH3NH3+][H2citrate][CH3NH2][H3citrate]

B is K = (KaKb)/Kw. Converting pKa and pKb to Ka and Kb gives Ka = 10−3.13 = 7.4 × 10−4 for citric acid and Kb = 10−3.34 = 4.6 × 10−4 for methylamine. Substituting these values into the equilibrium equation,

K=KaKbKw=(7.4×104)(4.6×104)1.01×1014=3.4×107

The value of pK can also be calculated directly by taking the negative logarithm of both sides of , which gives

pK = pKa + pKb − pKw = 3.13 + 3.34 − 14.00 = −7.53

Thus K = 10−(−7.53) = 3.4 × 107, in agreement with the earlier value. In either case, the K values show that the reaction of citric acid with the volatile, foul-smelling methylamine lies very far to the right, favoring the formation of a much less volatile salt with no odor. This is one reason a little lemon juice helps make less-than-fresh fish more appetizing.

Exercise

Dilute aqueous ammonia solution, often used as a cleaning agent, is also effective as a deodorizing agent. To see why, calculate the equilibrium constant for the reaction of aqueous ammonia with butyric acid (CH3CH2CH2CO2H), a particularly foul-smelling substance associated with the odor of rancid butter and smelly socks. The pKb of ammonia is 4.75, and the pKa of butyric acid is 4.83.

Answer: 2.6 × 104

Summary

If the concentration of one or more of the species in a solution of an acid or a base is determined experimentally, Ka and Kb can be calculated, and Ka, pKa, Kb, and pKb can be used to quantitatively describe the composition of solutions of acids and bases. The concentrations of all species present in solution can be determined, as can the pH of the solution and the percentage of the acid or base that is ionized. The equilibrium constant for the reaction of a weak acid with a weak base can be calculated from Ka (or pKa), Kb (or pKb), and Kw.

Key Takeaway

  • For a solution of a weak acid or a weak base, the percent ionization increases as the Ka or the Kb increases and as the analytical concentration decreases.

Key Equations

Percent ionization of acid

: [H+]CHA×100

Percent ionization of base

: [OH]CB×100

Equilibrium constant for reaction of a weak acid with a weak base

: K=KaKbKw

Conceptual Problems

  1. Explain why the analytical concentration (C) of H2SO4 is equal to [H2SO4] + [HSO4] + [SO42−].

  2. Write an expression for the analytical concentration (C) of H3PO4 in terms of the concentrations of the species actually present in solution.

  3. For relatively dilute solutions of a weak acid such as acetic acid (CH3CO2H), the concentration of undissociated acetic acid in solution is often assumed to be the same as the analytical concentration. Explain why this is a valid practice.

  4. How does dilution affect the percent ionization of a weak acid or a weak base?

  5. What is the relationship between the Ka of a weak acid and its percent ionization? Does a compound with a large pKa value have a higher or a lower percent ionization than a compound with a small pKa value (assuming the same analytical concentration in both cases)? Explain.

  6. For a dilute solution of a weak acid (HA), show that the pH of the solution can be approximated using the following equation (where CHA is the analytical concentration of the weak acid):

    pH=logKaCHA

    Under what conditions is this approximation valid?

Numerical Problems

  1. The pKa of NH3 is estimated to be 35. Its conjugate base, the amide ion (NH2), can be isolated as an alkali metal salt, such as sodium amide (NaNH2). Calculate the pH of a solution prepared by adding 0.100 mol of sodium amide to 1.00 L of water. Does the pH differ appreciably from the pH of a NaOH solution of the same concentration? Why or why not?

  2. Phenol is a topical anesthetic that has been used in throat lozenges to relieve sore throat pain. Describe in detail how you would prepare a 2.00 M solution of phenol (C6H5OH) in water; then write equations to show all the species present in the solution. What is the equilibrium constant expression for the reaction of phenol with water? Use the information in to calculate the pH of the phenol solution.

  3. Describe in detail how you would prepare a 1.50 M solution of methylamine in water; then write equations to show all the species present in the solution. What is the equilibrium constant expression for the reaction of methylamine with water? Use the information in to calculate the pH of the solution.

  4. A 0.200 M solution of diethylamine, a substance used in insecticides and fungicides, is only 3.9% ionized at 25°C. Write an equation showing the equilibrium reaction and then calculate the pKb of diethylamine. What is the pKa of its conjugate acid, the diethylammonium ion? What is the equilibrium constant expression for the reaction of diethylammonium chloride with water?

  5. A 1.00 M solution of fluoroacetic acid (FCH2CO2H) is 5% dissociated in water. What is the equilibrium constant expression for the dissociation reaction? Calculate the concentration of each species in solution and then calculate the pKa of FCH2CO2H.

  6. The pKa of 3-chlorobutanoic acid (CH3CHClCH2CO2H) is 4.05. What percentage is dissociated in a 1.0 M solution? Do you expect the pKa of butanoic acid to be greater than or less than the pKa of 3-chlorobutanoic acid? Why?

  7. The pKa of the ethylammonium ion (C2H5NH3+) is 10.64. What percentage of ethylamine is ionized in a 1.00 M solution of ethylamine?

  8. The pKa of Cl3CCO2H is 0.64. What is the pH of a 0.580 M solution? What percentage of the Cl3CCO2H is dissociated?

  9. The pH of a 0.150 M solution of aniline hydrochloride (C6H5NH3+Cl) is 2.70. What is the pKb of the conjugate base, aniline (C6H5NH2)? Do you expect the pKb of (CH3)2CHNH2 to be greater than or less than the pKb of C6H5NH2? Why?

  10. What is the pH of a 0.620 M solution of CH3NH3+Br if the pKb of CH3NH2 is 10.62?

  11. The pKb of 4-hydroxypyridine is 10.80 at 25°C. What is the pH of a 0.0250 M solution?

  12. The pKa values of formic acid and the methylammonium ion are 3.75 and 10.62, respectively. Calculate K for the following reaction:

    HCO2(aq)+CH3NH3+(aq)HCO2H(aq)+CH3NH2(aq)
  13. The pKa values of butanoic acid and the ammonium ion are 4.82 and 9.24, respectively. Calculate K for the following reaction:

    CH3CH2CH2CO2(aq)+NH4+(aq)CH3CH2CH2CO2H(aq)+NH3(aq)
  14. Use the information in to calculate the pH of a 0.0968 M solution of calcium formate.

  15. Calculate the pH of a 0.24 M solution of sodium lactate. The pKa of lactic acid is 3.86.

  16. Use the information in to determine the pH of a solution prepared by dissolving 750.0 mg of methylammonium chloride (CH3NH3+Cl) in enough water to make 150.0 mL of solution.

  17. Use the information in to determine the pH of a solution prepared by dissolving 855 mg of sodium nitrite (NaNO2) in enough water to make 100.0 mL of solution.

Answers

  1. pKb = 9.43; (CH3)2CHNH2 will be a stronger base and have a lower pKb; aniline is a weaker base because the lone pair on the nitrogen atom can be delocalized on the aromatic ring.

  2. 3.8 × 10−5

  3. 8.18

16.5 Acid–Base Titrations

Learning Objective

  1. To calculate the pH at any point in an acid–base titration.

In , you learned that in an acid–base titration, a buret is used to deliver measured volumes of an acid or a base solution of known concentration (the titrant) to a flask that contains a solution of a base or an acid, respectively, of unknown concentration (the unknown). If the concentration of the titrant is known, then the concentration of the unknown can be determined. The following discussion focuses on the pH changes that occur during an acid–base titration. Plotting the pH of the solution in the flask against the amount of acid or base added produces a titration curveA plot of the pH of the solution being titrated versus the amount of acid or base (of known concentration) added.. The shape of the curve provides important information about what is occurring in solution during the titration.

Titrations of Strong Acids and Bases

Part (a) of shows a plot of the pH as 0.20 M HCl is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as HCl is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of HCl (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M NaOH is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of NaOH as shown in part (b) in . As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid.

Figure 16.17 Solution pH as a Function of the Volume of a Strong Acid or a Strong Base Added to Distilled Water

(a) When 0.20 M HCl is added to 50.0 mL of distilled water, the pH rapidly decreases until it reaches a minimum at the pH of 0.20 M HCl. (b) Conversely, when 0.20 M NaOH is added to 50.0 mL of distilled water, the pH rapidly increases until it reaches a maximum at the pH of 0.20 M NaOH.

Suppose that we now add 0.20 M NaOH to 50.0 mL of a 0.10 M solution of HCl. Because HCl is a strong acid that is completely ionized in water, the initial [H+] is 0.10 M, and the initial pH is 1.00. Adding NaOH decreases the concentration of H+ because of the neutralization reaction: (OH+H+H2O) (in part (a) in ). Thus the pH of the solution increases gradually. Near the equivalence pointThe point in a titration where a stoichiometric amount of the titrant has been added., however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the H+ ions originally present have been consumed. (For more information on titrations and the equivalence point, see , .) For the titration of a monoprotic strong acid (HCl) with a monobasic strong base (NaOH), we can calculate the volume of base needed to reach the equivalence point from the following relationship:

Equation 16.50

moles of base= moles of acid(volume)b(molarity)b=(volume)a(molarity)aVbMb=VaMa

If 0.20 M NaOH is added to 50.0 mL of a 0.10 M solution of HCl, we solve for Vb:

Vb(0.20 M)=(0.0500 L)(0.10M)Vb=0.025 L=25 mL

Figure 16.18 The Titration of (a) a Strong Acid with a Strong Base and (b) a Strong Base with a Strong Acid

(a) As 0.20 M NaOH is slowly added to 50.0 mL of 0.10 M HCl, the pH increases slowly at first, then increases very rapidly as the equivalence point is approached, and finally increases slowly once more. (b) Conversely, as 0.20 M HCl is slowly added to 50.0 mL of 0.10 M NaOH, the pH decreases slowly at first, then decreases very rapidly as the equivalence point is approached, and finally decreases slowly once more.

At the equivalence point (when 25.0 mL of NaOH solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more NaOH produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M NaOH.

As shown in part (b) in , the titration of 50.0 mL of a 0.10 M solution of NaOH with 0.20 M HCl produces a titration curve that is nearly the mirror image of the titration curve in part (a) in . The pH is initially 13.00, and it slowly decreases as HCl is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M HCl.

The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities.

Note the Pattern

The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities.

Example 11

Calculate the pH of the solution after 24.90 mL of 0.200 M NaOH has been added to 50.00 mL of 0.100 M HCl.

Given: volumes and concentrations of strong base and acid

Asked for: pH

Strategy:

A Calculate the number of millimoles of H+ and OH to determine which, if either, is in excess after the neutralization reaction has occurred. If one species is in excess, calculate the amount that remains after the neutralization reaction.

B Determine the final volume of the solution. Calculate the concentration of the species in excess and convert this value to pH.

Solution:

A Because 0.100 mol/L is equivalent to 0.100 mmol/mL, the number of millimoles of H+ in 50.00 mL of 0.100 M HCl can be calculated as follows:

50.00 mL(0.100 mmol HClmL)=5.00 mmol HCl=5.00 mmol H+

The number of millimoles of NaOH added is as follows:

24.90 mL(0.200 mmol NaOHmL)=4.98 mmol NaOH=4.98 mmol OH

Thus H+ is in excess. To completely neutralize the acid requires the addition of 5.00 mmol of OH to the HCl solution. Because only 4.98 mmol of OH has been added, the amount of excess H+ is 5.00 mmol − 4.98 mmol = 0.02 mmol of H+.

B The final volume of the solution is 50.00 mL + 24.90 mL = 74.90 mL, so the final concentration of H+ is as follows:

[H+]=0.02 mmol H+74.90 mL=3×104 M

The pH is −log[H+] = −log(3 × 10−4) = 3.5, which is significantly less than the pH of 7.00 for a neutral solution.

Exercise

Calculate the pH of a solution prepared by adding 40.00 mL of 0.237 M HCl to 75.00 mL of a 0.133 M solution of NaOH.

Answer: 11.6

Titrations of Weak Acids and Bases

In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding Ka or Kb. As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned in , [H+] of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its pKa and its concentration. Because only a fraction of a weak acid dissociates, [H+] is less than [HA]. Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration. Part (a) in shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M NaOH superimposed on the curve for the titration of 0.100 M HCl shown in part (a) in . Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the HCl solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the HCl titration; the magnitude of the pH change at the equivalence point depends on the pKa of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess NaOH present, regardless of whether the acid is weak or strong.

Note the Pattern

The shape of the titration curve of a weak acid or weak base depends heavily on their identities and the Ka or Kb.

The titration curve in part (a) in was created by calculating the starting pH of the acetic acid solution before any NaOH is added and then calculating the pH of the solution after adding increasing volumes of NaOH. The procedure is illustrated in the following subsection and Example 12 for three points on the titration curve, using the pKa of acetic acid (4.76 at 25°C; Ka = 1.7 × 10−5).

Figure 16.19 The Titration of (a) a Weak Acid with a Strong Base and (b) a Weak Base with a Strong Acid

(a) As 0.200 M NaOH is slowly added to 50.0 mL of 0.100 M acetic acid, the pH increases slowly at first, then increases rapidly as the equivalence point is approached, and then again increases more slowly. The corresponding curve for the titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH is shown as a dashed line. (b) As 0.200 M HCl is slowly added to 50.0 mL of 0.100 M NH3, the pH decreases slowly at first, then decreases rapidly as the equivalence point is approached, and then again decreases more slowly. The corresponding curve for the titration of 50.0 mL of 0.100 M NaOH with 0.200 M HCl is shown as a dashed line.

Calculating the pH of a Solution of a Weak Acid or a Weak Base

As explained , if we know Ka or Kb and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a table of initial concentrations, changes in concentrations, and final concentrations. In this situation, the initial concentration of acetic acid is 0.100 M. If we define x as [H+] due to the dissociation of the acid, then the table of concentrations for the ionization of 0.100 M acetic acid is as follows:

CH3CO2H(aq)H+(aq)+CH3CO2(aq)
[CH3CO2H] [H+] [CH3CO2]
initial 0.100 1.00 × 10−7 0
change x +x +x
final (0.100 − x) x x

In this and all subsequent examples, we will ignore [H+] and [OH] due to the autoionization of water when calculating the final concentration. However, you should use and to check that this assumption is justified.

Inserting the expressions for the final concentrations into the equilibrium equation (and using approximations),

Ka=[H+][CH3CO2][CH3CO2H]=(x)(x)0.100xx20.100=1.74×105

Solving this equation gives x = [H+] = 1.32 × 10−3 M. Thus the pH of a 0.100 M solution of acetic acid is as follows:

pH = −log(1.32 × 10−3) = 2.879

Calculating the pH during the Titration of a Weak Acid or a Weak Base

Now consider what happens when we add 5.00 mL of 0.200 M NaOH to 50.00 mL of 0.100 M CH3CO2H (part (a) in ). Because the neutralization reaction proceeds to completion, all of the OH ions added will react with the acetic acid to generate acetate ion and water:

Equation 16.51

CH3CO2H(aq) + OH(aq) → CH3CO2(aq) + H2O(l)

All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the amounts of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation () to determine [H+] of the resulting solution.

Step 1: To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of CH3CO2H in the original solution and the amount of OH in the NaOH solution that was added. The acetic acid solution contained

50.00 mL(0.100 mmol CH3CO2HmL)=5.00 mmol CH3CO2H

The NaOH solution contained

5.00 mL(0.200 mmol NaOHmL)=1.00 mmol NaOH

Comparing the amounts shows that CH3CO2H is in excess. Because OH reacts with CH3CO2H in a 1:1 stoichiometry, the amount of excess CH3CO2H is as follows:

5.00 mmol CH3CO2H − 1.00 mmol OH = 4.00 mmol CH3CO2H

Each 1 mmol of OH reacts to produce 1 mmol of acetate ion, so the final amount of CH3CO2 is 1.00 mmol.

The stoichiometry of the reaction is summarized in the following table, which shows the numbers of moles of the various species, not their concentrations.

CH3CO2H(aq)+OH(aq)CH3CO2(aq)+H2O(l)
[CH3CO2H] [OH] [CH3CO2]
initial 5.00 mmol 1.00 mmol 0 mmol
change −1.00 mmol −1.00 mmol +1.00 mmol
final 4.00 mmol 0 mmol 1.00 mmol

This table gives the initial amount of acetate and the final amount of OH ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of CH3CO2 in equilibrium is insignificant compared to the amount of OH added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of OH, but the amount of OH due to the autoionization of water is insignificant compared to the amount of OH added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem.

Step 2: To calculate [H+] at equilibrium following the addition of NaOH, we must first calculate [CH3CO2H] and [CH3CO2] using the number of millimoles of each and the total volume of the solution at this point in the titration:

final volume=50.00 mL+5.00 mL=55.00 mL[CH3CO2H]=4.00 mmol CH3CO2H55.00 mL=7.27×102 M[CH3CO2]=1.00 mmol CH3CO255.00 mL=1.82×102 M

Knowing the concentrations of acetic acid and acetate ion at equilibrium and Ka for acetic acid (1.74 × 10−5), we can use to calculate [H+] at equilibrium:

Ka=[CH3CO2][H+][CH3CO2H] [H+]=Ka[CH3CO2H][CH3CO2]=(1.74×105)(7.27×102 M)1.82×102=6.95×105 M

Calculating −log[H+] gives pH = −log(6.95 × 10−5) = 4.158.

Comparing the titration curves for HCl and acetic acid in part (a) in , we see that adding the same amount (5.00 mL) of 0.200 M NaOH to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for HCl (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example 12, we calculate another point for constructing the titration curve of acetic acid.

Example 12

What is the pH of the solution after 25.00 mL of 0.200 M NaOH is added to 50.00 mL of 0.100 M acetic acid?

Given: volume and molarity of base and acid

Asked for: pH

Strategy:

A Write the balanced chemical equation for the reaction. Then calculate the initial numbers of millimoles of OH and CH3CO2H. Determine which species, if either, is present in excess.

B Tabulate the results showing initial numbers, changes, and final numbers of millimoles.

C If excess acetate is present after the reaction with OH, write the equation for the reaction of acetate with water. Use a tabular format to obtain the concentrations of all the species present.

D Calculate Kb using the relationship Kw = KaKb (). Calculate [OH] and use this to calculate the pH of the solution.

Solution:

A Ignoring the spectator ion (Na+), the equation for this reaction is as follows:

CH3CO2H(aq) + OH(aq) → CH3CO2(aq) + H2O(l)

The initial numbers of millimoles of OH and CH3CO2H are as follows:

25.00 mL(0.200 mmol OHmL)=5.00 mmol OH 50.00 mL(0.100 CH3CO2HmL)=5.00 mmol CH3CO2H

The number of millimoles of OH equals the number of millimoles of CH3CO2H, so neither species is present in excess.

B Because the number of millimoles of OH added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form.

CH3CO2H(aq)+OH(aq)CH3CO2(aq)+H2O(l)
[CH3CO2H] [OH] [CH3CO2]
initial 5.00 mmol 5.00 mmol 0 mmol
change −5.00 mmol −5.00 mmol +5.00 mmol
final 0 mmol 0 mmol 5.00 mmol

C Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate [H+] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction:

[CH3CO2]=5.00 mmol CH3CO2(50.00+25.00) mL=6.67×102 M

The equilibrium reaction of acetate with water is as follows:

CH3CO2(aq)+H2O(l)CH3CO2H(aq)+OH(aq)

The equilibrium constant for this reaction is Kb = Kw/Ka, where Ka is the acid ionization constant of acetic acid. We therefore define x as [OH] produced by the reaction of acetate with water. Here is the completed table of concentrations:

H2O(l)+CH3CO2(aq)CH3CO2H(aq)+OH(aq)
[CH3CO2] [CH3CO2H] [OH]
initial 0.0667 0 1.00 × 10−7
change x +x +x
final (0.0667 − x) x x

D Substituting the expressions for the final values from this table into ,

Kb=[CH3CO2H][OH][CH3CO2]=(x)(x)0.0667xx20.0667

We can obtain Kb by rearranging and substituting the known values:

Kb=KwKa=1.01×10141.74×105=5.80×1010=x20.0667

which we can solve to get x = 6.22 × 10−6. Thus [OH] = 6.22 × 10−6 M, and the pH of the final solution is 8.794 (part (a) in ). As expected for the titration of a weak acid, the pH at the equivalence point is greater than 7.00 because the product of the titration is a base, the acetate ion, which then reacts with water to produce OH.

Exercise

Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M HCl solution to 125.0 mL of a 0.150 M solution of ammonia. The pKb of ammonia is 4.75 at 25°C.

Answer: 9.23

As shown in part (b) in , the titration curve for NH3, a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is less than 7.00 because the titration produces an acid.

The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. illustrates the shape of titration curves as a function of the pKa or the pKb. As the acid or the base being titrated becomes weaker (its pKa or pKb becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point.

Figure 16.20 Effect of Acid or Base Strength on the Shape of Titration Curves

Unlike strong acids or bases, the shape of the titration curve for a weak acid or base depends on the pKa or pKb of the weak acid or base being titrated. (a) Solution pH as a function of the volume of 1.00 M NaOH added to 10.00 mL of 1.00 M solutions of weak acids with the indicated pKa values. (b) Solution pH as a function of the volume of 1.00 M HCl added to 10.00 mL of 1.00 M solutions of weak bases with the indicated pKb values. The shapes of the two sets of curves are essentially identical, but one is flipped vertically in relation to the other. Midpoints are indicated for the titration curves corresponding to pKa = 10 and pKb = 10.

One point in the titration of a weak acid or a weak base is particularly important: the midpointThe point in an acid–base titration at which exactly enough acid (or base) has been added to neutralize one-half of the base (or the acid) originally present: [HA]=[A]. of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in part (a) in and part (b) in for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A]. Recall from that the ionization constant for a weak acid is as follows:

Ka=[H3O+][A][HA]

If [HA] = [A], this reduces to Ka = [H3O+]. Taking the negative logarithm of both sides,

−log Ka = −log[H3O+]

From the definitions of pKa and pH, we see that this is identical to

Equation 16.52

pKa = pH

Thus the pH at the midpoint of the titration of a weak acid is equal to the pKaof the weak acid, as indicated in part (a) in for the weakest acid where we see that the midpoint for pKa = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration and the pKa (or the pKb) of a weak acid (or a weak base).

Note the Pattern

The pH at the midpoint of the titration of a weak acid is equal to the pKa of the weak acid.

Titrations of Polyprotic Acids or Bases

When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the pKa values are separated by at least three pKa units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid H3PO4 with NaOH is illustrated in and shows two well-defined steps: the first midpoint corresponds to pKa1, and the second midpoint corresponds to pKa2. Because HPO42− is such a weak acid, pKa3 has such a high value that the third step cannot be resolved using 0.100 M NaOH as the titrant.

Figure 16.21 Titration Curve for Phosphoric Acid (H3PO4), a Typical Polyprotic Acid

The curve for the titration of 25.0 mL of a 0.100 M H3PO4 solution with 0.100 M NaOH along with the species in solution at each Ka is shown. Note the two distinct equivalence points corresponding to deprotonation of H3PO4 at pH ≈ 4.6 and H2PO42− at pH ≈ 9.8. Because HPO42− is a very weak acid, the third equivalence point, at pH ≈ 13, is not well defined.

The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in . The initial pH is high, but as acid is added, the pH decreases in steps if the successive pKb values are well separated. lists the ionization constants and pKa values for some common polyprotic acids and bases.

Example 13

Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M NaOH solution to 100.0 mL of a 0.0510 M solution of oxalic acid (HO2CCO2H), a diprotic acid (abbreviated as H2ox). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion (O2CCO2, abbreviated ox2−).Oxalate salts are toxic for two reasons. First, oxalate salts of divalent cations such as Ca2+ are insoluble at neutral pH but soluble at low pH, as we shall see in . As a result, calcium oxalate dissolves in the dilute acid of the stomach, allowing oxalate to be absorbed and transported into cells, where it can react with calcium to form tiny calcium oxalate crystals that damage tissues. Second, oxalate forms stable complexes with metal ions, which can alter the distribution of metal ions in biological fluids.

Given: volume and concentration of acid and base

Asked for: pH

Strategy:

A Calculate the initial millimoles of the acid and the base. Use a tabular format to determine the amounts of all the species in solution.

B Calculate the concentrations of all the species in the final solution. Use to determine [H+] and convert this value to pH.

Solution:

A gives the pKa values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present:

100.0 mL(0.0510 mmol H2oxmL)=5.10 mmol H2ox 55.0 mL(0.120 mmol NaOHmL)=6.60 mmol NaOH

The strongest acid (H2ox) reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of OH to react with Hox, forming ox2− and H2O. The reactions can be written as follows:

H2ox5.10 mmol+OH6.60 mmolHox5.10 mmol+H2O5.10 mmol Hox5.10 mmol+OH1.50 mmolox21.50 mmol+H2O1.50 mmol

In tabular form,

H2ox OH Hox ox2−
initial 5.10 mmol 6.60 mmol 0 mmol 0 mmol
change (step 1) −5.10 mmol −5.10 mmol +5.10 mmol 0 mmol
final (step 1) 0 mmol 1.50 mmol 5.10 mmol 0 mmol
change (step 2) −1.50 mmol −1.50 mmol +1.50 mmol
final 0 mmol 0 mmol 3.60 mmol 1.50 mmol

B The equilibrium between the weak acid (Hox) and its conjugate base (ox2−) in the final solution is determined by the magnitude of the second ionization constant, Ka2 = 10−3.81 = 1.6 × 10−4. To calculate the pH of the solution, we need to know [H+], which is determined using exactly the same method as in the acetic acid titration in Example 12:

final volume of solution = 100.0 mL + 55.0 mL = 155.0 mL

Thus the concentrations of Hox and ox2− are as follows:

[Hox]=3.60 mmol Hox155.0 mL=2.32×102 M[ox2]=1.50 mmol ox2155.0 mL=9.68×103 M

We can now calculate [H+] at equilibrium using the following equation:

Ka2=[ox2][H+][Hox]

Rearranging this equation and substituting the values for the concentrations of Hox and ox2−,

[H+]=Ka2[Hox][ox2]=(1.6×104)(2.32×102)9.68×103=3.7×104 M

So

pH = −log[H+] = −log(3.7 ×10−4) = 3.43

This answer makes chemical sense because the pH is between the first and second pKa values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than pKa1), but we added only enough to titrate less than half of the second, less acidic proton, with pKa2. If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to pKa2.

Exercise

Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine (pKb1 = 4.27, pKb2 = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M HCl (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine.

Answer: 4.9

Indicators

In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicatorA compound added in small amounts to an acid–base titration to signal the equivalence point by changing color. is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful.

We can describe the chemistry of indicators by the following general equation:

where the protonated form is designated by HIn and the conjugate base by In. The ionization constant for the deprotonation of indicator HIn is as follows:

Equation 16.53

Kin=[H+][In][HIn]

The pKin (its pKa) determines the pH at which the indicator changes color.

Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH (). In all cases, though, a good indicator must have the following properties:

  • The color change must be easily detected.
  • The color change must be rapid.
  • The indicator molecule must not react with the substance being titrated.
  • To minimize errors, the indicator should have a pKin that is within one pH unit of the expected pH at the equivalence point of the titration.

Figure 16.22 Naturally Occurring pH Indicators in Red Cabbage Juice

Red cabbage juice contains a mixture of substances whose color depends on the pH. Each test tube contains a solution of red cabbage juice in water, but the pH of the solutions varies from pH = 2.0 (far left) to pH = 11.0 (far right). At pH = 7.0, the solution is blue.

Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values.

Figure 16.23 Some Common Acid–Base Indicators

Approximate colors are shown, along with pKin values and the pH range over which the color changes.

It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In increases, the color of the solution slowly changes from the characteristic color of HIn to that of In. As we will see in , the [In]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units.

We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in . This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M NaOH. The pH ranges over which two common indicators (methyl red, pKin = 5.0, and phenolphthalein, pKin = 9.5) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the HCl titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of NaOH has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the HCl titration, the phenolphthalein indicator will turn pink when about 50 mL of NaOH has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of NaOH will therefore cause the methyl red indicator to change color, resulting in a huge error.

Figure 16.24 Choosing the Correct Indicator for an Acid–Base Titration

The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M NaOH. Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pKin for methyl red (5.0) is very close to the pKa of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point.

In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pKin between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used.

The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container ().

Figure 16.25 pH Paper

pH paper contains a set of indicators that change color at different pH values. The approximate pH of a solution can be determined by simply dipping a paper strip into the solution and comparing the color to the standards provided.

Summary

The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component. The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. The pH at the midpoint, the point halfway on the titration curve to the equivalence point, is equal to the pKa of the weak acid or the pKb of the weak base. Thus titration methods can be used to determine both the concentration and the pKa (or the pKb) of a weak acid (or a weak base). Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself.

Key Takeaway

  • Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the pKa, and the pKb of the system.

Conceptual Problems

  1. Why is the portion of the titration curve that lies below the equivalence point of a solution of a weak acid displaced upward relative to the titration curve of a strong acid? How are the slopes of the curves different at the equivalence point? Why?

  2. Predict whether each solution will be neutral, basic, or acidic at the equivalence point of each titration.

    1. An aqueous solution of NaOH is titrated with 0.100 M HCl.
    2. An aqueous solution of ethylamine (CH3CH2NH2) is titrated with 0.150 M HNO3
    3. An aqueous solution of aniline hydrochloride (C6H5NH3+Cl) is titrated with 0.050 M KOH.
  3. The pKa values of phenol red, bromophenol blue, and phenolphthalein are 7.4, 4.1, and 9.5, respectively. Which indicator is best suited for each acid–base titration?

    1. titrating a solution of Ba(OH)2 with 0.100 M HCl
    2. titrating a solution of trimethylamine (Me3N) with 0.150 M HNO3
    3. titrating a solution of aniline hydrochloride (C6H5NH3+Cl) with 0.050 M KOH
  4. For the titration of any strong acid with any strong base, the pH at the equivalence point is 7.0. Why is this not usually the case in titrations of weak acids or weak bases?

  5. Why are the titration curves for a strong acid with a strong base and a weak acid with a strong base identical in shape above the equivalence points but not below?

  6. Describe what is occurring on a molecular level during the titration of a weak acid, such as acetic acid, with a strong base, such as NaOH, at the following points along the titration curve. Which of these points corresponds to pH = pKa?

    1. at the beginning of the titration
    2. at the midpoint of the titration
    3. at the equivalence point
    4. when excess titrant has been added
  7. On a molecular level, describe what is happening during the titration of a weak base, such as ammonia, with a strong acid, such as HCl, at the following points along the titration curve. Which of these points corresponds to pOH = pKb?

    1. at the beginning of the titration
    2. at the midpoint of the titration
    3. at the equivalence point
    4. when excess titrant has been added
  8. For the titration of a weak acid with a strong base, use the Ka expression to show that pH = pKa at the midpoint of the titration.

  9. Chemical indicators can be used to monitor pH rapidly and inexpensively. Nevertheless, electronic methods are generally preferred. Why?

  10. Why does adding ammonium chloride to a solution of ammonia in water decrease the pH of the solution?

  11. Given the equilibrium system CH3CO2H(aq) ⇌ CH3CO2(aq) + H+(aq), explain what happens to the position of the equilibrium and the pH in each case.

    1. Dilute HCl is added.
    2. Dilute NaOH is added.
    3. Solid sodium acetate is added.
  12. Given the equilibrium system CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH(aq), explain what happens to the position of the equilibrium and the pH in each case.

    1. Dilute HCl is added.
    2. Dilute NaOH is added.
    3. Solid CH3NH3+Cl is added.

Answer

    1. shifts to left; pH decreases
    2. shifts to right; pH increases
    3. shifts to left; pH increases

Numerical Problems

  1. Calculate the pH of each solution.

    1. A volume of 25.0 mL of 6.09 M HCl is added to 100.0 mL of distilled water
    2. A volume of 5.0 mL of 2.55 M NaOH is added to 75.0 mL of distilled water.
  2. What is the pH of a solution prepared by mixing 50.0 mL of 0.225 M HCl with 100.0 mL of a 0.184 M solution of NaOH?

  3. What volume of 0.50 M HCl is needed to completely neutralize 25.00 mL of 0.86 M NaOH?

  4. Calculate the final pH when each pair of solutions is mixed.

    1. 100 mL of 0.105 M HCl and 100 mL of 0.115 M sodium acetate
    2. 50 mL of 0.10 M HCl and 100 mL of 0.15 M sodium acetate
    3. 100 mL of 0.109 M acetic acid and 100 mL of 0.118 M NaOH
    4. 100 mL of 0.998 M acetic acid and 50.0 mL of 0.110 M NaOH
  5. Calculate the final pH when each pair of solutions is mixed.

    1. 100 mL of 0.983 M HCl and 100 mL of 0.102 M sodium fluoride
    2. 50.0 mL of 0.115 M HCl and 100 mL of 0.109 M sodium fluoride
    3. 100 mL of 0.106 M hydrofluoric acid and 50.0 mL of 0.996 M NaOH
    4. 100 mL of 0.107 M sodium acetate and 50.0 mL of 0.987 M acetic acid
  6. Calcium carbonate is a major contributor to the “hardness” of water. The amount of CaCO3 in a water sample can be determined by titrating the sample with an acid, such as HCl, which produces water and CO2. Write a balanced chemical equation for this reaction. Generate a plot of solution pH versus volume of 0.100 M HCl added for the titration of a solution of 250 mg of CaCO3 in 200.0 mL of water with 0.100 M HCl; assume that the HCl solution is added in 5.00 mL increments. What volume of HCl corresponds to the equivalence point?

  7. For a titration of 50.0 mL of 0.288 M NaOH, you would like to prepare a 0.200 M HCl solution. The only HCl solution available to you, however, is 12.0 M.

    1. How would you prepare 500 mL of a 0.200 M HCl solution?
    2. Approximately what volume of your 0.200 M HCl solution is needed to neutralize the NaOH solution?
    3. After completing the titration, you find that your “0.200 M” HCl solution is actually 0.187 M. What was the exact volume of titrant used in the neutralization?
  8. While titrating 50.0 mL of a 0.582 M solution of HCl with a solution labeled “0.500 M KOH,” you overshoot the endpoint. To correct the problem, you add 10.00 mL of the HCl solution to your flask and then carefully continue the titration. The total volume of titrant needed for neutralization is 71.9 mL.

    1. What is the actual molarity of your KOH solution?
    2. What volume of titrant was needed to neutralize 50.0 mL of the acid?
  9. Complete the following table and generate a titration curve showing the pH versus volume of added base for the titration of 50.0 mL of 0.288 M HCl with 0.321 M NaOH. Clearly indicate the equivalence point.

    Base Added (mL) 10.0 30.0 40.0 45.0 50.0 55.0 65.0 75.0
    pH
  10. The following data were obtained while titrating 25.0 mL of 0.156 M NaOH with a solution labeled “0.202 M HCl.” Plot the pH versus volume of titrant added. Then determine the equivalence point from your graph and calculate the exact concentration of your HCl solution.

    Volume of HCl (mL) 5 10 15 20 25 30 35
    pH 11.46 11.29 10.98 4.40 2.99 2.70 2.52
  11. Fill in the data for the titration of 50.0 mL of 0.241 M formic acid with 0.0982 M KOH. The pKa of formic acid is 3.75. What is the pH of the solution at the equivalence point?

    Volume of Base Added (mL) 0 5 10 15 20 25
    pH
  12. Glycine hydrochloride, which contains the fully protonated form of the amino acid glycine, has the following structure:

    It is a strong electrolyte that completely dissociates in water. Titration with base gives two equivalence points: the first corresponds to the deprotonation of the carboxylic acid group and the second to loss of the proton from the ammonium group. The corresponding equilibrium equations are as follows:

    +NH3CH2CO2H(aq)pKa1=2.3N+H3CH2CO2(aq)+H++NH3CH2CO2(aq)pKa2=9.6NH2CH2COO(aq)+H+
    1. Given 50.0 mL of solution that is 0.430 M glycine hydrochloride, how many milliliters of 0.150 M KOH are needed to fully deprotonate the carboxylic acid group?
    2. How many additional milliliters of KOH are needed to deprotonate the ammonium group?
    3. What is the pH of the solution at each equivalence point?
    4. How many milliliters of titrant are needed to obtain a solution in which glycine has no net electrical charge? The pH at which a molecule such as glycine has no net charge is its isoelectric point. What is the isoelectric point of glycine?
  13. What is the pH of a solution prepared by adding 38.2 mL of 0.197 M HCl to 150.0 mL of 0.242 M pyridine? The pKb of pyridine is 8.77.

  14. What is the pH of a solution prepared by adding 40.3 mL of 0.289 M NaOH to 150.0 mL of 0.564 M succinic acid (HO2CCH2CH2CO2H)? (For succinic acid, pKa1 = 4.21 and pKa2 = 5.64).

  15. Calculate the pH of a 0.15 M solution of malonic acid (HO2CCH2CO2H), whose pKa values are as follows: pKa1 = 2.85 and pKa2 = 5.70.

Answers

  1. 43 mL

    1. dilute 8.33 mL of 12.0 M HCl to 500.0 mL
    2. 72 mL
    3. 77.0 mL
  2. pH at equivalence point = 8.28

    Volume of Base Added (mL) 0 5 10 15 20 25
    pH 2.19 2.38 2.70 2.89 3.04 3.15
  3. 1.85

16.6 Buffers

Learning Objectives

  1. To understand how adding a common ion affects the position of an acid–base equilibrium.
  2. To know how to use the Henderson-Hasselbalch equation to calculate the pH of a buffer.

BuffersSolutions that maintain a relatively constant pH when an acid or a base is added. are solutions that maintain a relatively constant pH when an acid or a base is added. They therefore protect, or “buffer,” other molecules in solution from the effects of the added acid or base. Buffers contain either a weak acid (HA) and its conjugate base (A) or a weak base (B) and its conjugate acid (BH+), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH.

The Common Ion Effect

To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of H+). Le Châtelier’s principle can be used to predict the effect on the equilibrium position of the solution.

A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. Recall that the dissociation reaction of acetic acid is as follows:

Equation 16.54

CH3CO2H(aq)CH3CO2(aq)+H+(aq)

and the equilibrium constant expression is as follows:

Equation 16.55

Ka=[H+][CH3CO2][CH3CO2H]

Sodium acetate (CH3CO2Na) is a strong electrolyte that ionizes completely in aqueous solution to produce Na+ and CH3CO2 ions. If sodium acetate is added to a solution of acetic acid, Le Châtelier’s principle predicts that the equilibrium in will shift to the left, consuming some of the added CH3CO2 and some of the H+ ions originally present in solution:

+ CH3CO2CH3CO2H(aq)CH3CO2(aq)+H+(aq)

Because Na+ is a spectator ion, it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which [H+] is less than the initial value. Because [H+] has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding any base to a solution of a weak acid should increase the pH.

If we instead add a strong acid such as HCl to the system, [H+] increases. Once again the equilibrium is temporarily disturbed, but the excess H+ ions react with the conjugate base (CH3CO2), whether from the parent acid or sodium acetate, to drive the equilibrium to the left. The net result is a new equilibrium composition that has a lower [CH3CO2] than before. In both cases, only the equilibrium composition has changed; the ionization constant Kafor acetic acid remains the same. Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium, in this case CH3CO2, will therefore shift the equilibrium in the direction that reduces the concentration of the common ion. The shift in equilibrium is called the common ion effectThe shift in equilibrium that results when a strong electrolyte containing one ion in common with a reaction system that is at equilibrium is added to the system..

Note the Pattern

Adding a common ion to a system at equilibrium affects the equilibrium composition but not the ionization constant.

Example 14

In , we calculated that a 0.150 M solution of formic acid at 25°C (pKa = 3.75) has a pH of 2.28 and is 3.5% ionized.

  1. Is there a change to the pH of the solution if enough solid sodium formate is added to make the final formate concentration 0.100 M (assume that the formic acid concentration does not change)?
  2. What percentage of the formic acid is ionized if 0.200 M HCl is added to the system?

Given: solution concentration and pH, pKa, and percent ionization of acid; final concentration of conjugate base or strong acid added

Asked for: pH and percent ionization of formic acid

Strategy:

A Write a balanced equilibrium equation for the ionization equilibrium of formic acid. Tabulate the initial concentrations, the changes, and the final concentrations.

B Substitute the expressions for the final concentrations into the expression for Ka. Calculate [H+] and the pH of the solution.

C Construct a table of concentrations for the dissociation of formic acid. To determine the percent ionization, determine the anion concentration, divide it by the initial concentration of formic acid, and multiply the result by 100.

Solution:

  1. A Because sodium formate is a strong electrolyte, it ionizes completely in solution to give formate and sodium ions. The Na+ ions are spectator ions, so they can be ignored in the equilibrium equation. Because water is both a much weaker acid than formic acid and a much weaker base than formate, the acid–base properties of the solution are determined solely by the formic acid ionization equilibrium:

    HCO2H(aq)HCO2(aq)+H+(aq)

    The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated.

    HCO2H(aq)H+(aq)+HCO2(aq)
    [HCO2H] [H+] [HCO2]
    initial 0.150 1.00 × 10−7 0.100
    change x +x +x
    final (0.150 − x) x (0.100 + x)

    B We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so

    Ka=[H+][HCO2][HCO2H]=(x)(0.100+x)0.150x=x(0.100)0.150=103.75=1.8×104

    Rearranging and solving for x,

    x=(1.8×104)×0.150 M0.100 M=2.7×104=[H+]

    The value of x is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover, KaCHA = (1.8 × 10−4)(0.150) = 2.7 × 10−5, which is greater than 1.0 × 10−6, so again, our assumption is justified. The final pH is −log(2.7 × 10−4) = 3.57, compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Châtelier’s principle; the stress on the system has been relieved by the consumption of H+ ions, driving the equilibrium to the left.

  2. C Because HCl is a strong acid, it ionizes completely, and chloride is a spectator ion that can be neglected. Thus the only relevant acid–base equilibrium is again the dissociation of formic acid, and initially the concentration of formate is zero. We can construct a table of initial concentrations, changes in concentration, and final concentrations.

    HCO2H(aq)H+(aq)+HCO2(aq)
    [HCO2H] [H+] [HCO2]
    initial 0.150 0.200 0
    change x +x +x
    final (0.150 − x) (0.200 + x) x

    To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final [HCO2]. We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so

    Ka=[H+][HCO2][HCO2H]=(0.200+x)(x)0.150x=x(0.200)0.150=1.80×104

    Rearranging and solving for x,

    x=(1.80×104)×0.150 M0.200 M=1.35×104=[HCO2]

    Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows:

    percent ionization=1.35×104 M0.150 M×100=0.0900%

    Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Châtelier’s principle: adding H+ ions drives the dissociation equlibrium to the left.

Exercise

As you learned in Example 8, a 0.225 M solution of ethylamine (CH3CH2NH2, pKb = 3.19) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following:

  1. the pH of the solution if enough solid ethylamine hydrochloride (EtNH3Cl) is added to make the solution 0.100 M in EtNH3+
  2. the percentage of ethylamine that is ionized if enough solid NaOH is added to the original solution to give a final concentration of 0.050 M NaOH

Answer:

  1. 11.16
  2. 1.3%

Now let’s suppose we have a buffer solution that contains equimolar concentrations of a weak base (B) and its conjugate acid (BH+). The general equation for the ionization of a weak base is as follows:

Equation 16.56

B(aq)+H2O(l)BH+(aq)+OH(aq)

If the equilibrium constant for the reaction as written in is small, for example Kb = 10−5, then the equilibrium constant for the reverse reaction is very large: K = 1/Kb = 105. Adding a strong base such as OH to the solution therefore causes the equilibrium in to shift to the left, consuming the added OH. As a result, the OH ion concentration in solution remains relatively constant, and the pH of the solution changes very little. Le Châtelier’s principle predicts the same outcome: when the system is stressed by an increase in the OH ion concentration, the reaction will proceed to the left to counteract the stress.

If the pKb of the base is 5.0, the pKa of its conjugate acid is pKa = pKw − pKb = 14.0 – 5.0 = 9.0. Thus the equilibrium constant for ionization of the conjugate acid is even smaller than that for ionization of the base. The ionization reaction for the conjugate acid of a weak base is written as follows:

Equation 16.57

BH+(aq)+H2O(l)B(aq)+H3O+(aq)

Again, the equilibrium constant for the reverse of this reaction is very large: K = 1/Ka = 109. If a strong acid is added, it is neutralized by reaction with the base as the reaction in shifts to the left. As a result, the H+ ion concentration does not increase very much, and the pH changes only slightly. In effect, a buffer solution behaves somewhat like a sponge that can absorb H+ and OH ions, thereby preventing large changes in pH when appreciable amounts of strong acid or base are added to a solution.

Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacityThe amount of strong acid or strong base that a buffer solution can absorb before the pH changes dramatically., the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on K), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in , when NaOH is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH.

Figure 16.26 Effect of Buffer Concentration on the Capacity of a Buffer

A buffer maintains a relatively constant pH when acid or base is added to a solution. The addition of even tiny volumes of 0.10 M NaOH to 100.0 mL of distilled water results in a very large change in pH. As the concentration of a 50:50 mixture of sodium acetate/acetic acid buffer in the solution is increased from 0.010 M to 1.00 M, the change in the pH produced by the addition of the same volume of NaOH solution decreases steadily. For buffer concentrations of at least 0.500 M, the addition of even 25 mL of the NaOH solution results in only a relatively small change in pH.

Calculating the pH of a Buffer

The pH of a buffer can be calculated from the concentrations of the weak acid and the weak base used to prepare it, the concentration of the conjugate base and conjugate acid, and the pKa or pKb of the weak acid or weak base. The procedure is analogous to that used in Example 14 to calculate the pH of a solution containing known concentrations of formic acid and formate.

An alternative method frequently used to calculate the pH of a buffer solution is based on a rearrangement of the equilibrium equation for the dissociation of a weak acid. The simplified ionization reaction is HAH++A, for which the equilibrium constant expression is as follows:

Equation 16.58

Ka=[H+][A][HA]

This equation can be rearranged as follows:

Equation 16.59

[H+]=Ka[HA][A]

Taking the logarithm of both sides and multiplying both sides by −1,

Equation 16.60

log[H+]=log Kalog([HA][A])=logKa+log([A][HA])

Replacing the negative logarithms in ,

Equation 16.61

pH=pKa+log([A][HA])

or, more generally,

Equation 16.62

pH=pKa+log([base][acid])

and are both forms of the Henderson-Hasselbalch equationA rearranged version of the equilibrium constant expression that provides a direct way to calculate the pH of a buffer solution: pH = pKa + log([base]/[acid])., named after the two early-20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch equation may be limited to solutions whose concentrations are at least 100 times greater than their Ka values.

There are three special cases where the Henderson-Hasselbalch equation is easily interpreted without the need for calculations:

  1. [base] = [acid]. Under these conditions, [base]/[acid] = 1 in . Because log 1 = 0, pH = pKa, regardless of the actual concentrations of the acid and base. Recall from that this corresponds to the midpoint in the titration of a weak acid or a weak base.
  2. [base]/[acid] = 10. In , because log 10 = 1, pH = pKa + 1.
  3. [base]/[acid] = 100. In , because log 100 = 2, pH = pKa + 2.

Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = pKa − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit.

Note the Pattern

If [base] = [acid] for a buffer, then pH = pKa. Changing this ratio by a factor of 10 either way changes the pH by ±1 unit.

Example 15

What is the pH of a solution that contains

  1. 0.135 M HCO2H and 0.215 M HCO2Na? (The pKa of formic acid is 3.75.)
  2. 0.0135 M HCO2H and 0.0215 M HCO2Na?
  3. 0.119 M pyridine and 0.234 M pyridine hydrochloride? (The pKb of pyridine is 8.77.)

Given: concentration of acid, conjugate base, and pKa; concentration of base, conjugate acid, and pKb

Asked for: pH

Strategy:

Substitute values into either form of the Henderson-Hasselbalch equation ( or ) to calculate the pH.

Solution:

  1. According to the Henderson-Hasselbalch equation, the pH of a solution that contains both a weak acid and its conjugate base is pH = pKa + log([A]/[HA]). Inserting the given values into the equation,

    pH=3.75+log(0.2150.135)=3.75+log 1.593=3.95

    This result makes sense because the [A]/[HA] ratio is between 1 and 10, so the pH of the buffer must be between the pKa (3.75) and pKa + 1, or 4.75.

  2. This is identical to part (a), except for the concentrations of the acid and the conjugate base, which are 10 times lower. Inserting the concentrations into the Henderson-Hasselbalch equation,

    pH=3.75+log(0.02150.0135)=3.75+log 1.593=3.95 

    This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends only on the ratio of the concentrations of the conjugate base and the acid, not on the magnitude of the concentrations. Because the [A]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95).

  3. In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion (HPy+). We will therefore use , the more general form of the Henderson-Hasselbalch equation, in which “base” and “acid” refer to the appropriate species of the conjugate acid–base pair. We are given [base] = [Py] = 0.119 M and [acid] = [HPy+] = 0.234 M. We also are given pKb = 8.77 for pyridine, but we need pKa for the pyridinium ion. Recall from that the pKb of a weak base and the pKa of its conjugate acid are related: pKa + pKb = pKw. Thus pKa for the pyridinium ion is pKw − pKb = 14.00 − 8.77 = 5.23. Substituting this pKa value into the Henderson-Hasselbalch equation,

    pH=pKa+log([base][acid])=5.23+log(0.1190.234)=5.230.294=4.94

    Once again, this result makes sense: the [B]/[BH+] ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the pKa (5.23) and pKa − 1, or 4.23.

Exercise

What is the pH of a solution that contains

  1. 0.333 M benzoic acid and 0.252 M sodium benzoate?
  2. 0.050 M trimethylamine and 0.066 M trimethylamine hydrochloride?

The pKa of benzoic acid is 4.20, and the pKb of trimethylamine is also 4.20.

Answer:

  1. 4.08
  2. 9.68

The Henderson-Hasselbalch equation can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example 16.

Example 16

The buffer solution in Example 15 contained 0.135 M HCO2H and 0.215 M HCO2Na and had a pH of 3.95.

  1. What is the final pH if 5.00 mL of 1.00 M HCl are added to 100 mL of this solution?
  2. What is the final pH if 5.00 mL of 1.00 M NaOH are added?

Given: composition and pH of buffer; concentration and volume of added acid or base

Asked for: final pH

Strategy:

A Calculate the amounts of formic acid and formate present in the buffer solution using the procedure from Example 14. Then calculate the amount of acid or base added.

B Construct a table showing the amounts of all species after the neutralization reaction. Use the final volume of the solution to calculate the concentrations of all species. Finally, substitute the appropriate values into the Henderson-Hasselbalch equation () to obtain the pH.

Solution:

The added HCl (a strong acid) or NaOH (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction.

  1. A We begin by calculating the millimoles of formic acid and formate present in 100 mL of the initial pH 3.95 buffer:

    100 mL(0.135 mmol HCO2HmL)=13.5 mmol HCO2H 100 mL(0.215 mmol HCO2mL)=21.5 mmol HCO2

    The millimoles of H+ in 5.00 mL of 1.00 M HCl is as follows:

    5.00 mL(1.00 mmol H+mL)=5.00 mmol H+

    B Next, we construct a table of initial amounts, changes in amounts, and final amounts:

    HCO2(aq) + H+(aq) → HCO2H(aq)
    [HCO2] [H+] [HCO2H]
    initial 21.5 mmol 5.00 mmol 13.5 mmol
    change −5.00 mmol −5.00 mmol +5.00 mmol
    final 16.5 mmol ∼0 mmol 18.5 mmol

    The final amount of H+ in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final [H+] and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example 14 or the Henderson–Hasselbach equation. Because we have performed many equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch equation requires the concentrations of HCO2 and HCO2H, which can be calculated using the number of millimoles (n) of each and the total volume (VT). Substituting these values into the Henderson-Hasselbalch equation,

    pH=pKa+log([HCO2][HCO2H])=pKa+log(nHCO2/VfnHCO2H/Vf)=pKa+log(nHCO2nHCO2H)

    Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So

    pH=pKa+log(nHCO2nHCO2H)=3.75+log(16.5 mmol18.5 mmol)=3.750.050=3.70

    Once again, this result makes sense on two levels. First, the addition of HCl has decreased the pH from 3.95, as expected. Second, the ratio of HCO2 to HCO2H is slightly less than 1, so the pH should be between the pKa and pKa − 1.

  2. A The procedure for solving this part of the problem is exactly the same as that used in part (a). We have already calculated the numbers of millimoles of formic acid and formate in 100 mL of the initial pH 3.95 buffer: 13.5 mmol of HCO2H and 21.5 mmol of HCO2. The number of millimoles of OH in 5.00 mL of 1.00 M NaOH is as follows:

    5.00 mL(1.00 mmol OHmL)=5.00 mmol OH

    B With this information, we can construct a table of initial amounts, changes in amounts, and final amounts.

    HCO2H(aq) + OH(aq) → HCO2(aq) + H2O(l)
    [HCO2H] [OH] [HCO2]
    initial 13.5 mmol 5.00 mmol 21.5 mmol
    change −5.00 mmol −5.00 mmol +5.00 mmol
    final 8.5 mmol ∼0 mmol 26.5 mmol

    The final amount of OH in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both HCO2 and HCO2H into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels:

    pH=pKa+log(nHCO2nHCO2H)=3.75+log(26.5 mmol8.5 mmol)=3.75+0.494=4.24

    Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the pKa and pKa + 1, as expected for a solution with a HCO2/HCO2H ratio between 1 and 10.

Exercise

The buffer solution from Example 15 contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94.

  1. What is the final pH if 12.0 mL of 1.5 M NaOH are added to 250 mL of this solution?
  2. What is the final pH if 12.0 mL of 1.5 M HCl are added?

Answer:

  1. 5.30
  2. 4.42

Note the Pattern

Only the amounts (in moles or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch equation, not their concentrations.

Note the Pattern

The most effective buffers contain equal concentrations of an acid and its conjugate base.

The results obtained in Example 16 and its corresponding exercise demonstrate how little the pH of a well-chosen buffer solution changes despite the addition of a significant quantity of strong acid or strong base. Suppose we had added the same amount of HCl or NaOH solution to 100 mL of an unbuffered solution at pH 3.95 (corresponding to 1.1 × 10−4 M HCl). In this case, adding 5.00 mL of 1.00 M HCl would lower the final pH to 1.32 instead of 3.70, whereas adding 5.00 mL of 1.00 M NaOH would raise the final pH to 12.68 rather than 4.24. (Try verifying these values by doing the calculations yourself.) Thus the presence of a buffer significantly increases the ability of a solution to maintain an almost constant pH.

A buffer that contains approximately equal amounts of a weak acid and its conjugate base in solution is equally effective at neutralizing either added base or added acid. This is shown in for an acetic acid/sodium acetate buffer. Adding a given amount of strong acid shifts the system along the horizontal axis to the left, whereas adding the same amount of strong base shifts the system the same distance to the right. In either case, the change in the ratio of CH3CO2 to CH3CO2H from 1:1 reduces the buffer capacity of the solution.

Figure 16.27 Distribution Curve Showing the Fraction of Acetic Acid Molecules and Acetate Ions as a Function of pH in a Solution of Acetic Acid

The pH range over which the acetic acid/sodium acetate system is an effective buffer (the darker shaded region) corresponds to the region in which appreciable concentrations of both species are present (pH 3.76–5.76, corresponding to pH = pKa ± 1).

The Relationship between Titrations and Buffers

There is a strong correlation between the effectiveness of a buffer solution and the titration curves discussed in . Consider the schematic titration curve of a weak acid with a strong base shown in . As indicated by the labels, the region around pKa corresponds to the midpoint of the titration, when approximately half the weak acid has been neutralized. This portion of the titration curve corresponds to a buffer: it exhibits the smallest change in pH per increment of added strong base, as shown by the nearly horizontal nature of the curve in this region. The nearly flat portion of the curve extends only from approximately a pH value of 1 unit less than the pKa to approximately a pH value of 1 unit greater than the pKa, which is why buffer solutions usually have a pH that is within ±1 pH units of the pKa of the acid component of the buffer.

Figure 16.28 The Relationship between Titration Curves and Buffers

This schematic plot of pH for the titration of a weak acid with a strong base shows the nearly flat region of the titration curve around the midpoint, which corresponds to the formation of a buffer. At the lower left, the pH of the solution is determined by the equilibrium for dissociation of the weak acid; at the upper right, the pH is determined by the equilibrium for reaction of the conjugate base with water.

In the region of the titration curve at the lower left, before the midpoint, the acid–base properties of the solution are dominated by the equilibrium for dissociation of the weak acid, corresponding to Ka. In the region of the titration curve at the upper right, after the midpoint, the acid–base properties of the solution are dominated by the equilibrium for reaction of the conjugate base of the weak acid with water, corresponding to Kb. However, we can calculate either Ka or Kb from the other because they are related by Kw.

Blood: A Most Important Buffer

Metabolic processes produce large amounts of acids and bases, yet organisms are able to maintain an almost constant internal pH because their fluids contain buffers. This is not to say that the pH is uniform throughout all cells and tissues of a mammal. The internal pH of a red blood cell is about 7.2, but the pH of most other kinds of cells is lower, around 7.0. Even within a single cell, different compartments can have very different pH values. For example, one intracellular compartment in white blood cells has a pH of around 5.0.

Because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately pH 5.0 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the CO2/HCO3 system, which dominates the buffering action of blood plasma.

The acid–base equilibrium in the CO2/HCO3 buffer system is usually written as follows:

Equation 16.63

H2CO3(aq)H+(aq)+HCO3(aq)

with Ka = 4.5 × 10−7 and pKa = 6.35 at 25°C. In fact, is a grossly oversimplified version of the CO2/HCO3 system because a solution of CO2 in water contains only rather small amounts of H2CO3. Thus does not allow us to understand how blood is actually buffered, particularly at a physiological temperature of 37°C. As shown in , CO2 is in equilibrium with H2CO3, but the equilibrium lies far to the left, with an H2CO3/CO2 ratio less than 0.01 under most conditions:

Equation 16.64

CO2(aq)+H2O(l)H2CO3(aq)

with K′ = 4.0 × 10−3 at 37°C. The true pKa of carbonic acid at 37°C is therefore 3.70, not 6.35, corresponding to a Ka of 2.0 × 10−4, which makes it a much stronger acid than suggests. Adding and and canceling H2CO3 from both sides give the following overall equation for the reaction of CO2 with water to give a proton and the bicarbonate ion:

Equation 16.65

CO2(aq)+H2O(l)H2CO3(aq)K=4.0×103 (37°C)H2CO3(aq)H+(aq)+HCO3(aq)Ka=2.0×104 (37°CCO2(aq)+H2O(l)H+(aq)+HCO3(aq)K=8.0×107 (37°C)

The K value for the reaction in is the product of the true ionization constant for carbonic acid (Ka) and the equilibrium constant (K) for the reaction of CO2(aq) with water to give carbonic acid. The equilibrium equation for the reaction of CO2 with water to give bicarbonate and a proton is therefore

Equation 16.66

K=[H+][HCO3][CO2]=8.0×107

The presence of a gas in the equilibrium constant expression for a buffer is unusual. According to Henry’s law, [CO2]=kPCO2, where k is the Henry’s law constant for CO2, which is 3.0 × 10−5 M/mmHg at 37°C. (For more information about Henry’s law, see , .) Substituting this expression for [CO2] in ,

Equation 16.67

K=[H+][HCO3](3.0×105 M/mmHg)(PCO2)

where PCO2 is in mmHg. Taking the negative logarithm of both sides and rearranging,

Equation 16.68

pH=6.10+log([HCO3](3.0×105 M/mm Hg)(PCO2))

Thus the pH of the solution depends on both the CO2 pressure over the solution and [HCO3]. plots the relationship between pH and [HCO3] under physiological conditions for several different values of PCO2, with normal pH and [HCO3] values indicated by the dashed lines.

Figure 16.29 Buffering in Blood: pH versus [HCO3] Curves for Buffers with Different Values of PCO2

Only those combinations of pH and [HCO3] that lie on a given line are allowed for the particular value of PCO2 indicated. Normal values of blood plasma pH and [HCO3] are indicated by dashed lines.

According to , adding a strong acid to the CO2/HCO3 system causes [HCO3] to decrease as HCO3 is converted to CO2. Excess CO2 is released in the lungs and exhaled into the atmosphere, however, so there is essentially no change in PCO2. Because the change in [HCO3]/PCO2 is small, predicts that the change in pH will also be rather small. Conversely, if a strong base is added, the OH reacts with CO2 to form [HCO3], but CO2 is replenished by the body, again limiting the change in both [HCO3]/PCO2 and pH. The CO2/HCO3 buffer system is an example of an open system, in which the total concentration of the components of the buffer change to keep the pH at a nearly constant value.

If a passenger steps out of an airplane in Denver, Colorado, for example, the lower PCO2 at higher elevations (typically 31 mmHg at an elevation of 2000 m versus 40 mmHg at sea level) causes a shift to a new pH and [HCO3]. The increase in pH and decrease in [HCO3] in response to the decrease in PCO2 are responsible for the general malaise that many people experience at high altitudes. If their blood pH does not adjust rapidly, the condition can develop into the life-threatening phenomenon known as altitude sickness.

Summary

Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid (HA) and its conjugate weak base (A). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. The shift in equilibrium is called the common ion effect. Buffers are characterized by their pH range and buffer capacity. The useful pH range of a buffer depends strongly on the chemical properties of the conjugate weak acid–base pair used to prepare the buffer (the Ka or Kb), whereas its buffer capacity depends solely on the concentrations of the species in the solution. The pH of a buffer can be calculated using the Henderson-Hasselbalch equation, which is valid for solutions whose concentrations are at least 100 times greater than their Ka values. Because no single buffer system can effectively maintain a constant pH value over the physiological range of approximately 5 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the CO2/HCO3 system, which dominates the buffering action of blood plasma.

Key Equations

Henderson-Hasselbalch equation

: pH=pKa+log([A][HA])

: pH=pKa+log([base][acid])

Key Takeaway

  • The common ion effect allows solutions to act as buffers, whose pH can be calculated using the Henderson-Hasselbalch equation.

Conceptual Problems

  1. Explain why buffers are crucial for the proper functioning of biological systems.

  2. What is the role of a buffer in chemistry and biology? Is it correct to say that buffers prevent a change in [H3O+]? Explain your reasoning.

  3. Explain why the most effective buffers are those that contain approximately equal amounts of the weak acid and its conjugate base.

  4. Which region of the titration curve of a weak acid or a weak base corresponds to the region of the smallest change in pH per amount of added strong acid or strong base?

  5. If you were given a solution of sodium acetate, describe two ways you could convert the solution to a buffer.

  6. Why are buffers usually used only within approximately one pH unit of the pKa or pKb of the parent weak acid or base?

  7. The titration curve for a monoprotic acid can be divided into four regions: the starting point, the region around the midpoint of the titration, the equivalence point, and the region after the equivalence point. For which region would you use each approach to describe the behavior of the solution?

    1. a buffer
    2. a solution of a salt of a weak base
    3. a solution of a weak acid
    4. diluting a strong base
  8. Which of the following will produce a buffer solution? Explain your reasoning in each case.

    1. mixing 100 mL of 0.1 M HCl and 100 mL of 0.1 M sodium fluoride
    2. mixing 50 mL of 0.1 M HCl and 100 mL of 0.1 M sodium fluoride
    3. mixing 100 mL of 0.1 M hydrofluoric acid and 100 mL of 0.1 M HCl
    4. mixing 100 mL of 0.1 M hydrofluoric acid and 50 mL of 0.1 M NaOH
    5. mixing 100 mL of 0.1 M sodium fluoride and 50 mL of 0.1 M NaOH.
  9. Which of the following will produce a buffer solution? Explain your reasoning in each case.

    1. mixing 100 mL of 0.1 M HCl and 100 mL of 0.1 M sodium acetate
    2. mixing 50 mL of 0.1 M HCl and 100 mL of 0.1 M sodium acetate
    3. mixing 100 mL of 0.1 M acetic acid and 100 mL of 0.1 M NaOH
    4. mixing 100 mL of 0.1 M acetic acid and 50 mL of 0.1 M NaOH
    5. mixing 100 mL of 0.1 M sodium acetate and 50 mL of 0.1 M acetic acid
  10. Use the definition of Kb for a weak base to derive the following expression, which is analogous to the Henderson-Hasselbalch equation but for a weak base (B) rather than a weak acid (HA):

    pOH=pKblog([base][acid])
  11. Why do biological systems use overlapping buffer systems to maintain a constant pH?

  12. The CO2/HCO3 buffer system of blood has an effective pKa of approximately 6.1, yet the normal pH of blood is 7.4. Why is CO2/HCO3 an effective buffer when the pKa is more than 1 unit below the pH of blood? What happens to the pH of blood when the CO2 pressure increases? when the O2 pressure increases?

  13. Carbon dioxide produced during respiration is converted to carbonic acid (H2CO3). The pKa1 of carbonic acid is 6.35, and its pKa2 is 10.33. Write the equations corresponding to each pK value and predict the equilibrium position for each reaction.

Answer

    1. Not a buffer; the HCl completely neutralizes the sodium acetate to give acetic acid and NaCl(aq).
    2. Buffer; the HCl neutralizes only half of the sodium acetate to give a solution containing equal amounts of acetic acid and sodium acetate.
    3. Not a buffer; the NaOH completely neutralizes the acetic acid to give sodium acetate.
    4. Buffer; the NaOH neutralizes only half of the acetic acid to give a solution containing equal amounts of acetic acid and sodium acetate.
    5. Buffer; the solution will contain a 2:1 ratio of sodium acetate and acetic acid.

Numerical Problems

  1. Benzenesulfonic acid (pKa = 0.70) is synthesized by treating benzene with concentrated sulfuric acid. Calculate the following:

    1. the pH of a 0.286 M solution of benzenesulfonic acid
    2. the pH after adding enough sodium benzenesulfonate to give a final benzenesulfonate concentration of 0.100 M
  2. Phenol has a pKa of 9.99. Calculate the following:

    1. the pH of a 0.195 M solution
    2. the percent increase in the concentration of phenol after adding enough solid sodium phenoxide (the sodium salt of the conjugate base) to give a total phenoxide concentration of 0.100 M
  3. Salicylic acid is used in the synthesis of acetylsalicylic acid, or aspirin. One gram dissolves in 460 mL of water to create a saturated solution with a pH of 2.40.

    1. What is the Ka of salicylic acid?
    2. What is the final pH of a saturated solution that is also 0.238 M in sodium salicylate?
    3. What is the final pH if 10.00 mL of 0.100 M HCl are added to 150.0 mL of the buffered solution?
    4. What is the final pH if 10.00 mL of 0.100 M NaOH are added to 150.0 mL of the buffered solution?
  4. An intermediate used in the synthesis of perfumes is valeric acid, also called pentanoic acid. The pKa of pentanoic acid is 4.84 at 25°C.

    1. What is the pH of a 0.259 M solution of pentanoic acid?
    2. Sodium pentanoate is added to make a buffered solution. What is the pH of the solution if it is 0.210 M in sodium pentanoate?
    3. What is the final pH if 8.00 mL of 0.100 M HCl are added to 75.0 mL of the buffered solution?
    4. What is the final pH if 8.00 mL of 0.100 M NaOH are added to 75.0 mL of the buffered solution?

Answer

    1. 1.35 × 10−3
    2. 4.03
    3. 3.88
    4. 4.30

16.7 End-of-Chapter Material

Application Problems

    Problems marked with a ♦ involve multiple concepts.

  1. The analytical concentration of lactic acid in blood is generally less than 1.2 × 10−3 M, corresponding to the sum of [lactate] and [lactic acid]. During strenuous exercise, however, oxygen in the muscle tissue is depleted, and overproduction of lactic acid occurs. This leads to a condition known as lactic acidosis, which is characterized by elevated blood lactic acid levels (approximately 5.0 × 10−3 M). The pKa of lactic acid is 3.86.

    1. What is the actual lactic acid concentration under normal physiological conditions?
    2. What is the actual lactic acid concentration during lactic acidosis?
  2. When the internal temperature of a human reaches 105°F, immediate steps must be taken to prevent the person from having convulsions. At this temperature, Kw is approximately 2.94 × 10−14.

    1. Calculate the pKw and the pH and pOH of a neutral solution at 105°F.
    2. Is the pH greater than or less than that calculated in Exercise 1 for a neutral solution at a normal body temperature of 98.6°F?
  3. ♦ The compound diphenhydramine (DPH) is the active ingredient in a number of over-the-counter antihistamine medications used to treat the runny nose and watery eyes associated with hay fever and other allergies. DPH is a derivative of trimethylamine (one methyl group is replaced by a more complex organic “arm” containing two phenyl rings):

    The compound is sold as the water-soluble hydrochloride salt (DPH+Cl). A tablet of diphenhydramine hydrochloride contains 25.0 mg of the active ingredient. Calculate the pH of the solution if two tablets are dissolved in 100 mL of water. The pKb of diphenhydramine is 5.47, and the formula mass of diphenhydramine hydrochloride is 291.81 amu.

  4. ♦ Epinephrine, a secondary amine, is used to counter allergic reactions as well as to bring patients out of anesthesia and cardiac arrest. The pKb of epinephrine is 4.31. What is the percent ionization in a 0.280 M solution? What is the percent ionization after enough solid epinephrine hydrochloride is added to make the final epinephrineH+ concentration 0.982 M? What is the final pH of the solution?

  5. ♦ Fluoroacetic acid is a poison that has been used by ranchers in the western United States. The ranchers place the poison in the carcasses of dead animals to kill coyotes that feed on them; unfortunately, however, eagles and hawks are also killed in the process. How many milliliters of 0.0953 M Ca(OH)2 are needed to completely neutralize 50.0 mL of 0.262 M fluoroacetic acid solution (pKa = 2.59)? What is the initial pH of the solution? What is the pH of the solution at the equivalence point?

  6. Accidental ingestion of aspirin (acetylsalicylic acid) is probably the most common cause of childhood poisoning. Initially, salicylates stimulate the portion of the brain that controls breathing, resulting in hyperventilation (excessively intense breathing that lowers the PCO2 in the lungs). Subsequently, a potentially serious rebound effect occurs, as the salicylates are converted to a weak acid, salicylic acid, in the body. Starting with the normal values of PCO2 = 40 mmHg, pH = 7.40, and [HCO3] = 24 nM, show what happens during the initial phase of respiratory stimulation and the subsequent phase of acid production. Why is the rebound effect dangerous?

  7. Emphysema is a disease that reduces the efficiency of breathing. As a result, less CO2 is exchanged with the atmosphere. What effect will this have on blood pH, PCO2, and [HCO3]?

Answer

  1. 68.7 mL; 1.60; 7.85