Chapter 14 Chemical Kinetics

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".

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 Chapter 10 "Gases" 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. Figure 14.2 "The Effect of Temperature on Reaction Rates" 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:

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

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 (Figure 14.3 "The Effect of Surface Area on Reaction Rates"). 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.

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 Chapter 3 "Chemical Reactions" 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 (Figure 14.4 "The Effect of Catalysts on Reaction Rates"). 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.

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).

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.

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 $\left(t_0\right)$ and at the end of the interval $\left(t_1\right)\text{.}$ 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):

$$\begin{array}{cc}\hfill {\text{rate}}_{\left(t=0-2.0\text{ h}\right)}& =\frac{{\text{[salicyclic acid]}}_2-{\text{[salicyclic acid]}}_0}{2.0\text{ h}-0\text{ h}}\hfill \\ \hfill & =\frac{0.040\times {10}^{-3}\text{ M}-0\text{ M}}{2.0\text{ h}}=2.0\times {10}^{-5}\text{ M/h}\hfill \end{array}$$

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:

$$\begin{array}{cc}\hfill {\text{rate}}_{\left(t=0-2.0\text{ h}\right)}& =-\frac{{\text{[aspirin]}}_2-{\text{[aspirin]}}_0}{2.0\text{ h}-0\text{ h}}\hfill \\ \hfill & =-\frac{(5.51\times {10}^{-3}\text{ M)}-(5.55\times {10}^{-3}\text{ M)}}{2.0\text{ h}}\hfill \\ \hfill & =2\times {10}^{-5}\text{ M/h}\hfill \end{array}$$

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):

$$\begin{array}{cc}\hfill {\text{rate}}_{\left(t=200-300\text{h}\right)}& =\frac{{\text{[salicyclic acid]}}_{300}-{\text{[salicyclic acid]}}_{200}}{300\text{ h}-200\text{ h}}\hfill \\ \hfill & =\frac{(3.73\times {10}^{-3}\text{ M)}-(2.91\times {10}^{-3}\text{ M)}}{100\text{ h}}\hfill \\ \hfill & =8.2\times {10}^{-6}\text{ M/h}\hfill \end{array}$$

(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.)

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 $(\Delta t)\text{.}$ 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 $(\left[{\text{R}}_0\right])$ and the measured concentration of one or more reactants ([R]) after a given amount of time $(t)\text{.}$ 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

$$\text{rate}=-\frac{\Delta \text{[A]}}{\Delta t}=k{\text{[A]}}^m{\text{[B]}}^n$$

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

$$\text{rate}=-\frac{\Delta [{({\text{CH}}_3)}_3\text{CBr]}}{\Delta t}=k{[{({\text{CH}}_3)}_3\text{CBr]}}^m{[{\text{H}}_2\text{O]}}^n$$

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

$$\text{rate}=-\frac{\Delta [{\text{CH}}_3\text{Br]}}{\Delta t}=k\prime [{\text{CH}}_3\text{Br]}$$

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. $\begin{array}{l}\hfill 2\text{HI(g)}\stackrel{\text{Pt}}{\to }{\text{H}}_2\text{(g)}+{\text{I}}_2\text{(g)}\\ \hfill \text{rate}=-\frac{1}{2}\left(\frac{\Delta \text{[HI]}}{\Delta t}\right)=k{\text{[HI]}}^2\end{array}$
2. $\begin{array}{l}\hfill 2{\text{N}}_2\text{O(g)}\stackrel{\Delta }{\to }{\text{2N}}_2\text{(g)}+{\text{O}}_2\text{(g)}\\ \hfill \text{rate}=-\frac{1}{2}\left(\frac{\Delta [{\text{N}}_2\text{O]}}{\Delta t}\right)=k\end{array}$
3. $\begin{array}{l}\hfill \text{cyclopropane(g)}\text{ → }\text{propane(g)}\\ \hfill \text{rate}=-\frac{\Delta \text{[cyclopropane]}}{\Delta t}=k\text{[cyclopropane]}\end{array}$

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)]:

   $$k{\text{M}}^2=\frac{\text{M}}{\text{s}}k=\frac{\text{M/s}}{{\text{M}}^2}=\frac{1}{\text{M}\cdot \text{s}}={\text{M}}^{-1}\cdot {\text{s}}^{-1}$$

   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. $\begin{array}{cc}\hfill {\text{CH}}_{\text{3}}{\text{N=NCH}}_{\text{3}}(\text{g})\to {\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}(\text{g})+{\text{N}}_{\text{2}}(\text{g})& \text{rate}\text{ = }-\frac{\Delta [{\text{CH}}_3{\text{N=NCH}}_3]}{\Delta t}\\ \hfill & =k[{\text{CH}}_3{\text{N=NCH}}_3\text{]}\hfill \end{array}$
2. $\begin{array}{cc}\hfill 2{\text{NO}}_2\text{(g)}+{\text{F}}_2\text{(g)}\to {\text{2NO}}_2\text{F(g)}& \text{rate}\text{ = }-\frac{\Delta [{\text{F}}_2]}{\Delta t}=-\frac{1}{2}\left(\frac{\Delta [{\text{NO}}_2]}{\Delta t}\right)\\ \hfill & =k[{\text{NO}}_2][{\text{F}}_2]\hfill \end{array}$

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: $\text{rate}=\frac{\Delta \text{[B]}}{\Delta t}=-\frac{\Delta \text{[A]}}{\Delta t}$

general form of rate law when A and B are reactants

Equation 14.9: $\text{rate}=k{\text{[A]}}^m{\text{[B]}}^n$

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.

3. 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:

   $$\text{rate}=-\frac{\Delta \text{[A]}}{\Delta t}=\frac{1}{2}\left(\frac{\Delta \text{[C]}}{\Delta t}\right)$$

   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.

   $${\text{N}}_2{}^{+}+{\text{O}}_2\stackrel{k_1}{\to }{\text{N}}_2+{\text{O}}_2{}^{+}$$ $${\text{O}}_2{}^{+}+\text{O}\stackrel{k_2}{\to }{\text{O}}_2+{\text{O}}^{+}$$ $${\text{O}}^{+}+{\text{N}}_2\stackrel{k_3}{\to }{\text{NO}}^{+}+\text{N}$$

   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:

   $${\text{H}}_3{\text{AsO}}_4\text{(aq)}+3{\text{I}}^{-}\text{(aq)}+2{\text{H}}^{+}\text{(aq)}\underset{k_{\text{r}}}{\overset{k_{\text{f}}}{\rightleftharpoons }}{\text{H}}_3{\text{AsO}}_3\text{(aq)}+{\text{I}}_3{}^{-}\text{(aq)}+{\text{H}}_2\text{O(l)}$$

   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: k_f/k_r = [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

5. 298 s; 1270 s

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:

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

where [A]₀ is the initial concentration of reactant A. (Equation 14.16 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:

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:

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 Figure 14.8 "A Zeroth-Order Reaction", 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.

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 Figure 14.10 "The Catalyzed Oxidation of Ethanol"). 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").

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:

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:

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 Chapter 11 "Liquids", Section 11.9 "Essential Skills 6", discusses natural logarithms.) Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation 14.20 predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation 14.20 and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and t:

Because Equation 14.21 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 (Equation 14.19) or the integrated rate law (Equation 14.21) 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 (Figure 14.6) and the reaction of t-butyl bromide with water to give t-butanol (Equation 14.10). 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 Figure 14.11 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 Figure 14.11 have been studied extensively to find ways of maximizing the concentration of the active species.

The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Table 14.2 "Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C". 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 Table 14.2 "Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C" 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 Table 14.2 "Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C". For example, substituting the values for Experiment 3 into Equation 14.19,

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.

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 Figure 14.12 "The Hydrolysis of Cisplatin, a First-Order Reaction" 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 Figure 14.12 "The Hydrolysis of Cisplatin, a First-Order Reaction". 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 Figure 14.12 "The Hydrolysis of Cisplatin, a First-Order Reaction" for t = 100 min ([cisplatin] = 0.0086 M) and t = 1000 min ([cisplatin] = 0.0022 M),

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

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:

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:

Because Equation 14.23 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.

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 (Equation 14.22) or the integrated rate law (Equation 14.23).

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

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

This means that the reaction is second order in the monomer. Using Equation 14.22 and the data from any row in Table 14.3 "Rates of Reaction as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M", we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:

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 Figure 14.14 "Dimerization of a Monomeric Compound, a Second-Order Reaction". 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 Figure 14.14 "Dimerization of a Monomeric Compound, a Second-Order Reaction". 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 Figure 14.13 " " 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 Figure 14.16 "Properties of Reactions That Obey Zeroth-, First-, and Second-Order Rate Laws" in Section 14.4 "Using Graphs to Determine Rate Laws, Rate Constants, and Reaction Orders".

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:

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 Section 14.2 "Reaction Rates and Rate Laws", 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:

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 Table 14.4 "Rate Data for a Hypothetical Reaction of the Form A + B → Products". The general rate law for the reaction is given in Equation 14.25. 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".

Inserting the appropriate values from Table 14.4 "Rate Data for a Hypothetical Reaction of the Form A + B → Products",

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.

Substituting the appropriate values from Table 14.4 "Rate Data for a Hypothetical Reaction of the Form A + B → Products",

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

We can now calculate the rate constant by inserting the data from any row of Table 14.4 "Rate Data for a Hypothetical Reaction of the Form A + B → Products" into the experimentally determined rate law and solving for k. Using Experiment 2, we obtain

You should verify that using data from any other row of Table 14.4 "Rate Data for a Hypothetical Reaction of the Form A + B → Products" 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

$$k=\frac{\text{rate}}{{\text{[NO]}}^2[{\text{O}}_2]}=\frac{7.98\times {10}^{-3}\text{ M/s}}{{(0.0235\text{ M)}}^2(0.0125\text{ M)}}=1.16\times {10}^3{\text{ M}}^{-2}\cdot {\text{s}}^{-1}$$

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

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 t_1/2. 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 (Equation 14.21) to produce the following equation:

Substituting [A]₀/2 for [A] and t_1/2 for t (to indicate a half-life) into Equation 14.27 gives

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:

Thus, for a first-order reaction, each successive half-life is the same length of time, as shown in Figure 14.17 "The Half-Life of a First-Order Reaction", 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.

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.

Radioactive Decay Rates

As you learned in Chapter 1 "Introduction to Chemistry", 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: $A=-\Delta N/\Delta t.$ as the decrease in the number of the radioisotope’s nuclei per unit time:

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:

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 Equation 14.29 and Equation 14.30, 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 is the same as the equation for the reaction rate of a first-order reaction (Equation 14.19), 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 (Equation 14.31) or the integrated rate law:

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 Table 14.6 "Half-Lives and Applications of Some Radioactive Isotopes", along with some of their applications.

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:

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 (Figure 14.18 "Radiocarbon Dating"). 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).

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.

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:

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:

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.

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 (Table 14.7 "Common Types of Elementary Reactions and Their Rate Laws"). 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 Figure 14.19 "The Basis for Writing Rate Laws of Elementary Reactions". 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.

$$\begin{array}{ccc}\text{step 1}\hfill & {\text{NO}}_2+{\text{NO}}_2\stackrel{k_1}{\to }{\text{NO}}_3+\text{NO}& \text{rate}=k_1{[{\text{NO}}_2]}^2\text{ (predicted)}\\ \text{step 2}\hfill & \underset{\_}{{\text{NO}}_3+\text{CO}\stackrel{k_2}{\to }{\text{NO}}_2+{\text{CO}}_2}& \text{rate}=k_{\text{2}}{\text{[NO}}_{\text{3}}]\text{[CO] (predicted)}\\ \text{sum}\hfill & {\text{NO}}_2+\text{CO}\stackrel{k}{\to }\text{NO}+{\text{CO}}_2& \text{rate}=k{[{\text{NO}}_2]}^2\text{ (observed)}\end{array}$$

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.

$$\begin{array}{cc}\text{step }1\hfill & {\text{NO}}_2+{\text{NO}}_2\stackrel{k_1}{\to }{\text{N}}_2{\text{O}}_4\\ \text{step }2\hfill & \underset{\_}{{\text{N}}_2{\text{O}}_4+\text{CO}\stackrel{k_2}{\to }\text{NO}+{\text{NO}}_2+{\text{CO}}_2}\\ \text{sum}\hfill & {\text{NO}}_2+\text{CO}\stackrel{}{\to }\text{NO}+{\text{CO}}_2\end{array}$$

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:

$$\begin{array}{lll}\text{step }1\hfill & \text{ICl}+{\text{H}}_2\stackrel{k_1}{\to }\text{HCl}+\text{HI}\hfill & \text{rate}=k_1\text{[ICl]}[{\text{H}}_2]\text{ (slow)}\hfill \\ \text{step }2\hfill & \underset{\_}{\text{HI}+\text{ICl}\stackrel{k_2}{\to }\text{HCl}+{\text{I}}_2}\hfill & \text{rate}=k_2\text{[HI][ICl] (fast)}\hfill \\ \text{sum}\hfill & 2\text{ICl}+{\text{H}}_2\stackrel{}{\to }2\text{HCl}+{\text{I}}_2\hfill & \hfill \end{array}$$

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:

$$\begin{array}{ccc}\text{step 1}\hfill & \text{NO}+\text{NO}\stackrel{k_1}{\to }\stackrel{}{{\text{N}}_2{\text{O}}_2}& \text{(fast)}\\ \text{step 2}\hfill & {\text{N}}_2{\text{O}}_2+{\text{H}}_2\stackrel{k_2}{\to }\stackrel{}{{\text{N}}_2\text{O}+{\text{H}}_2\text{O}}& \text{(slow)}\\ \text{step }3\hfill & {\text{N}}_2\text{O}+{\text{H}}_2\stackrel{k_3}{\to }{\text{N}}_2+\stackrel{}{{\text{H}}_2\text{O}}& \text{(fast)}\end{array}$$

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):

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·):

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

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

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

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:

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

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:

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.