Chapter 10 – Chemical Kinetics

Introduction

To this point in our study of chemistry, we have been concerned only with the composition of the equilibrium mixture, not the length of time required to obtain it. However, the time required is also an important consideration. If a reaction proceeds too slowly, it will not be useful no matter how large the equilibrium constant, and if a reaction proceeds too rapidly, it may not be possible to control it. Thermodynamics allows us to predict the extent of a reaction, but it tells us nothing about the speed of the reaction or how it occurs. These latter two considerations are the domain of kinetics, the study of the rates and mechanisms of chemical reactions.

10.1. Reaction Rates

Introduction

Rates are changes in one quantity with respect to another, and reaction rates are measured as changes in concentration with respect to time. They are the most important of the kinetic parameters, and we begin our study of kinetics with a treatment of reaction rates.

Objectives

• •
  Express the relative rates of consumption of reactants and formation of products from the coefficients in a balanced chemical reaction.

10.1-1. Concentrations versus Time

It takes time for the concentrations of reactants and products to change from their initial to their equilibrium values, but once equilibrium is established, the concentrations no longer change. Kinetics studies the reaction during the time of change, while thermodynamics is concerned only with the initial and equilibrium conditions. Thus, the region of time in which the concentrations are changing is called the kinetic region, while the region after equilibrium is established is called the thermodynamic region. Consider the decomposition of 0.020 M HI at 700 °C, 2 HI(g) → H₂(g) + I₂(g). The concentrations of both reactants and products are shown in Figure 10.1a as a function of time in seconds. The concentration of HI decreases for about 10,000 seconds (about three hours) and then levels off at 0.015 M. During that time, the concentrations of H₂ and I₂ each rise from zero to about 0.0025 M. Once the concentrations are no longer changing, the system has reached equilibrium. Thus, approximately three hours are required to obtain an equilibrium mixture at this temperature. Δn_g = 0 for the reaction, so the thermodynamic equilibrium constant for the reaction is equal to K_c, so it can be determined from the equilibrium concentrations, If the reaction were extensive, 0.020 M HI would react to produce a mixture that was 0.010 M in each H₂ and I₂, and we now examine the reaction from the reverse direction by reacting 0.010 M H₂ with 0.010 M I₂ at 700 °C. The results are shown in Figure 10.1b. The HI concentration rises to its equilibrium value of 0.015 M, while the H₂ and I₂ concentrations each drop to their equilibrium value of 0.0025 M. Note that these are the same equilibrium concentrations obtained by the decomposition of HI shown in Figure 10.1a. We conclude that equilibrium concentrations, while dependent upon the initial concentrations, are independent of the direction from which they were obtained.

10.1-2. Average Rates

Rates are changes in one quantity with respect to another. For example, speed is the rate of change of position with respect to time. The average speed is determined by dividing the distance traveled (the change in position) by the time required to travel it. Thus, a person traveling 300 kilometers in 4 hours has an average speed of 75 km·h^–1 (~50 mph). The above is the average speed over the interval, but it is unlikely that the person traveled at a constant speed of 75 km·h^–1 for the entire trip. The instantaneous speed (the speed given by the speedometer at any instant) for such a trip might vary from 0 to over 100 km·h^–1.

10.1-3. Average Rates of Formation of HI

Reaction rates are the most important of the kinetic parameters. They are measured as changes in concentration with respect to time. Products are formed during a reaction, so we refer to their rate of formation. The average rate of formation of HI over a time interval Δt is given as the following. Consider the following three points in the plot of HI concentration versus time at 700 °C for the reaction
H₂ + I₂ → 2 HI shown in Figure 10.2.

time (s)	[HI] (M)
0	0.000
2000	0.010
5000	0.014

The average rate of formation of HI during the first 2000 s would be determined as and the average rate of formation in the interval 2000 to 5000 s would be Thus, the rate decreased by nearly 80%. Indeed, the rate of formation continues to decrease as it approaches 0 at equilibrium. Thus, rates of reaction vary with time.

10.1-4. Graphical Determination of a Rate of Formation

It is not the average rate of change that is important in kinetics studies. Rather it is the instantaneous rate of change; i.e., the rate at which the concentration is changing at a given time. The instantaneous rate at a given time is given by the slope of the tangent to the concentration versus time curve at that instant. We will assume that anytime a rate is given, that it is the instantaneous rate unless it is specified to be the average rate. We now determine the rate of formation of HI 2130 s into the reaction. Consider the graph shown in Figure 10.3, which shows the variation of the HI concentration with time in more detail than in Figure 10.1a. The rate of formation of HI at this point equals the slope of the tangent (shown as a green line). The slope is determined by taking two points on the tangent line and determining the change in concentration by the time interval between the two points. We use the two points at the ends of the line to determine the slope and therefore the rate of formation at that time as follows.

10.1-5. Graphical Determination of a Rate of Disappearance

Reactant concentrations decrease with time, so they are disappearing. Consequently, it is usually the rate of disappearance of reactants that is given. The rate of disappearance of H₂ is given as Note that the rate at which [H₂] is changing is negative, so its rate of disappearance is positive. The rates of disappearance of H₂ and I₂ at 2130 s into the reaction is determined graphically in Figure 10.4. Note that the slope of the tangent is negative, which indicates a disappearance. Thus, the rates of disappearance of H₂ and I₂ are each 1 × 10^–6 M·s^–1. Note that the rates of disappearance of H₂ and I₂ are one-half of the rate of formation of HI, consistent with the fact that 2 HI are formed for each H₂ and I₂ that react.

10.1-6. Rate of Reaction

The rate of formation of HI at any instant is exactly twice the rate of disappearance of H₂ or I₂ as dictated by the stoichiometry of the reaction (2 mol HI form for each mole of H₂ or I₂ that react). The rates of appearance of products and the rates of disappearance of reactants are often different, so what is meant by the term rate of reaction? To answer that question, we construct the reaction table to determine how much reaction takes place during some time interval Δt.

	H2	+	I2	→	2 HI
t	c		c		c'
Δt	–Δx		–Δx		+2Δx
t + Δt	c – Δx		c – Δx		c' + 2Δx

In the reaction table, Δx is the change in concentration of any species whose coefficient in the balanced chemical equation is unity. In the reaction above, Δx = –Δ[H₂] = –Δ[I₂] = Δ[HI]/2. The rate of reaction of the above reaction is defined as Although it is more common to talk about the rate of formation of a product or the rate of disappearance of a reactant, the rate of reaction is also used and care must be given to the differences caused by the coefficients in the balanced equation. In general, the rate of reaction is related to the rates of formation or disappearance as follows.

10.1-7. Exercise

10.1-8. Exercise

10.2. Rate Laws

Introduction

Rates are used to determine the rate law of a reaction, which shows how the rate varies with the concentrations of reactants.

Objectives

• •
  Determine the rate of a reaction given the rate law, the rate constant, and concentrations of reactants.
• •
  Determine the rate constant given the rate law and a rate at a given set of concentrations.

10.2-1. Definitions

The manner in which the rate varies at a given temperature is expressed by the rate law for the reaction. Rate laws can be very complicated, but they often take the form of a constant times the concentrations of the reactants each raised to some exponent. The exponent of a reactant's concentration is called the reactant order of that reactant, while the sum of the exponents of all reactants is referred to as the reaction order.

Reaction	Rate Law	Reaction and Reactant Orders
a    H2 + I2 → 2 HI	rate = ka[H2][I2]	A second order reaction that is first order in H2 and first order in I2.
b    2 HI → H2 + I2	rate = kb[HI]2	A second order reaction that is second order in HI.
c    2 N2O5 → 4 NO2 + O2	rate = kc[N2O5]	A first order reaction that is first order in N2O5

It is important to recognize the following. Thus, Reaction c is first order in N₂O₅ even though the stoichiometric coefficient of N₂O₅ is two. The rate constants k_a and k_b are called second-order rate constants, while k_c is a first-order rate constant. The reaction rate has units of concentration per unit time, but the units of the rate constant depend on the reaction order.

First Order	Second Order
kc = rate [N2O5] = M·s−1 M = s−1	kb = rate [HI]2 = M·s−1 M 2 = M −1·s−1

Thus, the units of a first order rate constant are s^–1 and the units of a second order rate constant are M^–1·s^–1.

10.2-2. Exercise

10.3. Determining Rate Laws

Introduction

The exponents of each of the concentrations in a rate law must be determined experimentally. In this section, we apply the method of initial rates and concentration versus time data to obtain rate laws.

Objectives

• •
  Use the method of initial rates to determine the rate law of a reaction.

10.3-1. Introduction

In the method of initial rates, average rates of reaction are determined at the beginning of the reaction. The time interval of the experiment must be small enough that only a small fraction of the reactants is consumed. This assures that the average rate is close to the instantaneous rate (Δt → 0). The method has the advantage that the concentrations are initial (makeup) concentrations, which means that they are predetermined. In addition, the initial concentrations of the products are zero, so the reverse reaction can be ignored. This is a real advantage when studying reactions where the forward and reverse reactions compete with one another at appreciable product concentrations.

10.3-2. Method

At least one experiment must be performed for each unknown in the rate law: one for the rate constant and one for each reactant order. For example, the rate law for the generic reaction, A + B → products has the form R = k[A]^a[B]^b, so a minimum of three rates must be measured using different initial concentrations of A and B in order to determine the values of k, a, and b. The three rates can be expressed as If all three experiments are carried out at the same temperature, all three rate constants are the same. Thus, the rate constant can be eliminated by dividing the rates by one another to produce the following two equations. We now have two equations in two unknowns that can be solved for a and b by taking the logarithms of both sides. However, the algebra can be simplified by experimental design. If the concentration of B is kept constant in Experiments 1 and 2 ([B]₁ = [B]₂), and the concentration of A is kept constant in Experiments 2 and 3 ([A]₂ = [A]₃) then the ratio of rates simplifies to the following. The ratios are all known, so the exponents can be determined with the use of logs as shown in Equation 10.3. The orders can often be determined by inspection in cases where the ratios are simple fractions. Most of the examples that we will deal with can be done by inspection (no logs required).

10.3-3. Decomposition of HI Exercise

10.3-4. Formation of HI Exercise

10.3-5. Exercise

First-order Kinetics

10.3-6. Integrated Rate Law

Consider the generic first order reaction A → products, which has the following rate law. The above can be rearranged as follows to get the [A] terms on the same side of the equation which can be solved by the following integration to produce the following two expressions that relate concentration and time. where [A]₀ is the concentration of A at t = 0, i.e., the initial concentration. Equation 10.4 shows that a plot of ln [A] versus time for a first order reaction is a straight line with a slope of –k and an intercept of ln [A]_o. This behavior can also be expressed as an exponential. Equation 10.5 shows that the concentration of a reactant drops exponentially with time. This behavior is called exponential decay. We conclude the following.

10.3-7. Determining a Concentration Exercise

10.3-8. First Order Behavior Exercise

10.3-9. Half-life Definition

The half-life, t_1/2, of a reaction is the time required for half of the existing reactant to disappear. Consider the graph shown in Figure 10.5. The initial concentration of A (azomethane) is 0.006 M, so the half-life is the time required for the concentration to drop to 0.003 M. As shown in the figure, this occurs after 4330 s, the half-life of the reaction. After another half-life (another 4330 s) the concentration has dropped to 0.0015 M, which is half of 0.003 M. After a third half-life the concentration is again halved. Thus the half-life of a first order reaction is constant. The concentration of a reactant after n half-lives is given by the following. Thus, after five half-lives (5 × 4330 = 21,650 s) the concentration of azomethane is (1/2)⁵(0.006 M) = 1.9 × 10^–4 M.

10.3-10. Half-life Equation

At the half-life, t = t_1/2 and [A] = [A]₀/2. Substitution of these half-life quantities into Equation 10.4

ln [A] = ln [A]₀ − kt

ln

[A]

[A]0

= −kt

yields Note that the identity ln(1/x) = –ln(x) was used. Solving for the half-life, The half-life of the azomethane decomposition can be determined from its first order rate constant determined in Exercise 10.8

(k = 1.60 × 10⁻⁴ s⁻¹)

with Equation 10.6 as follows which is the same value observed in the previous section. As shown in the following exercise, Equation 10.6 can also be used to determine the value of the first-order rate constant from a half-life.

10.3-11. Half-life Exercise

Second-order Kinetics

10.3-12. Integrated Rate Law

Consider the generic second order reaction 2 A → products, which has the following rate law. The above can be rearranged as follows to get the [A] terms on the same side of the equation, which can be solved by the following integration to produce the following expression that relates concentration and time.

10.3-13. Reaction Order Exercise

Another Look at the Relationship Between Equilibrium and Kinetics

10.3-14. Decomposition of HI Review

Recall that HI decomposes into H₂ and I₂, but it can also be formed from H₂ and I₂. If we represent these competing processes as 2 HI ⇌ H₂ + I₂ then the decomposition of HI is the forward reaction with a rate constant k_f, and the formation of HI is the reverse reaction with a rate constant k_r. We studied the competing reactions using the method of initial rates. The resulting rate laws were the following. The specific rate constants at 700 °C were found to be k_f = 1.8 × 10^–3 M^–1·s^–1 and k_r = 6.3 × 10^–2 M^–1·s^–1. The '2' in the rate laws is due to the coefficient of HI in the balanced equation.

10.3-15. Deviation from Second Order Kinetics

The decomposition of HI is second order in HI, so a plot of 1/[HI] versus time for the decomposition should be linear. As shown in Figure 10.6, the plot is linear at the beginning of the experiment, but it deviates from linearity after about 2000 s. The curvature of the line indicates that [HI] is greater than predicted by simple second order kinetics. The higher than expected concentration of HI results because it is being formed by the formation reaction. As the reaction proceeds, the decomposition slows because the concentration of HI drops, but the rate of formation increases because the concentrations of H₂ and I₂ are increasing. The net rate of disappearance of HI is equal to the rate at which it disappears in the decomposition minus the rate at which it forms in the formation. At equilibrium, the concentration of HI no longer changes because it is formed at the same rate that it disappears. Thus, at equilibrium, Gathering concentrations on one side and rate constants on the other, and substituting the known rate constants, we obtain the following: which is the equilibrium constant for the reaction 2 HI(g) ⇌ H₂(g) + I₂(g). We conclude that the equilibrium constant is equal to the ratio of the forward and reverse rate constants. The equilibrium concentrations in the decomposition of 0.020 M HI at 700 °C are [HI] = 0.015 M, and [H₂] = [I₂] = 0.0025 M. Using the known equilibrium concentrations and rate constants, we can determine the rates of formation and disappearance of HI as follows.

• •
  rate of formation = 2k_f[HI]² = 2(1.8 × 10³ M^–1·s^–1)(0.015 M)² = 8.0 × 10^–7 M/s
• •
  rate of disappearance = 2k_r[I₂][H₂] = 2(6.3 × 10^–2 M^–1·s^–1)(0.0025 M)(0.0025 M) = 8.0 × 10^–7 M/s

The reaction reaches equilibrium because HI is formed at the same rate that it is consumed. Thus, while the net concentrations are no longer changing at equilibrium, the reaction has not stopped.

10.4. Reaction Mechanisms and Rate Laws

Introduction

The rate law of a reaction gives us some clues about the steps involved in the reaction. In this section, we examine some simple examples.

Objectives

• •
  Identify the intermediates in a reaction mechanism and use an experimental rate law to identify the rate-determining step.
• •
  Determine the molecularity of an elementary step.
• •
  Determine if a mechanism is consistent with the rate law.

10.4-1. Elementary Reactions

Most reactions do not occur in a single step, but by a series of elementary steps called the reaction mechanism. Consider the decomposition of azomethane, which proceeds by a two-step mechanism.

• 1
  H₃C–N=N–CH₃ → 2 CH₃ + N₂
• 2
  2 CH₃ → C₂H₆

In the first step, the two N–C bonds break and the N≡N bond forms, and in the second, the highly reactive CH₃ groups quickly combine to form ethane. The individual steps of the mechanism are called elementary reactions because they each represent a single molecular event that results in a reaction. The two elementary reactions above combine to give the reaction mechanism for the decomposition of azomethane. The sum of the elementary processes must yield the net reaction. The two CH₃ (methyl) groups that are formed in the first step are consumed in the second. Since they appear in the mechanism, but not in the net reaction, the methyl groups are called intermediates. Intermediates typically are short-lived, but they can frequently be observed during reaction and sometimes even isolated. Indeed, evidence for a proposed intermediate is excellent support for a mechanism.

10.4-2. Molecularity

The number of molecules reacting in an elementary reaction is called the molecularity of the elementary reaction. There are only three molecularities.

Number of Particles	Molecularity
1	unimolecular
2	bimolecular
3	termolecular

Termolecular processes are very rare because the simultaneous collision of three reactant molecules is very rare. Processes with molecularities greater than three do not occur at all. Thus, the first step in the decomposition of azomethane is unimolecular while the second step is bimolecular.

10.4-3. The Rate Law of an Elementary Process

In order for two or three molecules to react, they must collide with one another. As a result, the rate of an elementary reaction is proportional to the collision frequency of the reactants. The collision frequency, which is the number of collisions between the reacting particles in a specified volume per unit time, can be shown to be directly proportional to the product of the molar concentrations of the colliding particles. Thus, we conclude the following.

The rate law of an elementary reaction is equal to a proportionality constant (the rate constant) times the product of the molar concentrations of the reactants.

This means we can write the rate law expression for elementary processes from their chemical equation. For example, Most reactions do not occur in a single step, so their rate laws cannot be determined from the overall balanced equation. They can be determined only by experiment.

10.4-4. Rate-Determining Step

The rate law of a reaction is a combination of the rate laws of the elementary reactions of which it is composed. Thus, mechanisms are proposed based on the observed rate law. A mechanism is acceptable only if the rate law derived from it agrees with the experimentally determined rate law. It is often the case that more than one mechanism yields the experimental rate law, so mechanisms are usually proposed, not proven. However, finding evidence for a proposed intermediate strongly supports a proposed mechanism and can sometimes prove that it is correct. Combining the rate laws of the elementary processes into a reaction rate law can be a formidable task for complicated reactions, so we restrict our discussion to a special case: reactions in which one step of the multi-step mechanism is much slower than any of the others. In this case,

• •
  the rate of reaction is determined by the rate of the slow step, which is called the rate-determining step (RDS), and the rate law of the reaction is the rate law of the RDS.

10.4-5. Rate-Determining Step Example

10.4-6. Intermediates and the Rate Law

The rate law of a reaction is based on the rate law of the rate-determining step, but the rate laws of reactions considered in this chapter involve the concentrations of only reactants—no products and no intermediates. Consequently, the concentrations of any intermediates that appear in the rate law of the RDS must be eliminated to obtain the experimental rate law. This is done by assuming that the rapid steps that precede the rate-determining step all reach equilibrium. Thus, the concentrations of any intermediate can be obtained in terms of the concentrations of the reactants and the equilibrium constants of the preceding reactions. This results in the following two properties of the rate law.

• •
  The rate constant of a reaction is composed of the rate constant of the RDS and equilibrium constants of previous steps.
• •
  The rate law of a reaction contains only those reacting molecules that appear in the RDS and steps prior to the RDS.

10.4-7. Determining a Rate Law from a Mechanism

Let us determine the rate law for the reaction with the following mechanism.

• 1
  NO + O₂ → NO₃
• 2
  NO + NO₃ → 2 NO₂

The rate law can contain only the concentrations of reactants in the net equation, so we begin by writing the net equation. NO₃ is an intermediate that is formed in Step 1 and consumed in Step 2, so the net equation is: Only NO and O₂ appear as reactants in the net equation, so only they can appear in the rate law. The rate law of the reaction is the rate law of the RDS. We will do this problem twice, assuming each step to be rate-determining. Remember that the rate law of an elementary process equals the rate constant times the concentrations of the reactants each raised to an exponent equal to their coefficients in the balanced equation. If Step 1 is RDS: NO + O₂ → NO₃ is the RDS, so the rate law of the RDS is rate = k[NO][O₂]. The rate law of the RDS contains the concentrations of reactants in the net equation only, so this result would be acceptable for the rate law of the reaction. If Step 2 is RDS: NO + NO₃ → 2 NO₂ is the RDS, so the rate law of the RDS is

rate = k' [NO][NO₃].

NO₃ is an intermediate, so it cannot appear in the rate law. To eliminate NO₃, we assume that the first (rapid) step reaches equilibrium to get an expression for [NO₃] in terms of reactant concentrations. The equilibrium constant expression for the first reaction is solving for the NO₃ concentration, we obtain Substitution of this expression into the rate law for the RDS, yields the following. Substituting k for

k'K

, we obtain the rate law expression. The result contains the concentrations of only reactants in the net equation, so it would be acceptable for the rate law of the reaction. Note that the rate law of the reaction is equal to the product of the rate constant of the RDS and the equilibrium constant of the previous step. We could arrive at the same rate laws by using the rule that the concentrations of only those molecules involved in the RDS and prior steps appear in the rate law. Thus, only one NO appears in the rate law if the first step is rate-determining because the second NO does not appear until the second step.

10.4-8. Rate Law from Mechanism Exercise

10.4-9. A Three-step Mechanism Exercise

10.5. The Effect of Temperature on Reaction Rates

Introduction

The rate of reaction depends upon both the reactant concentrations and the temperature. We have discussed the concentration dependence in the rate law, and we now treat the temperature dependence.

Prerequisites

• •
  CAMS 9.9 Activation Energy

Objectives

• •
  Use the Arrhenius equation to solve for a variable.
• •
  Sketch a potential energy profile for a reaction and interpret it in either the forward or reverse direction.
• •
  Define reaction coordinate, transition state, and activation energy.

10.5-1. Activation Energy

The rate of reaction depends upon both the reactant concentrations and the temperature. We have discussed the concentration dependence and now treat the temperature dependence. Consider the reaction between methyl iodide and hydroxide ion. The reaction is an elementary process, so it is first order in each reactant. In order for the reaction to occur, a hydroxide ion must collide with a CH₃I molecule, and it must collide between the three hydrogen atoms. As the distance between the hydroxide ion and the carbon atom decreases, the carbon-iodine bond stretches, and the three hydrogen atoms are pushed back, which forces the H–C–H angles to increase from 109°. As shown in Figure 10.7, these changes result in an energy increase, which reaches a maximum when the bond angle is 120° and the carbon is five-coordinate. The five-coordinate species (at the top of Figure 10.7) is called the transition state of the reaction because the reaction must proceed through this state to make the transition from reactants to products. The transition state can lead to or be formed from either side of the reaction (iodide ion can collide with CH₃OH to lead to the same transition state). The transition state cannot be isolated or directly observed. It is not an intermediate but rather a highly energetic species through which the reaction proceeds. The energy required to reach the transition state is called the activation energy. Figure 10.7 shows that the variation of energy along a complicated combination of C–I and C–O bond lengths and H–C–H bond angles is called the reaction coordinate. The activation energy of the forward reaction, E_a(f), is much less than that of the reverse reaction, E_a(r), because the energy of the reactants is greater than the energy of the products; i.e., the reaction is exothermic.

10.5-2. Activation Energy Factors

There are two factors that dictate whether a collision achieves the transition state: the steric factor and the energy factor. The steric factor indicates the fraction of collisions in which the reactants have the correct orientation to react. As shown in Figure 10.8, there are a number of ways in which CH₃I and OH^1– can collide, but only a fraction have the correct orientation to reach the transition state. The energy factor arises because even if the reactants have the correct orientation to react, they cannot do so unless there is sufficient energy in the collision to overcome the activation energy and achieve the transition state. The average energy available in a collision depends on the thermal energy (~RT), and the fraction of collisions with sufficient energy to overcome the activation energy is given by

e^−E_a/RT.

The rate constant represents the fraction of collisions that lead to the transition state, and it is the product of the steric and energy factors as shown in Equation 10.8. A, which is referred to as the pre-exponential, includes the steric factor. Equation 10.8, known as the Arrhenius equation, shows the temperature dependence of the rate constant. Thus, as the temperature increases, the thermal energy increases, and the fraction of collisions with sufficient energy to achieve the transition state increases. Increasing the temperature, however, increases the rate of both the forward and reverse reactions.

10.5-3. Arrhenius Plot

If we take the natural logarithm of both sides of Equation 10.8

k = Aexp−

Ea

RT

= Ae^−E_a/RT

, we can rewrite the Arrhenius equation as shown in Equation 10.9. A plot of ln k vs. 1/T is called an Arrhenius plot. If a reaction follows Arrhenius behavior, then the plot is linear with slope = –E_a/R and intercept = ln A. An Arrhenius plot is an excellent way to obtain the activation energy of a reaction.

10.5-4. Arrhenius Exercise

10.5-5. Using Only Two Temperatures

Determining the activation energy graphically is the best procedure, but it is often the case that only two data points are known. In these cases, we need consider only two temperatures: T₁ and T₂ and the two rate constants, k₁ and k₂, at those temperatures. In addition, only two temperatures are needed to determine a rate constant at one temperature given the activation energy and rate constant at another temperature. To get the expression for only two temperatures, apply Equation 10.9

ln k = ln A −

Ea

R

·

1

T

to each point and then take the difference to eliminate the ln A term and obtain the following. Equation 10.10 is an excellent way to determine a rate constant at one temperature given the rate constant at another temperature and the activation energy. However, we can also solve the expression for the activation energy, which allows us to determine the activation energy given the rate constants at two temperatures. However, care must always be taken when assuming a linear relationship with only two points. The activation energies derived from the above expression can deviate substantially from that determined from the best fit when the two points do not lie on the best line.

10.5-6. Activation Energy Exercise

10.5-7. Activation Energy from Ratio Exercise

10.6. Catalysis

Introduction

Catalysts increase the rate of reaction but are unchanged by it. They function by altering the path (mechanism) of the reaction so as to reduce the activation energy.

Prerequisites

• •
  CAMS 14.5 Metals as Catalysts

Objectives

• •
  Explain how the addition of a catalyst affects the reaction rate.

10.6-1. Review

Heterogeneous catalysts are in a different phase (usually solid) than the reactants. Look at the material listed as a prerequisite for this section for a discussion of the catalytic converter and the surface catalyzed hydrogenation of ethene to ethane. Homogeneous catalysts are in the same phase as the reaction. (See the discussion of the titanocene catalyzed polymerization of ethene discussed in the prerequisite.) Catalysts increase the rates of both the forward and reverse reactions, but they do not affect the equilibrium constant. Thus, they increase the rate at which equilibrium is established but not the amount of product that is formed. Catalysts not only increase the rate of reaction, they also lower the temperature required for reaction. Catalysts can also be very specific, increasing the rate of desired reactions without affecting the rates of undesired reactions. Enzymes are very specialized catalysts used by biological organisms.

10.6-2. Ozone's Role in the Stratosphere

High-energy ultraviolet (UV) light (E = hν) from the sun can damage living tissue because its energy is sufficient to break bonds in DNA, thus damaging genes. Fortunately, UV radiation is absorbed by ozone (O₃) in the stratosphere:

O₃ + hν → O₂ + O

O can go on to react with another O₂ to regenerate the ozone, which helps maintain the ozone level in the stratosphere. However, the oxygen atom can also react with an ozone molecule as shown below.

( Reaction 1 )

O₃ + O → 2 O₂

This reaction would deplete the ozone level, but it has a fairly high activation energy (17 kJ/mol) and few collisions in the stratosphere result in reaction. CFC's (chlorofluorocarbons), which are used as air conditioner refrigerants and in the production of plastics, are very stable and relatively unreactive. However, they absorb high-energy photons in the stratosphere to produce chlorine atoms. Consider the photochemical (caused by light) decomposition of CF₂Cl₂ (Freon-12):

CF₂Cl₂ + hν → CF₂Cl + Cl

The Cl atoms produced in reactions of this type can go on to catalyze the reaction between O and O₃. The behavior is discussed in the following section.

10.6-3. Ozone Depletion

The Cl atoms formed in the photochemical decomposition of CFC's can go on to catalyze the reaction between O and O₃ by the following two-step mechanism.

• 1
  O₃ + Cl → OCl + O₂
• 2
  OCl + O → Cl + O₂

The activation energy for Step 1 is only 2.1 kJ/mol, so it proceeds quite rapidly. The energy diagram for the reaction is given in the figure below. Summing Steps 1 and 2 yields Reaction 1 given in the previous section, with Cl forming OCl as an intermediate. Catalytic Cl lowers the activation energy for the reaction from 17 kJ/mol down to 2 kJ/mol and dramatically speeds the reaction, but it does not affect the initial or final states, so it does not affect the thermodynamics and equilibrium constant of the reaction.

10.7. Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.

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