Chapter 13. Chemical Kinetics

• Why do some chemical reactions proceed with lighting speed when others require days, months, or even years to produce detectable amounts of products?

• How do catalysts increase the rates of chemical reactions?• Why do small changes in temperature often have such

large effects on the cooking rate of food?• How does a study of the rate of a chemical reaction inform

us about the way in which molecules combine to form products?

• All of these questions involve chemical kinetics.

• There are many reactions whose equilibrium constants are known accurately, but whose rates and detailed reaction pathways remain poorly understood.

13.1 Rates of chemical Reactions• Measuring Reaction Rates

13.2 Rate Laws

Order of a Reaction

The order of a reaction does not have to be an integer; fractional powers are sometimes found.

• Example 13.1At elevated temperatures, HI reacts according to the chemical equation

2 HI(g) H2(g) + I2(g)At 443 C, the rate of the reaction increases with concentration of HI as

follows:[HI] (mol/L) 0.0050 0.010 0.020Rate (mol/L/s) 7.5x10-4 3.0x10-3 1.2x10-2

(a) Determine the order of the reaction and write the rate expression.(b) Calculate the rate constant, and give its units.(c) Calculate the reaction rate for a 0.0020 M concentration of HI.

• Overall reaction order

• Example 13.2Use the preceding rate expression to determine the effect of the following

changes on the rate of decomposition of H2PO2-(aq):

(a) Tripling the concentration of H2PO2-(aq) at constant pH.

(b) Changing the pH from 13 to 14 at a constant concentration of H2PO2

-(aq).

• Rate expressions that depend on more than one concentrationExample 13.3The reaction of NO(g) with O2(g) gives NO2(g):

2 NO(g) + O2(g) 2 NO2(g)From the dependence of the initial rate(-1/2d[NO]/dt) on the initial

concentration of NO and O2, determine the rate expression and the value of the rate constant.

[NO] (mol L-1) [O2] (mol L-1) Initial Rate (mol L-1 s-1)1.0 x 10-4 1.0 x 10-4 2.8 x 10-6

1.0 x 10-4 3.0 x 10-4 8.4 x 10-6

2.0 x 10-4 3.0 x 10-4 3.4 x 10-5

Integrated Rate Laws• Differential rate laws• Integrated rate laws: expresses the concentration of a

species directly as a function of the time.

First-Order Reactions

half-life: the time it takes for the original concentration to be reduce to half its value.

• Example 13.4(a) What is the rate constant k for the first-order decomposition of

N2O5(g) at 25 C if the half-life of N2O5(g) is 4.03 x 104 s?(b) What percentage of the N2O5 molecules will not have reacted after

one day?

• Second-Order Reactions

• Example 13.5The dimerization of tetrafluoroethylene (C2F4) to C4F8 is second order in the

reactant C2F4, and at 450 K its rate constant is k = 0.0448 L mol-1s-1. If the initial concentration of C2F4 is 0.100 mol L-1, what will its concentration be after 205 s?

13.3 Reaction Mechanisms• Many reaction do not occur in a single step, but rather

proceed through a series of steps. Each step is called an elementary reaction.

• Elementary reaction is directly caused by the collisions of atoms, ions, or molecules.

• The rate expression for an overall reaction cannot be derived from the stoichiometry of the balanced equation.

• On the other hand, the rate of an elementary reaction is directly proportional to the product of the concentrations of the reaction species, each raised to a power equal to its coefficient in the balanced elementary reaction.

Elementary ReactionsUnimolecular

Bimolecular

• Termolecular

Reaction Mechanisms• A reaction mechanism is detailed series of elementary

reactions, with their rates, that are combined to yield the overall reaction.

NO2 + CO NO + CO2

A mechanism:NO2 + NO2 NO3 + NO (slow)NO3 + CO NO2 + CO2 (fast)

Adding twoNO2 + NO2 + NO3 + CO NO3 + NO + NO2 + CO2

NO2 + CO NO + CO2

A reaction intermediate is a chemical species that is formed and consumed in the reaction but does not appear in the overall balanced chemical equaiton.

Here, ( ) is the reaction intermediate.

• Example 13.6Consider the following reaction mechanism:

Cl2 2 ClCl + CHCl3 HCl + CCl3

CCl3 + Cl CCl4

(a) What is the molecularity of each elementary step?(b) Write the overall equation for the reaction.(c) Identify the reaction intermediate(s).

Kinetics and Chemical Equilibrium

• The principle of detailed balance: at equilibrium, the rate of each elementary process is balanced by (equal to) the rate of its reverse process.

13.4 Reaction Mechanisms and Rate• In many reaction mechanisms, one step is significantly slower than all

the others; this step is called the rate-determining step.

• Example 13.7In basis aqueous solution the reaction

I- + OCl- Cl- + OI-

Follows a rate law that is consistent with the following mechanism:OCl-(aq) + H2O(l) HOCl(aq) + OH-(aq) (fast equilibrium)

I-(aq) + HOCl(aq) HOI(aq) + Cl-(aq) (slow)

OH-(aq) + HOI(aq) H2O(l) + OI(aq) (fast)

What rate law is predicted by this mechanism?

k1

k-1

k2

k3

• The rate law of the foregoing example showed an inverse dependence on the concentration of OH- ion. Such a form is often a clue that a rapid equilibrium occurs in the first step of a reaction, preceding the rate-determining step.

• Fractional orders of reaction provide a similar clue, as in the reaction of H2 and Br2 to form HBr.

H2 + Br2 2 HBrrate = kobs[H2][Br2]1/2

One reaction mechanism

Br2 + M Br + Br + M (fast equilibrium)

Br + H2 HBr + M (slow)

H + Br2 HBr + Br (fast)

k1

k-1

k2

k3

• Deducing a rate law from a proposed mechanism is relatively straightforward, but doing the reverse is much harder.

• In fact, several competing mechanisms often give rise to the same rate law, and only some independent type of measurement can help one to choose between them.

• A reaction mechanism can never be proved from an experimental rate law; it can only be disproved if it is inconsistent with the experimental behavior.

H2 + I2 2 HIrrate = kobs[H2][I2]

Sullivan: light : dramatic increase in the rate…..; inconsistent with the one-step mechanism.

• One such mechanism which can explain the observation…

The Steady-State Approximation• In some reaction mechanisms there is no single rate-determining step,

and the methods discussed so far cannot be used to predict the rate law.• In such cases, the steady-state approximation is helpful, in which it is

assumed that the concentration of reactive intermediates remain nearly constant through most of the reaction.

N2O5 + M N2O5* + M

N2O5* NO3 + NO2

NO3 + NO2 NO + NO2 + O2 (fast)

NO3 + NO 2 NO2 (fast)

k1

k-1

k2

k3

k4

This expression has two limiting cases:1. Low pressure: When [M] is small enough, k2 >> k-1[M] and we can

use the approximationrate = k1[N2O5][M]

This same result would be found by assuming the first step to berate-determining.

2. High pressure: When [M] is large enough, k-1[M] >> k2 and we can use the approximation

rate = (k1/k-1)k2[N2O5]This same result would be found by assuming the second step to be rate-determining.

Chain Reactions• A chain reaction is one that proceeds through a series of elementary

steps, some of which are repeated many times.

CH4(g) + F2(g) CH3F(g) + HF(g)

CH4 + F2 CH3 + HF + F (initiation)CH3 + F2 CH3F + F (propagation)CH4 + F CH3 + HF (propagation)CH3 + F + M CH3F + M (termination)

Branching chain reaction• The number of reaction intermediates keep increasing.

13.5 Effect of Temperature on Reaction Rates

• To consider the magnitude of the rate constant and the effect of temperature, it is necessary to establish the connection between collision rates and the rates of chemical reactions.

• Only gases, for which the kinetic theory of Chapter 4 is applicable, are considered here.

Gas-Phase Reaction Rate Constants• Collision frequency (chapter 4): ~109 s-1

• Time per collision: ~10-9 s• If every collision led to reaction, the reaction would be complete in

~10-9 s.Some reaction do proceed at almost this rate.An example: 2 CH3 C2H6

for which the observed rate constant is 1 x 106 L mol-1 s-1.If the initial pressure of CH3 were near 1 atm, then at 25 C the

concentration initially would be about 0.04 M. According to the second-order integrated rate law from Section 13.2, after a period of 10-9 s the concentration would have dropped to 0.02 M.

Much more commonly, however, reactions proceed at rates that are far lower-by factors of 1012 or more.

In 1889, Svante Arrhenius:

• The decomposition of hydroxylamine (NH2OH) in the presence of oxygen follows the rate law

-d[NH2OH]/dt = kobs[NH2OH][O2]Where kobs is 0.237 x 10-4 L mol-1 s-1 at 0 C and 2.64 x 10-4 L

mol-1 s-1 at 25 C. Calculate Ea and the factor A for this reaction.

Activation energy

The Reaction Coordinate and the Activated Complex

NO2(g) + CO(g) NO(g) + CO2(g)

• Rates of elementary reactions therefore increase with increasing temperature. This is not necessarily true for rates of overall reactions consisting of more than one elementary reaction.

• These sometimes have “negative activation energies”, meaning that the overall reaction rate slows at higher temperature.

2 NO(g) + O2(g) 2 NO2(g)rate = kobs[NO]2[O2]

NO + NO N2O2 (fast equilibrium)N2O2 + O2 2 NO2 (slow)

k1

k-1k2

k-2

13.7 Kinetics of Catalysis• A catalyst is a substance that takes part in a chemical

reaction and speeds it up but undergoes no permanent chemical change itself.

• Catalysis can be classified into two types: homogeneous and heterogeneous.

Homogeneous catalysisTl+(aq) + 2 Ce4+(aq) Tl3+(aq) + 2 Ce3+(aq)

Ag+ + Ce4+ Ag2+ + Ce3+ (fast)

Tl+ + Ag2+ Tl2+ + Ag+ (slow)

Tl2+ + Ce2+ Tl3+ + Ce3+ (fast)

k1

k-1

k2

k3

Heterogenous catalysisC2H4(g) + H2(g) C2H6(g)

• It is important to remember that a catalyst has no effect on the thermodynamics of the overall reaction.

• An inhibitor plays an opposite role to that of a catalyst.

Enzyme Catalysis

E + S ESES E + P

k1

k-1k2

• Steady State Approximation

• Michaelis-Menten equation.

When [S] is small, the more substrate is added, the greater is the rate of formation of product

As the amount of substrate becomes large, all the active sites of the enzyme molecules become bound to substrate and further addition of S does not affect the rate of the reaction.