Chapter 13 Chemical Kinetics

Daniel L. RegerScott R. GoodeDavid W. Ball

http://academic.cengage.com/chemistry/reger

Chapter 13 Chemical Kinetics

• Kinetics is the study of the rates of chemical reactions.• Rate is the change of concentration

(c) per unit time (t):

Kinetics

t

c

rate

• Square brackets are used to denote molar concentration.

• Rate is expressed either as rate of appearance of product or rate of disappearance of reactant.

• Rate has units of M/s or molar/s or mol/(L·s).

Rate of Reaction

ttt

c

reactantsproducts

Rate

Reaction Rate

• An average rate is a change in concentration measured over a non-zero time interval.

• Average rates are not very useful because they depend on the starting and ending times.

Average Rates

• The instantaneous rate of the reaction is equal to the slope of the line drawn tangent to the curve at time t.

These graphs show how to determine the instantaneous rate at 10.0 s.

Instantaneous Reaction Rate

• The relative rates of consumption of reactants and formation of products depend on the reaction stoichiometry.

• For the reaction2HBr (g) H2 (g) + Br2 (g)

two moles of HBr are consumed for every one mole of H2 that is formed so rate of change of [HBr] is double that of [H2].

Rate and Reaction Stoichiometry

tt

2HHBr

2

1

• For any reactiona A + b B c C + d D

the reaction rate is given by:

• Note the signs as well as the coefficients.

Rate and Reaction Stoichiometry

tdtc

tta

D1

C1

B

b

1A1rate

For the reaction 5H2O2 + 2MnO4

- + 6H+ → 2Mn2+ + 5O2 + 8H2Othe experimentally determined rate of disappearance of MnO4

- is 2.2 x 10-3 M/s.

(a) Calculate the reaction rate.

(b) What is the rate of appearance of O2?

Example Problem

• Experimental rate law: analysis of many experiments shows that the rate of a reaction is proportional to the product of the concentrations of the reactants raised to some power.• For a reaction aA + bB products

the rate law is the equation:rate = k[A]x[B]y

Relating Rate and Concentration

rate = k[A]x[B]y

• x and y are the orders of the reaction in [A] and [B] respectively.• The overall order of the reaction is

x + y.• x and y are usually small integers,

but may be zero, negative, or fractions.• k is the specific rate constant.

Rate Law

• The reaction orders are determined by noting the effect that changing the concentration of each reactant have on the rate.• The rate constant, k, is evaluated once

the orders in the rate law are known.


• Most often you will be given initial concentrations and rates and asked to determine the order, which is the exponent to which concentration is raised.• One way to learn to predict the order is

to predict the rate given the order and concentration and see the relationship.


Dependence of Rate on Order

[Concentration] [Rate]

First order rate law

rate = k[conc]1

1

2

3

1

2

3

Second order rate law

rate = k[conc]2

1

2

3

1

4

9

Zero order rate law rate = k[conc]0

1

2

3

1

1

1

• The initial rate method measures the time during which a known small fraction of the reactants are consumed.• Experiments are performed in which initial

concentrations of [A] and [B] are individually varied. • The time period of measurement is small

enough that the measured rate is approximately equal to the instantaneous rate.

Initial Rate Method

• Initial rates are given below for the reaction

F2 + 2ClO2 2FClO2

Determine the rate law and rate constant

Initial Rates

TrialInit. conc.

[F2], M

Init. conc.

[ClO2], MInit. Rate

Ms

1 0.10 0.010 1.2x10-3

2 0.10 0.040 4.8x10-3

3 0.20 0.010 2.4x10-3

• It is helpful to express concentrations and rates on a relative scale, by dividing the entries in each column by the smallest value.

Initial Rates

Trial [F2]Rel.

conc.[ClO2]

Rel. conc.

Initial rate

Rel. rate

1 0.10 1 0.010 1 1.2x10-3 1

2 0.10 1 0.040 4 1.8x10-3 4

3 0.20 2 0.010 1 2.4x10-3 2

• In trials 1 and 2, the concentration of F2 does not change, so the 4-fold change in [ClO2] causes the 4-fold change in rate; the order of ClO2 must be 1.

• In trials 1 and 3, [ClO2] is the same, so the doubling of rate was caused by doubling [F2]; therefore the order of F2 is also 1.

Initial Rates

• The rate law is first order in both [F2] and [ClO2].

rate = k[F2][ClO2]• Solve the equation for k, and substitute

the experimental concentrations and rates.

Initial Rates

11

3

22

s210100100

1021

ClOF

rate

MMM

sMk .

..

/.

• Write the rate law for the reaction given the following data:

2NO + 2H2 N2 + 2H2O

Test Your Skill

TrialInit conc. [NO], M

Init conc. [H2], M

Init. Rate,

MIs

1 0.00570 0.140 7.01x10-5

2 0.00570 0.280 1.40x10-4

3 0.0114 0.140 2.81x10-4

• A zero order rate law rate = k

means that the reaction rate is independent of reactant concentration.

Concentration-Time Dependence

• A plot of concentration vs. time yields a straight line

• A plot of rate vs. time yields a straight line with a slope of zero

Zero-Order Rate Laws

• First order rate law• Differential rate law

rate = k[R]• Integrated rate law


kt

kt

o

o

RInRIn

or

eRR

• A plot of concentration vs. time yields a curve.

• First order rate law: [R] = [R]o e -kt


• A plot of ln(concentration) vs. time yields a straight line.

• First order rate law: ln[R] = -kt + ln[R]o


• C12H22O11 + H2O C6H12O6 + C6H12O6 sucrose + water glucose + fructoseThe reaction is 1st order, k = 6.2 x 10-5 s-1. If [R]o = 0.40 M, what is [R] after 2 hr?

First Order Rate Law

• How long did it take for the concentration in the same experiment to drop to 0.30 M?

Test Your Skill

• Half-life, t½, is the time required for the initial concentration to decrease by ½.

Half-Life

• Half-life, t½, is the time required for the initial concentration to decrease by ½.• For a first order rate law, the half-life is

independent of the concentration.

kkkt

0.693ln2

1

1/2[R]

[R]ln

1

o

o

21

Half-Life

• k = 6.2 x 10-5 s-1 for the reaction C12H22O11 + H2O C6H12O6 + C6H12O6 Calculate the half-life.

Calculating Half-Life

• The age of objects that were once living can be found by 14C dating, because:• The concentration 14C is a constant in the

biosphere (the atmosphere and all living organisms).• When an organism dies, the 14C content

decreases with first order kinetics (t½ = 5730 years).• Scientists calculate the age of an object

from the concentration of 14C in a sample.

Radiocarbon Dating

• A sample of wood has 58% of the 14C originally present. What is the age of the wood sample?

Example: 14C Dating

• For a second order rate law rate = k[R]2

• A plot of 1/[R] vs. t is a straight line for a system described by second order kinetics.

kto[R]

1

[R]

1

Second Order Rate Law

• The reaction2NOCl 2NO + Cl2

obeys the rate lawrate = 0.020 M-1s-1 [NOCl]2

Calculate the concentration of NOCl after 30 minutes, when the initial concentration was 0.050 M.

Example: Second Order Rate Law

Given the experimental data for the decomposition of 1,3-pentadiene shown below, determine the order of the reaction.

Example: Order of Reaction

Example: Order of Reaction (cont.)

Review table 13.4 on page 534.

• Reactions proceed at faster rates at higher temperatures.

Influence of Temperature on k

• The reaction rate is proportional to the collision frequency, Z, the number of molecular collisions per second.• Z depends on the temperature and the

concentration of the colliding molecules.• Not all molecular collisions result in the

formation of products.

Collision Theory

• Activation energy (Ea): the minimum collision energy required for reaction to occur.• Activated complex: the highest energy

arrangement of atoms that occurs in the course of the reaction.

Collision Theory

NO + O3 [activated complex]* NO2 + O2

The Activated Complex

• The fraction of collisions with energy in excess of Ea is given by:

• The collision frequency is proportional to the concentrations of colliding species.

• The reaction rate is proportional to the rate of collisions time the fraction of collisions with energy in excess of Ea.

rate = Z × fr

Influence of Temperature on Kinetic Energy

RTEaf /er

Orientation of Reactants

• The steric factor, p, is a number between 0 and 1 that is needed to account for factors other than energy before a reaction can occur.• The reaction rate is proportional to the

steric factor times the collision frequency times the fraction of collisions with energy in excess of Ea:

rate = p x Z x fr

The Steric Factor

rate = p x Z x fr

Combine p and Zo into a term A:

Experiments show that

rate = k[colliding species], so

The Arrhenius Equation

RTEaep / species collidingZrate o

RTEaeA / species collidingrate

RTEaAek /

• Take the natural log of both sides of the equation:

• A plot of ln k vs. 1/T gives a straight line with a slope of -Ea/R and an intercept of ln A.

The Arrhenius EquationRTEaAek /

TR

EAk a 1

InIn

• Determine Ea for the reaction, 2NO2 2NO + O2

given the data:

Measuring Activation Energy

k (M-1·s-1) T (°C) In k 1/T (K-1)

0.003 500 -5.8 2.00x10-3

0.037 550 -3.30 1.82x10-3

0.291 500 -1.234 1.67x10-3

1.66 650 0.507 1.54x10-3

7.39 700 2.000 1.43x10-3

• Prepare a plot of ln k vs. 1/T

Solution

kJ 10141

J 10141

314810371

/slope

101.37slope

2

5

4

4

.

.

..

a

a

a

E

E

RE

• The rate of a reaction exactly doubles, when the temperature is changed from 25.0o C to 36.2o C. Calculate the activation energy for this reaction.

Example: Arrhenius Equation

• A catalyst is a substance that increases the reaction rate but is not consumed in the reaction.• A catalyst provides an alternate

reaction path with a lower activation energy.

Catalysis

• A homogeneous catalyst is one that is present in the same phase as the reactants.

• Bromide ion is a homogeneous catalyst for the decomposition of hydrogen peroxide.

2H2O2(aq) 2H2O(l) + O2(g)

• step 1:

H2O2(aq) + 2Br-(aq) + 2H+(aq) Br2(aq) + 2H2O(l)

• step 2:H2O2(aq) + Br2(aq) 2Br-(aq) + 2H+(aq) + O2(g)

Homogeneous Catalysis

• A heterogeneous catalyst is one that is present in a different phase from the reactants.• The gas phase reaction of hydrogen with

many organic compounds is catalyzed by solid platinum.

Heterogeneous Catalysis

• Enzymes are large molecules (macromolecules) which catalyze specific biochemical reactions.• Enzymes can increase the rates of

reactions by factors as large as 1014.• Enzymes are very specific in the

reactions they catalyze.• Enzymes are active under mild reaction

conditions.

Enzyme Catalysis

• A mechanism is a sequence of molecular-level steps that lead from reactants to products.• An elementary step is an equation that

describes an actual molecular level event.• The concentration dependence in the

rate law for an elementary step is given by the coefficients in the equation.

Reaction Mechanisms

• The molecularity of an elementary step is the number of reactant species involved in that step.• Most elementary steps are either

unimolecular (involving a single molecule) or bimolecular (collision of two species).

Molecularity

• The reaction2NO + O2 2NO2

is believed to occur by the following sequence of elementary steps:

2NO N2O2 bimolecular reactionN2O2 + O2 2NO2 bimolecular reaction

Elementary Steps

• The rate of an elementary step is proportional to the concentration of each reactant species raised to the power of its coefficient in the equation:• step 1: 2NO N2O2

rate1 = k1[NO]2

• step 2: N2O2 + O2 2NO2

rate2 = k2[N2O2][O2]

Rate Laws for Elementary Reactions

• Write the rate law for the elementary step

H2 + Cl H2Cl

Test Your Skill

• The overall rate of a multistep reaction is determined by its slowest step, called the rate-limiting step.• The rates of fast steps which follow the rate-

limiting step have no effect on the overall rate law.• The rates of fast steps that precede the

rate-limiting step usually affect the concentrations of the reactant species in the rate-determining step.

Rate-Limiting Steps

Complex Reaction Mechanisms• R P (2 steps)• R intermediates rate1 = k1[R]

• Intermediates P rate2 = k2[intermediates]

• If step 1 is slow, then it determines rate• If step 2 is slow, how can we measure

[intermediates]?• Many fast steps prior to slow step are fast

and reversible

Complex Reaction Mechanisms

• Fast reversible steps help with [intermediate]

• Consider the following reaction2NO + 2H2 N2 + 2H2O

• rate = k[NO]2[H2]

• step 1: 2NO N2O2 fast and reversible

• step 2: N2O2 + H2 N2O + H2O slow

• step 3: N2O + H2 N2 + H2O fast

Work on the board

• Consider the two-step reaction2NO + O2 2NO2

• step 1: 2NO N2O2

rate1 = k1[NO]2

• step 2: N2O2 + O2 2NO2

rate2 = k2[N2O2][O2]


• If the first step is the rate-limiting step, the rate law is:

rate = k1[NO]2

• If the first step is rapid and the second step is the rate limiting step, the rate law is:

rate = k2[N2O2][O2]


If the first step reaches equilibriumrate1 (forward) = rate-1 (reverse):

2

1

122

2212

1

1-1

2211

211

[NO]]O[N

]O[N[NO]

raterate

:equal are mequilibriu at rates the Because

]O[Nrate

[NO]rate

k

k

kk

k

k


• Substituting into the expression for step 2:rate = k2[N2O2][O2]

rate = k2 [NO]2[O2]

Combine all the rate constants:

• rate = k [NO]2[O2]

• The reaction is second order in NO and first order in O2.

k1

k -1


• For the reaction2NO2 + O3 N2O5 + O2

the experimentally determined rate law is rate= k[NO2][O3].

• Identify the rate limiting step in the proposed two-step mechanism:

NO2 + O3 NO3 + O2 step 1NO3 + NO2 N2O5 step 2

Test Your Skill

• H2 + I2 2HI rate = k[H2][I2]

• For many years this reaction was believed to occur as a single bimolecular step.

• From more recent data, a very different mechanism is likely. I2 2I Fast, reversible I + H2 H2I Fast, reversibleH2I + I 2HI Slow

• Both mechanisms give the same rate law.

The Hydrogen-Iodine Reaction

• Most enzymes follow the Michaelis-Menten mechanism.• You will see this again in

biochemistry!!

Enzyme Catalysis

Documents

Chapter 13 Chemical Kinetics