16-1

Chapter 4

Kinetics: Rates and Mechanisms

of Chemical Reactions

16-2

Kinetics: Rates and Mechanisms of Chemical Reactions

16.1 Factors That Influence Reaction Rate

16.2 Expressing the Reaction Rate

16.3 The Rate Law and Its Components

16.4 Integrated Rate Laws: Concentration Changes over Time

16.7 Reaction Mechanisms: Steps in the Overall Reaction

16.8 Catalysis: Speeding Up a Chemical Reaction

16.5 The Effect of Temperature on Reaction Rate

16.6 Explaining the Effects of Concentration and Temperature

16-3

Figure 16.1 Reaction rate: the central focus of chemical kinetics.

16-4

Figure 16.2 The wide range of reaction rates.

16-5

Factors That Influence Reaction Rate

Under a specific set of conditions, every reaction has its own

characteristic rate, which depends upon the chemical nature of

the reactants.

Four factors can be controlled during the reaction:

1. Concentration - molecules must collide to react;

2. Physical state - molecules must mix to collide;

3. Temperature - molecules must collide with enough energy to react;

4. The use of a catalyst.

16-6

Figure 16.3 The effect of surface area on reaction rate.

16-7

Figure 16.4 Collision energy and reaction rate.

16-8

Expressing the Reaction Rate

reaction rate - changes in the concentrations of reactants or

products per unit time

reactant concentrations decrease while product concentrations

increase

Rate of reaction = -

for A B

change in concentration of A

change in time

= -conc A2 - conc A1

t2 - t1

(conc A)-

t

16-9

Table 16.1 Concentration of O3 at Various Times in its

Reaction with C2H4 at 303 K

C2H4(g) + O3(g) C2H4O(g) + O2(g)

Time (s) Concentration of O3 (mol/L)

0.0

20.0

30.0

40.0

50.0

60.0

10.0

3.20x10-5

2.42x10-5

1.95x10-5

1.63x10-5

1.40x10-5

1.23x10-5

1.10x10-5

(conc A)-

t

16-10

Figure 16.5

The concentration of O3 vs. time during its reaction with C2H4.

C2H4(g) + O3(g) C2H4O(g) + O2(g)

- [C2H4]

t

rate =

- [O3]

t

=

16-11

Figure 16.6 Plots of [C2H4] and [O2] vs. time.

Tools of the

Laboratory

16-12

In general, for the reaction

aA + bB cC + dD

rate = 1

a- = -

[A]

t

1

b

[B]

t

1

c

[C]

t= +

1

d

[D]

t= +

The numerical value of the rate depends upon the substance that

serves as the reference. The rest is relative to the balanced

chemical equation.

16-13

Sample Problem 16.1

PLAN:

SOLUTION:

Expressing Rate in Terms of Changes in

Concentration with Time

PROBLEM: Because it has a nonpolluting product (water vapor), hydrogen

gas is used for fuel aboard the space shuttle and may be used

by earthbound engines in the near future.

2H2(g) + O2(g) 2H2O(g)

(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.

(b) When [O2] is decreasing at 0.23 mol/L*s, at what rate is [H2O]

increasing?

Choose [O2] as a point of reference since its coefficient is 1. For every

molecule of O2 which disappears, 2 molecules of H2 disappear and 2

molecules of H2O appear, so [O2] is disappearing at half the rate of

change of H2 and H2O.

-1

2

[H2]

t= -

[O2]

t= +

[H2O]

t

1

2

0.23 mol/L*s = +[H2O]

t

1

2= 0.46 mol/L*s

[H2O]

t

rate =(a)

[O2]

t- = -(b)

16-14

Sample Problem 16.2

SOLUTION:

Determining Reaction Order from Rate Laws

PROBLEM: For each of the following reactions, determine the reaction order

with respect to each reactant and the overall order from the

given rate law.

(a) 2NO(g) + O2(g) 2NO2(g); rate = k[NO]2[O2]

(b) CH3CHO(g) CH4(g) + CO(g); rate = k[CH3CHO]3/2

(c) H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) + 2H2O(l); rate = k[H2O2][I

-]

PLAN: Look at the rate law and not the coefficients of the chemical reaction.

(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.

(b) The reaction is 3/2 order in CH3CHO and 3/2 order overall.

(c) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+,

while being 2nd order overall.

16-15

Table 16.2 Initial Rates for a Series of Experiments in the

Reaction Between O2 and NO

Experiment

Initial Reactant

Concentrations (mol/L) Initial Rate

(mol/L*s)

1

2

3

4

5

O2 NO

1.10x10-2 1.30x10-2 3.21x10-3

1.10x10-2 3.90x10-2 28.8x10-3

2.20x10-2

1.10x10-2

3.30x10-2

1.30x10-2

2.60x10-2

1.30x10-2

6.40x10-3

12.8x10-3

9.60x10-3

2NO(g) + O2(g) 2NO2(g)

16-16

Determining Reaction Orders

Using initial rates -

Run a series of experiments, each of which starts with a different

set of reactant concentrations, and from each obtain an initial rate.

See Table 16.2 for data on the reaction

O2(g) + 2NO(g) 2NO2(g) rate = k [O2]m[NO]n

Compare 2 experiments in which the concentration of one reactant

varies and the concentration of the other reactant(s) remains constant.

k [O2]2m[NO]2

n

k [O2]1m[NO]1

n=

rate 2

rate 1 =

[O2]2m

[O2]1m

=

6.40x10-3 mol/L*s

3.21x10-3 mol/L*s

[O2]2

[O2]1

m

=1.10x10-2 mol/L

2.20x10-2 mol/Lm

; 2 = 2m m = 1

Do a similar calculation for the other reactant(s).

16-17

Sample Problem 16.3

PLAN:

SOLUTION:

Determining Reaction Orders from Initial Rate Data

PROBLEM: Many gaseous reactions occur in a car engine and exhaust

system. One of these is

NO2(g) + CO(g) NO(g) + CO2(g) rate = k[NO2]m[CO]n

Use the following data to determine the individual and overall reaction orders.

Experiment Initial Rate (mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L)

1

2

3

0.0050

0.080

0.0050

0.10

0.10

0.40

0.10

0.10

0.20

Solve for each reactant using the general rate law using the

method described previously.

rate = k [NO2]m[CO]n

First, choose two experiments in which [CO] remains

constant and the [NO2] varies.

16-18

Sample Problem 16.3 Determining Reaction Orders from Initial Rate Data

0.080

0.0050

rate 2

rate 1

[NO2] 2

[NO2] 1

m

=k [NO2]

m2[CO]n2

k [NO2]m

1 [CO]n1

=

0.40

0.10=

m

; 16 = 4m and m = 2

k [NO2]m

3[CO]n3

k [NO2]m

1 [CO]n1

[CO] 3

[CO] 1

n

=rate 3

rate 1=

0.0050

0.0050=

0.20

0.10

n

; 1 = 2n and n = 0

The reaction is

2nd order in NO2.

The reaction is

zero order in CO.

rate = k [NO2]2[CO]0 = k [NO2]

2

16-19

Table 16.3 Units of the Rate Constant k for Several Overall

Reaction Orders

Overall Reaction Order Units of k (t in seconds)

0 mol/L*s (or mol L-1 s-1)

1 1/s (or s-1)

2 L/mol*s (or L mol -1 s-1)

3 L2 / mol2 *s (or L2 mol-2 s-1)

16-20

Integrated Rate Laws

rate = -[A]

t= k [A]

rate = -[A]

t= k [A]0

rate = -[A]

t= k [A]2

first order rate equation

second order rate equation

zero order rate equation

ln[A]0

[A]t

= kt ln [A]0 = kt + ln [A]t

1

[A]t

1

[A]0- = kt

1

[A]t

1

[A]0+= kt

[A]t - [A]0 = - kt

16-21

Sample Problem 16.5

PLAN:

SOLUTION:

Determining the Reactant Concentration at a

Given Time

PROBLEM: At 1000oC, cyclobutane (C4H8) decomposes in a first-order

reaction, with the very high rate constant of 87 s-1, to two

molecules of ethylene (C2H4).

(a) If the initial C4H8 concentration is 2.00 M, what is the

concentration after 0.010 s?

(b) What fraction of C4H8 has decomposed in this time?

Find the [C4H8] at time, t, using the integrated rate law for a 1st order

reaction. Once that value is found, divide the amount decomposed by

the initial concentration.

; ln2.00

[C4H8]= (87 s-1)(0.010 s)

[C4H8] = 0.83 mol/L

ln[C4H8]0

[C4H8]t= kt

(a)

(b) [C4H8]0 - [C4H8]t

[C4H8]0

=2.00 M - 0.87 M

2.00 M= 0.58

16-22

Figure 16.7 Integrated rate laws and reaction orders.

ln[A]t = -kt + ln[A]0

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

[A]t = -kt + [A]0

16-23

Figure 16.8 Graphical determination of the reaction order for the

decomposition of N2O5.

16-24

Figure 16.9 A plot of [N2O5] vs. time for three half-lives.

t1/2 =

for a first-order process

ln 2

k

0.693

k=

16-25

Sample Problem 16.6

PLAN:

SOLUTION:

Using Molecular Scenes to Determine Half-Life

PROBLEM: The 1st order process of compound A (red) converting to

compound B (black) is depicted at 0.0 s and 30.0 s:

(a) Find the half-life, t1/2, of the reaction. (b) Calculate the rate constant, k.

(c) Draw a scene that represents the reaction mixture at 2.00 min.

Number of spheres represents concentration. Half-life is constant for

a 1st order reaction and the elapsed time when half the red spheres

turn black.

(a) At t = 0, 8 A and 0 B. At t = 30.0 s, 6 A and 2 B. For 4 A and

4 B, t = 60.0 s = t1/2.

(b) t1/2 = 0.693/k; k = 0.693/t1/2 = 0.693/60.0 s = 1.16 x 10-2 s-1

(c) The 2.00 min is 120. s or two half-lives represented by the

above scene.

(c)

16-26

Sample Problem 16.7

PLAN:

SOLUTION:

Determining the Half-Life of a First-Order Reaction

PROBLEM: Cyclopropane is the smallest cyclic hydrocarbon. Because its

60o bond angles allow poor orbital overlap, its bonds are weak.

As a result, it is thermally unstable and rearranges to propene at

1000oC via the following first-order reaction:

CH2

H2C CH2(g)

H3C CH CH2 (g)

The rate constant is 9.2 s-1, (a) What is the half-life of the reaction? (b) How

long does it take for the concentration of cyclopropane to reach one-quarter of

the initial value?

Use the half-life equation, t1/2 = 0.693

k, to find the half-life.

One-quarter of the initial value means two half-lives have passed.

t1/2 = 0.693/9.2 s-1 = 0.075 s(a) 2 t1/2 = 2(0.075 s) = 0.150 s(b)

16-27

Table 16.4 An Overview of Zero-Order, First-Order, and

Simple Second-Order Reactions

Zero Order First Order Second Order

Plot for straight line

Slope, y intercept

Half-life

Rate law rate = k rate = k[A] rate = k[A]2

Units for k mol/L*s 1/s L/mol*s

Integrated rate law in

straight-line form[A]t = -kt + [A]0 ln [A]t = -kt + ln [A]0 1/[A]t = kt + 1/[A]0

[A]t vs. t ln [A]t vs. t 1/[A]t = t

k, [A]0 -k, ln [A]0k, 1/[A]0

[A]0/2k ln 2/k 1/k[A]0

16-28

Figure 16.10 Dependence of the rate constant on temperature.

16-29

The Arrhenius Equation

The Effect of Temperature on Reaction Rate

RTEek

/aA

lnk = lnA - Ea/RT

lnk2

k1

=Ea

R-

1

T2

1

T1

-

where k is the kinetic rate constant at T

Ea is the activation energy

R is the energy gas constant

T is the Kelvin temperature

A is the collision frequency factor

16-30

Figure 16.11 Graphical determination of the activation energy.

ln k = (-Ea/R )(1/T) + lnA

16-31

Sample Problem 16.8

PLAN:

SOLUTION:

Determining the Energy of Activation

PROBLEM: The decomposition of hydrogen iodide,

2HI(g) H2(g) + I2(g)

has rate constants of 9.51x10-9 L/mol*s at 500. K and 1.10x10-5

L/mol*s at 600. K. Find Ea.

Use the modification of the Arrhenius equation to find Ea.

lnk2

k1

=Ea

-R

1

T2

1

T1

- Ea = - R lnk2

k1

1

T2

1

T1

-

-1

1

600 K

1

500 K-ln

1.10x10-5 L/mol*s

9..51x10-9 L/mol*s

Ea = - (8.314 J/mol*K)

Ea = 1.76x105 J/mol = 176 kJ/mol

16-32

Figure 16.12

Information sequence to determine the kinetic parameters of a reaction.

Series of plots

of concentra-

tion vs. time Initial

rates Reaction

ordersRate constant

(k) and actual

rate law

Integrated

rate law

(half-life,

t1/2)

Rate constant

and reaction

order

Activation

energy, Ea

Plots of

concentration

vs. time

Find k at

varied T

Determine slope

of tangent at t0 for

each plot

Compare initial

rates when [A]

changes and [B] is

held constant and

vice versa

Substitute initial rates,

orders, and concentrations

into general rate law:

rate = k [A]m[B]n

Use direct, ln or

inverse plot to

find order

Rearrange to

linear form and

graph

Find k at

varied T

16-33

Figure 16.13 The dependence of number of possible collisions

on the product of reactant concentrations.

A

A

B

B

A

A

B

B

A

4 collisions

Add another

molecule of A6 collisions

Add another

molecule of B

A

A

B

B

A B

Why concentrations Are

Multiplied in the Rate Law ?

16-34

Figure 16.14

The effect of temperature on the distribution of collision

energies.Energy required to

activate the molecules

into a state from which

reactant bonds can

change into product

bonds

The temperature rise

enlarges the fraction of

collisions with enough

energy to exceed the Ea

16-35

Table 16.5 The Effect of Ea and T on the Fraction (f) of Collisions

with Sufficient Energy to Allow Reaction

Ea (kJ/mol) f (at T = 298 K)

50 1.70x10-9

75 7.03x10-14

100 2.90x10-18

T f (at Ea = 50 kJ/mol)

25oC (298 K) 1.70x10-9

35oC (308 K) 3.29x10-9

45oC (318 K) 6.12x10-9

Increasing Ea

by 25 kJ/mol Decreased

16-36

Figure 16.15 Energy-level diagram for a reaction.

REACTANTS

PRODUCTS

ACTIVATED STATE

Colli

sio

n E

ne

rgy

Colli

sio

n E

ne

rgy

Ea (forward)

Ea (reverse)

The forward reaction is exothermic because the

reactants have more energy than the products.

16-37

Figure 16.16 An energy-level diagram of the fraction of collisions

exceeding Ea.

LargerSmaller

16-38

• In both reaction direction, a larger fraction of collision

exceeds the activation energy at the higher temperature, T2;

higher T increase reaction rate

16-39

Figure 16.17

The importance of molecular orientation to an effective collision.

NO + NO3 2 NO2

A is the frequency factor

A = pZ where Z is the collision frequency

p is the orientation probability factor

16-40

Figure 16.18

Nature of the transition state in the reaction between CH3Br and OH-.

CH3Br + OH- CH3OH + Br -

transition state or activated complex

16-41

Figure 16.19 Reaction energy diagram for the reaction of CH3Br and OH-.

Transition state/activated complex

Extremely unstable species

(has very high potential energy)

Forms only if the molecules collide

In an effective orientation and with

energy than Ea

Thus, the Ea is the quantity needed

To stretch and deform bonds in

order to reach the transition state

16-42Reaction progress

Po

ten

tia

l E

ne

rgy

Sample Problem 16.9

SOLUTION:

Drawing Reaction Energy Diagrams and

Transition States

PROBLEM: A key reaction in the upper atmosphere is

O3(g) + O(g) 2O2(g)

The Ea(fwd) is 19 kJ, and the Hrxn for the reaction is -392 kJ. Draw a

reaction energy diagram for this reaction, postulate a transition state, and

calculate Ea(rev).

PLAN: Consider the relationships among the reactants, products and

transition state. The reactants are at a higher energy level than the

products and the transition state is slightly higher than the

reactants.

O3+O

2O2

Ea= 19 kJ

Hrxn = -392 kJ

Ea(rev)= (392 + 19) kJ =

411kJ

OO

OO

breakingbond

formingbond

transition state

16-43

Table 16.6 Rate Laws for General Elementary Steps

Elementary Step Molecularity Rate Law

A product

2A product

A + B product

2A + B product

Unimolecular

Bimolecular

Bimolecular

Termolecular

Rate = [A]

Rate = k[A]2

Rate = k[A][B]

Rate = k[A]2[B]

REACTION MECHANISMS

16-44

Sample Problem 16.10

PLAN:

SOLUTION:

Determining Molecularity and Rate Laws for

Elementary Steps

PROBLEM: The following two reactions are proposed as elementary steps in

the mechanism of an overall reaction:

(1) NO2Cl(g) NO2(g) + Cl(g)

(2) NO2Cl(g) + Cl(g) NO2(g) + Cl2(g)

(a) Write the overall balanced equation.

(b) Determine the molecularity of each step.

(a) The overall equation is the sum of the steps.

(b) The molecularity is the sum of the reactant particles in the step.

2NO2Cl(g) 2NO2(g) + Cl2(g)

(c) Write the rate law for each step.

rate2 = k2[NO2Cl][Cl]

(1) NO2Cl(g) NO2(g) + Cl (g)

(2) NO2Cl(g) + Cl (g) NO2(g) + Cl2(g)

(a)

Step(1) is unimolecular.

Step(2) is bimolecular.

(b)

rate1 = k1[NO2Cl](c)

16-45

The Rate-Determining Step of a Reaction Mechanism

The overall rate of a reaction is related to the rate of the slowest, or

rate-determining step.

Correlating the Mechanism with the Rate Law

The elementary steps must add up to the overall equation.

The elementary steps must be physically reasonable.

The mechanism must correlate with the rate law.

16-46

Figure 16.21

Reaction energy diagram for the two-step reaction of NO2 and F2.

2NO2(g) + F2(g) 2NO2F(g)

Accepted mechanism

NO2(g) + F2(g) NO2F(g)

+ F(g) [slow]

NO2(g) + F(g) NO2F(g)

NO2(g) + NO2(g) + F2(g) +

F(g) NO2F(g) + NO2F(g)

+ F(g)

16-47

CATALYSTS

•Each catalyst has its own specific way of functioning.

•In general a catalyst lowers the energy of activation.

•Lowering the Ea increases the rate constant, k, and

thereby increases the rate of the reaction.

•A catalyst increases the rate of the forward AND the

reverse reactions.

•A catalyzed reaction yields the products more quickly,

but does not yield more product than the uncatalyzed

reaction.

•A catalyst lowers Ea by providing a different mechanism,

for the reaction through a new, lower energy pathway.

16-48

Figure 16.22

Reaction energy diagram of a catalyzed and an uncatalyzed process.

16-49

Figure 16.23

Mechanism for the catalyzed hydrolysis of an organic ester.

H++ R C

O

O

R'

R C

O

O

R'

Hfast R C

O

O

R'

H

R C

O

O

R'

H

R C

O

O

R'

H

resonance hybrid

resonance forms

R C

O

O

R'

H

H

O H R C

O

O

R'

H

H

O H

slow, rate-determining

step

R C

O

OH

R'

H+

O Hall fast

16-50

16-51

Figure 16.24 The metal-catalyzed hydrogenation of ethene.

H2C CH2 (g) + H2 (g) H3C CH3 (g)

16-52

Figure 16.24 The metal-catalyzed hydrogenation of ethene.

16-53

Tools of the Laboratory

Figure B16.1 Spectrometric monitoring of a reaction.

16-54

Figure B16.2

Figure B16.3

Conductometric monitoring

of a reaction.

Manometric monitoring of a

reaction.

Tools of the Laboratory

16-55

Figure B16.4

The widely separated amino acid groups that form an active site.

16-56

Figure B16.5 Two models of enzyme action.

16-57

Figure B16.6 The increasing size of the Antarctic ozone hole.

O2 2O

O + O3 2O2 (ozone breakdown)

O + O2 O3 (ozone formation)