General Chemistry

Principles and Modern Applications Petrucci Harwood Herring 8th Edition

Chapter 15: Chemical Kinetics

Philip Dutton University of Windsor, Canada N9B 3P4
Prentice-Hall 2002

Contents
15-1 15-2 15-3 15-4 15-5 15-6 15-7 The Rate of a Chemical Reaction Measuring Reaction Rates Effect of Concentration on Reaction Rates: The Rate Law Zero-Order Reactions First-Order Reactions Second-Order Reactions Reaction Kinetics: A Summary

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 2 of 55

Contents
15-8 15-9 15-10 15-11 Theoretical Models for Chemical Kinetics The Effect of Temperature on Reaction Rates Reaction Mechanisms Catalysis Focus On Combustion and Explosions

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 3 of 55

15-1 The Rate of a Chemical Reaction

Rate of change of concentration with time.
2 Fe3+(aq) + Sn2+ 2 Fe2+(aq) + Sn4+(aq) t = 38.5 s t = 38.5 s [Fe2+] = 0.0010 M [Fe2+] = (0.0010 0) M

Rate of formation of Fe2+=

[Fe2+]
t =

0.0010 M 38.5 s

= 2.610-5 M s-1

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 4 of 55

Rates of Chemical Reaction

2 Fe3+(aq) + Sn2+ 2 Fe2+(aq) + Sn4+(aq)

1 [Fe3+] [Sn4+] 1 [Fe2+] = = t t 2 t 2

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 5 of 55

General Rate of Reaction

aA+bBcC+dD Rate of reaction = rate of disappearance of reactants

1 [B] 1 [A] ==b t a t

= rate of appearance of products

1 [D] 1 [C] = = d t c t
Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 6 of 55

15-2 Measuring Reaction Rates

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

2 MnO4-(aq) + 5 H2O2(aq) + 6 H+

2 Mn2+ + 8 H2O(l) + 5 O2(g)

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 7 of 55

Example 15-2
Determining and Using an Initial Rate of Reaction.
H2O2(aq) H2O(l) + O2(g)

-(-2.32 M / 1360 s) = 1.7 10-3 M s-1

Rate =

-[H2O2] t

-(-1.7 M / 2600 s) = 6 10-4 M s-1

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 8 of 55

Example 15-2
What is the concentration at 100s? [H2O2]i = 2.32 M Rate = 1.7 10-3 M s-1 = - [H2O2] t

-[H2O2] = -([H2O2]f - [H2O2]i) = 1.7 10-3 M s-1 t

[H2O2]100 s 2.32 M = -1.7 10-3 M s-1 100 s

[H2O2]100 s = 2.32 M - 0.17 M = 2.17 M

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 9 of 55

15-3 Effect of Concentration on Reaction Rates: The Rate Law

a A + b B . g G + h H . Rate of reaction = k [A]m[B]n . Rate constant = k Overall order of reaction = m + n + .

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 10 of 55

Example 15-3

Method of Initial Rates

Establishing the Order of a reaction by the Method of Initial Rates. Use the data provided establish the order of the reaction with respect to HgCl2 and C2O22- and also the overall order of the reaction.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 11 of 55

Example 15-3
Notice that concentration changes between reactions are by a factor of 2.
Write and take ratios of rate laws taking this into account.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 12 of 55

Example 15-3
R3 = k[HgCl2]3m[C2O42-]3n R2 = k[HgCl2]2m[C2O42-]2n = k(2[HgCl2]3)m[C2O42-]3n k(2[HgCl2]3)m[C2O42-]3n R2 = R3 k[HgCl2]3m[C2O42-]3n k2m[HgCl2]3m[C2O42-]3n R2 2mR3 = = = 2.0 m 2n R3 k[HgCl2]3 [C2O4 ]3 R3 2m = 2.0 therefore m = 1.0
Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 13 of 55

Example 15-3
R2 = k[HgCl2]21[C2O42-]2n = k(0.105)(0.30)n R1 = k[HgCl2]11[C2O42-]1n = k(0.105)(0.15)n k(0.105)(0.30)n R2 = R1 k(0.105)(0.15)n R2 = R1 (0.30)n 7.110-5 n = 2 = = 3.94 n -5 (0.15) 1.810 2n = 3.98 therefore n = 2.0
Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 14 of 55

Example 15-3

1 R2 = k[HgCl2]2 [C2O42-]22

First order +

Second order

= Third Order

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 15 of 55

15-4 Zero-Order Reactions

A products Rrxn = k [A]0

Rrxn = k
[k] = mol L-1 s-1

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 16 of 55

Integrated Rate Law

-[A] t
Move to the

= k

-d[A]
dt

infinitesimal

= k

And integrate from 0 to time t d[A] -

[A]0 [A]t

= k dt
0

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

Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 17 of 55

15-5 First-Order Reactions

H2O2(aq) H2O(l) + O2(g) d[H2O2 ] = -k [H2O2] dt [k] = s-1 = - k dt
0 t

[A]0

[A]t

d[H2O2 ]

[H2O2]

ln

[A]t [A]0

= -kt

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

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 18 of 55

First-Order Reactions

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 19 of 55

Half-Life
t is the time taken for one-half of a reactant to be consumed.
ln [A]t [A]0 ln

= -kt

[A]0 = -kt [A]0

- ln 2 = -kt ln 2 0.693 t = = k k
Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 20 of 55

Half-Life
ButOOBut(g) 2 CH3CO(g) + C2H4(g)

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 21 of 55

Some Typical First-Order Processes

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 22 of 55

15-6 Second-Order Reactions

Rate law where sum of exponents m + n + = 2.
A products d[A] dt

= -k[A]2

[k] = M-1 s-1 = L mol-1 s-1

= - k dt
0 t

[A]2
[A]0

[A]t

d[A]

1 1 = kt + [A]t [A]0
Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 23 of 55

Second-Order Reaction

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 24 of 55

Pseudo First-Order Reactions

Simplify the kinetics of complex reactions Rate laws become easier to work with.
CH3CO2C2H5 + H2O CH3CO2H + C2H5OH

If the concentration of water does not change appreciably during the reaction.
Rate law appears to be first order.

Typically hold one or more reactants constant by using high concentrations and low concentrations of the reactants under study.
Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 25 of 55

Testing for a Rate Law

Plot [A] vs t.

Plot ln[A] vs t.

Plot 1/[A] vs t.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 26 of 55

15-7 Reaction Kinetics: A Summary

Calculate the rate of a reaction from a known rate law using:
Rate of reaction = k [A]m[B]n .

Determine the instantaneous rate of the reaction by:

Finding the slope of the tangent line of [A] vs t or,
Evaluate [A]/t, with a short t interval.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 27 of 55

Summary of Kinetics
Determine the order of reaction by:
Using the method of initial rates. Find the graph that yields a straight line. Test for the half-life to find first order reactions. Substitute data into integrated rate laws to find the rate law that gives a consistent value of k.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 28 of 55

Summary of Kinetics
Find the rate constant k by:
Determining the slope of a straight line graph.

Evaluating k with the integrated rate law.

Measuring the half life of first-order reactions.

Find reactant concentrations or times for certain conditions using the integrated rate law after determining k.
Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 29 of 55

15-8 Theoretical Models for Chemical Kinetics

Collision Theory Kinetic-Molecular theory can be used to calculate the collision frequency.
In gases 1030 collisions per second. If each collision produced a reaction, the rate would be about 106 M s-1. Actual rates are on the order of 104 M s-1. Still a very rapid rate. Only a fraction of collisions yield a reaction.
Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 30 of 55

Activation Energy
For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s).

Activation Energy is:

The minimum energy above the average kinetic energy that molecules must bring to their collisions for a chemical reaction to occur.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 31 of 55

Activation Energy

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 32 of 55

Kinetic Energy

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 33 of 55

Collision Theory
If activation barrier is high, only a few molecules have sufficient kinetic energy and the reaction is slower.

As temperature increases, reaction rate increases.

Orientation of molecules may be important.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 34 of 55

Collision Theory

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 35 of 55

Transition State Theory

The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 36 of 55

15-9 Effect of Temperature on Reaction Rates

Svante Arrhenius demonstrated that many rate constants vary with temperature according to the equation: k = Ae-Ea/RT ln k =
-Ea 1 R T

+ ln A

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 37 of 55

Arrhenius Plot
N2O5(CCl4) N2O4(CCl4) + O2(g)

-Ea R

= -1.2104 K

-Ea = 1.0102 kJ mol-1

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 38 of 55

Arrhenius Equation
k = Ae-Ea/RT ln k =

-Ea 1
R

+ ln A

ln k2 ln k1 =

-Ea 1 R

1 E a + ln A - ln A T2 R T1 1 T1

ln

k1 k2

-Ea 1 R T2

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 39 of 55

15-10 Reaction Mechanisms

A step-by-step description of a chemical reaction. Each step is called an elementary process.
Any molecular event that significantly alters a molecules energy of geometry or produces a new molecule.

Reaction mechanism must be consistent with:

Stoichiometry for the overall reaction. The experimentally determined rate law.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 40 of 55

Elementary Processes
Unimolecular or bimolecular. Exponents for concentration terms are the same as the stoichiometric factors for the elementary process. Elementary processes are reversible. Intermediates are produced in one elementary process and consumed in another. One elementary step is usually slower than all the others and is known as the rate determining step.
Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 41 of 55

A Rate Determining Step

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 42 of 55

Slow Step Followed by a Fast Step

H2(g) + 2 ICl(g) I2(g) + 2 HCl(g) d[P] dt = k[H2][ICl]

Postulate a mechanism:
H2(g) + ICl(g)
slow

d[HI] HI(g) + HCl(g) dt d[I2] dt d[P] dt

= k[H2][ICl]
= k[HI][ICl]

HI(g) + ICl(g) fast I2(g) + HCl(g)

H2(g) + 2 ICl(g) I2(g) + 2 HCl(g)

= k[H2][ICl]

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 43 of 55

Slow Step Followed by a Fast Step

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 44 of 55

Fast Reversible Step Followed by a Slow Step

2NO(g) + O2(g) 2 NO2(g) Postulate a mechanism:
fast

d[P] dt

= -kobs[NO2]2[O2]

2NO(g) K=

k
-1

k1

N2O2(g)
[N2O2] [NO]
k2

[N2O2] =

k1
k-1

[NO]2 = K [NO]2

k1 k-1

slow

N2O2(g) + O2(g)

2NO2(g)

d[NO2] dt

= k2[N2O2][O2]

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

Prentice-Hall 2002 General Chemistry: Chapter 15

d[I2]
dt

= k2

k1
k-1

[NO]2[O2]

Slide 45 of 55

The Steady State Approximation

2NO(g) 2NO(g) k1 k-1 N2O2(g) N2O2(g) k3 2NO2(g) 2NO(g) N2O2(g) k1 k2 N2O2(g) 2NO(g) k3 2NO2(g)

N2O2(g) + O2(g)

N2O2(g) + O2(g)

d[NO2] dt d[N2O2] dt

= k3[N2O2][O2]

= k1[NO]2 k2[N2O2] k3[N2O2][O2] = 0

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 46 of 55

The Steady State Approximation

d[N2O2]
dt = k1[NO]2 k2[N2O2] k3[N2O2][O2] = 0 k1[NO]2 = [N2O2](k2 + k3[O2]) k1[NO]2

[N2O2] =

(k2 + k3[O2])
k1k3[NO]2[O2] (k2 + k3[O2])
Slide 47 of 55

d[NO2] dt
Prentice-Hall 2002

= k3[N2O2][O2] =

General Chemistry: Chapter 15

Kinetic Consequences of Assumptions

2NO(g) k1 k2 N2O2(g) 2NO(g) k3 2NO2(g)

d[NO2] dt

k1k3[NO]2[O2]

N2O2(g)

(k2 + k3[O2])
N2O2(g) + O2(g)

Let k2 << k3 Or Let k2 >> k3

d[NO2]

dt
d[NO2] dt

k1k3[NO]2[O2]
( k3[O2]) k1k3[NO]2[O2] ( k2)

= k1[NO]2 k1k3 k2

[NO]2[O2]

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 48 of 55

11-5 Catalysis
Alternative reaction pathway of lower energy. Homogeneous catalysis.
All species in the reaction are in solution.

Heterogeneous catalysis.
The catalyst is in the solid state. Reactants from gas or solution phase are adsorbed. Active sites on the catalytic surface are important.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 49 of 55

11-5 Catalysis

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 50 of 55

Catalysis on a Surface

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 51 of 55

Enzyme Catalysis

E + S ES
k-1

k1

ES E + P

k2

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 52 of 55

Saturation Kinetics
E + S ES E + P
k-1 k1 k2

d[P] dt d[P] dt

= k2[ES]

= k1[E][S] k-1[ES] k2[ES]= 0 k1[E][S] = (k-1+k2 )[ES] [E] = [E]0 [ES]

k1[S]([E]0 [ES]) = (k-1+k2 )[ES] [ES] = k1[E]0 [S]

(k-1+k2 ) + k1[S]

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 53 of 55

Michaelis-Menten

d[P] dt d[P] dt = k2[E]0 d[P] dt d[P] dt

k1k2[E]0 [S]

(k-1+k2 ) + k1[S]
= k2[E]0 [S]

(k-1+k2 ) + [S]
k1

d[P] dt

k2

KM

[E]0 [S]

k2[E]0 [S]
KM + [S]

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 54 of 55

Chapter 15 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.

Choose a variety of problems from the text as examples.

Practice good techniques and get coaching from people who have been here before.

Prentice-Hall 2002

General Chemistry: Chapter 15

Slide 55 of 55