Atkins’ Physical ChemistryEighth Edition

Chapter 22 – Lecture 2

The Rates of Chemical Reactions

Copyright © 2006 by Peter Atkins and Julio de Paula

Peter Atkins • Julio de Paula

Integrated Rate Laws

Relation Between Concentration and Time

• Rate law provides rate as a function of concentration

• Need relationship between concentration and time :

• i.e., How do we determine the concentration of areactant at some specific time?

Fig. 22.7 Exponential decay of the reactant, 2nd-order

kt]A[

1]A[

1

o

Plot of [A]/[A]o vs. t [A]o

Rearranges to:

oo [A] kt11

[A][A]

Summary of the Kinetics of Reactions

Order Rate LawConcentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

ln[A] = ln[A]o - kt

1[A]

=1

[A]o

+ kt

[A] = [A]o - kt

t½ln 2k

=

t½ =[A]o

2k

t½ =1

k[A]o

rate = k [A][B] t½ =1

k[A]o

2

Generally, as T increases, so does the reaction rate

k is temperature dependent

Temperature affectsthe rate of chemiluminescence

in light sticks

Dependence of rate constant on temperature

In a chemical rxn, bonds are broken and new bonds are formed

Molecules can only react if they collide with each other

Furthermore, molecules must collide with the correct orientation and with enough energy

Cl + NOCl Cl2 + NO

Activation Energy

An energy barrier

Reactant

Activatedcomplex

Product

Transition state

Activation energy (Ea) ≡ minimum amount of energy required to initiate a chemical reaction

Transition state ≡ very short-lived and cannot be removed from reaction mixture

methyl isonitrile acetonitrile

Rank the following series of reactions from slowest to fastest

• The lower the Ea the faster the reaction

• Slowest to fastest: (2) < (3) < (1)

ExothermicExothermic Endothermic

How does a molecule acquire sufficient energyto overcome the activation barrier?

Effect of temperature on distribution of kinetic energies

Fraction ofmolecules with

E ≥ Ea

R = 8.314 J/(mol ∙ K)T = kelvin temperatureg ≡ degeneracy of state

Maxwell–Boltzmann Distributions

RTaE

eg

g

N

N

oo

Fraction of molecules with E ≥ Ea

e.g., suppose Ea = 100 kJ/mol at T = 300 K:

= 3.9 x 10-18 (implies very few energetic molecules)

at T = 310 K: = 1.4 x 10-17 (about 3.6 times more molecules)

RTaE

eg

g

N

N

oo

R = 8.314 J/(mol ∙ K)T = kelvin temperatureg ≡ degeneracy of state

oN

N

oN

N

Temperature Dependence of the Rate ConstantTemperature Dependence of the Rate Constant

k = A • exp(− Ea/RT )

Ea ≡ activation energy (J/mol)

R ≡ gas constant (8.314 J/K•mol)

T ≡ kelvin temperature

A ≡ frequency factor

ln k = −Ea

R1T

+ ln A

(Arrhenius equation)

y = mx +b

Fig. 22.10 Arrhenius equation plot of ln k versus 1/T

RTE

A lnk ln a

A lnT1

RE

k ln a

Rearranges to:

bx my

k1 = A • exp(−Ea/RT1)

The Arrehenius equation can be used to relate

rate constants k1 and k2 at temperatures T1 and T2.

21

21a

2

1

TTTT

RE

kk

ln

k2 = A • exp(−Ea/RT2)combine to give:

Better as an estimate, since slope depends only on two points

Fig. 22.12 Potential energy profile for exothermic reaction

RT/EaAek Note: this rate constant givesthe rate of successful collisions

Transition state