Chemical Kinetics

Chapter 14

The Rate Law

Rate law – description of the effect of concentration on rate

aA + bB cC + dD

Rate = k [A]x[B]y

reaction is xth order in A

reaction is yth order in B

reaction is (x +y)th order overall

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2][ClO2]

Rate Laws

• Rate laws always determined experimentally

• Reaction order always defined in terms of reactant (not product) concentrations

• Order of a reactant is not related to the stoichiometric coefficients

1

Concentration and Initial Rates

NH4+(aq) + NO2

−(aq) N2(g) + 2 H2O(l)

Double [NH4+] with [NO2

−] constant: x = 1Rate doubles

Assume: rate = k [NH4+]x[NO2

−]y

Double [NO2−] with [NH4

+] constant: Rate doubles y = 1

Therefore rate law is rate = k [NH4+][NO2−]

Table14.2

Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8

2- (aq) + 3I- (aq) 2SO42- (aq) + I3

- (aq)

Experiment [S2O82-] [I-]

Initial Rate (M/s)

1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

3 0.16 0.017 2.2 x 10-4

rate = k [S2O82-]x[I-]y

Double [I-], rate doubles (experiment 1 & 2)

y = 1

Double [S2O82-], rate doubles (experiment 2 & 3)

x = 1

k = rate

[S2O82-][I-]

=2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

rate = k [S2O82-][I-]

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?

First-Order Reactions

A product rate = −[A]t

rate = k [A]

k = rate[A]

= 1/s or s-1M/sM

=[A]t

= k [A]−

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

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

Integrated

rate law

t

0

A

A

dt k[A]

d[A]

o

kt [A]

[A] ln

o

Which now integrates to:

Consider the process in which methyl isonitrile is converted to acetonitrile

CH3CN

Fig 14.7(a) Data collected for this reaction at 198.9 °C.

First-Order Reactions

CH3NC

Plot of ln P vs time, results in a straight line Therefore,

Process is first-orderk is the negative of the slope: 5.1 10-5 s−1

First-Order Reactions

Fig 14.7

orange red-brown

Decomposition of N2O5

2 N2O5 (in CCl4) → 4 NO2 (g) + O2 (g)

→

Decomposition of N2O5

2 N2O5 (in CCl4) → 4 NO2 (g) + O2 (g)

Plot of ln [N2O5] vs time

Linear plot indicates: 1st order

Second-Order Reactions

A product rate = −[A]t

rate = k [A]2

[A]t

= k [A]2−

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

1[A]

=1

[A]o

+ kt

One type:

A + B product

Second type:

rate = k [A][B]

Initial

rate law:

Combining the two rate expressions:

Which now integrates to: Integrated

rate law

The decomposition of NO2 at 300°C is described by the equation

NO2 (g) NO (g) + 1/2 O2 (g)

plot of ln[NO2] vs time:

Second-Order Reactions

plot of 1/[NO2] vs time:Fig 14.8

Half-life = time required for one-half of a reactant to react

t½ = t when [A] = [A]0/2

Because [A] at t1/2 is one-half of the original [A],

[A]t = 0.5 [A]0

1st order t1/2 = 0.693/k

2nd order

Fig 14.9

1st order rxn

t½ =1

k[A]o

Half-Life

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

t½ln2k

=

t½ =[A]o

2k

t½ =1

k[A]o