Objectives:

1.Write the rate law for zero order, 1st order and 2nd order reaction

2. Define half-life.3. Draw the respective graphs for the different

order reactions4. Solve quantitative problems.

Integrated Rate Law

Zero Order Reaction

A zero order reaction is a reaction independent of the concentration of reactant.

A product

The rate law is given byrate = k[A]0

rate = k

[A] M

rate

Integrated rate equations

- d[A] = k dt Using calculus, - d[A] = kdt - ∫d [A] = k ∫dt - [A] = kt + c substituting t=0, [A] = [A]0

- [A]0 = k(0)+c c = - [A]0

[A]0 -[A] = kt

Unit of k for zero order reaction M s-1

Graphs for zero order reaction

[A]o – [A] = kt

[A] = -k t + [A]o[A]

t

[A]o

t

[A]o – [A]

y = mx + c

y = m x + c

[A]

rate

Half-life (t½)

Half life (t½) is the time required for the concentration of a reactant to decrease to half of its initial value.

zero order reactionSubstituting t = t1/2 , and [A] = [A]0 into the zeroorder reaction, gives 2 [A]0 - [A] = kt [A]0 – [A]0 = kt1/2

2 Solving for t1/2 gives t1/2 = [A]0 2k

From the rate law, rate = k[A] To obtain the units of k k = rate [A] unit k = M s-1 M = s-1

First Order Reactions

A first order reaction is a reaction whereby its rate depends on the concentration of reactant raised to the first power.

Rate = k[A]y = mx + c

Rate

Ms-1

[A] ,M

For first order reaction,

- d[A]dt

= k[A]

- d[A][A]

= k dt

- d[A][A]

= k dt∫ ∫

- ln [A] = kt + c

substituting

t = 0, [A]=[A]0

- ln [A]0 = k(0) + c c = ln[A]0

-ln [A] = kt – ln[A]0

ln[A]0

[A] = kt

Rate = k[A]

Characteristic graphs of 1st order reaction

ln[A]

ln[A]o

t

ln[A]o

[A]

t

ln[A]o – ln[A] =kt

[A]

tln

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

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

The reaction 2A B is first order with respect A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?

ln[A]o – ln[A] = kt

kt = ln[A]0 – ln[A]

t =ln[A]0 – ln[A]

k= 66 s

[A]0 = 0.88 M[A] = 0.14 M

ln[A]0

[A]k

=ln

0.88 M0.14 M

2.8 x 10-2 s-1=

Example

Example Decomposition of H2O2 (aq) is first order, given that k

= 3.66 x 10-3 s-1 and [H2O2 ]o = 0.882 M, determinea) the time at which [H2O2] = 0.600 Mb) the [H2O2 ] after 225 s.

Solution :

a) ktln[H2O2]0

[H2O2]=

[H2O2]=

=ln 0.8820.600

3.66 x 10-3 s-1 x t

ln 1.47 = 3.66 x 10-3 s-1 x t

t =ln 1.473.66 x 10-3

= 105.26 s

b)ln

[H2O2]0

[H2O2]=

[H2O2]= kt

ln0.882[H2O2]

=[H2O2]

= 3.66 x 10-3s-1x 225 s

[H2O2] = 0.387 M

Exercise,

The conversion of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7 X 10-4 s-1 at 500°C.

CH2

CH2 CH2 CH3-CH=CH2

a) If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 minute. (0.18 M)

b) How long will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (13 min)

The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.

t=t½ when [A] = [A]0

2

ln[A]0

[A]0/2k

=t½ln2k

=0.693

k=

What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?

t½ln2k

=0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

How do you know decomposition is first order?units of k (s-1)

Half-life of a first-order reaction

A product

First-order reaction

No. of half-lives [A]o = 8 M

1

2

3

4

4

2

1

1/2

t½ln2k

=

Example The decomposition of ethane C2H6 to methyl radicals is a first order reaction with a rate constant of 5.36 x 10-4 s-1 at 700o C. C2H6 (g) → 2 CH3 (g)Calculate the half life of the reaction in minutes.Solution

t1/2 =ln 2

k

=0.693

5.36 x 10-4

= 1.29 x 103 s= 21.5 min

Problem 2

What is the half-life of a compound if 75% of a given sample of the compound decomposes in 60 min? Assume it is first-order reactions kinetics.

(t1/2=30 min)

ln [A0]/[A] = kt ln [A0]/[A] = kt

t = 60 mint = 60 min[A0] = 1[A0] = 1[A] = 0.25 [A] = 0.25

Second Order ReactionsA second order reaction is a reaction which rate depends on the concentration of one reactant raised to the second power or on the concentration of two different reactants each raised to the first power.

Example A productWhere

Rate = - d[A]dt

= k[A]2

To obtain the units of k

k =rate[A]2

= M/sM2

= M-1 s-1

Using calculus, the following expression can be obtained

1[A]

1-[A]0

= + kt

Unit k

Characteristic graphs for second order reaction

rate

[A]

Rate = k [A]2

rate

[A]2

Graphs for second order reaction

[A]

t

[A]1 + kt=

[A]o

1

1/[A] M-1

1/[A]o

t

1/[A] – 1/[A]o

t

22ndnd –order reaction, r = k[A] –order reaction, r = k[A]22

If [A] doubles,If [A] doubles, rr22 = k (2[A]) = k (2[A])22

= k ( 4 [A]= k ( 4 [A]22 ) ) = 4 k [A]= 4 k [A]22

= 4 r= 4 r

R will increase by 4 times if [A] doublesR will increase by 4 times if [A] doubles

Half life of a second order reaction

1[A]

1[A]0

= + kt

Substituting t = t1/2[A]= [A]o

1[A]0

1[A]0

= + kt1/2

t1/2= 1k[A]0

2

2

Detemination of half-life using graph for second order reaction[A]0

[A]0/2

[A]0/4

[A]0/8

tx 2x 4x

t1/2= 1k[A]0

ExampleIodine atoms combine to form molecular iodine inthe gaseous phase I (g) + I (g) I2(g)

This reaction is a second order reaction , with the rate constant of 7.0 x 109 M-1 s-1

If the initial concentration of iodine was 0.086 M, i) calculate it’s concentration after 2 min.ii) calculate the half life of the reaction if the initial concentration of iodine is 0.06 M and 0.42 M respectively.

Solution :

i)1[A]

1[A]0

= + kt

1[A]

1[0.086]

= + (7.0 x 109 x 2 x 60 )

= 8.4 x 1011

[A] = 1.190 x 10-12 M

ii) [I2] = 0.06 M

t1/2= 1k[A]0

=1

7.0 x 109 x 0.060

= 2.4 x 10-10 s

[I2] = 0.42 M

t1/2= 1k[A]0

7.0 x 109 x 0.042

1=

= 3.4 x 10-10 s

Example,

The following results were obtained from an experimental investigation on dissociation of dinitrogen pentoxide at 45oC

N2O5(g) 2 NO2(g) + ½ O2(g)

time, t/min 0 10 20 30 40 50 60

[N2O5] x 10-4 M 176 124 93 71 53 39 29

Plot graph of [N2O5] vs time, determine

i) The order of the reaction

ii) the rate constant k

Using graph

Solution :

[N2O5] x 10-4 /M

Time ( min)

180

160

140

80

120100

8060

4020

10 20 30 40 50 60 70

Based on the above graph, Time taken for concentration of N2O5 to change from 176 x 10-4 M to 88 x 10-4 M is 20 minTime taken for concentration of N2O5 to change from 88 x 10-4 M to 44 x 10-4 M is also 20 minThe half life for the reaction is a constant and does not depend on the initial concentration of N2O5

Thus, the above reaction is first order

i)

ii) k =ln2

20 min = 0.03 min-1

Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions

Order Rate LawConcentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

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

1[A]

=1

[A]0+ kt

[A] = [A]0 - kt

t½ln2k

=

t½ = [A]0

2k

t½ = 1k[A]0

Zero order 1st order 2nd orderA product A product A product

r = k [A]0 r = k [A]1 r = k [A]2

[A]

r

[A]

r

[A]2

r

[A]

rUnit k = M s-1 Unit k = s-1 Unit k = M-1 s-1

Integrated rate law Integrated rate law Integrated rate law

[A]0 – [A] = kt ln([A]0 / [A]) = kt 1/[A] – 1/[A]0 = kt

t

[A]

[A]0

t

[A]0 - [A]t

[A]

t

[A]

t

ln[A]

ln[A]0

t

1/[A]

1/[A]0

t

ln([A]0 / [A])

t

1/[A] – 1/[A]0

t1/2 = [A]0/2k

t1/2 = ln2/k

t1/2 = 1/k[A]0