RATE LAW

11.1

Objectives:

5. calculate the value and determine the units of rate constants

At the end of the lesson the students should be able to:

1. define rate law and write the rate equation2. define the order of reaction and the rate constant3. calculate the order with respect to a certain reactant from experimental data4. determine the overall order of a reaction from experimental data

The Rate LawThe rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.

aA + bB cC + dD

Rate = k [A]x[B]y

reaction order is x with respect to A

reaction order is y with respect to B

Overall reaction order is (x + y) The exponents x, y, The exponents x, y, …… can be integers, can be integers, fractions or decimal or negative values.fractions or decimal or negative values.

kk is called is called rate constantrate constant

Rate α [reactant]

Rate Law

• The values of x and y can only be determined experimentally.

• Reaction order is usually defined in terms of reactant concentrations.

• The order of a reactant is not related to the stoichiometric coefficients of the reactants in the balanced chemical equation.

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

rate = k [F2][ClO2]1

The units of rate constant, k

A Products

Rate, r = k [A]x

i) The reaction is zero order

Rate = k [A]0

Rate = k

unit k =unit rate= mol L-1 s-1 or M s-1

ii) First orderRate = k [A]1

Unit k =M s-1

M

= s-1

iii) Second orderRate = k [A]2

[A]2k =

rate

=M s-1

M2= M-1 s-1Unit k

[A]k =

rate

Example :

S2O82- + 3I- 2SO4

2- + I3-

The above reaction is first order with respect to iodide ions and to thiosulphate ions.

a) Write the rate of equation for the reaction.

b) What is the unit of rate constant, k?

Solution :

a) Rate = k [S2O82- ]1[I-]1

b) Rate = k [S2O82- ]1[I-]1

k = rate[S2O8

2- ]1[I-]1= Ms-1

M2= M-1 s-1Unit k

The order of reaction

For reaction

A Products

Rate = k [A]x

i) If x = 0

Rate is not dependent on [A]

Rate = k [A]0

Rate = k

Therefore this reaction is zero order with respect to A

ii) If x= 1Rate = k [A]1

Assume [A]i = 1.0M

Rate = k (1.0M)

If the [A] is doubled from 1.0M to 2.0M,

Rate = k (2.0M)

= 2k(1.0M)hence

Rate = 2k[A]

Doubling the [A] will double the rate of reaction.

Therefore this reaction is first order with respect to A

iii) If x = 2Rate = k[A]2

Assume [A]i = 1.0 M

Rate = k (1.0 M)2

If the [A] is doubled from 1.0 M to 2.0 M,

Rate = k (2.0 M)2

= 4k(1.0 M)

henceRate = 4k[A]

Doubling [A], the rate will increase by a factor of 4.

Therefore the reaction is second order with respect to A

Example

Determining Reaction Order from Rate Law

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]

overall reaction order = 3

The reaction order respect to NO : 2

The reaction order respect to O2 : 1

Solution:

The reaction order with respect to CH3CHO : 3/2

The reaction order (overall) : 3/2

Solution:

The reaction of order with respect to H2O2 : 1

The reaction of order with respect to I- : 1

and zero order in H+, while overall order is 2.

(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-]

Experiment

Initial Reactant Concentrations (molL-1) Initial Rate

(M s-1)O2 NO

1

2

3

4

5

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

Determination of the orders of reaction rate;

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

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.

rate2 k [O2]2m[NO]2

n

k [O2]1m[NO]1

n=rate1

=[O2]2

m

[O2]1m

=[O2]2

[O2]1

m

6.40x10-3Ms-1

3.21x10-3Ms-1

=

1.10x10-2mol/L

2.20x10-2mol/L m

;2 = 2m , m = 1

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

Solution:

The reaction is first order with respect to O2

To find the order with respect to NO, we compare experiment 3 and 1, in which [O2] is held constant and [NO] is doubled:

=k [O2]3

m [NO]3n

k [O2]1m [NO]1

n

Rate 3

Rate 1

[NO]3=[NO]1

n

12.8 x 10-3Ms-1

3.21 x 10-3Ms-1=

2.60 x 10-2mol/L1.30 x 10-2mol/L

n

4 = 2n n = 2;

The reaction is second order with respect to NO

Thus the rate law is :

Rate = k[O2][NO]2

The results of the kinetic studies are given below.

Exercise:

exp [ClO2]

M

[OH-]M

Initial rate, Ms-1

1 0.0421 0.0185 8.21 × 1 0-3

2 0.0522 0.0185 1.26 × 1 0-2

3 0.0421 0.0285 1.26 × 1 0-2

ClO2(aq) + 2OH- (aq) → products

a) Explain what is meant by the order of reaction.b) Reffering to the data determine (i) rate law /rate equation (ii) rate constant, k (iii) the reaction rate if the concentration of both ClO2 and OH- = 0.05 M

i) When [A] is doubled, rate also doubles. But doubling the [B] has no effect on rate.

ii) When [A] is increased 3x, rate increases 3x, and increasing of [B] 3x causes the rate to increase 9x.

iii) Reducing [A] by half has no effect on the rate, but reducing [B] by half causes the rate to be half the value of the initial rate.

Exercise:

A + B → C

Write rate law for this equation,

Exercise:

Many gaseous reactions occur in a car engine and exhaustsystem. One of the gas reaction is given below.

NO2(g) + CO(g) NO(g) + CO2(g)

Rate = k [NO2]m[CO]n

Experiment Initial Rate(Ms-1) Initial [NO2](M) Initial [CO](M)

1 0.0050 0.10 0.10

2 0.080 0.40 0.10

3 0.0050 0.10 0.20

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