Consider the following data for the reaction A + B → Products.

Run

Initial

Initial

Initial rate (mol s^–1)

concentration

concentration

1

0.10 M

0.1 M

2.1 x 10^–3

2

0.20 M

1.0 M

8.4 x 10^–3

3

0.20 M

2.0 M

8.4 x 10^–3

Determine the order of reaction with respect to A and with respect to B and the overall order of reaction.

Solution

The rate law may be expressed as

Rate = k {[A]}^{p} {[B]}^{q}

Comparing experiments 2 and 3

(Rate)_{2} = k {[0.2]}^{p} [1.0]^{q} = 8.4 \times 10^{- 3} . . . (i) (Rate)_{3} = k [0.2]^{p} {[2.0]}^{q} = 8.4 \times 10^{- 3} . . . (ii)

Dividing eqn. by (ii) by (i),

\frac{(Rate)_{3}}{(Rate)_{2}} = \frac{k {[0.2]}^{p} {[2.0]}^{q}}{k {[0.2]}^{p} {[1.0]}^{q}} = \frac{8.4 \times 10^{- 3}}{8.4 \times 10^{- 3}}

{[2]}^{q} = {[2]}^{0} or q = 0 .

Comparing experiments (i) and (ii)

(Rate)_{2} = k {[0.20]}^{p} {[1.0]}^{q} = 8.4 \times 10^{- 3} . . . (iii) (Rate)_{1} = k {[0.10]}^{p} {[1.0]}^{q} = 2.1 \times 10^{- 3} . . . (iv)

Dividing eqn. (iii) by (iv),

\frac{(Rate)_{2}}{(Rate)_{1}} = \frac{k {[0.20]}^{p} {[1.0]}^{q}}{k [0.10]^{p} [1.0]^{q}} = \frac{8.4 \times 10^{- 3}}{2.7 \times 10^{- 3}} = 4 {[2]}^{q} = {[2]}^{2} or q = 2 .

Order with respect to A = 2.
Order with respet to B = 0.
Overall order of reaction = 2.

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