In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below:

A/mol L^–1

0.20

0.20

0.40

B/mol L^–1

0.30

0.10

0.05

r/mol L^–1S^–1

5.07 x 10^–5

5.07 x 10^–5

1.43 x 10^–4

What is the order of the reaction with respect of A and B?

Solution

In a reaction A and B, Let order of reaction w.r.t. A is x and w.r.t. B is y. Then the rate of reaction can be written as

rate = k[A]^x [B]^yFrom given table data, 1 and 2 we can write

5.07 x 10^–5 = k[10.20]^x [0.30]^y ...(i)
5.07 x 10^–5 = k[0.20]^x [0.10]^y ...(ii)

Dividing (ii) by (i), we get

$\frac{5.07 \times 10^{- 5}}{5.07 \times 10^{- 5}} = \frac{k {[0.20]}^{x} {[0.10]}^{y}}{k [0.20]^{x} {[0.30]}^{y}}$

or $1 = {(\frac{0.10}{0.30})}^{y} or y = 0$

From given table data, 2 and 3 we can write

5.07 \times 10^{- 5} = k {[0.20]}^{x} {[0.20]}^{y} = k {[0.20]}^{y} \times 1 [∵ y = 1] . . . (iii) 7.06 \times 10^{- 5} = k [0.20]^{x} [0.05]^{y} = k {[0.40]}^{x} \times 1 . . . (iv)

Dividing (iv) by (iii), we get

\frac{7.60 \times 10^{- 5}}{5.07 \times 10^{- 5}} = \frac{k {[0.40]}^{x}}{k {[0.20]}^{x}} = {[\frac{0.4}{0.2}]}^{x} = (2)^{2}

(2)^{x} = 3 / 2 = 1.5

x = 0.5

Thus the order of reaction w.r.t. A is

\frac{1}{2}

and w.r.t. B is zero.

For Daily Free Study Material Join wiredfaculty

Chemical Kinetics

Some More Questions From Chemical Kinetics Chapter

What will be the effect of temperature on rate constant?

A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction.

The rate law for the decomposition of N₂O₅ is rate = k[N₂O₅]. What is the significance of k in this equation?

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

A/mol L^–1	0.20	0.20	0.40
B/mol L^–1	0.30	0.10	0.05
r/mol L^–1S^–1	5.07 x 10^–5	5.07 x 10^–5	1.43 x 10^–4