Question

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
r0/mol L−1 s−1	5.07 × 10−5	5.07 × 10−5	1.43 × 10−4

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

Answer 1

Suppose the order of the reaction with respect to A is m and with respect to B is n

∴ r₀ = k[A]^m [B]ⁿ

For the given data, we have

5.07 × 10⁻⁵ = k (0.20)^m × (0.30)ⁿ   ...(i)

5.07 × 10⁻⁵ = k (0.20)^m × (0.10)ⁿ   ...(ii)

1.43 × 10⁻⁴ = k (0.40)^m × (0.05)ⁿ   ...(iii)

Dividing eq. (i) by eq. (ii), we have

`(5.07 xx 10^-5)/(5.07 xx 10^-5) = (k (0.20)^m xx (0.30)^n)/(k (0.20)^m xx (0.10)^n)`

or 1 = (3)ⁿ

or n = 0

Dividing eq. (iii) by eq. (ii) and putting the value of n, we get

`(1.43 xx 10^-4)/(5.07 xx 10^-5) = (k (0.40)^m xx (0.05)^0)/(k (0.20)^m xx (0.10)^0)`

or, `(1.43 xx 10^-4)/(5.07 xx 10^-5)` = 2^m

or (2)^m = 2.82

or, m log₁₀ 2 = log₁₀ 2.82 = 0.45

or, m = `0.45/(log_10 2)`

= `0.45/0.3010`

= 1.49

= 1.50

Hence, order with respect to A = 1.5 and order with respect to 8 = 0.