Question

For the reaction \[\ce{A + B <=>}\] Product, the following data was obtained.

Experiment number	Initial conception of [A] (mol L−1)	Initial concentration of [B] (mol L−1)	Initial Rate (mol L−1 minL−1)
1	0.15	0.15	9.6 × 10−2
2	0.30	0.15	3.84 × 10−1
3	0.15	0.30	1.92 × 10−1
4	0.30	0.30	7.68 × 10−1

Calculate the following:

1. The overall order of the reaction
2. The rate law equation
3. The value of rate constant

Answer 1

Let the rate of equation be:

Rate = K[A]^x [B}^y

In experiment 1

Rate = K[0.15]^x [0.15]^y = 9.6 × 10⁻² ...(1)

In experiment 2

Rate = K[0.30]^x [0.15]^y = 3.84 × 10⁻¹ ...(2)

In experiment 3

Rate = K[0.15]^x [0.30]^y = 1.92 × 10⁻¹ ...(3)

In experiment 4

Rate = K[0.30]^x [0.30]^y = 7.68 × 10⁻¹ ...(4)

Dividing eq. (2) by eq. (1)

`(3.84 × 10^-1)/(9.6 × 10^-2) = (K[0.30]^x[0.15]^y)/(K[0.15]^x[0.15]^y)`

4= [2]^x

x = 2

Thus, the order of reaction with respect to [A] is 2.

Dividing eq. (3) by eq. (1),

`(1.92 × 10^-1)/(9.6 × 10^-2) = (K[0.15]^x[0.30]^y)/(K[0.15]^x[0.15]^y)`

2 = [2]^y

y = 1

Thus, the order of reaction with respect to [B] is 1.

1. The overall order of reaction is 2 + 1 = 3
2. The rate law equation
   Rate = K[A]² [B]¹
3. Rate constant (K)
   From eq. (1),
   Rate = K[0.15]² [0.15]¹
   = 9.6 × 10⁻²
   K = `(9.6 × 10^-2)/([0.15]^2[0.15]^1)`
   = 28.44 mol⁻² L² min⁻¹

Thus, the value of the rate constant is 28.44 mol⁻² L² min⁻¹.