Question

For the reaction A + B → product, the rate of reaction is 3.6 * 10 ^ - 2 mol dm ^ - 3 sec^ -1. When [A] = 0.2 mol dm ^ - 3 and [B] = 0.1 mol dm ^ - 3, find the rate constant of the reaction if it is second order with respect to both reactants.

• A. 18 mol ^ - 3 dm ^ 9 sec^ -1
• B. 90 mol ^ - 3 dm ^ 9 sec^ -1
• C. 36 mol ^ - 3 dm ^ 9 sec^ -1
• D. 72 mol ^ - 3 dm ^ 9 sec^ -1

Answer 1

Answer: (B)

90 mol ^ - 3 dm ^ 9 sec^ -1

Answer 2

The rate law for the reaction is:

$\text{Rate} = k[A]^2[B]^2$

Substitute the given values:

$0.036 \text{ mol dm}^{-3} \text{ sec}^{-1} = k (0.2)^2 (0.1)^2$

Calculate the concentrations:

$(0.2)^2 = 0.04$

$(0.1)^2 = 0.01$

Therefore:

$k = \frac{0.036}{0.04 \times 0.01} = \frac{0.036}{0.0004} = 90$