Question

A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled, (ii) reduced to half?

Hint:

For second-order reactions, the rate is proportional to the square of the concentration. Doubling the concentration increases the rate by a factor of 4, while halving it decreases the rate by a factor of 4.

Answer 1

To solve the problem, we need to determine how the rate of a second-order reaction is affected when the concentration of the reactant is (i) doubled and (ii) reduced to half.

1. Understand the Rate Law:
For a second-order reaction with respect to a reactant A, the rate law is \( \text{Rate} = k [A]^2 \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.

2. Case (i) - Concentration Doubled:
If the concentration is doubled, \( [A] \) becomes \( 2[A] \). The new rate is:

\( \text{Rate}_{\text{new}} = k (2[A])^2 = k \cdot 4[A]^2 = 4 \times \text{Rate}_{\text{initial}} \).
The rate increases by a factor of 4.

3. Case (ii) - Concentration Reduced to Half:
If the concentration is reduced to half, \( [A] \) becomes \( \frac{1}{2}[A] \). The new rate is:

\( \text{Rate}_{\text{new}} = k \left(\frac{1}{2}[A]\right)^2 = k \cdot \frac{1}{4}[A]^2 = \frac{1}{4} \times \text{Rate}_{\text{initial}} \).
The rate decreases to one-fourth of the initial rate.

Final Answer:
(i) The rate increases by a factor of 4 when the concentration is doubled.
(ii) The rate decreases to one-fourth when the concentration is reduced to half.