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

This analysis investigates the impact on the rate of a second-order reaction when the reactant concentration is altered.

1. Understanding the Rate Law:
For a second-order reaction involving reactant A, the rate is defined by the equation \( \text{Rate} = k [A]^2 \), where \( k \) represents the rate constant and \( [A] \) denotes the reactant concentration.

2. Scenario (i) - Doubled Concentration:
When the concentration of A is doubled, it becomes \( 2[A] \). The resultant rate is calculated as:

\( \text{Rate}_{\text{new}} = k (2[A])^2 = k \cdot 4[A]^2 = 4 \times \text{Rate}_{\text{initial}} \).
This indicates a fourfold increase in the reaction rate.

3. Scenario (ii) - Halved Concentration:
When the concentration of A is halved, it becomes \( \frac{1}{2}[A] \). The new rate is determined by:

\( \text{Rate}_{\text{new}} = k \left(\frac{1}{2}[A]\right)^2 = k \cdot \frac{1}{4}[A]^2 = \frac{1}{4} \times \text{Rate}_{\text{initial}} \).
Consequently, the reaction rate is reduced to one-fourth of its initial value.

Summary of Findings:
(i) Doubling the reactant concentration results in a quadrupling of the reaction rate.
(ii) Reducing the reactant concentration by half leads to a decrease in the reaction rate to one-fourth of its original magnitude.