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

Step 1: Understanding Second-Order Reactions. For a second-order reaction, the rate law is given by: \[ \text{Rate} = k [A]^2 \] where \( [A] \) is the concentration of the reactant and \( k \) is the rate constant.
Step 2: Effect of Concentration Change.
(i) If the concentration of \( A \) is doubled, the rate will increase by a factor of \( 2^2 = 4 \). This is because the rate is proportional to the square of the concentration. \[ \text{New Rate} = k (2[A])^2 = 4k [A]^2 \]
(ii) If the concentration of \( A \) is reduced to half, the rate will decrease by a factor of \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \). \[ \text{New Rate} = k \left(\frac{1}{2}[A]\right)^2 = \frac{1}{4} k [A]^2 \] Thus, doubling the concentration quadruples the rate, while halving the concentration reduces the rate to one-quarter.