Question

A reaction follows second order kinetics. How is the rate of reaction affected if the concentration of the reactant is reduced to half? Choose the correct value from the following:

• A. four times
• B. eight times
• C. \( \frac{1}{4} \) of the original value
• D. three times

Hint:

For second-order reactions, the rate depends on the square of the concentration. Reducing the concentration by half reduces the rate to \( \frac{1}{4} \) of its original value.

Answer 1

The Correct Option is C

Step 1: For a second-order reaction, the rate law is:
\[ \text{Rate} = k [A]^2 \] Step 2: If the concentration of the reactant \([A]\) is reduced to half:
\[ \text{New Rate} = k \left( \frac{[A]}{2} \right)^2 = k [A]^2 \times \frac{1}{4} \] Step 3: So, the new rate is one-fourth of the original rate.