Question

The rate of the chemical reaction between substance A and B is found to follow the rate law. rate = \( k[A]^2[B] \), where k is the rate constant. The concentration of A is reduced to half of its original value. To make the reaction occur at 50% of its original rate, the concentration of B should be

• A. decreased by \( \frac{1}{4} \)
• B. halved
• C. kept constant
• D. doubled

Hint:

In rate laws, when the concentration of a reactant is changed, you must adjust the concentration of other reactants accordingly to maintain the reaction rate.

Answer 1

The Correct Option is D

The rate law is given by \( \text{rate} = k[A]^2[B] \), where \( k \) is the rate constant, and \( [A] \) and \( [B] \) are the concentrations of reactants A and B, respectively.

When the concentration of A is halved, the term \( [A]^2 \) decreases by a factor of 4 because the square of a halved concentration is \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \). This causes the rate of the reaction to become \( \frac{1}{4} \) of the original rate.

To maintain the same rate, the concentration of B must be increased to compensate for the decrease in \( [A]^2 \). Since the rate is directly proportional to \( [B] \), the concentration of B must be doubled to counteract the reduction in \( [A]^2 \) and keep the rate constant.

Therefore, to maintain the same rate, the concentration of B must be doubled.