Question

Rate law for the reaction \(\text{aA} + \text{bB} \rightarrow \text{cC} + \text{dD}\) is \(r = k[A][B]\), the rate of reaction doubles if

• A. Concentration of both A and B are doubled.
• B. Concentration of A is doubled and concentration of B is kept constant.
• C. Concentration of B is doubled and concentration of A is halved.
• D. Concentration of A is kept constant and concentration of B is halved.

Hint:

In rate law \(r = k[A][B]\), doubling one reactant doubles the rate, while doubling both makes the rate four times.

Answer 1

The Correct Option is B

Concept:
Given rate law: \[ r = k[A][B] \] This means:
• first order with respect to A
• first order with respect to B So the rate depends directly on both concentrations.

Step 1: Check option (A).
If both \([A]\) and \([B]\) are doubled: \[ r' = k(2[A])(2[B]) = 4k[A][B] \] So rate becomes four times, not double.

Step 2: Check option (B).
If \([A]\) is doubled and \([B]\) is constant: \[ r' = k(2[A])([B]) = 2k[A][B] \] So rate becomes double.

Step 3: Check option (C) and (D).
For option (C): \[ r' = k\left(\frac{[A]}{2}\right)(2[B]) = k[A][B] \] Rate remains same. For option (D): \[ r' = k[A]\left(\frac{[B]}{2}\right) = \frac{1}{2}k[A][B] \] Rate becomes half. Hence, the correct answer is:
\[ \boxed{(B)\ \text{Concentration of A is doubled and concentration of B is kept constant}} \]