Question

The rate of reaction \(\text{A}+\text{B}\rightarrow \text{P}\) is \(4\times 10^{-2} \text{mol dm}^{-3} \text{s}^{-1}\)
When \([A] = 0.2 \text{mole dm}^{-3}\) and \([B] = 0.1 \text{mole dm}^{-3}\), What is the rate constant of reaction, if it is first order with respect to A and second order with respect to B ?

• A. \(10 \text{mol}^{-2}\text{dm}^6 \text{s}^{-1}\)
• B. \(20 \text{mol}^{-2}\text{dm}^6 \text{s}^{-1}\)
• C. \(25 \text{mol}^{-2}\text{dm}^6 \text{s}^{-1}\)
• D. \(40 \text{mol}^{-2}\text{dm}^6 \text{s}^{-1}\)

Hint:

Rate = k [A][B]^2, so k = rate / ([A][B]^2).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The rate law with order 1 in A and order 2 in B is \(r = k[A][B]^2\). The overall order is 3, so the unit of k is \(\text{mol}^{-2}\text{dm}^{6}\text{s}^{-1}\).

Step 2: Detailed Explanation
Given \(r = 4 \times 10^{-2}\), \([A] = 0.2\), \([B] = 0.1\).
\[ k = \frac{r}{[A][B]^2} = \frac{4 \times 10^{-2}}{0.2 \times (0.1)^2} \]
\[ k = \frac{4 \times 10^{-2}}{0.2 \times 0.01} = \frac{4 \times 10^{-2}}{2 \times 10^{-3}} = 20 \]
So \(k = 20 \text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}\). Option (A), 10, comes from forgetting to square [B] properly, and (C), 25, from a wrong denominator.

Final Answer:
The rate constant is 20 \(\text{mol}^{-2}\text{dm}^6\text{s}^{-1}\), option (B). \[ \boxed{20 \text{ mol}^{-2}\text{dm}^6\text{s}^{-1}} \]