Question

The rate of the reaction \(2A+3B\rightarrow 2C+D\) is \(6\times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\), when \([A] = [B] = 0.3\text{ mol dm}^{-3}\). If the reaction is of first order for A and zeroth order for B, then find the rate constant.

• A. \(1\times 10^{-3}\text{ s}^{-1}\)
• B. \(2\times 10^{-3}\text{ s}^{-1}\)
• C. \(3\times 10^{-3}\text{ s}^{-1}\)
• D. \(4\times 10^{-3}\text{ s}^{-1}\)

Hint:

Rate = k[A], since B is zero order.

Answer 1

The Correct Option is B

Step 1: Idea:
Order is found from experiment and is given here. Use it directly.

Step 2: Divide:
$k = \text{rate}/[A] = (6\times 10^{-4})/(0.3) = 2\times 10^{-3}$.

Step 3: Unit check:
Rate has unit mol dm$^{-3}$ s$^{-1}$ and $[A]$ has mol dm$^{-3}$, so $k$ has unit s$^{-1}$.

Final Answer:
The B concentration does not matter in the rate law. \[ \boxed{B} \]