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: Understanding the Concept:
The rate law is written from the given orders: first order in A and zeroth order in B. Zero order in B means B does not appear in the rate expression.

Step 2: Rate law:
\[ \text{Rate} = k[A]^1[B]^0 = k[A] \]

Step 3: Substitute:
\[ 6\times 10^{-4} = k\times 0.3 \]
\[ k = \frac{6\times 10^{-4}}{0.3} = 2\times 10^{-3}\ \text{s}^{-1} \]

Step 4: Check units:
For a first order reaction the units of \(k\) are \(s^{-1}\), which matches. The stoichiometric coefficients (2A and 3B) do not enter the rate law unless the reaction is elementary. So the answer is (B).

Final Answer:
k = rate / [A] = 2e-3 per second. \[ \boxed{2\times 10^{-3}\ \text{s}^{-1}} \]