Question

A reaction \( A_2 + B \to \) Products, involves the following mechanism: \[ A_2 \rightleftharpoons 2A \text{ (fast)} \quad \text{(A being the intermediate)} \] \[ A + B \xrightarrow{k_2} \text{ Products (slow)}. \] The rate law consistent to this mechanism is:

• A. rate = \( k[A_2][B] \)
• B. rate = \( k[A_2]^2[B] \)
• C. rate = \( k[A_2]^{1/2}[B] \)
• D. rate = \( k[A_2][B]^2 \)

Hint:

When an intermediate is involved, express its concentration using the equilibrium constant for its formation.

Answer 1

The Correct Option is C

In the given mechanism, \( A_2 \) is in equilibrium and is an intermediate, so the rate law for the overall reaction depends on the slow step. Since \( A_2 \) is decomposed into 2A in the fast step, the rate is proportional to \( [A_2]^{1/2} \) due to its equilibrium concentration. Therefore, the rate law is \( k[A_2]^{1/2}[B] \).
Step 2: Conclusion.
The correct answer is (C), rate = \( k[A_2]^{1/2}[B] \).