JEE Main 2025 · 29 January · morning · Single correct · +4 / −1
The reaction A₂ + B₂ → 2AB follows the mechanism:
A₂ ⇌(k₁) A + A (fast)
A + B₂ →(k₂) AB + B (slow)
A + B → AB (fast)
The overall order of the reaction is:
- (a) 3
- (b) 2.5
- (c) 1.5
- (d) 2
Answer and solution
Answer (c) 1.5
Solution
Chemistry ma’am · 3 steps
The slow step sets the rate, but it contains A, so swap A out using the fast equilibrium before it.
- Slow step. The slowest step is rate-determining, so rate = k₂[A][B₂]. But A is an intermediate. It does not appear in the overall equation, so its concentration cannot stay in the final rate law.
- Fast equilibrium. The first step reaches equilibrium quickly, so K = k₁/k₋₁ = [A]²/[A₂]. Solving for the intermediate gives [A] = K^(1/2)[A₂]^(1/2).
Rate = k₂ K^(1/2)[A₂]^(1/2)[B₂]
- So. Add the powers: 1/2 + 1 = 3/2. The overall order is 1.5, option (c).
The idea to keep. Write the rate from the slow step, then replace every intermediate using the fast equilibrium before it. A 2:1 split like A₂ ⇌ 2A always brings in a square root.
Answer from the NTA final answer key. Solution by PrepWiser.