Exams › JEE Main › Chemistry
For the reaction
Cl₂(aq) + H₂S(aq) → S(s) + 2H^+(aq) + 2Cl^-(aq),
the experimentally observed rate law is
rate = k[Cl₂][H₂S].
Which mechanism below is consistent with this rate law?
- A. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁻ + HS⁻ (slow)
Cl⁻ + HS⁻ → H⁺ + Cl⁻ + S (fast)
- B. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁻ + HS⁻ (fast)
Cl⁻ + HS⁻ → H⁺ + Cl⁻ + S (slow)
- C. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁻ + HS⁻ (slow)
Cl⁻ + HS⁻ → H⁺ + Cl⁻ + S (slow)
- D. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁻ + HS⁻ (fast)
Cl⁻ + HS⁻ → H⁺ + Cl⁻ + S (fast)
Correct answer: A. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁻ + HS⁻ (slow)
Cl⁻ + HS⁻ → H⁺ + Cl⁻ + S (fast)
Solution
Observed rate = k[Cl2][H2S] is first order in each reactant, so the slow step must be the bimolecular reaction Cl2 + H2S. That is the mechanism whose FIRST step (Cl2 + H2S -> H+ + 2Cl- + HS-) is slow, giving rate = k[Cl2][H2S].
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