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Chemical Kinetics and Nuclear Chemistry question

2010 · Shift 0 · Q19
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Chemical Kinetics and Nuclear Chemistry question

2010 · Shift 0 · Q19

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
Consider the reaction : Cl2Cl_2Cl2​(aq) + H2SH_2SH2​S(aq) → SSS(s) + 2H+H^+H+ (aq) + 2Cl−Cl^-Cl− (aq) The rate equation for this reaction is rate = k [Cl2Cl_2Cl2​] [H2SH_2SH2​S] Which of these mechanisms is/are consistent with this rate equation? (A) Cl2Cl_2Cl2​ + H2SH_2SH2​S →\to→ H+H^+H+ + Cl−Cl^-Cl− + Cl+Cl^+Cl+ + HS−HS^-HS− (slow) Cl+Cl^+Cl+ + HS−HS^-HS−→\to→ H+H^+H+ + Cl−Cl^-Cl− + SSS (fast) (B) H2SH_2SH2​S ⇔\Leftrightarrow⇔ H+H^+H+ + HS−HS^-HS− (fast equilibrium) Cl2Cl_2Cl2​ + HS−HS^-HS−→\to→ 2Cl−Cl^-Cl− + H+H^+H+ + SSS (slow)
  1. A
    B only
  2. B
    Both A and B
  3. C
    Neither A nor B
  4. D
    A only
View written solutionFree

Correct answer: D

  1. Given reaction and observed rate law

The overall reaction is

Cl2(aq)+H2S(aq)→S(s)+2H+(aq)+2Cl−(aq)Cl_2(aq)+H_2S(aq) \to S(s)+2H^+(aq)+2Cl^-(aq)Cl2​(aq)+H2​S(aq)→S(s)+2H+(aq)+2Cl−(aq)

with experimental rate law

rate=k[Cl2][H2S].\text{rate}=k[Cl_2][H_2S].rate=k[Cl2​][H2​S].

We must check which proposed mechanism gives this rate law.

  1. Check mechanism (A)

Mechanism (A):

  1. Slow step:
Cl2+H2S→H++Cl−+Cl++HS−Cl_2+H_2S \to H^+ + Cl^- + Cl^+ + HS^- Cl2​+H2​S→H++Cl−+Cl++HS−
  1. Fast step:
Cl++HS−→H++Cl−+SCl^+ + HS^- \to H^+ + Cl^- + SCl++HS−→H++Cl−+S

Rate from mechanism (A)

Since the first step is slow, it is the rate-determining step. So,

rate=k[Cl2][H2S].\text{rate}=k[Cl_2][H_2S].rate=k[Cl2​][H2​S].

This exactly matches the given rate law.

Check overall reaction

Adding the two steps:

  • Cl+Cl^+Cl+ and HS−HS^-HS− cancel.

So net reaction becomes

Cl2+H2S→2H++2Cl−+SCl_2+H_2S \to 2H^+ +2Cl^- +SCl2​+H2​S→2H++2Cl−+S

which is the given overall reaction.

Hence, mechanism (A) is consistent.

  1. Check mechanism (B)

Mechanism (B):

  1. Fast equilibrium:
H2S⇌H++HS−H_2S \rightleftharpoons H^+ + HS^-H2​S⇌H++HS−
  1. Slow step:
Cl2+HS−→2Cl−+H++SCl_2+HS^- \to 2Cl^- +H^+ +SCl2​+HS−→2Cl−+H++S

Rate from slow step

Since step 2 is slow,

rate=k[Cl2][HS−].\text{rate}=k[Cl_2][HS^-].rate=k[Cl2​][HS−].

But HS−HS^-HS− is an intermediate, so express it in terms of H2SH_2SH2​S. From the fast equilibrium,

K=[H+][HS−][H2S]K=\frac{[H^+][HS^-]}{[H_2S]}K=[H2​S][H+][HS−]​

Thus,

[HS−]=K[H2S][H+].[HS^-]=K\frac{[H_2S]}{[H^+]}.[HS−]=K[H+][H2​S]​.

Substitute into rate law:

rate=k[Cl2](K[H2S][H+])\text{rate}=k[Cl_2]\left(K\frac{[H_2S]}{[H^+] }\right)rate=k[Cl2​](K[H+][H2​S]​)

or

rate=k′[Cl2][H2S][H+]−1.\text{rate}=k'[Cl_2][H_2S][H^+]^{-1}.rate=k′[Cl2​][H2​S][H+]−1.

This is not equal to the observed rate law

k[Cl2][H2S].k[Cl_2][H_2S].k[Cl2​][H2​S].

It has an extra factor of [H+]−1[H^+]^{-1}[H+]−1.

Hence, mechanism (B) is not consistent.

  1. Evaluate options
  • (A) consistent
  • (B) not consistent

Therefore, the correct choice is:

D: A only\boxed{\text{D: A only}}D: A only​
  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So, the answers agree.

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