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Question #44810

Difficulty: 8/10 | 🟢 Free

The Question

The decomposition of a substrate SS in the presence of a homogeneous catalyst CC follows the mechanism:

\ceS+C<=>[k1][k−1]I−>[k2]P+C\ce{S + C <=>[k_1][k_{-1}] I ->[k_2] P + C}

The rate constants at 298 K298\text{ K} are given as k1=2×107 M−1s−1k_1 = 2 \times 10^7 \, \text{M}^{-1}\text{s}^{-1}, k−1=105 s−1k_{-1} = 10^5 \, \text{s}^{-1}, and k2=105 s−1k_2 = 10^5 \, \text{s}^{-1}. The initial concentration of the catalyst is [C]0=10−6 M[C]_0 = 10^{-6} \, \text{M} and the initial concentration of the substrate is [S]0=0.1 M[S]_0 = 0.1 \, \text{M}. Applying the steady-state approximation for the intermediate II, what is the time required for the concentration of SS to reduce to 0.05 M0.05 \, \text{M}?

(Assume the reaction volume remains constant and no products are initially present.)

Options

  • A.

    (0.5+0.1ln⁡2) s(0.5 + 0.1 \ln 2) \, \text{s}

  • B.

    (0.1ln⁡2) s(0.1 \ln 2) \, \text{s}

  • C.

    0.5 s0.5 \, \text{s}

  • D.

    (0.5−0.1ln⁡2) s(0.5 - 0.1 \ln 2) \, \text{s}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

First Order Half-Life Independence|Rate Law Derivation from Mechanism

Why This Question Is Tricky

Common Mistakes

This question tests the transition regime of saturation kinetics (Michaelis-Menten type).

Students often blindly apply the first-order half-life formula (t1/2=ln⁡2/kefft_{1/2} = \ln 2 / k_{eff}) or the zero-order half-life formula (t1/2=[S]0/2Vmaxt_{1/2} = [S]_0 / 2V_{max}).

The trap is that at [S]0=0.1 M[S]_0 = 0.1 \, \text{M} and KM=0.01 MK_M = 0.01 \, \text{M}, the reaction is neither purely zero-order nor purely first-order throughout the half-life period.

It starts predominantly zero-order, but the first-order term (KMln⁡2K_M \ln 2) is highly significant.

Failing to integrate the full rate law derived from the Steady-State Approximation leads to selecting the pure zero-order (0.5 s) or pure first-order (0.1ln⁡20.1 \ln 2 s) distractors.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterChemical Kinetics - Class 12
ConceptFirst Order Half-Life Independence
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree