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Question #44813

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The Question

An optically pure sample of $(+)-2$ -iodooctane ( $0.10\text{ M}$ ) is treated with a large excess of radioactive $\ce{Na^{131}I}$ in acetone at $298\text{ K}$ . The isotopic exchange proceeds via an $\text{S}_\text{N}2$ mechanism with $100\%$ inversion of configuration. The pseudo-first-order half-life for the isotopic exchange process is measured to be $40\text{ minutes}$ . Calculate the time required for the observed optical rotation of the solution to drop to $12.5\%$ of its initial value.

Options

A.
$60\text{ minutes}$
B.
$120\text{ minutes}$
C.
$240\text{ minutes}$
D.
$80\text{ minutes}$

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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|Optical Activity and Enantiomers

Why This Question Is Tricky

Common Mistakes

The core trap tests the distinction between the rate of a chemical reaction (isotopic exchange) and the rate of change of a physical property (optical rotation).

Students often assume the half-life of the chemical concentration decay is identical to the half-life of the optical rotation decay.

However, because each inversion event not only consumes a $(+)$ -enantiomer but also produces a $(-)$ -enantiomer, the difference in concentration $([+] - [-])$ decreases twice as fast as the chemical exchange occurs.

Failing to account for the factor of 2 in $k_{rac} = 2k_{inv}$ leads to the incorrect assumption that $t_{1/2}(\alpha) = 40\text{ min}$ , resulting in the trap answer of $120\text{ minutes}$ .

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Kinetics - Class 12
Concept	First Order Half-Life Independence
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free