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Question #44794

Difficulty: 8/10 | 🟢 Free

The Question

A $0.15 \text{ M}$ aqueous solution of pure $\alpha$ -D-glucopyranose is prepared at $298 \text{ K}$ . The specific rotations $[\alpha]_D^{25}$ of pure $\alpha$ -D-glucopyranose and $\beta$ -D-glucopyranose are $+110^\circ$ and $+20^\circ$ , respectively. Upon standing, the solution undergoes mutarotation to reach an equilibrium specific rotation of $+50^\circ$ . The mutarotation process follows reversible pseudo-first-order kinetics. If the specific rotation of the solution drops to $+80^\circ$ exactly $23.1 \text{ min}$ after preparation, what is the initial instantaneous rate of formation of the $\beta$ -anomer? (Given: $\ln 2 = 0.693$ )

Options

A.
$3.0 \times 10^{-3} \text{ M min}^{-1}$
B.
$4.5 \times 10^{-3} \text{ M min}^{-1}$
C.
$1.5 \times 10^{-3} \text{ M min}^{-1}$
D.
$2.2 \times 10^{-3} \text{ M min}^{-1}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Instantaneous vs. Average Rate|D-Glucose Anomeric Equilibrium

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between the observed rate constant ( $k_{obs}$ ) of a reversible first-order reaction and the actual forward rate constant ( $k_f$ ).

A common trap is assuming mutarotation is a simple $A \to B$ first-order process, leading students to incorrectly use $k_{obs}$ as the forward rate constant (yielding Option B).

Furthermore, it requires correctly applying the optical rotation data to find both the equilibrium constant and the pseudo-first-order rate constant, integrating thermodynamic and kinetic principles.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Kinetics - Class 12
Concept	Instantaneous vs. Average Rate
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free