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Question #44246

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The Question

In an aqueous solution at $298\text{ K}$ , D-glucose undergoes mutarotation to establish an equilibrium between its $\alpha$ and $\beta$ anomers. The standard Gibbs free energy change for the conversion $\alpha\text{-D-glucopyranose} \rightleftharpoons \beta\text{-D-glucopyranose}$ is $\Delta G^\circ = -RT \ln 2$ . This solution is rapidly flash-frozen to $0\text{ K}$ , forming a kinetically trapped, completely random solid solution of the two anomers without any phase separation. Assuming no other sources of disorder, what is the residual molar entropy ( $S_{m,0}$ ) of this solid at $0\text{ K}$ ?

Options

A.
$R \left( \ln 3 - \frac{2}{3} \ln 2 \right)$
B.
$0$
C.
$R \ln 2$
D.
$R \left( \frac{1}{3} \ln 3 + \frac{2}{3} \ln 2 \right)$

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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

Third Law Application|D-Glucose Anomeric Equilibrium

Why This Question Is Tricky

Common Mistakes

This problem tests the limits of the Third Law of Thermodynamics.

Students often assume that the entropy of any solid at $0\text{ K}$ is strictly zero.

However, when an equilibrium mixture of isomers (like the anomeric $\alpha$ / $\beta$ forms of D-glucose) is flash-frozen, the molecules are kinetically trapped in a random solid solution.

This frozen-in configurational disorder results in a non-zero residual entropy, calculated via statistical thermodynamics.

The problem elegantly bridges macroscopic equilibrium ( $\Delta G^\circ$ ) with microscopic microstates ( $S = k_B \ln W$ ).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Thermodynamics - Class 11 Part 1
Concept	Third Law Application
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free