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Question #44521

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The Question

Two volatile liquids, \ce{A} and \ce{B}, form a minimum boiling azeotrope at $T_{az} = 324 \text{ K}$ and $1 \text{ atm}$ . The azeotropic mixture has a mole fraction of \ce{A}, $x_A = 0.6$ . The pure liquids \ce{A} and \ce{B} have boiling points $T_A = 340 \text{ K}$ and $T_B = 390 \text{ K}$ at $1 \text{ atm}$ , respectively. The enthalpies of vaporization of pure \ce{A} and pure \ce{B} are $30.6 \text{ kJ mol}^{-1}$ and $35.1 \text{ kJ mol}^{-1}$ , respectively, and can be assumed to be independent of temperature. The enthalpy of mixing to form $1 \text{ mole}$ of the liquid azeotrope from pure liquids \ce{A} and \ce{B} at $324 \text{ K}$ is $\Delta H_{\text{mix}} = +1.62 \text{ kJ mol}^{-1}$ . Assuming the vapor phase behaves as an ideal gas mixture, what is the entropy of vaporization ( $\Delta S_{\text{vap}}$ ) of $1 \text{ mole}$ of the azeotropic mixture at its boiling point?

Options

A.
$90 \text{ J K}^{-1} \text{ mol}^{-1}$
B.
$95 \text{ J K}^{-1} \text{ mol}^{-1}$
C.
$100 \text{ J K}^{-1} \text{ mol}^{-1}$
D.
$105 \text{ J K}^{-1} \text{ mol}^{-1}$

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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

Azeotrope Boiling Point Logic|Entropy of Phase Change|Spontaneity Criteria|Molecular Mass Abnormalities

Why This Question Is Tricky

Common Mistakes

The question tests the thermodynamic understanding of non-ideal solutions forming azeotropes.

Students often fall into the trap of assuming that the enthalpy of vaporization of a mixture is simply the mole-fraction weighted average of the pure components (ideal behavior).

However, for a minimum boiling azeotrope, the positive deviation from Raoult's Law implies weaker intermolecular forces and an endothermic enthalpy of mixing ( $\Delta H_{\text{mix}} > 0$ ).

To vaporize the azeotrope, one must account for the energy already absorbed during mixing.

The hidden variable is failing to construct the correct thermodynamic cycle (Hess's Law) and incorrectly applying or ignoring $\Delta H_{\text{mix}}$ , leading to erroneous entropy values.

Distractor A uses the ideal boiling point, Distractor C ignores the enthalpy of mixing, and Distractor D adds the enthalpy of mixing instead of subtracting it.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Azeotrope Boiling Point Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free