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Question #44248

Difficulty: 8/10 | 🟢 Free

The Question

A physical chemistry student is tasked with determining the boiling point of an ideal binary liquid solution containing a non-volatile solute \ce{B} in a volatile solvent \ce{A} (mole fraction of \ce{A}, $x_A = 0.30$ ). For pure solvent \ce{A}, the normal boiling point is $T^\circ_b = 360 \text{ K}$ and the enthalpy of vaporization is $\Delta H^\circ_{vap} = 36.0 \text{ kJ mol}^{-1}$ . Assume $\Delta H^\circ_{vap}$ is independent of temperature.

The student determines the partial molar entropy of \ce{A} in the liquid solution at various temperatures by integrating $\frac{C_p}{T} dT$ from $0 \text{ K}$ . Assuming the mixture forms a perfect glass at $0 \text{ K}$ , the student incorrectly applies the Third Law of Thermodynamics ( $S_{0\text{K}} = 0$ ) to the mixture, ignoring any residual configurational entropy.

Using this flawed absolute entropy and the correct standard absolute entropy of \ce{A(g)} (obtained from statistical mechanics), the student calculates a boiling point for the solution, $T_{calc}$ . Recognizing the Third Law violation, the professor calculates the true boiling point, $T_{true}$ .

Given $R \ln(0.30) = -10.0 \text{ J K}^{-1} \text{ mol}^{-1}$ , determine the values of $T_{calc}$ and $T_{true}$ .

Options

A.
$T_{calc} = 360 \text{ K}, \quad T_{true} = 400 \text{ K}$
B.
$T_{calc} = 400 \text{ K}, \quad T_{true} = 360 \text{ K}$
C.
$T_{calc} = 360 \text{ K}, \quad T_{true} = 327 \text{ K}$
D.
$T_{calc} = 327 \text{ K}, \quad T_{true} = 400 \text{ K}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Third Law Application|Boiling Point Elevation (Solvent vs Solution)

Why This Question Is Tricky

Common Mistakes

This problem tests the boundary conditions of the Third Law of Thermodynamics and its connection to colligative properties.

Students often mechanically apply $S(0)=0$ without realizing that for non-crystalline systems (like a frozen glassy solution), residual configurational entropy exists.

The trap demonstrates a profound thermodynamic equivalence: ignoring the residual entropy of mixing at 0 K mathematically cancels out the entropy of mixing at higher temperatures, leading to the erroneous prediction that the solution has the same boiling point as the pure solvent (zero boiling point elevation).

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