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Question #44226

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The Question

Consider two independent systems at $298 \text{ K}$ :

\textbf{System I:} $x$ moles of an ideal monoatomic gas undergoes adiabatic free expansion from a volume $V$ to $10V$ . The entropy change of the universe for this process is $\Delta S_I$ .

\textbf{System II:} A galvanic cell is constructed as follows: $\ce{Pt | H2 (g, 1 bar) | HA (aq, 0.1 M) || HCl (aq, 0.1 M) | H2 (g, 1 bar) | Pt}$ where $\ce{HA}$ is a weak monoprotic acid with $K_a = 10^{-5}$ at $298 \text{ K}$ . The standard enthalpy of dissociation of $\ce{HA}$ is $\Delta H^\circ_{diss} = 17.115 \text{ kJ mol}^{-1}$ . The entropy change of the cell reaction ( $\Delta S_{rxn} = nF \left(\frac{\partial E}{\partial T}\right)_P$ ) per mole of $\ce{H2}$ oxidized is $\Delta S_{II}$ .

Given that $\Delta S_I = \Delta S_{II}$ , and assuming ideal behavior for all solutions and gases, what is the value of $x$ and the initial cell potential $E_{cell}$ at $298 \text{ K}$ ?

(Given: $\frac{2.303 RT}{F} = 0.059 \text{ V}$ and $2.303 RT = 5.705 \text{ kJ mol}^{-1}$ at $298 \text{ K}$ )

Options

A.
$x = 1$ , $E_{cell} = 0.118 \text{ V}$
B.
$x = 4$ , $E_{cell} = 0.118 \text{ V}$
C.
$x = 8$ , $E_{cell} = 0.236 \text{ V}$
D.
$x = 2$ , $E_{cell} = 0.236 \text{ V}$

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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

Adiabatic Free Expansion|Concentration Cells|Nernst Equation Concentration Dependence

Why This Question Is Tricky

Common Mistakes

The question tests the thermodynamic understanding of concentration cells involving temperature-dependent equilibria.

Students often memorize $\Delta S_{rxn} = nR \ln(C_2/C_1)$ for concentration cells, assuming it depends only on concentrations.

However, when a weak acid is involved, the concentration of $\ce{H+}$ itself depends on temperature via the dissociation constant $K_a$ .

By applying the Nernst equation and differentiating with respect to temperature, the hidden variable—the enthalpy of dissociation ( $\Delta H^\circ_{diss}$ )—emerges.

Ignoring this term leads to the trap answer $x=4$ .

Additionally, it tests the concept of adiabatic free expansion where $\Delta S_{univ}$ is solely the system's entropy change.

Quick Facts About This Question

Question Information

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