Start Learning Free

Before You Begin

Question #44218

Difficulty: 8/10 | 🟢 Free

The Question

Consider a hydrogen-oxygen fuel cell operating reversibly and isothermally at 298 K298 \text{ K} and 1 bar1 \text{ bar}. The overall cell reaction is: \ceH2(g)+1/2O2(g)−>H2O(l)\ce{H2(g) + 1/2O2(g) -> H2O(l)} Given the standard thermodynamic data at 298 K298 \text{ K}: ΔfH∘(\ceH2O,l)=−285.8 kJ mol−1\Delta_f H^\circ(\ce{H2O, l}) = -285.8 \text{ kJ mol}^{-1} ΔfS∘(\ceH2O,l)=−163.3 J K−1 mol−1\Delta_f S^\circ(\ce{H2O, l}) = -163.3 \text{ J K}^{-1} \text{ mol}^{-1} R=8.314 J K−1 mol−1R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} Assume ideal gas behavior and negligible volume for liquid water. Which of the following represents the correct values of heat exchanged (qq) and the change in internal energy (ΔU\Delta U) for the consumption of 1 mole1 \text{ mole} of \ceH2(g)\ce{H2(g)}?

Options

  • A.

    q=−48.66 kJ mol−1q = -48.66 \text{ kJ mol}^{-1}, ΔU=−282.08 kJ mol−1\Delta U = -282.08 \text{ kJ mol}^{-1}

  • B.

    q=−285.80 kJ mol−1q = -285.80 \text{ kJ mol}^{-1}, ΔU=−282.08 kJ mol−1\Delta U = -282.08 \text{ kJ mol}^{-1}

  • C.

    q=−48.66 kJ mol−1q = -48.66 \text{ kJ mol}^{-1}, ΔU=0 kJ mol−1\Delta U = 0 \text{ kJ mol}^{-1}

  • D.

    q=−237.14 kJ mol−1q = -237.14 \text{ kJ mol}^{-1}, ΔU=−285.80 kJ mol−1\Delta U = -285.80 \text{ kJ mol}^{-1}

Check Your Answer

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

First Law Consistency|Fuel Cell Efficiency|Electrode Potential Sign Conventions

Why This Question Is Tricky

Common Mistakes

The problem tests the application of the First Law of Thermodynamics to electrochemical systems involving non-PV work.

Students often fall into two traps: (1) Assuming ΔU=0\Delta U = 0 because the process is isothermal, confusing ideal gas expansion with a chemical reaction. (2) Assuming q=ΔHq = \Delta H at constant pressure, forgetting that this relation only holds when non-PV work (like electrical work in a fuel cell) is zero.

By forcing the calculation of qq in a reversible fuel cell, students must recognize that qrev=TΔSq_{rev} = T\Delta S and ΔU\Delta U depends on ΔH\Delta H and ΔngRT\Delta n_g RT.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptFirst Law Consistency
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree