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Question #44225

Difficulty: 8/10 | 🟢 Free

The Question

One mole of liquid \ce{X2} at 500 K500\text{ K} and 1 atm1\text{ atm} is placed in an insulated, constant-pressure chamber. A catalyst is introduced, triggering the endothermic decomposition and phase change: \ceX2(l)−>2X(g)\ce{X2(l) -> 2X(g)} The reaction proceeds via a highly reactive steady-state intermediate \ce{X^*(g)}, ensuring the process is strictly irreversible. The reaction goes to completion. Given the following thermodynamic parameters at 500 K500\text{ K}: ΔHrxn∘=10.0 kJ mol−1\Delta H^\circ_{\text{rxn}} = 10.0\text{ kJ mol}^{-1} Sm∘(\ceX2,l)=100 J K−1mol−1S^\circ_{m}(\ce{X2}, l) = 100\text{ J K}^{-1}\text{mol}^{-1} Sm∘(\ceX,g)=100 J K−1mol−1S^\circ_{m}(\ce{X}, g) = 100\text{ J K}^{-1}\text{mol}^{-1} Cp,m(\ceX,g)=25 J K−1mol−1C_{p,m}(\ce{X}, g) = 25\text{ J K}^{-1}\text{mol}^{-1} (assumed independent of temperature) Assume the heat capacity of the chamber is negligible. What is the total entropy change of the universe (ΔSuniv\Delta S_{\text{univ}}) for this process? (Take ln⁡(0.6)=−0.51\ln(0.6) = -0.51)

Options

  • A.

    0 J K^{-1}

  • B.

    20.0 J K^{-1}

  • C.

    74.5 J K^{-1}

  • D.

    125.5 J K^{-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

Heat Capacity Ratio|Entropy of Phase Change|Steady State Approximation

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to navigate non-equilibrium thermodynamics during a phase change.

The presence of a steady-state intermediate makes the phase change strictly irreversible.

A lethal trap is assuming that because the process involves a phase change, ΔSsys=ΔH/T\Delta S_{\text{sys}} = \Delta H/T (yielding Option B: 20 J K−120\text{ J K}^{-1}), or that an adiabatic process implies ΔSsys=0\Delta S_{\text{sys}} = 0 (Option A).

Students must recognize that for irreversible adiabatic processes, ΔSuniv>0\Delta S_{\text{univ}} > 0 and must be calculated via a reversible thermodynamic path that accounts for the changing heat capacity of the system as the gaseous mole fraction increases.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptHeat Capacity Ratio
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree