JEE Main · 2021 · Shift-IImediumTHERMO-071

The incorrect expression among the following is: (1) For isothermal process wreversible = -nRTVfVi (2) K = H - T SRT…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

The incorrect expression among the following is:

(1) For isothermal process wreversible=−nRTln⁡VfViw_{\text{reversible}} = -nRT\ln\frac{V_f}{V_i} (2) ln⁡K=ΔH∘−TΔS∘RT\ln K = \frac{\Delta H^\circ - T\Delta S^\circ}{RT} (3) ΔGSystemΔSTotal=−T\frac{\Delta G_{\text{System}}}{\Delta S_{\text{Total}}} = -T (at constant P) (4) K=e−ΔG∘/RTK = e^{-\Delta G^\circ/RT}

Options
  1. a

    For isothermal process: wrev=−nRTln⁡VfViw_{\text{rev}}=-nRT\ln\frac{V_f}{V_i}

  2. b

    ln⁡K=ΔH∘−TΔS∘RT\ln K=\frac{\Delta H^\circ-T\Delta S^\circ}{RT}

    ✓
  3. c

    ΔGSystemΔSTotal=−T\frac{\Delta G_{\text{System}}}{\Delta S_{\text{Total}}}=-T (at constant P)

  4. d

    K=e−ΔG∘/RTK=e^{-\Delta G^\circ/RT}

Correct Answerb

ln⁡K=ΔH∘−TΔS∘RT\ln K=\frac{\Delta H^\circ-T\Delta S^\circ}{RT}

Detailed Solution

🧠 ln⁡K=−ΔG∘/RT\ln K = -\Delta G^\circ / RT — the sign matters Starting from ΔG∘=ΔH∘−TΔS∘=−RTln⁡K\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -RT\ln K: ln⁡K=−(ΔH∘−TΔS∘)RT=TΔS∘−ΔH∘RT\ln K = \frac{-(\Delta H^\circ - T\Delta S^\circ)}{RT} = \frac{T\Delta S^\circ - \Delta H^\circ}{RT}

Option (2) writes ln⁡K=ΔH∘−TΔS∘RT\ln K = \frac{\Delta H^\circ - T\Delta S^\circ}{RT} — the sign in the numerator is flipped. That makes it incorrect.

🗺️ Checking the rest (1) wrev=−nRTln⁡(Vf/Vi)w_{\text{rev}} = -nRT\ln(V_f/V_i) — correct isothermal reversible work. ✓

(3) ΔGsystem/ΔStotal=−T\Delta G_{\text{system}}/\Delta S_{\text{total}} = -T: since ΔG=−TΔStotal\Delta G = -T\Delta S_{\text{total}} at constant PP, dividing both sides by ΔStotal\Delta S_{\text{total}} gives −T-T. ✓

(4) K=e−ΔG∘/RTK = e^{-\Delta G^\circ/RT}: rearranging ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln K gives this directly. ✓

Only expression (2) = option (b) is wrong.

⚠️ Trap: The minus sign in ln⁡K=−ΔG∘/RT\ln K = -\Delta G^\circ/RT trips many students. A negative ΔG∘\Delta G^\circ gives a positive ln⁡K>0\ln K > 0, i.e., K>1K > 1 — which makes sense for a spontaneous reaction.

Answer: (b)\boxed{\text{Answer: (b)}}

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