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JEE Main 2023 · Chemistry · Chemical Thermodynamics · medium

For a reaction Δ H^°=+30 kJ mol^-1 and Δ S^°=+100 J K^-1 mol^-1 (both independent of temperature). The equilibrium constant at 600 K is closest to…

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For a reaction ΔH∘=+30 kJ mol−1\Delta H^\circ=+30\ \mathrm{kJ\,mol^{-1}} and ΔS∘=+100 J K−1 mol−1\Delta S^\circ=+100\ \mathrm{J\,K^{-1}\,mol^{-1}} (both independent of temperature). The equilibrium constant at 600 K600\ \mathrm{K} is closest to (R=8.314 J K−1 mol−1R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}})
  1. A. 4.1×1024.1\times10^{2}
  2. B. 2.4×10−32.4\times10^{-3}
  3. C. 1.6×1011.6\times10^{1}
  4. D. 6.06.0
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Answer: (A) 4.1×1024.1\times10^{2}

ΔG∘=ΔH∘−TΔS∘=30000−600(100)=−30000 J mol−1\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ=30000-600(100)=-30000\ \mathrm{J\,mol^{-1}}. Then ln⁡K=−ΔG∘/RT=30000/(8.314×600)=6.014\ln K=-\Delta G^\circ/RT=30000/(8.314\times600)=6.014, so K=e6.014≈4.1×102K=e^{6.014}\approx4.1\times10^{2}. Note the reaction is non-spontaneous below T=ΔH/ΔS=300T=\Delta H/\Delta S=300 K, so at 600600 K it is spontaneous and K>1K>1. Forgetting to convert kJ to J gives the nonsensical K≈e−0.006K\approx e^{-0.006} type answers.
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