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JEE Main 2019 · Chemistry · Chemical Thermodynamics · easy

For a reaction A → B, Δ H^° = +40 kJ mol^-1 and Δ S^° = +120 J K^-1 mol^-1, both independent of temperature. Taking R = 8.314 J K^-1 mol^-1, the…

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For a reaction A→B\mathrm{A \to B}, ΔH∘=+40 kJ mol−1\Delta H^\circ = +40\ \mathrm{kJ\,mol^{-1}} and ΔS∘=+120 J K−1 mol−1\Delta S^\circ = +120\ \mathrm{J\,K^{-1}\,mol^{-1}}, both independent of temperature. Taking R=8.314 J K−1 mol−1R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}, the equilibrium constant at 300 K300\ \mathrm{K} is
  1. A. 1.01.0
  2. B. 5.05.0
  3. C. 0.800.80
  4. D. 0.200.20
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Answer: (D) 0.200.20

ΔG∘=ΔH∘−TΔS∘=40000−300(120)=4000 J mol−1\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 40000 - 300(120) = 4000\ \mathrm{J\,mol^{-1}}. Then ln⁡K=−ΔG∘/RT=−4000/(8.314×300)=−1.604\ln K = -\Delta G^\circ/RT = -4000/(8.314\times300) = -1.604, so K=e−1.604=0.20K = e^{-1.604} = 0.20. ΔG∘>0\Delta G^\circ>0 means K<1K<1; the value 5.05.0 arises from forgetting the minus sign ; the reaction becomes spontaneous in the standard state only above T=ΔH/ΔS=333 KT = \Delta H/\Delta S = 333\ \mathrm{K}.
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