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JEE Main 2017 · Chemistry · Chemical Thermodynamics · medium

For a reaction A→ B, Δ H=+24 kJ mol^-1 and Δ S=+80 J K^-1mol^-1, both independent of temperature. The equilibrium constant is exactly 1 at a certain…

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For a reaction A→B\mathrm{A\to B}, ΔH=+24 kJ mol−1\Delta H=+24\ \mathrm{kJ\,mol^{-1}} and ΔS=+80 J K−1mol−1\Delta S=+80\ \mathrm{J\,K^{-1}mol^{-1}}, both independent of temperature. The equilibrium constant is exactly 11 at a certain temperature T0T_0. The equilibrium constant at 2T02T_0 is nearly (R=8.314 J K−1mol−1R=8.314\ \mathrm{J\,K^{-1}mol^{-1}})
  1. A. 5.65.6
  2. B. 1.2×1021.2\times10^{2}
  3. C. 1818
  4. D. 2.4×1032.4\times10^{3}
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Answer: (B) 1.2×1021.2\times10^{2}

K=1K=1 means ΔG=0\Delta G=0, so T0=ΔH/ΔS=24000/80=300T_0=\Delta H/\Delta S=24000/80=300 K. At 2T0=6002T_0=600 K: ΔG=24000−600×80=−24000 J mol−1\Delta G=24000-600\times80=-24000\ \mathrm{J\,mol^{-1}}. Then ln⁡K=24000/(8.314×600)=4.81\ln K=24000/(8.314\times600)=4.81 and K=e4.81≈1.2×102K=e^{4.81}\approx1.2\times10^{2}. Using ΔG\Delta G at 300300 K (zero) would wrongly give K=1K=1.
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