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JEE Main 2021 · Chemistry · Chemical Thermodynamics · medium

For the gas-phase reaction N2O4(g)rightleftharpoons 2NO2(g), Δ H^°=+58.0 kJ mol^-1 and Δ S^°=+176 J K^-1 mol^-1 (both assumed temperature…

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For the gas-phase reaction N2O4(g)⇌2NO2(g)\mathrm{N_2O_4}(g)\rightleftharpoons 2\mathrm{NO_2}(g), ΔH∘=+58.0 kJ mol−1\Delta H^\circ=+58.0\ \mathrm{kJ\,mol^{-1}} and ΔS∘=+176 J K−1 mol−1\Delta S^\circ=+176\ \mathrm{J\,K^{-1}\,mol^{-1}} (both assumed temperature independent). The value of KpK_p at 298 K298\ \mathrm{K} is closest to (R=8.314 J K−1 mol−1R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}, standard pressure 11 bar)
  1. A. 0.110.11
  2. B. 9.49.4
  3. C. 1.1×10−21.1\times10^{-2}
  4. D. 2.3×10−42.3\times10^{-4}
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Answer: (A) 0.110.11

First get ΔG∘=ΔH∘−TΔS∘=58000−298×176=5552 J mol−1\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ=58000-298\times176=5552\ \mathrm{J\,mol^{-1}}. Then ln⁡Kp=−ΔG∘RT=−55528.314×298=−2.241\ln K_p=-\dfrac{\Delta G^\circ}{RT}=-\dfrac{5552}{8.314\times298}=-2.241, so Kp=e−2.241=0.106≈0.11K_p=e^{-2.241}=0.106\approx0.11. The value 9.49.4 results from dropping the minus sign; 1.1×10−21.1\times10^{-2} results from a factor-of-ten slip in the exponent. Since ΔG∘>0\Delta G^\circ>0, Kp<1K_p<1 must hold, which already eliminates 9.49.4.
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