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JEE Main 2022 · Chemistry · Chemical Thermodynamics · medium

For N2(g)+3H2(g)rightleftharpoons2NH3(g), Δ H^°=-92 kJ and Δ S^°=-200 J K^-1 (both assumed temperature independent). The equilibrium constant at 500…

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For N2(g)+3H2(g)⇌2NH3(g)\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}, ΔH∘=−92 kJ\Delta H^\circ=-92\ \mathrm{kJ} and ΔS∘=−200 J K−1\Delta S^\circ=-200\ \mathrm{J\,K^{-1}} (both assumed temperature independent). The equilibrium constant at 500 K500\ \mathrm K (R=8.314 J mol−1K−1R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}) is about
  1. A. 1.5×10−11.5\times10^{-1}
  2. B. 6.96.9
  3. C. 2.1×10−22.1\times10^{-2}
  4. D. 4.1×1094.1\times10^{9}
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Answer: (A) 1.5×10−11.5\times10^{-1}

ΔG∘=ΔH∘−TΔS∘=−92000−500(−200)=+8000 J\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ=-92000-500(-200)=+8000\ \mathrm J. Then ln⁡K=−ΔG∘/RT=−8000/(8.314×500)=−1.925\ln K=-\Delta G^\circ/RT=-8000/(8.314\times500)=-1.925, so K=e−1.925=0.146≈1.5×10−1K=e^{-1.925}=0.146\approx1.5\times10^{-1}. The positive ΔG∘\Delta G^\circ at this temperature means the reaction is not spontaneous under standard conditions. Ignoring the entropy term would give K≈4×109K\approx4\times10^9; a sign slip gives 6.96.9.
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