JEE Main 2020 · Chemistry · Chemical Thermodynamics · medium

For the reaction N2(g)+3H2(g)rightleftharpoons2NH3(g), Δ H^°=-92.0 kJ and Δ S^°=-198 J K^-1, both assumed independent of temperature. Using R=8.314 J…

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For the reaction

\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}

\Delta H^\circ=-92.0\ \mathrm{kJ}

and

\Delta S^\circ=-198\ \mathrm{J\,K^{-1}}

, both assumed independent of temperature. Using

R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}

and

2.303RT

with

T=450\ \mathrm{K}

, the value of

\log_{10}K

450\ \mathrm{K}

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Answer: (A) $0.34$

\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ=-92000-450(-198)=-92000+89100=-2900\ \mathrm{J}

. Then

\log K=-\dfrac{\Delta G^\circ}{2.303RT}=\dfrac{2900}{2.303\times8.314\times450}=0.337\approx0.34

. Using

\Delta H

alone would give

10.7

in magnitude, and the reaction is only marginally spontaneous near

T=\Delta H/\Delta S\approx465

K, where

K=1

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