JEE Main 2018 · Chemistry · Chemical Thermodynamics · medium

For N2O4(g)rightleftharpoons 2NO2(g), Δ H^°=57.2 kJ mol^-1 and Δ S^°=175.7 J mol^-1K^-1, both assumed independent of temperature. The equilibrium…

Single correct option

For

\mathrm{N_2O_4(g)\rightleftharpoons 2NO_2(g)}

\Delta H^\circ=57.2\ \mathrm{kJ\,mol^{-1}}

and

\Delta S^\circ=175.7\ \mathrm{J\,mol^{-1}K^{-1}}

, both assumed independent of temperature. The equilibrium constant

K_p

(standard state

1

bar) at

400\ \mathrm{K}

is closest to (

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

)

Show answer and solution

Answer: (D) $51$

\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ=57200-400\times175.7=-13080\ \mathrm{J\,mol^{-1}}

. Then

\ln K=-\dfrac{\Delta G^\circ}{RT}=\dfrac{13080}{8.314\times400}=3.933

, so

K=e^{3.933}=51

. (The reaction becomes spontaneous standard-state above

T=\Delta H/\Delta S=326

K, so

K>1

400

K is expected;

0.0196

is the reciprocal, obtained by using the wrong sign of

\Delta G^\circ

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