JEE Advanced 2022 Paper 2 · Chemistry · Chemical Thermodynamics · medium

For the gas-phase reaction N2(g)+3H2(g)rightleftharpoons2NH3(g) assume Δ H^°=-92.0 kJ mol^-1 and Δ S^°=-198.0 J K^-1 mol^-1, both independent of…

One or more options correct

For the gas-phase reaction

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

assume

\Delta H^\circ=-92.0\ \mathrm{kJ\,mol^{-1}}

and

\Delta S^\circ=-198.0\ \mathrm{J\,K^{-1}\,mol^{-1}}

, both independent of temperature, and

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

. Choose the correct statement(s).

A. The reaction is spontaneous under standard conditions at every temperature.
B. At $298\ \mathrm{K}$ , $\Delta G^\circ\approx-33\ \mathrm{kJ\,mol^{-1}}$ .
C. The reaction is spontaneous under standard conditions only above about $465\ \mathrm{K}$ .
D. At $500\ \mathrm{K}$ , the equilibrium constant $K_p$ (standard-state basis) is greater than $1$ .

Show answer and solution

Answer: B

Use

\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ

.
(A) False. Both

\Delta H^\circ

and

\Delta S^\circ

are negative, so

\Delta G^\circ

becomes positive at high temperature.
(B) True.

\Delta G^\circ(298)=-92.0-298\times(-0.198)=-92.0+59.0=-33.0\ \mathrm{kJ\,mol^{-1}}

.
(C) False.

\Delta G^\circ<0

requires

T<\Delta H^\circ/\Delta S^\circ=92000/198=464.6

K, so the reaction is spontaneous below, not above, about 465 K.
(D) False. At

500

\Delta G^\circ=-92.0+99.0=+7.0\ \mathrm{kJ\,mol^{-1}}

and

K_p=\exp(-7000/(8.314\times500))=0.186<1

(computed numerically). At 298 K, by comparison,

K_p=e^{13.3}\approx6\times10^5

.
Only (B) is correct.

Original question written in the JEE Advanced 2022 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake