Data given for the following reaction is as follows : ${\mathrm{FeO}}_{(s)}+{C}_{(\mathrm{graphite})}\longrightarrow {\mathrm{Fe}}_{(s)}+{\mathrm{CO}}_{(g)}$ Substance ${\Delta }_{f}H^{\circ}({\mathrm{kJmol}}^{-1})$ $\Delta S^{\circ}({\mathrm{Jmol}}^{-1}{K}^{-1})$ ${\mathrm{FeO}}_{(s)}$ $-266.3$ $57.49$ ${C}_{(\mathrm{graphite})}$ $0$ $5.74$ ${\mathrm{Fe}}_{(s)}$ $0$ $27.28$ ${\mathrm{CO}}_{(g)}$ $-110.5$ $197.6$ The minimum temperature in $K$ at which the reaction becomes spontaneous is_____.(Integer answer)

JEE Main 2021 — Chemistry Physical Chemistry

2021integerhard

Data given for the following reaction is as follows :

${\mathrm{FeO}}_{(s)}+{C}_{(\mathrm{graphite})}\longrightarrow {\mathrm{Fe}}_{(s)}+{\mathrm{CO}}_{(g)}$

Substance	${\Delta }_{f}H^{\circ}({\mathrm{kJmol}}^{-1})$	$\Delta S^{\circ}({\mathrm{Jmol}}^{-1}{K}^{-1})$
${\mathrm{FeO}}_{(s)}$	$-266.3$	$57.49$
${C}_{(\mathrm{graphite})}$	$0$	$5.74$
${\mathrm{Fe}}_{(s)}$	$0$	$27.28$
${\mathrm{CO}}_{(g)}$	$-110.5$	$197.6$

The minimum temperature in

K

at which the reaction becomes spontaneous is_____.(Integer answer)

Official previous-year question

Held on 27 Aug 2021 · Verified 6 Jul 2026.

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Solution

${T}_{\mathrm{min}}=(\frac{\Delta ^{\circ}H}{\Delta ^{\circ}S})$

${\Delta }^{o}{H}_{\mathrm{rxn}}=[{\Delta }_{f}^{^{\circ}}H({\mathrm{Fe}}_{(s)})+{\Delta }_{f}^{^{\circ}}H({\mathrm{CO}}_{(g)})]-[{\Delta }_{f}^{^{\circ}}H({\mathrm{FeO}}_{(s)})+{\Delta }_{f}^{^{\circ}}H({C}_{(\mathrm{graphite})})]$

$=[0-110.5]-[-266.3+0]$

$=155.8\mathrm{kJ}/\mathrm{mol}$

$\Delta ^{\circ}{S}_{\mathrm{rxn}}=[\Delta ^{\circ}S({\mathrm{Fe}}_{(s)})+\Delta ^{\circ}S({\mathrm{CO}}_{(g)})]-[\Delta ^{\circ}S({\mathrm{FeO}}_{(s)})+\Delta ^{\circ}S({C}_{(\mathrm{graphite})})]$

$=[27.28+197.6]-[57.49+5.74]$

$=161.65J/\mathrm{mol}.K$

${T}_{\mathrm{min}}=\frac{155.8\times {10}^{3}J/\mathrm{mol}}{161.65J/\mathrm{mol}-K}=963.8K$

$\approx 964K$ (nearest integer)

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Solution

${T}_{\mathrm{min}}=(\frac{\Delta ^{\circ}H}{\Delta ^{\circ}S})$

$=[0-110.5]-[-266.3+0]$

$=155.8\mathrm{kJ}/\mathrm{mol}$

$\Delta ^{\circ}{S}_{\mathrm{rxn}}=[\Delta ^{\circ}S({\mathrm{Fe}}_{(s)})+\Delta ^{\circ}S({\mathrm{CO}}_{(g)})]-[\Delta ^{\circ}S({\mathrm{FeO}}_{(s)})+\Delta ^{\circ}S({C}_{(\mathrm{graphite})})]$

$=[27.28+197.6]-[57.49+5.74]$

$=161.65J/\mathrm{mol}.K$

${T}_{\mathrm{min}}=\frac{155.8\times {10}^{3}J/\mathrm{mol}}{161.65J/\mathrm{mol}-K}=963.8K$

$\approx 964K$ (nearest integer)

JEE Main 2021 — Chemistry Physical Chemistry

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JEE Main 2021 — Chemistry Physical Chemistry

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JEE Main Physical Chemistry in other years

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