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(A) For a reaction to be spontaneous,$\Delta G < 0$. Since $\Delta G = \Delta H - T\Delta S$,the reaction becomes spontaneous when $\Delta H - T\De

Vedclass · Accepted Answer

$964$

Vedclass · Answer

(A) For a reaction to be spontaneous,$\Delta G < 0$. Since $\Delta G = \Delta H - T\Delta S$,the reaction becomes spontaneous when $\Delta H - T\Delta S < 0$,or $T > \frac{\Delta H}{\Delta S}$.
First,calculate $\Delta H^{\circ}_{rxn}$:
$\Delta H^{\circ}_{rxn} = [\Delta H^{\circ}_{f}(Fe) + \Delta H^{\circ}_{f}(CO)] - [\Delta H^{\circ}_{f}(FeO) + \Delta H^{\circ}_{f}(C_{(	ext{graphite})})]$
$= [0 + (-110.5)] - [-266.3 + 0] = 155.8 \ 	ext{kJ mol}^{-1} = 155800 \ 	ext{J mol}^{-1}$.
Next,calculate $\Delta S^{\circ}_{rxn}$:
$\Delta S^{\circ}_{rxn} = [S^{\circ}(Fe) + S^{\circ}(CO)] - [S^{\circ}(FeO) + S^{\circ}(C_{(	ext{graphite})})]$
$= [27.28 + 197.6] - [57.49 + 5.74] = 224.88 - 63.23 = 161.65 \ 	ext{J mol}^{-1} 	ext{K}^{-1}$.
The minimum temperature $T$ is given by:
$T = \frac{\Delta H^{\circ}_{rxn}}{\Delta S^{\circ}_{rxn}} = \frac{155800 \ 	ext{J mol}^{-1}}{161.65 \ 	ext{J mol}^{-1} 	ext{K}^{-1}} \approx 963.81 \ 	ext{K}$.
Rounding to the nearest integer,we get $964 \ 	ext{K}$.

Substance	$\Delta H^{\circ} \text{ (kJ mol}^{-1})$	$\Delta S^{\circ} \text{ (J mol}^{-1} \text{ K}^{-1})$
$FeO_{(s)}$	$-266.3$	$57.49$
$C_{(\text{graphite})}$	$0$	$5.74$
$Fe_{(s)}$	$0$	$27.28$
$CO_{(g)}$	$-110.5$	$197.6$

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