Question

Given the following data:

Substances	$\Delta H^{\circ }$(kJ/ mol)	$S^{\circ }$ (J/mol K)	$\Delta G^{\circ }$ (kJ/mol)
FeO(s)	-266.3	57.49	-245 .12
C(Graphite)	0	5.74	0
Fe(s)	0	27.28	0
CO(g)	-110.5	197.6	-137.15

Determine at what temperature the following reaction is spontaneous? $FeO\left(s\right)+C(Graphite)\to Fe\left(s\right)+CO\left(g\right)$

• A. $298K$
• B. $668K$
• C. $964K$
• D. $\Delta G^{\circ }is+ve$, hence the reaction will never be spontaneous

Answer 1

Answer: (C)

$\Delta G=\Delta H-T\Delta S$
$\Delta H=\Sigma \Delta H_p-\Sigma \Delta H_R$
$\Delta H=-110.5+266.3=155.8kJ$
$\Delta S=\Sigma S_p-\Sigma S_R$
$\Delta S=197.6+27.28-\left(57.5+5.74\right)=161.64J/molK$
For reaction to be spontaneous, $\Delta G<0$
$\Delta H-T\Delta S<0T>\frac{\Delta H}{\Delta S}=\frac{155800}{161.64}=964K$