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

Answer 1

Answer: (C)

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