Given the following data:
Substance
$Δ H^{\circ}$
(kJ/mol)
$S ° (J / molK)$
$Δ G^{\circ}$
(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?
$FeO (s) + C (graphite) ⟶ Fe (s) + CO (g)$

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$Δ G^{\circ}$ is +ve, hence the reaction will never be spontaneous

668 K

964 K

298 K

answer is C.

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Detailed Solution

Gibbs free energy for the given case,
$\begin{array}{l} Δ G^{\circ} = - 137.15 + 245.12 = + 107.97 \\ Δ H^{\circ} = - 110.5 + 266.3 = + 155.8 \\ Δ S^{\circ} = 27.28 + 197.6 - 5.74 - 57.49 = + 161.65 \\ T = \frac{Δ H^{\circ}}{Δ S^{\circ}} = \frac{155.8 \times 10^{3}}{161.65} = 964 K \end{array}$

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