Question

Given the following data :

Substance	$\Delta$H0 (kJ/mol)	S0(J/mol K)	$\Delta$G0 (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)$\longrightarrow$Fe(s) + CO(g)

• A. 298 K
• B. 668 K
• C. 964 K
• D. DG° is +ve, hence the reaction will never be spontaneous

Answer 1

Answer: (C)

964 K

Answer 2

$\Delta$G = $\Delta$H – T$\Delta$S

$\Delta$H = $\sum_{}^{}{\Delta H_{P} -}\sum_{}^{}{\Delta H_{R}}$

$\Delta$H = – 110.5 + 266.3 = 155.8 KJ

$\Delta$S = $\sum_{}^{}{S_{P} - \sum_{}^{}S_{R}}$

$\Delta$S = 197.6 + 27.28 – (57.5 + 5.74) = 161.64 J/mole K

For reaction to be spontaneous, $\Delta$G < 0

$\Delta$H – T$\Delta$S < 0 $\Rightarrow$ T > $\frac{\Delta H}{\Delta S}$ = $\frac{155800}{161.64}$= 964 K