Given

Fe₂O_3(s) + 3CO_(g) $$ \to $$ 2Fe_(s) + 3CO_2(g); $$\Delta $$H = $$-$$ 26.8 kJ .....(1)

FeO_(s) + CO(g) $$ \to $$  Fe_(s) + CO_2(g); $$\Delta $$H = $$-$$ 16.5 kJ .....(2)

Fe₂O_3(s) + CO_(g) $$ \to $$  2FeO_(s) + CO_2(g), $$\Delta $$H = ? ....(3)

Equation (3) can be calculated as :

(1) - 2(2)

$$ \therefore $$ $$\Delta $$H = –26.8 + 33.0 = +6.2 kJ

We know, from Gibb's equation,

$$\Delta $$G = $$\Delta $$H – T$$\Delta $$S

When $$\Delta $$G = 0, $$\Delta $$H = T$$\Delta $$S

$$ \therefore $$ $$T = {{\Delta H} \over {\Delta S}}$$ = $${{40.63 \times {{10}^3}} \over {108.8}}$$ = 373.4 K

Since the ideal gas expands spontaneously into vacuum,, P_ext = 0.

$$ \therefore $$ Work done is also zero.

When K_p > Q, rate of forward reaction > rate of backward reaction.

$$ \therefore $$ Reaction is spontaneous.

When $$\Delta $$G^o < RT ln Q, $$\Delta $$G^o is positive, reverse reaction is feasible, thus reaction is non spontaneous.

When K_p = Q, rate of forward reaction > rate of backward reaction.

$$ \therefore $$ Reaction is in equilibrium.

When T$$\Delta $$S > $$\Delta $$H, $$\Delta $$G will be negative only when $$\Delta $$H = +ve.

$$ \therefore $$ Reaction is spontaneous and endothermic.