Question 16.6: CALCULATING ΔG° FOR A REACTION FROM ΔH° AND ΔS° Iron metal i...

(a) The following values of ΔH°_{f} and S° are found in Appendix B:

So we have

ΔH° = [2 ΔH°_{f}(Fe) + 3 ΔH°_{f}(CO_{2})] – [ΔH°_{f}(Fe_{2}O_{3}) + 3 ΔH°_{f}(CO)]

= [(2 mol)(0 kJ/mol) + (3 mol)(-393.5 kJ/mol)]

– [(1 mol)(-824.2 kJ/mol) + (3 mol)(-110.5 kJ/mol)]

ΔH° = -24.8 kJ

and

ΔS° = [2 S°(Fe) + 3 S°(CO_{2})] – [S°(Fe_{2}O_{3}) + 3 S°(CO)]

= \left[(2  mol)\left(27.3  \frac{J}{K  ·  mol}\right)  +  (3  mol)\left(213.6 \frac{J}{K  ·  mol}\right)\right]

– \left[(1  mol)\left(87.4 \frac{J}{K  ·  mol}\right)  +  (3  mol)\left(197.6 \frac{J}{K  ·  mol}\right)\right]

ΔS° = +15.2 J/K    or    0.0152 kJ/K

Therefore,

ΔG° = ΔH° – TΔS°

= (-24.8 kJ) – (298 K)(0.0152 kJ/K)

ΔG° = -29.3 kJ

(b) Because ΔG° is negative, the reaction is spontaneous at 25 °C. This means that a mixture of Fe_2O_{3}(s),  CO(g),  Fe(s),  and  CO_{2}(g), with each gas at a partial pressure of 1 atm, will react at 25 °C to produce more iron metal.
(c) Because ΔH° is negative and ΔS° is positive, ΔG° will be negative at all temperatures. The forward reaction is therefore spontaneous at all temperatures, and the reverse reaction does not become spontaneous at higher temperatures.