Question 17.4: Consider the oxidation of iron metal: 4 Fe(s) + 3 O2(g) → 2 ......
Consider the oxidation of iron metal:
4 Fe(s) + 3 O_{2}(g)\longrightarrow 2 Fe_{2}O_{3}(s)Determine the sign of ΔS_{total}, and decide whether the reaction is spontaneous at 25 °C.
STRATEGY
To determine the sign of ΔS_{total} = ΔS_{sys} + ΔS_{surr}, we need to calculate the values of ΔS_{sys} and ΔS_{surr}. The entropy change in the system equals the standard entropy of reaction and can be calculated using the standard molar entropies. To obtain ΔS_{surr} = -ΔH°/T, calculate ΔH° for the reaction from standard heats of formation (Section 9.9).
| IDENTIFY | |
| Known | Unknown |
| Values of ΔH°_{f} and ΔS° for reactants and products (Appendix B) | ΔS_{total} |
=(2 mol)(87.4\frac{J}{K \cdot mol})-[(4 mol)(27.3\frac{J}{K \cdot mol})+(3 mol)(205.0\frac{J}{K \cdot mol})]
= – 549.5 J/K
ΔH° = 2 ΔH°_{f}(Fe_{2}O_{3}) – [4 ΔH°_{f}(Fe) + 3 ΔH°_{f}(O_{2})]Because ΔH°_{f} = 0 for elements and ΔH°_{f} = -824.2 kJ/mol for Fe_{2}O_{3}, ΔH° for the reaction is
ΔH° = 2 ΔH°_{f}(Fe_{2}O_{3}) = (2 mol)(-824.2 kJ/mol) = -1648.4 kJ
Therefore, at 25 °C = 298.15 K,
ΔS_{surr}=\frac{-ΔH}{T}=\frac{-(-1,648,400 J)}{298.15 K}=5529 J/KΔS_{total} = ΔS_{sys} + ΔS_{surr} = -549.5 J/K + 5529 J/K = 4980 J/K
Because the total entropy change is positive, the reaction is spontaneous under standard-state conditions at 25 °C.