Question 16.4: DETERMINING WHETHER A REACTION IS SPONTANEOUS Consider the o...
DETERMINING WHETHER A REACTION IS SPONTANEOUS
Consider the oxidation of iron metal:
4 Fe(s) + 3 O_{2}(g) → 2 Fe_{2}O_{3}(s)
By determining the sign of ΔS_{total}, show 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 in Table 16.1. To obtain ΔS_{surr} = -ΔH°/T , calculate ΔH° for the reaction from standard enthalpies of formation (Section 8.9).
TABLE 16.1 Standard Molar Entropies for Some Common Substances at 25 °C
| Substance Formula S°[J/(K · mol)] | Substance Formula S°[J/(K · mol)] |
| Gases
Acetylene C_{2}H_{2} 200.8 Ammonia NH_{3} 192.3 Carbon dioxide CO_{2} 213.6 Carbon monoxide CO 197.6 Ethylene C_{2}H_{4} 219.5 Hydrogen H_{2} 130.6 Methane CH_{4} 186.2 Nitrogen N_{2} 191.5 Nitrogen dioxide NO_{2} 240.0 Dinitrogen tetroxide N_{2}O_{4} 304.3 Oxygen O_{2} 205.0 |
Liquids
Acetic acid CH_{3}CO_{2}H 160 Ethanol CH_{3}CH_{2}OH 161 Methanol CH_{3}OH 127 Water H_{2}O 69.9 Solids Calcium carbonate CaCO_{3} 91.7 Calcium oxide CaO 38.1 Diamond C 2.4 Graphite C 5.7 Iron Fe 27.3 Iron(III) oxide Fe_{2}O_{3} 87.4 |
= (2 mol)\left(87.4 \frac{J}{K · mol} \right) – \left[(4 mol)\left(27.3 \frac{J}{K · mol}\right) + (3 mol)\left(205.0 \frac{J}{K · mol}\right)\right]
= -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} (Appendix B), 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/K
ΔS_{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 standardstate conditions at 25 °C.
BALLPARK CHECK
Since the reaction consumes 3 mol of gas, ΔS_{sys} is negative. Because the oxidation (burning) of iron metal is highly exothermic, ΔS_{surr} = -ΔH°/T is positive and very large. The value ΔS_{surr} of is greater than the absolute value of ΔS_{sys} , and so ΔS_{total} is positive, in agreement with the solution.