Burdge/Overby, Chemistry: Atoms First, 2e Ch14 | Page 22
SEC TION 14.5 ? Predicting Spontaneity
591
Worked Example 14.6 demonstrates the calculation of standard free-energy changes.
Worked Example 14.6
Calculate the standard free-energy changes for the following reactions at 25°C:
(a) CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
2Mg(s) + O2(g)
(b) 2MgO(s)
Strategy Look up the ?G ° values for the reactants and products in each equation, and use
f
Equation 14.12 to solve for ?G° .
rxn
Setup From Appendix 2, we have the following values: ?G °[CH4(g)] = –50.8 kJ/mol,
f
?G °[CO2(g)] = –394.4 kJ/mol, ?G °[H2O(l)] = –237.2 kJ/mol, and ?G °[MgO(s)] = –569.6 kJ/mol.
f
f
f
All the other substances are elements in their standard states and have, by definition, ?G ° = 0.
f
Solution?
(a) ?G° = (?G °[CO2(g)] + 2?G °[H2O(l)]) ? (?G °[CH4(g)] + 2?G °[O2(g)])
rxn
f
f
f
f
= [(–394.4 kJ/mol) + (2)(–237.2 kJ/mol)] ? [(–50.8 kJ/mol) + (2)(0 kJ/mol)]
= –818.0 kJ/mol
(b) ?G ° = (2?G °[Mg(s)] + ?G °[O2(g)]) ? (2?G °[MgO(s)])
rxn
f
f
f
= [(2)(0 kJ/mol) + (0 kJ/mol)] ? [(2)(–569.6 kJ/mol)]
= 1139 kJ/mol
Think About It?
Note that, like standard enthalpies of formation (?H°), standard free energies of formation (?G°) depend
f
f
on the state of matter. Using water as an example, ?G°[H2O(l)] = –237.2 kJ/mol and ?G°[H2O(g)] =
f
f
–228.6 kJ/mol. Always double-check to make sure you have selected the right value from the table.
Practice Problem A t t e m p t ? Calculate the standard free-energy changes for the following
reactions at 25°C:
(a) H2(g) + Br2(l)
(b) 2C2H6(g) + 7O2(g)
2HBr(g)
4CO2(g) + 6H2 ?
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