Calculating DGorxn
EXAMPLE: Combustion of acetylene
C2H2(g) + 5/2 O2(g) ? 2 CO2(g) + H2O(g)
From standard enthalpies of formation: DHorxn = -1238 kJ
From standard molar entropies: DSorxn = - 0.0974 kJ/K
Calculate DGorxn from DGo = ?Ho - T?So
DGorxn = -1238 kJ - (298 K)(-0.0974 kJ/K)
Reaction is product-favored in spite of negative DSorxn. Reaction is “enthalpy driven”