We have already calculated ΔS° for the reaction CaCO3() → CaO () + CO2 (g) as 160.6 J/K. We can find ΔH° for the reaction in a similar manner:
ΔH° = (1 mol)[ΔH° for CaO(*)] + (1 mol)[ΔH° for CO2(g)] – (1 mol)[ΔH° for CaCO3(*)] =
(1 mol) (–635.3 kJ/mol) + (1 mol) (–393.5 kJ/mol) – (1 mol) (–1207.0 kJ/mol)
= 178.2 kJ
Now we use Equation 23.6 to find the value of ΔG°:
ΔG° = ΔH° – TΔS° = 178.2 kJ – 298.15 K ×160.6 JK×1 kJx1000 J
= 130.3 kJ
Notice that we used 298.15 K, or 25°C as the value of T. This is thermodynamic standard temperature. Also note how the ΔS° value was in joules per kelvin, while the ΔH° value was in kJ.
