Answer:
a) ΔG° = 132.5 kJ
b) ΔG° = 13.68 kJ
c) ΔG° = -58.58 kJ
Explanation:
The values of the enthalpy of formation(H°f) and the entropy (S°) for the substances in the reaction are:
CaCO₃(s): H°f = -1206.9 kJ/mol S° = 92.9 J/K.mol
CaO(s): H°f = -635.09 kJ/mol S° = 39.75 J/K.mol
CO₂(g): H°f = -393.51 kJ/mol S° = 213.74 J/K.mol
ΔH° = ∑n*H°f,products - ∑n*H°f,reactants (where n is the number of moles)
ΔH° = (-393.51 + (-635.09)) - (-1206.9) = 178.3 kJ
ΔS° = ∑n*S°,products - ∑n*S°,reactants
ΔS° = (213.74 + 39.75) - 92.9 = 160.59 J/K = 0.1606 kJ/K
ΔG° = ΔH° - TΔS°
a) ΔG° = 178.3 - 285*0.1606 = 132.5 kJ
b) ΔG° = 178.3 - 1025*0.1606 = 13.68 kJ
c) ΔG° = 178.3 - 1475*0.1606 = -58.58 kJ
