Consider the following reaction:CaCO3 (s) →CaO (s) + CO2 (g). Estimate ΔG∘ for this reaction at each of the following temperatures. (Assume that ΔH∘ and ΔS∘ do not change too much within the given temperature range.) A. 285 KΔG∘ = B. 1025 KΔG∘=C. 1475 KΔG∘ =

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