Question

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.) 295K 1010K 1450K

Answer 1

Given:

Hof(CaCO3(s)) = -1206.92 KJ/mol

Hof(CaO(s)) = -635.09 KJ/mol

Hof(CO2(g)) = -393.509 KJ/mol

Balanced chemical equation is:

CaCO3(s) ---> CaO(s) + CO2(g)

ΔHo rxn = 1*Hof(CaO(s)) + 1*Hof(CO2(g)) - 1*Hof( CaCO3(s))

ΔHo rxn = 1*(-635.09) + 1*(-393.509) - 1*(-1206.92)

ΔHo rxn = 178.321 KJ

Given:

Sof(CaCO3(s)) = 92.9 J/mol.K

Sof(CaO(s)) = 39.75 J/mol.K

Sof(CO2(g)) = 213.74 J/mol.K

Balanced chemical equation is:

CaCO3(s) ---> CaO(s) + CO2(g)

ΔSo rxn = 1*Sof(CaO(s)) + 1*Sof(CO2(g)) - 1*Sof( CaCO3(s))

ΔSo rxn = 1*(39.75) + 1*(213.74) - 1*(92.9)

ΔSo rxn = 160.59 J/K

1)

So we have:

ΔHo rxn = 178.321 KJ

ΔSo rxn = 160.59 J/K

= 0.16059 KJ/K

T = 295 K

Now use:

ΔGo rxn = ΔHo rxn - T*ΔSo rxn

= 178.321 - 295.0*0.16059

= 131 KJ

Answer: 131 KJ

2)

Now use:

ΔGo rxn = ΔHo rxn - T*ΔSo rxn

= 178.321 - 1010*0.16059

= 16.1 KJ

Answer: 16.1 KJ

3)

Now use:

ΔGo rxn = ΔHo rxn - T*ΔSo rxn

= 178.321 - 1450*0.16059

= -54.5 KJ

Answer: -54.5 KJ