Question

1. Calculate the standard reaction entropy, enthalpy and Gibbs free energy for the following reactions

a) N2(g) + NO2(g) ---> NO(g) + N2O(g) and

b) 2F2(g) +2H2O(l) ---> 4HF(aq) + O2(g)

Answer 1

1)

a)

Given:

Hof(N2(g)) = 0.0 KJ/mol

Hof(NO2(g)) = 33.18 KJ/mol

Hof(NO(g)) = 90.25 KJ/mol

Hof(N2O(g)) = 82.05 KJ/mol

Balanced chemical equation is:

N2(g) + NO2(g) ---> NO(g) + N2O(g)

ΔHo rxn = 1*Hof(NO(g)) + 1*Hof(N2O(g)) - 1*Hof( N2(g)) - 1*Hof(NO2(g))

ΔHo rxn = 1*(90.25) + 1*(82.05) - 1*(0.0) - 1*(33.18)

ΔHo rxn = 139 KJ/mol

Given:

Sof(N2(g)) = 191.61 J/mol.K

Sof(NO2(g)) = 240.06 J/mol.K

Sof(NO(g)) = 210.76 J/mol.K

Sof(N2O(g)) = 219.85 J/mol.K

Balanced chemical equation is:

N2(g) + NO2(g) ---> NO(g) + N2O(g)

ΔSo rxn = 1*Sof(NO(g)) + 1*Sof(N2O(g)) - 1*Sof( N2(g)) - 1*Sof(NO2(g))

ΔSo rxn = 1*(210.76) + 1*(219.85) - 1*(191.61) - 1*(240.06)

ΔSo rxn = -1.06 J/mol.K

Given:

Gof(N2(g)) = 0.0 KJ/mol

Gof(NO2(g)) = 51.31 KJ/mol

Gof(NO(g)) = 86.55 KJ/mol

Gof(N2O(g)) = 104.2 KJ/mol

Balanced chemical equation is:

N2(g) + NO2(g) ---> NO(g) + N2O(g)

ΔGo rxn = 1*Gof(NO(g)) + 1*Gof(N2O(g)) - 1*Gof( N2(g)) - 1*Gof(NO2(g))

ΔGo rxn = 1*(86.55) + 1*(104.2) - 1*(0.0) - 1*(51.31)

ΔGo rxn = 139.44 KJ/mol

ΔHo rxn = 139 KJ/mol

ΔSo rxn = -1.06 J/mol.K

ΔGo rxn = 139.44 KJ/mol

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