Question

Knowing that the ∆G for the combustion of liquid ethanol (CH3CH2OH(l)) at the constant temperature of 302.1K and standard pressure is -1328.4 kJ/mol, derive the standard molar entropy of liquid ethanol. Give your answer in J K-1 mol-1. Use the following values, assuming that ∆H and ∆S are temperature-independent within the range considered: ∆_fH^o(CO2(g)) = -398.8 kJ/mol ∆_fH^o(H2O(l)) = -288.3 kJ/mol ∆_fH^o(Ethanol(l)) = -276.6 kJ/mol Sm^o(CO2(g)) = 211.1 J K-1 mol-1 Sm^o(H2O(l)) = 72.4 J K-1 mol-1 Sm^o(O2(g))= 202.8 J K-1 mol-1

Answer 1

The reaction is

C₂H₅OH + 3O₂ - - - - - - > 2CO₂ + 3H₂O

ΔH_reaction = ΣΔH_f (products) - ΣΔH_f(reactants)

=2 ΔH_f (CO₂) + 3 ΔH_f (H₂O) - ΔH_f (C₂H₅OH) - 3 ΔH_f (O₂)

= 2x (-398.8 kJ/mol) + 3 x (-288.3 kJ/mol) - (- 276.6 kJ/mol) - 0 = - 1385.9 kJ/mol

We know

ΔG_reaction = ΔH_reaction - TΔS_reaction

ΔS_reaction = (ΔH_reaction - ΔG_reaction ) / T

= (-1385.9 kJ/mol - (-1328.4 kJ/mol)) / 302.1K

= - 0.1903 kJ/mol K = - 190.3 J/mol

ΔS_reaction = ΣS_mo ( products) - ΣS_mo(reactants)

=2 S_mo (CO₂) +3 S_mo (H₂O) - S_mo(C₂H₅OH) - 3 S_mo(O₂ )

-190.3 J/mol K = 2x211.1 J/mol K + 3x72.4 J/mol K - S_mo(C₂H₅OH) - 3x202.8 J/mol K

So S_mo(C₂H₅OH) = - 221.3 J/mol K