Answer:
ΔfG°(C₆H₆(g)) = 129.7kJ/mol
Explanation:
Bringing out the parameters mentioned in the question;
Vapor pressure = 94.4 mm of Hg
The vaporization reaction is given as;
C₆H₆(l) ⇄ C₆H₆(g)
Equilibrium in terms of activities is given by:
K = a(C₆H₆(g)) / a(C₆H₆(l))
Activity of pure substances is one:
a(C₆H₆(l)) = 1
Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions
K = p(C₆H₆(g)) / p°
Where p° = 1atm = 760mmHg standard pressure
We now have;
K = 94mmHg / 760mmHg = 0.12421
Gibbs free energy is given as;
ΔG = - R·T·ln(K)
where R = gas constant = 8.314472J/molK
So ΔG° of vaporization of benzene is:
ΔvG° = - 8.314472 · 298.15 · ln(0.12421)
ΔvG° = 5171J/mol = 5.2kJ/mol
Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:
ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))
Hence:
ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))
ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol
ΔfG°(C₆H₆(g)) = 129.7kJ/mol
