Answer:
ΔfG°(C₆H₆(g)) = 129.7kJ/mol
Explanation:
Identifying the given parameters from the question;
Vapor pressure = 94.4 mm of Hg
The reaction for vaporization is expressed as;
C₆H₆(l) ⇄ C₆H₆(g)
The equilibrium in terms of activities can be defined as:
K = a(C₆H₆(g)) / a(C₆H₆(l))
The activity for pure substances equals one:
a(C₆H₆(l)) = 1
For an ideal gas phase, activity is approximated as the ratio of partial pressure to total pressure. Under standard conditions:
K = p(C₆H₆(g)) / p°
Where p° = 1atm = 760mmHg is the standard pressure
Thus, we find;
K = 94mmHg / 760mmHg = 0.12421
The formula for Gibbs free energy is:
ΔG = - R·T·ln(K)
Here, R represents the gas constant = 8.314472J/molK
Consequently, the ΔG° for the vaporization of benzene is calculated as:
ΔvG° = - 8.314472 · 298.15 · ln(0.12421)
ΔvG° = 5171J/mol = 5.2kJ/mol
The change in Gibbs free energy for the reaction is determined by the difference between the Gibbs free energy of formation of the products and reactants:
ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))
<pThus:
ΔfG°(C₆H₆(g)) = ΔvG° + ΔfG°(C₆H₆(l))
ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol
ΔfG°(C₆H₆(g)) = 129.7kJ/mol
