Question

The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at 298 K is 124.5 kJ/mol. What is the standard molar Gibbs free energy of formation of gaseous benzene at 298 K

Answer 1

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

Answer 2

The formation of gaseous benzene at 298 K has the standard Gibbs free energy value of 129.7 kJ/mol.

What are vapor pressure and Gibbs free energy?

Vapor pressure is the pressure exerted by the substance in the vapor phase. The given vapor pressure of benzene is 94.4 mm Hg.

The Gibbs free energy is the randomness and the free energy possessed by the compound. The Gibbs free energy of liquid benzene   ($\Delta G_l$) is 124.5 kJ/mol.

The pressure of liquid benzene is calculated as:

The equilibrium constant (K) is given as:

$K=\rm \dfrac{Vapor \;pressure}{Liquid\;pressure}$

Benzene is assumed in an ideal state. The pressure of the liquid in the ideal state is 760 mm Hg.

Substituting the values for the equilibrium constant:

$K=\dfrac{94}{760}\\\\ K=0.124$

The Gibbs free energy ($\Delta G$) for the compound is given as:

$\Delta G=RT\;\rm ln K$

The gas constant has the value of $R=8.314\;\rm J/mol.K$

The temperature of the liquid is, $T=298\;\rm K$

Substituting the values for the free energy of vaporization of benzene is:

$\Delta G_v=8.314\;\times\;298\;\times\;ln\;0.124\\\Delta G_v=\rm 5171\;J/mol\\\Delta \textit G_v=5.2\;kJ/mol$

The Gibbs free energy for the formation of gaseous benzene is given as the sum of the vaporization and the liquid formation. The Gibbs free energy for the formation of gaseous benzene ($\Delta G_f$) is given as:

$\Delta G_f=\Delta G_v+\Delta G_l\\\Delta G_f=\rm 5.2\;kJ/mol+124.5\;kJ/mol\\\Delta \textit{G}_\textit f=129.7\;kJ/mol$

The standard Gibbs free energy for the formation of gaseous benzene at 298 K is 129.7 kJ/mol.

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