In: Chemistry
Consider the evaporation of methanol at 25.0°C
CH3OH(l) --> CH3OH(g)
Find ΔG°rxn at 25.0°C if the nonstandard pressure of CH3OH = 150.0 mmHg
The given reaction is
CH3OH (l) <======> CH3OH (g)
The equilibrium constant is given as
Kp = PCH3OH where the P term denotes the partial pressure of CH3OH vapor.
Since only CH3OH vapor is present, we shall assume PCH3OH = Ptot = P = 150 mm Hg.
Convert P to atm as below P = 150 mm Hg = (150 mm Hg)*(1 atm/760 mm Hg) = 0.1974 atm (1 atm = 760 mm Hg).
Therefore, Kp = P when P is expressed in atm; Note that Kp must be dimensionless. Therefore, Kp = 0.1974.
Now use the equation ΔG0rxn = -R*T*ln P where T = 25⁰C = 298 K; therefore,
ΔG0 = -(8.314 J/mol.K)*(298 K)*ln (0.1974) = -(2477.572 J/mol)*(-1.6225) = 4019.8606 J/mol = (4019.8606 J/mol)*(1 kJ/1000 J) = 4.0198 kJ/mol ≈ 4.02 kJ/mol (ans).