Question

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

Answer 1

The given reaction is

CH₃OH (l) <======> CH₃OH (g)

The equilibrium constant is given as

K_p = P_CH3OH where the P term denotes the partial pressure of CH₃OH vapor.

Since only CH₃OH vapor is present, we shall assume P_CH3OH = P_tot = 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, K_p = P when P is expressed in atm; Note that K_p must be dimensionless. Therefore, K_p = 0.1974.

Now use the equation ΔG⁰_rxn = -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).