Question

Calculate the standard Helmholtz energy of formation, ΔfA, of CH3OH(l) at 298 K from the standard Gibbs energy of formation and the assumption that H2 and O2 are perfect gases.

Answer 1

The reaction can be written as

C (graphite, s) + 2 H₂ (g) + ½ O₂ (g) ----à CH₃OH (l); Δ_fG⁰ = -166.2 kJ/mol; Δ_fH⁰ = -238.6 kJ/mol and S⁰ = 127 J/mol.K

The Helmholtz free energy of formation is given by

Δ_fA⁰ = Δ_fU⁰ – TS⁰

The reaction takes place under standard conditions and involves H₂ and O₂ gases, both of which behave ideally. The temperature remains constant and hence, Δ_fU⁰ = 0 (internal energy change for an ideal gas is zero).

Therefore, Δ_fA⁰ = -TS⁰ = -(298 K)*(127 J/mol.K) = -37846 J/mol = (-37846 J/mol)*(1 kJ/1000 J) = -37.846 kJ/mol (ans).