For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species.?

For the reaction

C2H6(g)+H2(g)↽−−⇀2CH4(g)

 

the standard change in Gibbs free energy is Δ𝐺°=−32.8 kJ/mol . What is ΔG for this reaction at 298 K when the partial pressures are 𝑃C2H6=0.100 atm , 𝑃H2=0.450 atm , and 𝑃CH4=0.700 atm ?

1 Answer

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  • 6 months ago

    C₂H₆(g) + H₂(g) ⇌ 2CH₄(g)

    ΔG

    = ΔG° + RT ln{𝑃CH₄²/(𝑃C₂H₆ × 𝑃H₂)}

    = -32.8 + (8.314 × 10⁻³) × 298 × ln{0.700²/(0.100 × 0.450) kJ/mol

    = -26.9 kJ/mol

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