Question

How does partial pressure affect Gibbs free energy?

Answer 1

If you increase the partial pressure of a product gas, `ΔG` becomes more positive. If you increase the partial pressure of a reactant gas, `ΔG` becomes more negative.

The equation for Gibbs free energy is

`ΔG = ΔG^o + RTlnQ`

For a gas phase reaction of the type

aA(g) + bB(g) ⇌ cC(g) + dD(g),

`Q_"P" = (P_"C"^"c"P_"D"^"d")/(P_"A"^"a"P_"B"^"b")`, so

`ΔG = ΔG^o +RTln((P_"C"^"c"P_"D"^"d")/(P_"A"^"a"P_"B"^"b"))`

This shows that if you increase the partial pressure of a product gas, `ΔG` becomes more positive.

If you increase the partial pressure of a reactant gas, `ΔG` becomes more negative.

EXAMPLE

For the reaction

2A(g) + 2B(g) ⇌ C(g) + 3D(g)

at 25 °C, the equilibrium partial pressures are `P_A` = 6.30 atm, `P_B"` = 7.20 atm, `P_C` = 6.40 atm, and `P_D` = 9.10 atm. What is the standard Gibbs free energy change for this reaction at 25 °C.

At equilibrium, `ΔG` = 0, so

`ΔG^o = -RTln((P_"C"P_"D"^3)/(P_"A"^2P_"B"^2))` =

-8.314 J·K⁻¹mol⁻¹ × 298 K × `ln((6.40 ×9.10^3)/(6.30^2 × 7.20^2))` =

-2478 J·K⁻¹mol⁻¹ × ln2.34 = -2110 J/mol = -2.11 kJ/mol