# How does partial pressure affect Gibbs free energy?

##### 1 Answer
Aug 17, 2014

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