How does partial pressure affect Gibbs free energy?

1 Answer
Ernest Z.
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

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