How does partial pressure affect Gibbs free energy?

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Answer 1 · 2015-12-17T22:16:36

$DeltaG=DeltaG^@+RTln(((P_(NH_3))^2)/((P_(N_2))(P_(H_2))^3))$

Consider the following reaction:

$N_2(g) +3H_2(g) rightleftharpoons 2NH_3(g)$

The free energy of this equilibrium could be found by the following expression:

$DeltaG=DeltaG^@+RTlnK$

where $K$ is the equilibrium constant that is given by:

$K=((P_(NH_3))^2)/((P_(N_2))(P_(H_2))^3)$

Therefore, the Gibbs free energy is related to the partial pressure by:

$DeltaG=DeltaG^@+RTln(((P_(NH_3))^2)/((P_(N_2))(P_(H_2))^3))$

Here is a video that explains the origin on this expression in details:

Thermodynamics | Free Energy, Pressure & Equilibrium.