How does partial pressure affect Gibbs free energy?

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How does partial pressure affect Gibbs free energy?

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Answer 1 · 2015-12-17T22:16:36

$Δ G = Δ G^{\circ} + R T ln (\frac{{(P_{N H_{3}})}_{N H_{3}}^{2}}{(P_{N_{2}}) {(P_{H_{2}})}_{H_{2}}^{3}})$ $DeltaG=DeltaG^@+RTln(((P_(NH_3))^2)/((P_(N_2))(P_(H_2))^3))$

Explanation:

Consider the following reaction:

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ $N_2(g) +3H_2(g) rightleftharpoons 2NH_3(g)$

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

$Δ G = Δ G^{\circ} + R T ln K$ $DeltaG=DeltaG^@+RTlnK$

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

$K = \frac{{(P_{N H_{3}})}_{N H_{3}}^{2}}{(P_{N_{2}}) {(P_{H_{2}})}_{H_{2}}^{3}}$ $K=((P_(NH_3))^2)/((P_(N_2))(P_(H_2))^3)$

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

$Δ G = Δ G^{\circ} + R T ln (\frac{{(P_{N H_{3}})}_{N H_{3}}^{2}}{(P_{N_{2}}) {(P_{H_{2}})}_{H_{2}}^{3}})$ $DeltaG=DeltaG^@+RTln(((P_(NH_3))^2)/((P_(N_2))(P_(H_2))^3))$

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How does partial pressure affect Gibbs free energy?

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