# How does partial pressure affect Gibbs free energy?

##### 1 Answer
Dec 17, 2015

$\Delta G = \Delta {G}^{\circ} + R T \ln \left(\frac{{\left({P}_{N {H}_{3}}\right)}^{2}}{\left({P}_{{N}_{2}}\right) {\left({P}_{{H}_{2}}\right)}^{3}}\right)$

#### Explanation:

Consider the following reaction:

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s 2 N {H}_{3} \left(g\right)$

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

$\Delta G = \Delta {G}^{\circ} + R T \ln K$

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

$K = \frac{{\left({P}_{N {H}_{3}}\right)}^{2}}{\left({P}_{{N}_{2}}\right) {\left({P}_{{H}_{2}}\right)}^{3}}$

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

$\Delta G = \Delta {G}^{\circ} + R T \ln \left(\frac{{\left({P}_{N {H}_{3}}\right)}^{2}}{\left({P}_{{N}_{2}}\right) {\left({P}_{{H}_{2}}\right)}^{3}}\right)$

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

Thermodynamics | Free Energy, Pressure & Equilibrium.