# In a reaction (at equilibrium) that makes more moles of gas than it consumes, what is the effect of increasing the pressure?

Feb 9, 2017

Well, if you make more mols of gas...

$a A \left(g\right) \to b B \left(g\right)$

where $b > a$ and these are stoichiometric coefficients.

Write the equilibrium constant.

${K}_{P} = {P}_{B}^{b} / \left({P}_{A}^{a}\right)$

As $b > a$, let $b = a + c$, where $c$ is any positive constant. Then:

${K}_{P} = {P}_{B}^{a + c} / \left({P}_{A}^{a}\right)$

$= {\left({P}_{B} / {P}_{A}\right)}^{a} {P}_{B}^{c}$

If you increase the total pressure, you must increase the partial pressures.

Even if the mols of $B$ are equal to the mols of $A$ (instead of larger), even though the partial pressures increase by the same amount, the ${P}_{B}^{c}$ out front causes the ${K}_{P}$ to increase overall.

Therefore, the equilibrium is skewed towards the products, and by Le Chatelier's principle, it wishes to undo the imbalance and shift towards the reactants.