Editing Le Chatelier's principle (section)

===Effect of change in volume===
Changing the volume of the system changes the partial pressures of the products and reactants and can affect the equilibrium concentrations.  With a pressure increase due to a decrease in volume, the side of the equilibrium with fewer moles is more favorable<ref name="Atkins-1993-p114">{{harvnb|Atkins|1993|p=114}}.</ref> and with a pressure decrease due to an increase in volume, the side with more moles is more favorable.  There is no effect on a reaction where the number of moles of gas is the same on each side of the chemical equation.

Considering the reaction of nitrogen gas with hydrogen gas to form ammonia:

:{{underset|4 moles|N<sub>2</sub> + 3 H<sub>2</sub>}} ⇌ {{underset|2 moles|2 NH<sub>3</sub>}}&nbsp;&nbsp;&nbsp;&nbsp;ΔH = −92kJ mol<sup>−1</sup>

Note the number of [[mole (unit)|moles]] of gas on the left-hand side and the number of moles of gas on the right-hand side. When the volume of the system is changed, the partial pressures of the gases change. If we were to decrease pressure by increasing volume, the equilibrium of the above reaction will shift to the left, because the reactant side has a greater number of moles than does the product side. The system tries to counteract the decrease in partial pressure of gas molecules by shifting to the side that exerts greater pressure. Similarly, if we were to increase pressure by decreasing volume, the equilibrium shifts to the right, counteracting the pressure increase by shifting to the side with fewer moles of gas that exert less pressure. If the volume is increased because there are more moles of gas on the reactant side, this change is more significant in the denominator of the [[equilibrium constant]] expression, causing a shift in equilibrium.
<!-- If we take the above reaction at [[standard conditions for temperature and pressure]] (STP), <math>K_c</math> would be as follow:

:<math>K_c=\frac{{[NH_3]} ^2} {{[N_2]}^1 {[H_2]}^3}</math>
::<math>=\frac{{(12)} ^2} {{(4)}^1 {(2)}^3}</math>
::<math>=1.125</math>

If we double the pressure of the above situation, by halving the volume of both sides then <math>K_c</math> would now be as follow:

:<math>K_c=\frac{{[NH_3]} ^2} {{[N_2]}^1 {[H_2]}^3}</math>
::<math>=\frac{{(6)} ^2} {{(2)}^1 {(1)}^3}</math>
::<math>=18</math> -->