Editing Le Chatelier's principle (section)

==Chemistry==
===Effect of change in concentration===
Changing the concentration of a chemical will shift the equilibrium to the side that would counter that change in concentration. The chemical system will attempt to partly oppose the change that affected the original state of equilibrium. In turn, the rate of reaction, extent, and yield of products will be altered corresponding to the impact on the system.

This can be illustrated by the equilibrium of [[carbon monoxide]] and [[hydrogen]] gas, reacting to form [[methanol]]. 

:[[Carbon|C]][[Oxygen|O]] + 2 H<sub>2</sub> ⇌ CH<sub>3</sub>OH

Suppose we were to increase the concentration of CO in the system. Using Le Chatelier's principle, we can predict that the concentration of methanol will increase, decreasing the total change in CO. If we are to add a species to the overall reaction, the reaction will favor the side opposing the addition of the species. Likewise, the subtraction of a species would cause the reaction to "fill the gap" and favor the side where the species was reduced. This observation is supported by the [[collision theory]]. As the concentration of CO is increased, the frequency of successful collisions of that reactant would increase also, allowing for an increase in forward reaction, and generation of the product. Even if the desired product is not [[thermodynamic]]ally favored, the end-product can be obtained if it is continuously removed from the [[Solution (chemistry)|solution]].

The effect of a change in concentration is often exploited synthetically for [[condensation reaction]]s (i.e., reactions that extrude water) that are equilibrium processes (e.g., the formation of an [[ester]] from carboxylic acid and alcohol or an [[imine]] from an amine and aldehyde).  This can be achieved by physically sequestering water, by adding desiccants like anhydrous magnesium sulfate or [[molecular sieve]]s, or by continuous removal of water by distillation, often facilitated by a [[Dean–Stark apparatus|Dean-Stark apparatus]].

===Effect of change in temperature===
[[File:NO2-N2O4.jpg|thumb|The [[reversible reaction]] 2NO<sub>2</sub>(g) ⇌ N<sub>2</sub>O<sub>4</sub>(g) is exothermic, so the equilibrium position can be shifted by changing the temperature.<br/>When heat is removed and the temperature decreases, the reaction shifts to the right and the flask turns colorless due to an increase in N<sub>2</sub>O<sub>4</sub>. This demonstrates Le Chatelier's principle: the equilibrium shifts in the direction that releases energy.<br/>When heat is added and the temperature increases, the reaction shifts to the left and the flask turns reddish brown due to an increase in NO<sub>2</sub>, in accordance with Le Chatelier's principle.]]
The effect of changing the temperature in the equilibrium can be made clear by 1) incorporating heat as either a reactant or a product, and 2) assuming that an increase in temperature increases the heat content of a system. When the reaction is [[exothermic]] (Δ''H'' is negative and energy is released), heat is included as a product, and when the reaction is [[endothermic]] (Δ''H'' is positive and energy is consumed), heat is included as a reactant. Hence, whether increasing or decreasing the temperature would favor the forward or the reverse reaction can be determined by applying the same principle as with concentration changes.

Take, for example, the [[reversible reaction]] of [[nitrogen]] gas with [[hydrogen]] gas to form [[ammonia]]:

:N<sub>2</sub>(g) + 3 H<sub>2</sub>(g) ⇌ 2 NH<sub>3</sub>(g)&nbsp;&nbsp;&nbsp;&nbsp;Δ''H'' = −92 [[Kilojoule|kJ]] mol<sup>−1</sup>

Because this reaction is exothermic, it produces heat:

:N<sub>2</sub>(g) + 3 H<sub>2</sub>(g) ⇌ 2 NH<sub>3</sub>(g) + ''heat''

If the temperature were increased, the heat content of the system would increase, so the system would consume some of that heat by shifting the equilibrium to the left, thereby producing less ammonia. More ammonia would be produced if the reaction were run at a lower temperature, but a lower temperature also lowers the rate of the process, so, in practice (the [[Haber process]]) the temperature is set at a compromise value that allows [[ammonia]] to be made at a reasonable rate with an equilibrium concentration that is not too unfavorable.

In [[exothermic reaction]]s, an increase in temperature decreases the [[equilibrium constant]], ''K'', whereas in [[endothermic reaction]]s, an increase in temperature increases ''K''.

Le Chatelier's principle applied to changes in concentration or pressure can be understood by giving ''K'' a constant value. The effect of temperature on equilibria, however, involves a change in the equilibrium constant. The dependence of ''K'' on temperature is determined by the sign of Δ''H''. The theoretical basis of this dependence is given by the [[Van 't Hoff equation]].

===Effect of change in pressure===
The equilibrium concentrations of the products and reactants do not directly depend on the [[Total pressure (fluids)|total pressure]] of the system. They may depend on the [[partial pressure]] of the products and reactants, but if the number of moles of gaseous reactants is equal to the number of moles of gaseous products, pressure has no effect on equilibrium.

Changing total pressure by adding an inert gas at constant volume does not affect the equilibrium concentrations (see {{slink||Effect of adding an inert gas}}).

Changing total pressure by changing the volume of the system changes the partial pressures of the products and reactants and can affect the equilibrium concentrations (see {{slink||Effect of change in volume}} below).

===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> -->

===Effect of adding an inert gas===
{{See also|#Effect of change in pressure}}
An [[inert gas]] (or [[noble gas]]), such as [[helium]], is one that does not react with other elements or compounds.  Adding an inert gas into a gas-phase equilibrium at constant volume does not result in a shift.<ref name=Atkins-1993-p114/> This is because the addition of a non-reactive gas does not change the equilibrium equation, as the inert gas appears on both sides of the chemical reaction equation.  For example, if A and B react to form C and D, but X does not participate in the reaction: <chem>\mathit{a}A{} + \mathit{b}B{} + \mathit{x}X <=> \mathit{c}C{} + \mathit{d}D{} + \mathit{x}X</chem>. While it is true that the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant; rather, it is a change in partial pressures that will cause a shift in the equilibrium. If, however, the volume is allowed to increase in the process, the partial pressures of all gases would be decreased resulting in a shift towards the side with the greater number of moles of gas. The shift will never occur on the side with fewer moles of gas. It is also known as Le Chatelier's postulate.

===Effect of a catalyst===
A [[catalyst]] increases the rate of a reaction without being consumed in the reaction.  The use of a catalyst does not affect the position and composition of the equilibrium of a reaction, because both the forward and backward reactions are sped up by the same factor.

For example, consider the Haber process for the synthesis of ammonia (NH<sub>3</sub>):

:N<sub>2</sub> + 3 H<sub>2</sub> ⇌ 2 NH<sub>3</sub>

In the above reaction, iron (Fe) and molybdenum (Mo) will function as catalysts if present. They will accelerate any reactions, but they do not affect the state of the equilibrium.