# General Chemistry/Chemical Equilibria/Le Chatelier's Principle

← Chemical Equilibria/Equilibrium ·Chemical Equilibria/Acid-Base Equilibrium →

## Le Chatelier's Principle

Le Chatelier's Principle states that when a system that is in dynamic equilibrium is disrupted in some way, the system will respond with chemical or physical changes to restore a new equilibrium state.
There are several changes that can effect the equilibrium position of a system:
* Concentration
* Pressure/Volume
* Temperature

### Concentration

If the concentrations in a system are changed, Le Chatelier's Principle predicts that the equilibrium position will shift to minimize the change.
$CO(g)+H_{2}O(g)\rightleftharpoons CO_{2}(g)+H_{2}(g)$
For example, let us say that the reaction above is at equilibrium. Adding more reactants ($CO(g)$ and $H_{2}O(g)$) will disturb the equilibrium system because it raises the reactant concentration. The system will then produce more products ($CO_{2}(g)$ and $H_{2}(g)$) and will decrease the reactant concentration in order to obtain equilibrium again. The system is now at a new equilibrium position, and the change created has been minimized.
The above example shows that increasing the reactants in an equilibrium system favors the products because the system produces more products and reduces the amount of reactants. The table below shows the response of an equilibrium system to changes in concentration:
| Change in Concentration | System response |
| Increase reactant | Favors products |
| Decrease reactant | Favors reactants |
| Increase product | Favors reactants |
| Decrease product | Favors products |

### Pressure/Volume

If some the substances in a system are gas, changing their partial pressure is the same as changing their concentrations.
If the volume is increased, the overall pressure decreases (and vice versa). Consider the reaction A + B → 3C. There are three moles of products for every two moles of reactants. The product side has higher pressure than the reactant side. So, if pressure is increased (or volume decreased), equilibrium will shift towards the lower pressure side. Products will be consumed (by the reverse reaction) and reactants will form.
If pressure is decreased (or volume increased), equilibrium will shift towards the higher pressure side.

### Temperature

Changes in concentration, pressure, and volume affect the equilibrium position, but the equilibrium constant K$_{eq}$ is unchanged. These changes can be calculated using the equilibrium expression and known values of K$_{eq}$ and concentrations.
Temperature, however, does change the value of K$_{eq}$. When given a value of K, you will also be given a temperature because K is dependent on the temperature.
An increase in temperature will favor the endothermic (heat-absorbing) side of a reaction. A decrease in temperature will favor the exothermic (heat-releasing) side of a reaction.
For example, the following reaction is very exothermic:
$2{\hbox{H}}_{2}+{\hbox{O}}_{2}\to 2{\hbox{H}}_{2}{\hbox{O}}$
Because it is exothermic, you can think of it like this:
$2{\hbox{H}}_{2}+{\hbox{O}}_{2}\to 2{\hbox{H}}_{2}{\hbox{O}}+heat$
At room temperature, K$_{eq}$ for this reaction is 3.2 x 10$^{81}$. If this reaction happens at a very high temperature, what will happen? Less product will form (or maybe the reverse reaction will occur if the temperature is high enough) and K$_{eq}$ will be smaller. What will happen if this reaction occurs at a very low temperature? More product will form and K$_{eq}$ will be larger.