The last change for Le Chatelier is to change the temperature. We can understand this by looking at the enthalpy change for a reaction and thinking about the reaction as either producing or consuming "heat" as a result of the reaction.
For an exothermic reaction
\[{\rm reactants \rightleftharpoons products + heat}\]
So "heat" is a product. If you raise the temperature (by inputting energy in the form of heat) then the reaction will "counteract" this by shifting to the reactant side. The converse is true of endothermic reactions
\[{\rm reactant + heat \rightleftharpoons products} \]
Now adding heat is like adding a reactant, the reaction shifts towards the products.
Why is this?
The reason here is quite different than for the previous examples since what is changing is the value of the equilibrium constant. The value of the equilibrium constant depends on temperature for two reasons.
\[\Delta G_{\rm r}^{\circ} = -RT\ln K\]
There is a factor of the temperature in the relationship between the standard free energy and K. In addition, the standard free energy depends on temperature.
\[\Delta G_{\rm r}^{\circ} =\Delta H_{\rm r}^{\circ} -T\Delta S_{\rm r}^{\circ}\]
combining these two we see that
\[-RT\ln K = \Delta H_{\rm r}^{\circ} -T\Delta S_{\rm r}^{\circ}\]
\[\ln K = {\Delta H_{\rm r}^{\circ} \over -RT} + {\Delta S_{\rm r}^{\circ} \over R}\]
\[ K = {exp\left({-\Delta H_{\rm r}^{\circ} \over RT}\right) \times constant}\]
So we can see that \(K\) (or the \(\ln K\)) has a temperature dependence that depends on the value of \(\Delta H_{\rm r}^{\circ}\) (its the term with the \(T\) in the resulting formula). The constant in the expression depends on the value of \(\Delta S_{\rm r}^{\circ}\).
So \(K\) has a strong dependence on temperature. Moreover, the sign of the change depends on the sign of \(\Delta H_{\rm r}^{\circ}\). If the standard reaction enthalpy is positive (endothermic) then as \(T\) increases \(K\) increase (more products at equilibrium than before). For exothermic reactions, as \(T\) increases \(K\) decreases (more reactants at equilibrium than at the previous temperature).
Often this is misinterpreted that endothermic reactions require high temperatures to be spontaneous. It is simply that at higher temperatures, endothermic reactions shift to the right.
This provides a lot of useful insight into typical chemical processes. Often it can be hard to look at a reaction and know if it is endothermic or exothermic. But bond breaking requires energy (endothermic) and bond formation releases energy (exothermic). So very often decomposition reactions where one compound breaks apart into two (or more) typically have more bond breaking than forming and tend to be endothermic. So for example, if you have rust (iron oxide \(Fe_2O_3\)) and you'd like to make iron metal
\[{\rm 2Fe_2O_3(s) \rightleftharpoons 4Fe(s) + 3O_2(g)} \]
At room temperature the equilibrium constant for this reaction is extremely small. So the equilibrium is essentially all reactants. However, this reaction is endothermic (like most decompositions). So if you want to increase the value of K you can raise the temperature. Thus to make rust into iron spontaneously, you can get the temperature very high and the equilibrium will shift to the right. This is the process of "smelting" iron ore.
The temperature dependence of K has one final application. For many reactions, it can be very difficult to measure if they are endothermic or exothermic (since the amounts of materials might be extremely small). However, we can often measure concentrations. So there are situations in which we measure the temperature dependence of \(K\) to determine \(\Delta H_{\rm r}^{\circ}\). In this way we can try to put a number to \(\Delta H_{\rm r}^{\circ}\) either by measuring heats (calorimetry) or concentrations (equilibria).
A video explaining the relationship of Temperature Change and LeChatelier's Principle. Which way will the reaction shift when temperature is increased? decreased?© 2013 mccord/vandenbout/labrake