# Why does changing the temperature shift the equilibrium?

I know that raising the temperature in a reversible chemical reaction causes the equilibrium to shift to the endothermic side.

I know that $$\Delta G = \Delta H − T\Delta S$$, but I don't know how to put them together to show why the equilibrium would shift.

To summarise, I already know what happens when, and I'm interested in why.

I think the best way to think of equilibria intuitively is in terms of rates of reaction. At equilibrium, the forward and the reverse reactions are happening at the same rate.

If you increase the temperature, what happens to the rates of the forward and reverse reactions?

Using the Arrhenius equation, $$k = A\exp\left(\frac{-E_\mathrm{a}}{RT}\right),$$ you can see that as temperature increases, the rate will increase. The amount it increases depends on the activation energy $$E_\mathrm{a}$$.

The activation energy on the endothermic side of the reaction will always be larger than the exothermic side. This is because the transition state is always higher in energy than either of the reactants/products, and the reactants for the endothermic reaction are by definition lower in energy than the products! This means that the endothermic reaction will be sped up more than the exothermic reaction. This means the reaction is now out of equilibrium and the endothermic reaction will happen until the reaction is back in equilibrium again.

• But when the temperature increases the equilbrium constant approaches 1 indepedent if the reaction is exergonic or endergonic. This seems counterintuitive. We would expect in higher temperatures the formation of the most stable compound. But why when the temperature goes high also the unstable compound is formed? Thanks in advance. Jan 6 '20 at 22:09
• @adosar At high enough temperature the activation energy becomes irrelevant. This means that the most stable compound doesn't matter. At high enough temperature everything is equally stable! You get the highest entropy combinations of products.
– Nick
Jan 9 '20 at 17:58

Suppose you have some reaction $$\ce{A <=> B}.$$

The equilibrium constant for the reaction is $$K$$ and the Gibbs free energy change is $$\Delta G$$. The equilibrium constant is $$K = \frac{[\ce{B}]}{[\ce{A}]}.$$

Increasing value of $$K$$ shifts the equilibrium towards the right, i.e. more $$\ce{B}$$, and reducing the value shifts it to the left, i.e. more $$\ce{A}$$. The equilibrium constant and the Gibbs free energy are related by $$K = \exp\left(-\frac{\Delta G}{RT}\right)$$

Putting in the expression you give for $$\Delta G$$ gives us \begin{align} K &= \exp\left(-\frac{\Delta H - T\Delta S}{RT}\right) \\ &= \exp\left(-\frac{\Delta H}{RT}\right)\, \exp\left(\frac{\Delta S}{R}\right). \end{align}

If we assume the variation of $$\Delta H$$ and $$\Delta S$$ is small we can ignore the entropy term because it's a constant, and we get $$K \propto \exp\left(-\frac{\Delta H}{RT}\right).$$

If the reaction is endothermic $$\Delta H$$ is positive, so we have the exponential of a negative number and this is less than one. If we increase the temperature we decrease $$\Delta H/(RT)$$ and the exponential increases. So increasing the temperature makes the equilibrium coefficient bigger, i.e. it drives the endothermic reaction.

To understand why this happens consider what happens when we make the temperature very high. If you do this the value of $$\Delta H/(RT)$$ approaches zero and the exponential approaches unity, i.e. the concentrations of $$\ce{A}$$ and $$\ce{B}$$ are the same. What we've done is make the thermal energy much greater than the activation energy in either direction, so the rates of the forward and backward reactions are the same.