Effect of temperature on equilibrium

What is the effect of temperature on equilibrium?

The change in temperature of a chemical reaction at equilibrium results in a shift in position of the equilibrium that opposes the temperature change. For example, the increase in temperature of an endothermic reaction encourages the reaction to occur because the increased energy is absorbed to facilitate the reaction, thereby lowering the temperature.

Quantitatively, the effect of temperature on a chemical reaction system is best explained using the van’t Hoff equation (see below for derivation):

\frac{dlnK}{d\frac{1}{T}}=-\frac{\Delta_rH^o}{R}\; \; \; \; \; \; \; \; 20

where ΔrHo is the enthalpy of a reaction at standard conditions, K is the equilibrium constant, T is the temperature of the system and R is the gas constant.

Eq20 can be rewritten as:

ln\left ( \frac{K_f}{K_i} \right )=-\frac{\Delta_rH^o }{R}\left ( \frac{1}{T_f}-\frac{1}{T_i} \right )\: \: \: \: \: \: \: \:\: 20a

where the subscripts f and i represent ‘final’ and ‘initial’ respectively.

Let’s consider an increase in T of an endothermic reaction (ΔrHo > 0). The RHS of eq20a under such conditions is positive, which means that Kf > Ki, i.e. the equilibrium constant increases with an increase in T for an endothermic reaction. Therefore, unlike concentration and pressure, temperature affects the value of K. Nevertheless, Le Chatelier’s principle still applies, as illustrated by the following example:

N_2(g)+O_x(g)\rightleftharpoons 2NO(g)\; \; \; \; \; \; \; \Delta_rH^o=+180\: kJmol^{-1}

K=\frac{p_{NO}^{\; \; \; \; \; \; 2}}{p_{N_2}\, p_{O_2}}\; \; \; \; \; \; \; \; 21

If K increases as a result of an increase in T, the numerator of eq21 must increase relative to the denominator. This means that the position of the equilibrium shifts to the right, which is consistent with the definition of Le Chatelier’s principle.

Using the same logic and eq20a, we can describe the effect of temperature on the position of an exothermic reaction’s equilibrium.

In summary,



Equilibrium position







Derive the van’t Hoff equation.


Substituting the definition of the standard Gibbs energy of reaction ΔrGo = ΔrHorSo in the Gibbs energy equilibrium constant relation ΔrGo = –RTlnK, we have

lnK=-\frac{\Delta_r H^o}{RT}+\frac{\Delta_rS^o}{R}

If we assume that ΔrHo and ΔrSo are constant for a reaction over a range of temperature, the change of lnK with respect to the change in temperature is:

\frac{dlnK}{dT}=\frac{\Delta_r H^o}{RT^2}

This is the called the van’t Hoff equation, which can also be written as:

\frac{dlnK}{d\frac{1}{T}}=-\frac{\Delta_r H^o}{R}



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