D22.4 Effect of Temperature on Equilibrium

Recall that:

ΔrG° = ΔrH° – TΔrS°

Therefore:

RT(lnK°) = ΔrH° – TΔrS°

This equation can be used to calculate K° at different temperatures, if we assume that ΔrH° and ΔrS° for a reaction have the same values at all temperatures. This is a good, but not perfect, assumption, and we will use it in this course unless specified otherwise. It is not a good assumption if there is a phase change for a reactant or a product within the temperature range of interest.

Dividing both sides of the equation by “-RT” gives:

lnK° =  -\dfrac{{\Delta}_rH^{\circ}}{R}\left(\dfrac{1}{T}\right) +  \dfrac{{\Delta}_rS^{\circ}}{R}
y = mx + b

A plot of lnK° vs.  \frac{1}{T} is called a van’t Hoff plot. The graph has slope =  -\frac{{\Delta}_rH^{\circ}}{R} and y-intercept =  \frac{{\Delta}_rS^{\circ}}{R} . If the concentrations of reactants and products are measured at various temperatures so that K° can be calculated at each temperature, both the reaction entropy change and enthalpy change can be obtained from a van’t Hoff plot.

Figure: Van’t Hoff plots. The reaction N2 + O2 = 2 NO has ΔrH° = 180.5 kJ/mol. For this endothermic reaction, as T increases (1/T decreases) the equilibrium constant increases. The reaction N2 + 3 H2 = 2 NH3 has ΔrH° = −92.2 kJ/mol; for this exothermic reaction, as T increases (1/T decreases) the equilibrium constant decreases.

Based on the equation for the van’t Hoff plot, an exothermic reaction (ΔrH° < 0) has K° decreasing with increasing temperature, and an endothermic reaction (ΔrH° > 0) has K° increasing with increasing temperature. This shows that the magnitude of ΔrH° will dictate how rapidly changes as a function of temperature.

For example, suppose that K°1 and K°2 are the equilibrium constants for a reaction at temperatures T1 and T2, respectively:

 \text{ln}K_1^{\circ} = -\dfrac{{\Delta}_{\text{r}}H^{\circ}}{R}\left(\dfrac{1}{T_1}\right)\;+\;\dfrac{{\Delta}_{\text{r}}S^{\circ}}{R}
 \text{ln}K_2^{\circ} = -\dfrac{{\Delta}_{\text{r}}H^{\circ}}{R}\left(\dfrac{1}{T_2}\right)\;+\;\dfrac{{\Delta}_{\text{r}}S^{\circ}}{R}

Subtracting the two equations yields:

 \begin{array}{rcl}\text{ln}K_2^{\circ}\;-\;\text{ln}K_1^{\circ} &=& \left(-\dfrac{{\Delta}_{\text{r}}H^{\circ}}{R}\left(\dfrac{1}{T_2}\right)\;+\;\dfrac{{\Delta}_{\text{r}}S^{\circ}}{R}\right) - \left(-\dfrac{{\Delta}_{\text{r}}H^{\circ}}{R}\left(\dfrac{1}{T_1}\right)\;+\;\dfrac{{\Delta}_{\text{r}}S^{\circ}}{R}\right)\\[1em] \text{ln}\left(\dfrac{K_2^{\circ}}{K_1^{\circ}}\right) &=& -\dfrac{{\Delta}_{\text{r}}H^{\circ}}{R}\left(\dfrac{1}{T_2}\;-\;\dfrac{1}{T_1}\right)\\[1em]&=& \dfrac{{\Delta}_{\text{r}}H^{\circ}}{R}\left(\dfrac{1}{T_1}\;-\;\dfrac{1}{T_2}\right) \end{array}

Thus calculating ΔrH° from tabulated ΔfH° values and measuring the equilibrium constant at one temperature allows us to calculate the equilibrium constant at any other temperature (assuming that ΔrH° and ΔrS° are independent of temperature).

Exercise: Temperature Dependence of the Equilibrium Constant

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Chemistry 109 Fall 2021 by John Moore, Jia Zhou, and Etienne Garand is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.