# 10.6: Temperature Dependence of Equilibrium Constants - the van’t Hoff Equation

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- Contributed by Patrick Fleming
- Assistant Professor (Chemistry and Biochemistry) at California State University East Bay

The value of \(K_p\) is independent of pressure, although the composition of a system at equilibrium may be very much dependent on pressure. Temperature dependence is another matter. Because the value of \(\Delta G_{rxm}^o\) is dependent on temperature, the value of \(K_p\) is as well. The form of the temperature dependence can be taken from the definition of the Gibbs function. At constant temperature and pressure

\[ \dfrac{\Delta G^o_{T_2}}{T_2} - \dfrac{\Delta G^o_{T_1}}{T_1} = \Delta H^o \left(\dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \]

Substituting

\[ \Delta G^o = -RT \ln K\]

For the two values of \(\Delta G_{}^o\) and using the appropriate temperatures, yields

\[ \dfrac{-R{T_2} \ln K_2}{T_2} - \dfrac{-R{T_1} \ln K_1}{T_1} = \Delta H^o \left(\dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \]

And simplifying the expression so that only terms involving \(K\) are on the left and all other terms are on the right results in the **van ’t Hoff equation**, which describes the temperature dependence of the equilibrium constant.

\[ \ln \left(\dfrac{\ K_2}{\ K_1}\right) = - \dfrac{\Delta H^o}{R} \left(\dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \label{vH}\]

Because of the assumptions made in the derivation of the Gibbs-Helmholtz equation, this relationship only holds if \(\Delta H^o\) is independent of temperature over the range being considered. This expression also suggests that a plot of \(\ln(K)\) as a function of \(1/T\) should produce a straight line with a slope equal to \(–\Delta H^o/R\). Such a plot is known as a **van ’t Hoff plot**, and can be used to determine the reaction enthalpy.

Example \(\PageIndex{1}\)

A certain reaction has a value of \(K_p = 0.0260\) at 25 °C and \(\Delta H_{rxm}^o = 32.4 \,kJ/mol\). Calculate the value of \(K_p\) at 37 °C.

**Solution**

This is a job for the van ’t Hoff equation!

- T
_{1}= 298 K - T
_{2 }= 310 K - \(\Delta H_{rxm}^o = 32.4 \,kJ/mol\)
- K
_{1}= 0.0260 - K
_{2}= ?

So Equation \ref{vH} becomes

\[ \begin{align*} \ln \left( \dfrac{\ K_2}{0.0260} \right) &= - \dfrac{32400 \,J/mol}{8.314 \,K/(mol \,K)} \left(\dfrac{1}{310\, K} - \dfrac{1}{298 \,K} \right) \\[4pt] K_2 &= 0.0431 \end{align*}\]

Note: the value of \(K_2\) **increased **with **increasing **temperature, which is what is expected for an **endothermic **reaction. An increase in temperature should result in an increase of product formation in the equilibrium mixture. But unlike a change in pressure, a change in temperature actually leads to a change in the value of the equilibrium constant!

Example \(\PageIndex{2}\)

Given the following average bond enthalpies for \(\ce{P-Cl}\) and \(\ce{Cl-Cl}\) bonds, predict whether or not an increase in temperature will lead to a larger or smaller degree of dissociation for the reaction

\[\ce{PCl_5 \rightleftharpoons PCl_3 + Cl_2} \nonumber\]

X-Y | D(X-Y) (kJ/mol) |
---|---|

P-Cl |
326 |

Cl-Cl |
240 |

**Solution**

The estimated reaction enthalpy is given by the total energy expended breaking bonds minus the energy recovered by the formation of bonds. Since this reaction involves breaking two P-Cl bonds (costing 652 kJ/mol) and the formation of one Cl-Cl bond (recovering 240 kJ/mol), it is clear that the reaction is endothermic (by approximately 412 kJ/mol). As such, an increase in temperature should increase the value of the equilibrium constant, causing the degree of dissociation to be increased at the higher temperature.

## Contributors and Attributions

Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)