We have seen that the equilibrium constant is related to the standard Gibbs energy change by
\[ \Delta G^\circ = -RT\ln K \]This relationship shows how the equilibrium constant is connected to thermodynamics, but it does not directly reveal how the equilibrium constant changes with temperature.
To answer that question, we begin with the Gibbs-Helmholtz equation:
\[ \left( \frac{\partial (\Delta G^\circ/T)} {\partial T} \right)_p = -\frac{\Delta H^\circ}{T^2} \]Substituting
\[ \Delta G^\circ = -RT\ln K \]into the Gibbs-Helmholtz equation and simplifying yields
\[ \frac{d(\ln K)} {dT} = \frac{\Delta H^\circ} {RT^2} \]This differential form of the van't Hoff equation relates the temperature dependence of the equilibrium constant to the reaction enthalpy.
If the reaction enthalpy is approximately constant over the temperature range of interest, the differential equation can be integrated to give
\[ \ln\!\left( \frac{K_2}{K_1} \right) = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]This equation allows equilibrium constants measured at one temperature to be used to predict equilibrium constants at another temperature.
Notice the similarity to the Clausius-Clapeyron equation:
\[ \ln\!\left( \frac{p_2}{p_1} \right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]In both cases, the temperature dependence is governed by an enthalpy change.
The sign of the reaction enthalpy determines how the equilibrium constant changes with temperature.
For an endothermic reaction,
\[ \Delta H^\circ > 0 \]increasing the temperature increases the equilibrium constant.
Consequently, product formation becomes more favorable at higher temperatures.
For an exothermic reaction,
\[ \Delta H^\circ < 0 \]increasing the temperature decreases the equilibrium constant.
In this case, reactants become more favorable as the temperature rises.
These conclusions are entirely consistent with Le Chatelier's principle. Increasing the temperature favors the direction that absorbs heat.
The integrated van't Hoff equation can be rearranged into the linear form
\[ \ln K = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T} \right) + C \]This equation has the form of a straight line:
\[ y=mx+b \]Therefore, a plot of
\[ \ln K \]versus
\[ \frac{1}{T} \]should be approximately linear.
The slope of the line is
\[ -\frac{\Delta H^\circ}{R} \]allowing the reaction enthalpy to be determined directly from equilibrium measurements made at different temperatures.
Big picture: The van't Hoff equation connects the temperature dependence of equilibrium constants to the reaction enthalpy. It provides the equilibrium analogue of the Clausius-Clapeyron equation and explains why temperature changes can alter both the equilibrium composition and the value of the equilibrium constant itself.
A reaction has an equilibrium constant
\[ K_1=0.0250 \]at 298 K and a standard reaction enthalpy
\[ \Delta H^\circ = 32.5\ \text{kJ mol}^{-1} \]Assuming that \(\Delta H^\circ\) is independent of temperature, calculate the equilibrium constant at 350 K.
The integrated van't Hoff equation is
\[ \ln\!\left( \frac{K_2}{K_1} \right) = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]The known values are
\[ K_1=0.0250 \] \[ T_1=298\ \text{K} \] \[ T_2=350\ \text{K} \] \[ \Delta H^\circ = 32.5\times10^3\ \text{J mol}^{-1} \]Substituting into the van't Hoff equation gives
\[ \ln\!\left( \frac{K_2}{0.0250} \right) = -\frac{32500} {8.314} \left( \frac{1}{350} - \frac{1}{298} \right) \]Evaluating the temperature term:
\[ \left( \frac{1}{350} - \frac{1}{298} \right) = -4.99\times10^{-4}\ \text{K}^{-1} \]Therefore,
\[ \ln\!\left( \frac{K_2}{0.0250} \right) = 1.95 \]Exponentiating both sides:
\[ \frac{K_2}{0.0250} = e^{1.95} = 7.03 \] \[ K_2 = (0.0250)(7.03) \] \[ K_2 = 0.176 \]Thus,
\[ \boxed{K_2=0.176} \]Physical interpretation: The reaction is endothermic because \(\Delta H^\circ\) is positive. As the temperature increases, the equilibrium constant increases from 0.0250 to 0.176, indicating that product formation becomes more favorable at the higher temperature. This behavior is consistent with Le Chatelier's principle, which predicts that increasing the temperature favors the direction that absorbs heat.
Four quantities are given. Calculate the missing quantity using the integrated van't Hoff equation.
\[ \ln\!\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]| Quantity | Value |
|---|---|
| \(K_1\) | |
| \(K_2\) | |
| \(T_1\) | |
| \(T_2\) | |
| \(\Delta H^\circ\) | |
| Your answer | |
Big picture: The van't Hoff equation connects chemical equilibrium to thermodynamics by relating changes in the equilibrium constant to the reaction enthalpy. It explains how temperature influences equilibrium composition and provides a powerful method for extracting thermodynamic information from equilibrium measurements.