CRE: Temperature Dependence of the Equilibrium Constant

The Equilibrium Constant as a Function of Temperature

Since the equilibrium constant is a function of $T$ only (we already saw how changes in $P$ affect equilibrium), we would like to determine:

$\frac{d\ln[K]}{dT} = \frac{d}{dT}\left [\frac{-\Delta g_{rxn}^o}{RT} \right ] $

taking the derivative of the quotient on the right gives

$\frac{d\ln[K]}{dT} = \frac{-1}{RT}\frac{d[\Delta g_{rxn}^o]}{dT} + \frac{\Delta g_{rxn}^o}{RT^2} $

Using the fundamental property relations, we have shown that

$\left ( \frac{\partial g}{\partial T} \right )_P = -s$ so $\left ( \frac{\partial \Delta g_{rxn}^o}{\partial T} \right )_P = -\Delta s_{rxn}^o$

combining these expressions yields

$\frac{d\ln[K]}{dT} = \frac{\Delta s_{rxn}^o}{RT} + \frac{\Delta g_{rxn}^o}{RT^2} $

which, using the definition of $g \equiv h - Ts$ can be simplified to

$\frac{d\ln[K]}{dT} = \frac{\Delta h_{rxn}^o}{RT^2} $

whereupon integration gives

$\ln[K] = \int \frac{\Delta h_{rxn}^o}{RT^2} dT$

When $\Delta h_{rxn}^o \ne \mathcal{F} (T)$

When the heat of reaction is independent of temperature, we can pull it out of the integral and evaluate the indefinite integral to give

$\ln[K] = -\frac{\Delta h_{rxn}^o}{RT} + C$

As with the Clausius-Clapeyron equation, we can eliminate the integration constant by using a reference value (typically at 298K$) and write$

$\ln[\frac{K}{K_o}] = -\frac{\Delta h_{rxn}^o}{R} \left [ \frac{1}{T}-\frac{1}{T_o} \right ]$

Note:

When we have an exothermic reaction ($\Delta h_{rxn}^o<0$) this means that $K$ will decrease with increasing $T$. Alternatively, when we have an endothermic reaction ($\Delta h_{rxn}^o<0$) $K$ will increase with increasing $T$.

When $\Delta h_{rxn}^o = \mathcal{F} (T)$

When the heat of reaction is not independent of temperature, we need to formally account for the temperature dependence inside the integral.

This is done by recalling that:

$\Delta h_{rxn}^o = \sum_i \nu_i\Delta h_{f_i}^o$

Since $\Delta h_{f_i}^o$ data is typically available at 298K, we can write

$\Delta h_{rxn}^o = \sum_i \nu_i\left [\Delta h_{f_{i_{298K}}}^o+\int_{298K}^T cp_{i} dT \right ]$

and plugging the resultant function of temperature into the $\ln[K]$ integral.

Outcome:

Use thermochemical data to calculate the equilibrium constant and its dependence on temperature

Test Yourself:

Calculate the equilibrium constant for the following vapor-phase reaction occurring at 250C:

$C_2H_4(g)+H_2O(g) \Longleftrightarrow C_2H_5OH(g)$