### 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:

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