# How do I relate equilibrium constants to temperature change to find the enthalpy of reaction?

Jun 20, 2015

(I moved this into Energy Change in Reactions since the varying $K$ or $k$ value is coupled with varying temperatures, and molecules move at different speeds at different temperatures, and so are differently energetic. Thus, $\Delta T$ is proportional to $\Delta E$.)

There's a useful equation we can use.

$\ln \left(\frac{{K}_{p 2}}{{K}_{p 1}}\right) = - \frac{\Delta {H}_{R}}{R} \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]$

and its variation:

$\ln \left(\frac{{K}_{c 1}}{{K}_{c 2}}\right) = - \frac{\Delta {H}_{R}}{R} \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]$

where $R = 8.314472 \cdot {10}^{- 3} \frac{k J}{m o l \cdot K}$ and $\Delta {H}_{R}$ is the enthalpy of reaction.

You may also have seen another variation with kinetics:

$\ln \left(\frac{{k}_{2}}{{k}_{1}}\right) = - \frac{{E}_{a}}{R} \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]$

which is very similar, and just contains the rate constants ${k}_{i}$ and the activation energy ${E}_{a}$, and everything else is the same. Anyways:

You can use both values of $K$ (or $k$), both values of $T$, and $R$ to solve for $\Delta {H}_{R}$. If the result is negative, the reaction is exothermic, and vice versa.

You should have seen something like this plot before, which is represented by the above first or second equation ($\ln K$ vs. $\frac{1}{T}$):

This is derived from the Van't Hoff Equation, $\frac{\mathrm{dl} n K}{\mathrm{dT}} = \frac{\Delta H}{R {T}^{2}}$, if you were curious. Just multiply over $\mathrm{dT}$ and integrate the function from ${T}_{1}$ to ${T}_{2}$. This works well as long as the temperature range is small enough such that $\Delta H$ varies linearly with temperature in that range.