Is there a way to derive a modified van 't Hoff equation if we can't make the assumption that $\Delta H_\mathrm{r}$ doesn't change?

We know that $\mathrm{d} \ln K/\mathrm{d}T = \Delta H_\mathrm{r}/(RT^2)$ and $\Delta H_\mathrm{r} = \Delta H_\mathrm{r}^\circ + \int C_p\,\mathrm dT$

Is there a way to integrate the $C_p$ term along with the $T^2$?


1 Answer 1


There is. Suppose that at 298 K $$\Delta G^○_{298} = -RT\ln{K_{p\,298}} = \Delta H^○_{298} - T\Delta S^○_{298},$$ and if both $\Delta H^○$ and $\Delta S^○$ depend on $T$, $$-RT\ln{K_p} = \Delta H^○_{298} + \int_{298}^T \Delta C_pdT - T\Delta S^○_{298} - T\int_{298}^T \frac{\Delta C_p}{T}dT.$$ Then both sides of it can be divided by $T$ and knowing that $\int udv = uv - \int vdu$, where in our case $v = 1/T$ and $u = \int_{298}^T \Delta C_pdT$, we can "simplify" the last equation: $$-R\ln{K_p} = \frac{\Delta H^○_{298}}{T} - \Delta S^○_{298} - \int_{298}^T \frac{\int_{298}^T \Delta C_pdT}{T^2} dT.$$ I'll not endeavor to differentiate that with respect to $T$, but now you can probably figure out the rest. Since $C_p$ is usually a not-that-complex polynomial of $T$, the math should be rather straightforward.


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