# Is ln(K2/K1) = (∆h°rxn/R)(1/T1 - 1/T2) even when ∆h°rxn varies strongly with temperature?

In all of my physical chemistry books I find the same expression for estimating the equilibrium constant of a reaction at a non-standard temperature. The derivation starts with $R \ln K_\text{eq} = -\Delta g^\circ/T = -(\Delta h^\circ/T - \Delta s^\circ)$. Then one takes the temperature derivative of each side assuming that $\Delta h^\circ$ and $\Delta s^\circ$ can both be considered constants, which eliminates the entropy term. The equation $\mathrm{d}\ln K = (1/R)(\Delta h^\circ/T^2)\mathrm{d}T$ is then integrated between $T_1$ and $T_2$ to give $\ln(K_2/K_1) = (\Delta h^\circ/R)(1/T_1 - 1/T_2)$.

Here's the question: $\Delta h_\text{r}^\circ$ is really NOT a constant, but a function of temperature. Assuming constant heat capacities, $\Delta h_\text{r}^\circ$ at $T = \Delta h_\text{r}^\circ + \Delta c_p(T - T_0)$. Likewise $\Delta s_\text{r}^\circ$ at $T = \Delta s_\text{r}^\circ + \Delta c_p \ln(T/T_0)$. Where $\Delta c_p$ is the difference in the heat capacity of the products and the reactants.

I have tried to insert these expressions into the typical derivation but am getting wonky results. Can someone help me with the derivation or point me to a source where $\ln K$ vs. $T$ is derived with temperature-sensitive reaction enthalpies and entropies.

Edit 4/19/2015: Would it be more effective to invoke $\mathrm{d}\Delta g = \Delta v\mathrm{d}P - \Delta s\mathrm{d}T$ which at constant pressure, and in the case of free energy of reaction, gives $\mathrm{d}\Delta g_\text{r}^\circ/\mathrm{d}T = \Delta s_\text{r}^\circ + \Delta c_p \ln(T/T_0)$ ?

Thanks in advance for any help.

• Why wonky results? What do you obtain? – user1420303 Apr 19 '15 at 19:50
• – J. LS Apr 20 '15 at 12:47

For each species, the quantities $g^0$, $s^0$, and $h^0$ are defined as the free energy, the entropy, and the enthalpy of the pure species, respectively, at 1 atm. pressure and temperature T. So, for constant pressure of 1 atm., $d\Delta g^0=\Delta s^0dT$. But, $\Delta s^0=\frac{\Delta g^0-\Delta h^0}{T}$. So $\frac{d\Delta g^0}{dT}=\frac{\Delta g^0-\Delta h^0}{T}$
So, $\frac{d(\Delta g^0/T)}{d{(1/T)}}=\Delta h^0$
• I'm getting that the expression comes out as $\frac{d(\Delta g^o/T)}{d{(1/T)}}=\Delta h^o$ without the $/T$. – jbhunter Apr 20 '15 at 23:07
• Found an answer to my original question in Castellan's 1971 book Physical Chemistry (equation 11-59). Starting with $d(ln K)/dT = \Delta h_r^o/RT^2$ expand $\Delta h_r^o$ as a polynomial series $\Delta h_o^o + AT + BT^2 + ...$ then integrate. That was a lot simpler than I expected. Thanks everyone for the ideas that put me on the right track. – jbhunter Apr 20 '15 at 23:14
• Thanks for spotting my math error. I corrected it. In my equations, the superscript 0 signifies at 1 atm pressure. However, the standard heat of reaction $\Delta h^0$ is a function of temperature (as your latter post indicates). The terms in your polyn0mial are obtained from the standard heat of reaction at 298 K and the temperature-dependent heat capacity variations of the reactants and products (by integration). – Chet Miller Oct 26 '15 at 10:10