While I was reading about the usefulness of the quantity $\Delta H$, I found that it can be used to calculate the how the equilibrium constant varies with temperature. How can this be done?

Does it agree with the predictions of Le Chatelier's principle (that for an exothermic reaction, increasing the temperature disfavours product formation, and vice versa)?

  • 2
    $\begingroup$ In this answer of mine you can find a derivation of the formula for the equilibrium constant which gives you its temperature dependence. $\endgroup$
    – Philipp
    Jun 3 '15 at 11:10

The equation linking $\Delta H^\circ$ and $K$ is called the van 't Hoff equation. Since Philipp's comment on your question already links to a very thorough discussion of where the equation $\Delta G^\circ = -RT\ln{K}$ comes from, I won't repeat it.

The definition of the Gibbs free energy, $G$, is $G = H - TS$. Using $\mathrm dG = V\,\mathrm dp - S\,\mathrm dT$ we obtain the Maxwell relation

$$\left(\frac{\partial G}{\partial T}\right) = -S$$

and hence the Gibbs–Helmholtz equation (derivation here)

$$\left(\frac{\partial (G/T)}{\partial T}\right) = -\frac{H}{T^2} \quad \Leftrightarrow \quad \left(\frac{\partial (\Delta G^\circ/T)}{\partial T}\right) = -\frac{\Delta H^\circ}{T^2}$$

Since $\ln K = -\Delta G^\circ/RT$, we have

$$\frac{\mathrm d(\ln{K})}{\mathrm dT} = -\frac{1}{R}\frac{\mathrm d}{\mathrm dT}\left(\frac{\Delta G^\circ}{T}\right) = \frac{\Delta H^\circ}{RT^2}$$

This is the differential form of the van 't Hoff equation; it's not the most useful thing to us though because it only tells you the slope of a plot of $\ln{K}$ against $T$ at one given point. We usually separate the variables and integrate with respect to both sides:

$$\int_{\ln{K_1}}^{\ln{K_2}}\!\mathrm d(\ln{K}) = \int_{T_1}^{T_2}\!\frac{\Delta H^\circ}{RT^2}\,\mathrm dT$$

$$\ln{K_2} - \ln{K_1} = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2} \right) $$

So, if you know the equilibrium constant $K_1$ at a certain temperature $T_1$ and you want to find the equilibrium constant $K_2$ at a different temperature $T_2$, you can just plug in your values into the equation and solve for $K_2$.

Note that this equation supports what you know of Le Chatelier's principle; if the reaction is exothermic, $\Delta H^\circ < 0$, and if you increase the temperature from $T_1$ to $T_2 > T_1$ then $(1/T_1 - 1/T_2) > 0$. The RHS of the equation is therefore negative, and that means that $\ln{K_2} < \ln{K_1} \Rightarrow K_2 < K_1$ which implies that the equilibrium position has shifted to the left.

Note that the last step (the integration) makes the assumption that $\Delta H^\circ$ is a constant over the temperature range $T_1$ to $T_2$. Note that this is, in general, not true but if the temperature range isn't too huge you will get pretty accurate results from the use of this equation.

  • $\begingroup$ The change in enthalpy $\Delta H^\circ$ refers to a standard state (a specific pressure). So $\Delta H^\circ$ depends also on temperature. How do we know that if a reaction is endothermic at specific conditions $(T_1,p^\circ)$ would also be endothermic in different conditions $(T_2,p^\circ)$, so we can apply Le Chatelier's principle? $\endgroup$
    – ado sar
    May 8 '20 at 18:08
  • $\begingroup$ @adosar you need to find the dependence of $\Delta H$ on temperature. It depends on the heat capacities of products and reactants. A full explanation is far too long for a comment, but look up Kirchhoff's law. Atkins' textbook will have a section on it. There is a brief mention of it at chemistry.stackexchange.com/questions/39620/… $\endgroup$
    – orthocresol
    May 8 '20 at 18:25
  • $\begingroup$ Thanks. I will check it out. $\endgroup$
    – ado sar
    May 8 '20 at 18:41

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