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While reading about equilibrium, I read that the equilibrium constant, $K_\mathrm{c}$, changes with temperature due to the forward and backward reactions having different activation energies. Is it possible for a reaction to have the same activation energy for the forward and backward reactions, such that $K_\mathrm{c}$ does not change when temperature changes? If so, please provide an example.

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The van't Hoff equation describes how the equilibrium constant changes with temperature:

$$\ln \frac{K_2}{K_1} = −\Delta H_R^\circ (\frac{1}{T_2} − \frac{1}{T_1})$$

So if the reaction enthalpy is zero for a given temperature interval, the equilibrium constant will not change. This also means that the activation energy in either direction is the same, at least if entropy is not considered (the typical diagram explaining activation energy is for a single molecule with potential energy as the y-axis, enthalpy as the difference in energy between reactant and product, and activation energies as the difference between reactant or product and the activated complex).

If so, please provide an example.

Take the transfer of protons across a membrane. Equilibrium is reached when the concentrations on either side are equal. A change in temperature will not change the equilibrium constant (which is 1, with a reaction enthalpy of zero).

Another example would be a reaction that turns one enantiomer into the other, e.g. D-alanine to L-alanine. In the absence of chiral co-solvents or solvents or other unusual circumstances, again K = 1 and enthalpy = zero over a range of temperatures.

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  • $\begingroup$ Nice answer (+1)! I was thinking of something like hydroxide ion taking a proton from water, yielding water and a hydroxide ion, but that is too rinky dink! $\endgroup$ – Ed V Jul 28 at 21:12
  • $\begingroup$ @EdV If you want to do a self-exchange reaction such as that one, you have to used isotopic labeling to keep reactant and product apart. $\endgroup$ – Karsten Theis Jul 28 at 21:39
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    $\begingroup$ Sure, but if you simply do not care, then both sides are the same and we know it must happen, as can be proven with isotopic labeling. But this is just too rinky dink, in any event! $\endgroup$ – Ed V Jul 28 at 21:44

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