4
$\begingroup$

Maybe this is a silly question, but it's not one I've been able to answer to my own satisfaction. Increasing the temperature of a reaction favours the endothermic direction, because it makes it relatively more likely for particles to collide with enough energy to make the activation energy. Yes?

But doesn't lowering the activation energy have almost the same effect on the relative likelihood of particles colliding at least that energetically? Someone on a related question about whether catalysts shift the position of equilibrium said:

The energy difference between the reactants and products is unchanged by catalysis

But isn't that equally true for increasing the temperature?

$\endgroup$
3
$\begingroup$

Your presumption in the first paragraph is incorrect. The fact that more particles collide with enough energy to make the activation energy with an increase in temperature means that the rate of reaction increases as temperature goes up, but that doesn't affect the equilibrium of the reaction.

Instead, the effect on temperature has to do with the comparative free energy of the two sides of the equilibrium. The controlling factor in where the equilibrium lies is the difference in free energy between the reactants and products (the ΔG). In fact, the equilibrium constant has a very simple relationship to the free energy: $\ce{\Delta G = -RT ln K }$ ... or $\ce{K = e^{-\frac{\Delta G}{RT} }}$. The ΔG is made up of two components - the enthalpy and the entropy: $\ce{\Delta G = \Delta H -T\Delta S}$, where T is the (absolute) temperature. The side of the reaction which is favored is the one with the lowest (most negative) free energy.

So when we substitute, we get the following relationship for the temperature dependence of the equilibrium constant, in terms of entropy and enthalpy:

$$ K = e^{-\frac{\Delta H}{RT} } \cdot e^{\frac{\Delta S}{R} } $$

The temperature dependence is only in the enthalpy ($\ce{\Delta H}$). Enthalpy, by the way, is what we normally think of as endothermic/exothermic. Roughly, reactions with a positive $\ce{\Delta H}$ absorb heat and are thus endothermic, and those with a negative $\ce{\Delta H}$ give off heat and are thus exothermic.

So imagine we have an endothermic reaction (positive $\ce{\Delta H}$), which means that the exponent for enthalpy in the above equation is negative. When we increase the temperature, that exponent gets smaller in magnitude (closer to zero), which means it becomes less negative. This causes an increase in $e^x$, which means that K gets bigger, which means the equilibrium shifts in the forward (endothermic) reaction. Likewise, if you have the reaction written in the exothermic direction, $\ce{\Delta H}$ is negative, the exponent is positive, an increase in temperature causes the exponent to become less positive, causing a decrease in the $e^x$, which results in a decrease of K -- the equilibrium now shifts in the reverse (endothermic) direction.

So while the statement that increasing temperatures means that the equilibrium shifts in the endothermic direction is true, it has nothing to do with the activation energy of the reaction. The arguments above are from a purely thermodynamic standpoint, and don't involve any consideration about the mechanism of reaction or the activation energy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.