Impact of temperature on an Energy profile diagram

What happens to the energy profile diagram if you increase the temperature of the system?


From my understanding, I think the whole diagram would shift upwards, as the reactants and products will both have more energy at this higher temperature. However I'm not too sure what would happen at the peak. Would that also increase? Or would it remain the same, if so doesn't this mean increasing temperature technically reduces the activation energy of the reaction?


ie in the following diagram, if black is the original energy profile diagram, at a higher temperature, will it look like red or blue?
[![enter image description here](https://i.stack.imgur.com/4Q4xc.jpg)](https://i.stack.imgur.com/4Q4xc.jpg)




$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)\_{\!#3}}$


Let's suppose that the reaction is taking place at constant pressure (for constant volume, see the last paragraph). In that case, the most likely state functions to be represented on the vertical axis are H and G. Then:


$$\pd{G}{T}{p} = -S$$
$$\pd{H}{T}{p} = C\_p$$


From this we can see that the vertical coordinate can't be G. That's because your diagram has the dependent variable increase with temperature. But, since $S>0$ (always), $\pd{G}{T}{p} <0$ (always).


Now, what about the sign of $C\_p$? Well, $C\_p=\frac{\text{đ}q\_p}{dT}$. If you flow heat into a system ($\text{đ}q > 0$) without making any other changes, the temperature will always increase. And vice versa if heat flows out of a system. Thus the signs of $\text{đ}q\_p$ and $dT$ will always be the same $ \implies C\_p > 0 \implies \pd{H}{T}{p} >0$.


Thus, by elimination (and based on our starting assumption), we conclude the vertical coordinate is H.


In that case, we would expect the temperature-dependence of the enthalpies of the reactants, activation complex, and products to be relatively small—the main effect of increasing temperature would be to increase the rates of the forward and reverse reactions, by increasing the proportion of reactants and products able to pass over the activation barrier, as well as the collision frequency.


However, to the extent that you do increase the temperature enough to see a temperature-dependence in enthalpies, it would increase the enthalpies of everything—the reactants, products, and activation complex. Thus, qualitatively, the blue curve would be correct.


I say "qualitatively" because the heat capacities of the reactants, products, and activation complex will all be different, and thus the extent to which their respective enthalpies are raised by an increase in temperature will likewise be different.


Note that the same argument would apply if we were at constant volume instead of constant pressure, in which case the most likely state functions to use for the vertical axis would be either U (the internal energy) or A (the Helmholtz free energy). And here the only one of these two whose value could increase with temperature (at constant volume) would be U.


$$\pd{A}{T}{V} = -S < 0$$
$$\pd{U}{T}{V} = C\_V > 0$$


$\endgroup$


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A reaction coordinate diagram has a horizontal axis that represents the progression of the reaction and a vertical axis for free energy. Temperature has little effect on the diagram.


Where temperature mostly comes into play is interpreting what a given difference in free energy means. In general, the relationship of interest is between free energy and either a ratio of concentrations or, indirectly, the rate of a reaction.


In such a relationship, we see a term that looks like $-\frac{\Delta G}{kT}$.


This basically means that for any of these relationships, at higher temperature, the effect of having different free energies is less. For example, rates are going to increase at higher temperatures (the height of the barrier is less important); and equilibria will approach equal preference for both sides (the intrinsic preference of the equilibrium matters less).


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