# 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? • A diagram seems to be missing. Energy can mean electronic energy, kinetic energy etc. Evidently it is a reaction schematic you are referring to. The activation energy depends on the energy of the intermediate wrt reactants. Apr 6, 2022 at 6:15
• @AdnanAL-Amleh, this is incorrect. Apr 8, 2022 at 22:46
• @AdnanAL-Amleh That article is designed for beginning students, and thus paints a simplified picture in which the T-dependence of activation energy is ignored. This simplified picture makes sense for beginning students, particularly since the T-dependence of the activation energy is small. However, the picture the OP drew explicitly introduces a possible T-dependence in the energies of reactants, products, and activated complex. Hence this question, intentionally or not, goes beyond that simplified picture. E.g., suppose you arbitrarily set the energy of reactants as 10 and the activated ... Apr 9, 2022 at 3:38
• ..complex as 20. Then the activation energy in the forward direction is 20 – 10 = 10. Now suppose we increase T, and suppose that increases the energy of the reactants by 5 (to 10 + 5 = 15) and of the activated complex by 4 (to 20 + 4 = 24). Then the activation energy in the forward direction is now 24 - 15 = 9. I.e., because the increase in T affects the energy of the reactants and the activated complex differently, the activation energy (which is the difference between them) changes as you change T. But, as I note in my answer, this is typically a small effect. Apr 9, 2022 at 4:30
• @theorist:Thank you for remediating my misconception, that's I taught my students all the time that Activation energy independent of temperature but dependent on catalysis as written in the text . Apr 10, 2022 at 1:21

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

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.

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

$$\endgroup$$

• @KarstenTheis Thanks, I've added an explanation for why heat capacity must also be positive. Regarding activation energy/enthalpy: Yes, that restates what I said in my last paragraph. Consequently, the activation enthalpy (at constant p) can either increase or decrease, since it's measured relative to the enthalpies of either the reactants or products (depending on which direction you're considering). But, at constant p, the enthalpy of the activation complex (as opposed to the activation enthalpy) always increases with temperature. Apr 8, 2022 at 22:21
• I really thought I have a quite solid grasp on this, but I don't get it. I do not understand any of your argument. I do not expect the OP to understand it either. Apr 8, 2022 at 22:51
• Why should it necessarily be enthalpy and not internal energy (not that it affects any conclusions)? [genuine question; not trying to be accusatory, because I'm sure I've seen these things plotted with just pure energies] Apr 8, 2022 at 23:41
• @orthocresol I added a paragraph at the end that should address your question. Please let me know if it doesn't. Apr 9, 2022 at 2:09
• @Martin-マーチン I'm not sure what you're trying to say here. Are you asking me to help you understand my answer, or just complaining that my answer is not understandable? I don't wish to presume here, but I gather that Karsten and orthocresol understood it, and I would think with your background you'd have no problem understanding it as well. Besides, it's not that complicated. So I guess you could say that I am confused by your confusion :). Apr 13, 2022 at 6:45

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).

• @Maurice : Vertical axis represent free energy or enthalpy? Apr 8, 2022 at 0:16
• @Adnan AL-Amieh. Why do you ask me about the axis ? I am not the author of the question. Apr 8, 2022 at 7:53
• If the vertical axis is free energy, the difference between the product and reactant (ie the $\Delta G$ of the reaction) is temperature dependent, even assuming standard concentrations (as is normal in a reaction diagram). It is also common to use enthalpy for the vertical axis, which varies much less with temperature. Apr 8, 2022 at 12:48
• @Andrew That's a good point. In principle those values are not truly constant as functions of temperature...
– Zhe
Apr 8, 2022 at 15:04
• @Andrew Interestingly, if the choices are blue or red, and assuming we're at constant p, the vertical axis cannot be G. Apr 8, 2022 at 22:27