@curious
While I've built some potential energy surfaces and estimated reaction barriers by DFT, I've never tried to calculate actual reaction rates from this information. Because there are other factors involved beyond what the DFT calculation will predict. Although, you can get a qualitative idea from the shape of the potential energy surface what vibrational modes need to be activated to lead to a reaction. Still, once the number of degrees of freedom increases beyond two, it becomes hard to picture a meaningful true reaction coordinate (the limitation being, of course, that the most dimensions you can plot on a piece of paper in a single graph is three).
@arkantos
Ok -
When you increase the temperature, two things happen. First, equilibrium is reached faster. Second, the equilibrium constant changes.
Your reaction is endothermic*, and thus at T1, you can see that equilibrium tends to favor reactants to products, as expected. When you increase the temperature to T2 (blue), not only does the overall reaction speed up (reach equilibrium faster), but it also begins to favor the products a little more than what was observed at T1 (still more R than P, but the ratio of R to P is less). What this essentially means is that when you increase the temperature from T1 to T2, both the forward and backward reaction rates increase, but the backward reaction rate increases LESS.
Why?
The transition state energy for both reactions is the same, which means the activation barrier for the reverse reaction is smaller than that of the forward direction. So generally speaking the reverse reaction (exothermic) happens faster than the forward (endothermic) reaction because at a given temperature, the probability of colliding product molecules having enough energy to go back to reactants exceeds the probability of the colliding reactants having enough energy to go forward to products.
But recall that as the temperature increases, the effect of temperature on any reaction rate approaches an assymptotic limit for reasons explained above. In our equilibrium where we have two reactions going on at one time, as the temperature increases the reaction with the smaller reaction barrier (the reverse, exothermic reaction) is closer to this "diffusion limited case" than tht reaction with the larger reaction barrier (the forward, endothermic reaction). Therefore increasing the temperature speeds up the endothermic reaction to a greater extent than the backward, exothermic reaction, and thus increasing the temperature will favor the formation of products in an overall endothermic equilibrium process. (The reverse is true for an overall exothermic process - increasing temperature will favor reactants in this case, although it's just a matter of bookkeeping at this point. Since we're talking about an equilibrium, there's always both an exothermic and endothermic process going on simultaneously. All that changes is which one is "forward" and which is "backward", an arbitrary distinction governed typically by what materials are initially put in the reaction flask.)
As you can guess, the exact opposite occurs when you lower the temperature. This will favor an exothermic reaction, and in your endothermic reaction will tend to favor formation of reactants, which is what your red plot shows.
My guess is that your professor was alluding to this effect when he/she said that increasing temperature reduces the rate of an exothermic reaction. Probably the professor just misspoke. Increasing temperature doesn't reduce the rate of an exothermic reaction. The rate still increases: it just increases less than the (simultaneous) endothermic counterpart, so that the endothermic products are - relative to the lower temperature reaction - favored.
Note that this reasoning really only works for small temperature changes. If the temperature changes a lot, then the actual value of ΔH begins to change by significant degree, and the situation is more complicated.
*Here we run into the annoying endothermic/endergonic issue again. And again we'll ignore it!