1
$\begingroup$

I'm trying to explain how a change in temperature favours an endothermic/exothermic reaction from a collision theory point of view, but I'm not really familiar with the concept just yet. For an increase in temperature, can I say that:

An increase in temperature leads to a greater frequency of collisions between reactants and products, leading to an increased rate of reaction. Since the rate of the endothermic reaction increases more than the exothermic reaction, the equilibrium shifts towards the reactants/products.

If this is right, I'm struggling to see why the endothermic reaction rate will increase more with this logic. The exothermic reaction has a lower activation energy, so shouldn't its rate of reaction increase more than endothermic since it has a lower energy requirement to complete the reaction? Unless the increase in particles with enough energy for the exothermic reaction is comparatively lower to the increase in particles with enough energy for the endothermic reaction (aka proportion)?

$\endgroup$
  • 1
    $\begingroup$ You cannot just explain the kinetics with thermodynamics and vice versa. Your proposal seems to work qualitatively in one direction, but it leaves out e.g. activation energy, so it can never give you a correct picture of the kinetics. $\endgroup$ – Karl Jan 14 at 9:05
1
$\begingroup$

If you look at the graph of the reactivity with relation to temperature it increases proportionally until it reaches its activation energy and then it proceeds to decrease. For exothermic reactions since the activation energy is met early the reactivity decreases when more temperature is provided. While for endothermic reactions since they require a higher activation energy combined with the fact that they absorb some of the energy for the reaction to take place, we can draw the conclusion that increase in temperature favours endothermic reactions.

$\endgroup$
  • $\begingroup$ I think I kinda of understand what you're trying to say. Do you have a link with a graph perhaps that I can look at? I tried looking at the Arrhenius equation on wikipedia, but that just shows an exponential graph. Not sure if that's the right path to go down. $\endgroup$ – CipherBot Jan 15 at 11:09

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.