# How to explain that an increase in temperature favours an endothermic reaction?

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)?

• 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.
– Karl
Jan 14 '20 at 9:05

## 2 Answers

I believe that collision theory in itself wouldn't be enough to explain this behavior. You can explain it by having Arrhenius equation and the activation energy in mind. What you got wrong was the effect of temperature on reaction rates depending on the activation energies. Actually, the speed constant of a reaction with higher activation energy will be more significantly affected by variation in temperature. This can be verified easily by manipulation of the Arrhenius equation. Apply ln to both sides to obtain:

$$\ln{k}=\ln{A}-\frac{E_a}{RT}$$

The graph of ln(k) as a function of (1/T) is a straight line with slope's magnitude directly proportional to activation energy.

Now, assuming that the endothermic reaction has higher activation energy than its inverse, it will be more affected by, say, an increase in temperature. While both reactions will become faster, the speed constant of the endothermic reaction will increase to a greater degree and that will shift the equilibrium favoring the products of the endothermic reaction. If the temperature is decreased, the endothermic reaction will be slowed down to a greater degree, favoring the products of the exothermic reaction.

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.

• 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. Jan 15 '20 at 11:09