# 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, 2020 at 9:05

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, 2020 at 11:09

Here is an answer to that question, according to a final exam's marking guidelines at a particular school:

As the temperature increases, the average kinetic energy of the particles increases. This means that an increasing proportion of the particles has energy equal to, or greater than, the activation energy for the reaction. In terms of collision theory, a change in temperature has a greater impact on the rate of a reaction with a large $$E_\mathrm{a}$$ than on one with a small $$E_\mathrm{a}$$. As an example, raising the temperature from $$\pu{20 ^\circ C}$$ to $$\pu{30 ^\circ C}$$ for a reaction with an activation energy of $$\pu{60 kJ mol-1}$$ increases the fraction of collisions with enough energy to react by 97%. The same increase in temperature for a reaction with an activation energy of only $$\pu{42 kJ mol-1}$$ produces only a 75% increase in the fraction of collisions with enough energy to react. For an endothermic reaction, the activation energy is greater for the forward reaction than for the reverse exothermic reaction. The rate of the endothermic reaction is increased more than the rate of the backward reaction in response to a change in temperature. Hence, an increase in temperature will always favour the reaction that cools the reaction vessel (the endothermic reaction).

Suppose a hypothetical reaction A + B <--> C + D, where left to right is exothermic and right to left is endothermic.

To understand why endothermic rate is more impacted with increase in temperature you must consider the Arrhenius equation:

k = Ae^(-Ea/RT)

• k = rate constant
• A = frequency factor (or per-exponential factor) --> this factor varies only slightly with temeprature.
• e^(-Ea/RT) = fraction of molecules that have energies equal to or more than the activation energy.

To understand why temperature favours endothermic reaction you need to graph y = e^(-Ea/RT).

The x represents Temperature. The y represents frequency of molecules with energies equal to or above the activation energy. The green curve is for exothermic (forward) reaction where Ea is small, and the blue curve represents the endothermic (reverse) reaction where Ea is high.

You can see that as temperature increases the green curve plateaus the blue curve continues to rise. Eventually the blue curve and green curve will be almost on top of one another. This indicates that given a temperature high enough both the forward and reverse rates will have similar values for their rate constant k.

Whats interesting to note here is that for the exothermic curve (green) the rate constant rises very rapidly with a small temperature increase and then any increase beyond that has no significant impact. For the endothermic reaction (blue), however, the rate constant rises steadily and the higher the temperature goes the more it increases. Because both the curves eventually plateau as you extrapolate out the exothermic rate will never be less than the endothermic rate, they will just become really close to one another.

From a molecular standpoint A and B already have a good proportion of their molecules close to the forward Ea, and so any further increase in temperature won't make a significant difference to the amount of molecules that can reach the transition state. For C and D the majority of the molecules have low energy compared to the reverse Ea, and so increasing temperature will drastically increase the proportion of molecules that can reach the transition state.

Effect on Equilibrium Constant Keq:

So we've established why increasing temperature increase the endothermic reaction more than it increase the exothermic reaction rate. This means we will now have more A and B forming than we did at the lower temperature and more C and D disappearing.

Keq = [C][D] / [A][B]

This will shift Keq to a lower value. If we decrease the temperature the endothermic rate (reverse rate) will decrease faster than the exothermic rate (forward rate) and so [C] and [D] will rapidly form and [A] and [B] will be used up, shifting Keq back to a higher value.

I know that was a long answer but hopefully it helps understand why reactions behave the way they do under temperature change.

• You might want to update your post with chemistry and mathematics markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. Jan 10 at 21:49

(A = B + energy) is the total equality for a chemical reaction; Mass-Energy is conserved. At equilibrium the chemical potentials of products and reactants are equal. Since the equilibrium is dynamic this means that the rates of the forward and reverse reactions, when the products, reactants and energy content are contained, are equal. kf [A] = kr [B] [energy], [B][Energy]/[A] =K[equilibrium]= kf/kr. Keq expressed thusly is independent of temperature. [Normal Keq and rate constants ignore energy and are functions of temperature.] This means an increase of heat content, an increase in temperature, means a decrease in [products] and increase in the [reactants]. An increase in temperature changes the equilibrium in the endothermic direction.

This explanation uses the First Law and makes the combination of concentration and inherent energy content of the reactants and products paramount. The kinetic theory of matter is implicit; the chemical activity of the molecules depends on their concentrations and their energy content. It makes no mention of activation energy, which is good, because the activation energy concept depends on mechanism; this approach by starting at equilibrium is independent of mechanism. A second apparent contradiction is a phase change at constant pressure of a pure substance. Here both activities are defined as 1.; a change in energy will change the potential of both phases. The result is phase change to eliminate the change in energy by changing the amounts of each phase maintaining constant chemical activities, T, and P. The reaction in one direction is energy driven in the other entropy driven.