[thepackage]

Temperature increases the kinetic energy of particles, hence a higher proportion of particles have enough energy to satisfy the activation energy.
Doesn’t this mean both the forward and backward reactions are equally favoured in equilibrium? since particles from both the backward and forward reactions have higher kinetic energy.

I sort of understand the reasoning on why endothermic reactions would be favoured, but why doesn't the scenario i've described above also occur?
Also, if the endothermic reaction is favoured, won't this mean that the surrounding temperature will decrease until equilibrium is established again?

[Borek]

Sounds like you are confusing kinetics with equilibrium. They are completely independent. Kinetics tells you how fast you will get to the equilibrium, but doesn't tell where the equilibrium lies.

[Corribus]

Doesn’t this mean both the forward and backward reactions are equally favoured in equilibrium?

No, the forward reaction is favored more as temperature is increased.

The reason this is the case is rather complicated and derives from statistical mechanics. A hand-wavy explanation goes something like this.

Consider for sake of argument a reaction A + B C + D.  In an endothermic reaction, the products C and D have higher potential energy than the reactants A and B. This means that products have a smaller reaction barrier going back to reactants than the reactants have to go to the products.  However, since both reactants and products are at the same temperature, reactants and products have the same kinetic energy. This means that* a collision between a C and a D is more likely to result in a conversion to an A and a B than vice-versa - since C and D have a smaller hill to climb. Equilibrium favors the reactants in an endothermic reaction.

Now raise the temperature. Since reactants and products will still be at the same (but elevated) temperature, they all still have the same kinetic energy. Since C and D were already close to the top of the reaction barrier (by virtue of their higher potential energy), the additional boost from the temperature increase does not help them as much. Contrast to A and B, which had much further to go. In this case the boost helps them much more. This isn't necessarily an intuitive result, but perhaps an analogy helps. Imagine two groups of people separated by a wall, the same number of people on each side. Each group of people is trying to climb the wall to get to the other side. The ground on one side of the wall is five feet higher than the other. From the low-lying ground, the top of the wall is 10 feet high, and from the high-lying ground, the wall is only 5 feet high. Very few people can get over the wall from the low-lying side because it's hard to scale a 10 foot wall, but a lot of people can get over the wall from the high-lying side because here the wall is only 5 feet high. On average, more people will end up on the low-lying side. Now suppose you arrange for tall step stools of equal height to be put on each side of the wall. Which side does this help out more? Obviously it helps the people on the low-lying side more. The people on the high-lying side could already easily scale the wall - excepting the really short people, most of them don't need a step-stool. In this situation, people who have a higher wall to climb will benefit more. So in comparison to the situation without a step stool, more people will end up on the higher ground side of the wall. Equilibrium has shifted because of the step-stool to favor distribution at the higher elevation.

Mathematically, the reason it works out this way for molecules is the exponential relationship between rate, temperature, and the activation energy.

If you assume an Arrhenius relationship, the forward rate is

[tex]k_1=A e^{-E_1/k_b T}[/tex]

(k_b is the Boltzmann constant; T is the temperature) and the backward rate is

[tex]k_{-1}=A e^{-E_{-1}/k_b T}[/tex]

This means the Equilibrium constant:

[tex]K=\frac{k_1}{k_{-1}}=\frac{e^{-E_1/k_b T}}{e^{-E_{-1}/k_b T}}=e^{\frac{E_{-1}-E_{1}}{k_b T}}[/tex]

Assuming the pre-exponential A terms are the same in the forward and backward direction. We can pause here for a second and see if that if E₁ > E_-1 (that is, the activation energy for the forward direction is greater than that of the backward direction), then the factor above the exponential term is negative and equilibrium constant is less than 1 - the reaction is endothermic. Likewise, if E₁ < E_-1, the factor above the exponential is positive and the equilibrium constant is greater than 1 – the reaction is exothermic. All this squares with our expectations and the definitions of exothermic and endothermic reactions.

Now, we define the difference in the activation energies as ΔE, where ΔE = E_-1 - E₁. At some lower temperature T_low, we have the equilibrium constant is:
[tex] K_{low}=e^{\frac{\Delta E}{k_b T_{low}}} [/tex]
Likewise at a higher temperature,
[tex] K_{high}=e^{\frac{\Delta E}{k_b T_{high}}} [/tex]
Therefore
[tex] K_{high} - K_{low} =e^{\frac{\Delta E}{k_b T_{high}}} - e^{\frac{\Delta E}{k_b T_{low}}} [/tex]
Using the properties of exponential functions, you should be able to convince yourself that if ΔE < 0 (that is, the reaction is endothermic because the reaction barrier for the reverse process is smaller than the reaction barrier for the forward process), then the right-hand side of this equation must be positive, and therefore
[tex] K_{high} > K_{low}[/tex]
I.e., the equilibrium constant is greater at higher temperature -> the reaction favors the forward direction as the temperature is increased.
On the other hand, if ΔE > 0 (that is, the reaction is exothermic because the reaction barrier for the reverse process is larger than the reaction barrier for the forward process), then the right-hand side of the above equation is negative , which implies
[tex] K_{low} > K_{high}[/tex]
I.e., the equilibrium constant is greater at lower temperature -> the reaction favors the backward direction as the temperature increases.

*This is assuming all other factors that influence reaction events are the same in both directions (that is, the Arrhenius A factors are identical).