equilibrium constant and temperature (1 Viewer)

[Masaken]

I understand that with increasing temperature in exothermic reactions, the equilibrium constant decreases, while in endothermic the constant increases. But why is that the case (like what does the increasing temperature do to the products/reactants of the reaction causing the constant to increase/decrease, I feel like it's related to the mod 4 concepts + collision theory but I'm not so sure)? I've tried searching it up but I haven't found a definitive answer

 

[Vall]

I think researching "le chatelier's principle" will help. (I don't remember enough chem to give a proper answer lol)

 

[Masaken]

I get the principle itself, but I have no idea how it relates to endothermic and exothermic reactions (because the value of Keq either increases or decreases based on whether the reaction is endo/exo)

 

[someth1ng]

The full explanation refers to the Van 't Hoff equation (don't think this is HSC).

However, it comes down to this:

• Increasing temperature will increase the reaction rate for both the forward AND reverse reactions.
• The activation energy of the reverse and forward reactions are rarely equal. The temperature will not have the same impact on each reaction rate.
• Reactions with larger activation energy require more energy to proceed, so the increasing temperature will benefit that reaction more.
• For exothermic reactions, the forward reaction has lower activation energy than the reverse reaction, so the increased temperature will favor the reverse reaction more, reducing the equilibrium constant. The opposite is true for endothermic reactions.

Hope this helps.

 

[Masaken]

Ohhh I get it now, thanks very much to you both

 

[wizzkids]

This topic is never explained properly; I sympathise with your doubts and confusion. Let's start with temperature. What is temperature? Temperature is a measure of the average kinetic energy of particles. The Maxwell-Boltzmann distribution function describes how the kinetic energy of particles behaves when the average temperature is changed. See the figure below. Notice how the change of temperature from 300K to 600K lowers the peak, but spreads the kinetic energies to much higher values.

Next, we need to bring in the potential energy/reaction co-ordinate model, as mentioned by @someth1ng. Only molecules that possess more than the required activation energy E are candidates for overcoming the activation barrier. Notice how, as the temperature increases, the M-B Distribution function allocates more of the available kinetic energy to the upper "tail" of the distribution. Now, reactions that have a high activation energy are modelled by the vertical line E and reactions with a lower activation energy are modelled by the vertical line E. Compare the dark blue curve with the violet curve. Notice how the fraction of molecules that have E doubles when the temperature changes from 300K to 600K, but the fraction of molecules that have E quadruples when the temperature changes from 300K to 600K. Lastly, remember that all reaction rates increase when temperature increases, but reactions that have a high E increase more than reactions that have a lower value of E. Therefore, we conclude that the equilibrium constant K changes if the temperature changes. It shifts in the endothermic direction if temperature increases, and so if the forward reaction is endothermic, then K increases.