Le Châtelier’s principle dictates that the reaction should be less favourable (K will decrease), but let’s say the reaction has positive entropy—shouldn’t ∆G decrease?
And vice versa. For an endothermic reaction with negative entropy, how does the reaction become more favorable (incr. K) and less spontaneous (increase ∆G) with temperature?
In: Chemistry
So, you have a chemical reaction A + B –> C + D + heat. It’s exothermic.
The reverse reaction will also be happening: C + D + heat –> A + B. This one is endothermic.
Raising the temperature will increase the reaction rates of *both* of these. But not by the same amount. So one direction is favoured over the other.
Le Chatelier’s principle says “if we provide more of C, D or heat, we get more of C + D + heat –> A + B happening. Or, if we provide more of A or B, we get more of A + B –> C + D + heat happening”.
So, as you noted, Le Chatelier’s principle says “raising the temperature favours the endothermic reaction”.
In terms of molecules, what’s happening is this:
* You have a mixture of A, B, C and D.
* Occasionally, an A and a B collide. If they collide hard enough, they overcome the activation energy, and you get a C + D plus some excess energy.
* Also, occasionally, a C and a D collide. If they collide hard enough, they overcome the activation energy, and you get an A + B. But the activation energy barrier is a bigger barrier this way round, because the C and D also must provide all the extra energy that gets stored in the A + B as chemical potential energy.
* So for A+B -> C+D+heat to happen, the A and the B only need to have so much energy, but for the reverse to happen, the C and the D need a chunk more energy.
* Now we raise the temperature. Every collision now has more energy. For the A+B, well, they often had enough energy before, so the new high temperature doesn’t make much difference. For C+D, however, they often did not have enough energy to even form A+B, now it’s much more likely they do. So although both reactions speed up, C+D+heat -> A+B speeds up more, ie, the endothermic reaction is favoured.
It becomes less spontaneous as an increase of temperature, makes it harder to decrease disorder. Imagine it like trying to stack blocks on a table that’s being shook. The shaking is the temperature of the reaction. The more shaking/temperature, the harder it is to decrease disorder/stack blocks.
Here’s a more succinct answer: if heat is a product, raising the temperature is like adding the product, which makes the reaction less favorable. But if the reaction is endothermic, heat is a reactant, so adding heat will push it along.
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