If ∆ G = ∆ H – T ∆ S, then why does increasing temperature for an exothermic reaction make it less favorable?

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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

3 Answers

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.