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

Question

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

Answer 1

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.