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

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.