Textbook 7.1 [ENDORSED]

[Sunny Wu 3A]

In some reactions, two or more different products can be formed. If the product formed by the fastest reaction predominates, the reaction is considered to be under "kinetic control." If the dominant product is the most thermodynamically stable, the reaction is considered to be under "thermodynamic control."
b) Does kinetic control predominate at low or high temperatures?

I looked up kinetic vs thermodynamic control, and the answer was explained with this graph:

Why is it that a lower activation energy corresponds to a higher free energy of the products? The answer to part b is that kinetic control predominates at lower temperatures.

Edit: After thinking about it I think that this graph corresponds to reaction pathways where one pathway is thermodynamically favored and the other one is kinetically favored, and only under these conditions will you have thermodynamic/kinetic control. In other reactions where for one pathway the Ea and the free energy of the products are both lower, then there is no control of one vs the other and that pathway will always be favored at any temperature. If anyone could confirm/refute this thank you :)

[Chem_Mod]

The reason why kinetic control would predominate at low temperatures is because of activation energies. If you look at the graphs that you have, the green line has a lower activation energy barrier. You can think of a lower activation energy as occurring "faster", so as you said, because the product formed by the green line is due to the "fastest reaction", a lower temperature will more likely favor the green reaction because it requires less of an energy input to occur. Now for why the lower activation energy corresponds to a higher free energy, I think it's easier to explain the converse. The blue line is "thermodynamically favored" because while it is slower than the green reaction, it is able to form a product that is lower in free energy. We must remember that a lower free energy corresponds to a more stable product. Therefore, this reaction is thermodynamically favored because while it requires a greater energy input and is a slower reaction, it forms a more stable product.