Would this reaction be product favored at higher or lower temperatures?

Question

Would this reaction favor the products at low temperatures, higher temperatures, or both? If possible please explain a bit how you know. Thanks

Mg(s) 1/2 O2(g) ====> MgO(s) Delta H = -601.70 kJ

Answer 1

Delta H is negative, which means that the reaction is exothermic.  For exothermic reactions, lower temperature favors the products.  Think of heat as a product in the reaction.  If you lower temperature. you take away heat.  LeChatelier's principle applies, and the reaction shifts towards making more heat, which is the products side..

Answer 2

Delta G= Delta H -T * Delta S

a negative delta G means that the reaction is spontaneous (ie that products are favored)

Delta S is the change in entropy, and as you can see, the products side is solid (more order and lower entropy) than the gaseous reactant side, so Delta S is negative

so w/ Delta G=-601.7-T*(something negative),

This means you have a negative number + T*(a positive number) (cuz of the double negative)

And since T>0 always, if you're trying to get the reaction to favor products, you want Delta to be as small as possible, you want T to be as low as possible