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For a reaction that is exothermic and proceeds with an increase in entropy, how will increasing temperature affect ΔG?
ΔG = ΔH - TΔS
ΔH is negative because it's exothermic and the question states ΔS is positive. ΔG is negative.
If you increase T, ΔG will become more negative and the rxn is more spontaneous (this is the correct answer)
But if Heat is a product (the rxn is exothermic), wouldn't increasing the T (i.e, adding more heat) shift the rxn towards the endothermic?
ΔG becoming more negative but the rxn shifting to the reverse is contradictory. Which happens?
ΔG = ΔH - TΔS
ΔH is negative because it's exothermic and the question states ΔS is positive. ΔG is negative.
If you increase T, ΔG will become more negative and the rxn is more spontaneous (this is the correct answer)
But if Heat is a product (the rxn is exothermic), wouldn't increasing the T (i.e, adding more heat) shift the rxn towards the endothermic?
ΔG becoming more negative but the rxn shifting to the reverse is contradictory. Which happens?