Which of the following is true for a particular reaction if ∆G° is -40.0 kJ/mol at 290 K and –20.0 kJ/mol at 390 K?

Question

∆H > 0 , ∆S > 0

∆H < 0 , ∆S < 0

∆H > 0 , ∆S < 0

∆H < 0 , ∆S > 0

None of these

Answer 1

∆H < 0 , ∆S < 0

ΔG = ∆H -T ∆S

If ∆H is negative and ∆S is negative, -T∆S would be positive. As T increases, -T∆S becomes even more positive, which would make ΔG less negative.

For example, let's assume ∆H is -10000 kj/mole and ∆S is -20 kj/mole then at 290K, ΔG = -10000 -(290 x -20) = - 4200 kj/mole, but if T = 390K then ΔG = -2200 kj/mole.