Question

Calculate the change in Gibbs free energy for each of the following sets of ΔH∘rxn, ΔS∘rxn, and T. (Assume that all reactants and products are in their standard states.)

Part A: ΔH∘rxn=−80.kJ, ΔS∘rxn=−154J/K, T=292K Express your answer as an integer.

Part B: ΔH∘rxn=−80.kJ, ΔS∘rxn=−154J/K, T=851K Express your answer as an integer.

Part C: ΔH∘rxn=80.kJ, ΔS∘rxn=−154J/K, T=292K Express your answer as an integer.

Part D: ΔH∘rxn=−80.kJ, ΔS∘rxn=154J/K, T=392K Express your answer as an integer.

Answer 1

The formula to calculate change in Gibbs free energy G⁰ is given by G⁰ = H - TS⁰

But one thing that we have to keep in mind is to match the units like we have to put all the values in J or KJ to make the calculation easy and to get the precise answer because if we don't do so we can get wrong answers.

We can also tell if the reaction is spontaneous or non spontaneous by knowing the value of gibbs free energy. If its value comes out to be negative then the reaction is spontaneous and if its positive the the reaction is non spontaneous but it its value is 0 then the reaction is in equilibrium.

Part A

Since change in entropy is in J/K, let's convert change in enthalpy in J too

1 KJ = 1000 J, -80 kJ = -80,000 J

G⁰ = -80,000 - (292 × -154) = -80,000 - (-44968) = -80,000 + 44,968 = -35,032 J

If we want to write it in kJ, we'd have to simply divide it by 1000 which would give

G⁰ = -35.032 kJ

The value of gibbs energy is negative. Therefore, the reaction is spontaneous.

Part B :- similar to Part A

G⁰ = - 80,000 - (851 × -154) = -80,000 + 131,054 = 51,054

G⁰ = 51.054 kJ

The value is positive, so the reaction is non spontaneous

Part C :-

G⁰ = 80,000 - (292 × -154) = 80,000 + 44,968 = 124968 J

G⁰ = 124.968 kJ

Part D :-

G⁰ = -80,000 - (392 × 154) = -80,000 - 60, 368 = - 140,368 J

G⁰ = -140.36 kJ

Tge value is, so the reaction is spontaneous.

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