What is the value of Δ G ∘ at 500 K . (in kJ )

Δ G ∘ = Δ H ∘ − T Δ S ∘ = − 50 × 1 0 3 + 500 × 125 = 12.5 kJ

Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIGibb's Free EnergyMedium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

Suppose a chemical reaction is spontaneous below 400 K and non-spontaneous above 400 K under standard conditions. $Δ H^{\circ}$ and $Δ S^{\circ}$ of the reaction do not vary with temperature, and $Δ H^{\circ} = - 50 kJ$ .

What is the value of $Δ G^{\circ}$ at 500 K . (in kJ )

Options:

Answer:

Solution:

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = - 50 \times 1 0^{3} + 500 \times 125 = 12.5 kJ$

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Gibb's Free Energy

⏱ 2mℹ️ Source: QB

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