What is the value of Δ S ∘ at 400 K . (in J / K mole)

Δ G ∘ = Δ H ∘ − T Δ S ∘ = 0 Δ S ∘ = T Δ H = 400 − 50 × 1 0 3 = − 125 ( J / K )

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ChemistryThermodynamics - IICalculation of DSMedium2 minQB

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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 $Δ S^{\circ}$ at 400 K . (in $J / K$ mole)

Options:

Answer:

Solution:

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = 0$ $Δ S^{\circ} = \frac{Δ H}{T} = \frac{- 50 \times 1 0 ^{3}}{400} = - 125 (J / K)$

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Calculation of DS

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