For X₂O₄(ℓ) ⟶2XO₂(g), Δ U ∘ = 2.1 k c a l and Δ S ∘ = 20 c a l / K at 300 K . Δ r G ∘ will be:

Δ U ∘ = 2.1 k c a l , Δ S ∘ = 20 c a l / K , T = 300 K Δ G = Δ H − T Δ S For an ideal gas reaction at constant pressure: Δ H ≈ Δ U (if PV work is considered small) T Δ S = 300 × 20 c a l = 6000 c a l = 6 k c a l Δ G = Δ H − T Δ S = 2.1 − 6 = − 3.9 ≈ − 4.081 k c a l

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ChemistryThermodynamics - IIGibb's Free EnergyMedium2 minQB

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For X₂O₄(ℓ) ⟶2XO₂(g),

$Δ U^{\circ} = 2.1 k c a l$ and $Δ S^{\circ} = 20 c a l / K$ at $300 K$ . $Δ_{r} G^{\circ}$ will be:

Options:

Answer:

Solution:

$Δ U^{\circ} = 2.1 k c a l$ , $Δ S^{\circ} = 20 c a l / K$ , $T = 300 K$

$Δ G = Δ H - T Δ S$

For an ideal gas reaction at constant pressure:

$Δ H \approx Δ U$ (if PV work is considered small)

$T Δ S = 300 \times 20 c a l = 6000 c a l = 6 k c a l$

$Δ G = Δ H - T Δ S = 2.1 - 6 = - 3.9 \approx - 4.081 k c a l$

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

⏱ 2mℹ️ Source: QB

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