For the reaction: X 2 O 4 l → 2X O 2 g ΔU = 2 .1 kcal, ΔS = 20 cal K − 1 300 K Hence Δ G is

X 2 O 4 l → 2 X O 2 g ; Δ n g = 2 - 0 = 2 Δ H = Δ U + Δ n g R T = 2.1 + 2 × 2 1000 × 300 ΔH = 3 .3 kcal ΔG = ΔH − T ⋅ ΔS = 3 .3 − 300 × 20 1000 ; ΔG = - 2 .7 kcal

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Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIAdiabatic Process/PolytropicMedium2 minPYQ_2020

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For the reaction: $X_{2} O_{4} (l) \to 2X O_{2} (g)$
$ΔU = 2 .1 kcal, ΔS = {20 cal K}^{- 1} 300 K$
Hence $Δ G$ is

Options:

Answer:

Solution:

$X_{2} O_{4} (l) \to 2 X O_{2} (g); Δ n_{g} = 2 - 0 = 2$
$Δ H = Δ U + Δ n_{g} R T = 2.1 + 2 \times \frac{2}{1000} \times 300$
$ΔH = 3 .3 kcal$
$ΔG = ΔH - T \cdot ΔS = 3 .3 - 300 × \frac{20}{1000}; ΔG = - 2 .7 kcal$

Stream:NTA_ABHYASSubject:ChemistryTopic:Thermodynamics - IISubtopic:Adiabatic Process/Polytropic

⏱ 2mℹ️ Source: PYQ_2020

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