Standard entropies of x 2 , y 2 and x y 3 are 60, 40 and 50 J K - 1 m o l - 1 respectively. For the reaction 1 2 x 2 + 3 2 y 2 ⇌ x y 3 . Δ H = - 30 k J to be at equilibrium, the temperature should be

Δ S = S x y 3 - 1 2 S x 2 - 3 2 S y 2 Δ S = 50 - 1 2 × 60 - 3 2 × 40 = - 40 J K - 1 m o l - 1 At equilibrium Δ G = 0 Δ G = Δ H - T Δ S 0 = - 30 - ( - 40 × 1 0 - 3 × T ) T = 750 K

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ChemistryThermodynamics - IFirst law of ThermodynamicsEasy2 minPYQ_2020

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Standard entropies of $x_{2}, y_{2}$ and $x y_{3}$ are 60, 40 and 50 J $K^{- 1} m o l^{- 1}$ respectively. For the reaction $\frac{1}{2} x_{2} + \frac{3}{2} y_{2} ⇌ x y_{3} . Δ H = - 30 k J$ to be at equilibrium, the temperature should be

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Answer:

Solution:

$Δ S = S_{x y_{3}} - \frac{1}{2} S_{x_{2}} - \frac{3}{2} S_{y_{2}}$
$Δ S = 50 - \frac{1}{2} \times 60 - \frac{3}{2} \times 40$
$= - 40$ J $K^{- 1} m o l^{- 1}$
At equilibrium $Δ G = 0$
$Δ G = Δ H - T Δ S$
$0 = - 30 - (- 40 \times 1 0^{- 3} \times T)$
$T = 750 K$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics

⏱ 2mℹ️ Source: PYQ_2020

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