Standard entropies of X 2 , Y 2 and XY 3 are 60 , 30 and 50 JK − 1 mol − 1 respectively. For the reaction 1 2 X 2 + 3 2 Y 2 ⇌ XY 3 , ΔH = − 30 kJ to be at equilibrium, the temperature should be:

ΔS for the reaction 1 2 X 2 + 3 2 Y 2 ⇌ XY 3 ΔS = 50 − 1 2 × 60 + 3 2 × 30 = − 25 J For equilibrium ΔG = 0 = ΔH − TΔS T = ΔH ΔS = − 30000 − 25 = 1200 K

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ChemistryThermodynamics - IIAdiabatic Process/PolytropicMedium2 minPYQ_2020

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Standard entropies of $X_{2}, Y_{2}$ and ${XY}_{3}$ are $60, 30$ and $50 {JK}^{- 1} {mol}^{- 1}$ respectively. For the reaction $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ {XY}_{3},$ $ΔH = - 30 kJ$ to be at equilibrium, the temperature should be:

Options:

Answer:

Solution:

$ΔS$ for the reaction $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ {XY}_{3}$

$ΔS = 50 - (\frac{1}{2} × 60 + \frac{3}{2} × 30) = - 25 J$

For equilibrium $ΔG = 0 = ΔH - TΔS$

$T = \frac{ΔH}{ΔS} = \frac{- 30000}{- 25} = 1200 K$

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

⏱ 2mℹ️ Source: PYQ_2020

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