Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IFirst law of ThermodynamicsEasy2 minPYQ_2020
ChemistryEasysingle choice

AsO2(l) is cooled at 1 atm pressure, it freezes to form solid I at 54.5 K. At a lower temperature, solid I rearrange to solid II, which has a different crystal structure. Thermal measurements show that for the phase transition solid I to slid II,ΔH=-743.1 Jmol-1andΔS=-17.0 JK-1mol-1. At what temperature are solids I and II in equilibrium?

Options:

Answer:
C
Solution:

Since,ΔG=ΔH-TΔS
At equilibriumΔG=0
∴ΔH=TΔS
GivenΔH=-743.1 J/mol
ΔS=-17.0 JK-1mol-1
∴-743.1-17.0=T
T=43.7 K
At this temperature solid, I and II are in equilibrium.

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics
⏱ 2mℹ️ Source: PYQ_2020

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