Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IIFree Energy and EquilibriumEasy2 minQB
ChemistryEasynumerical
The standard entropy change (in cal/K-mol) for the reaction: A ⇌ B, if the value of ΔH° = 4.5 kcal/mol and KC = e⁻⁵ at 300 K.
Answer:
5.00
Solution:
ΔG° = −RT ln KC = −(2)(300)(−5) = +3000 cal mol⁻¹ = 3 kcal mol⁻¹. ΔS° = (ΔH° − ΔG°)/T = (4.5 − 3.0)×10³/300 = 1500/300 = 5 cal K⁻¹ mol⁻¹.
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Free Energy and Equilibrium
⏱ 2mℹ️ Source: QB
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