Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IIFree Energy and EquilibriumMedium2 minQB
ChemistryMediumnumerical
The standard entropy change (in cal/K-mol) for the reaction: X ⇌ Y, if the value of ΔH° = 7.5 kcal/mol and KC = e⁻¹⁰ at 300 K.
Answer:
5.00
Solution:
ΔG° = −RT ln KC = −(2 cal K⁻¹ mol⁻¹)(300 K)(−10) = +6000 cal mol⁻¹ = 6 kcal mol⁻¹. ΔS° = (ΔH° − ΔG°)/T = (7.5 − 6.0)×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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