Enthalpy change for the transition of liquid water to steam is 41 kJ mol − 1 at 10 0 ∘ C . The entropy change for the above process is

Δ S t = T △ H τ = ( 373 41000 ) J K − 1 m o l − 1 = 110 J K − 1 m o l − 1

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ChemistryThermodynamics - IIFree Energy and EquilibriumMedium2 minQB

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Enthalpy change for the transition of liquid water to steam is $41 kJ mol^{- 1}$ at $10 0^{\circ} C$ . The entropy change for the above process is

Answer:

Solution:

$Δ S_{t} = \frac{△ H _{τ}}{T} = (\frac{41000}{373}) J K^{- 1} m o l^{- 1} = 110 J K^{- 1} m o l^{- 1}$

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Free Energy and Equilibrium

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