If the enthalpy change for the transition of liquid water to steam is 30 kJ mol − 1 at 2 7 ∘ C , the entropy change for the process would beNEET 2011 (Screening)

Δ G ∘ = Δ H ∘ − T Δ S ∘ Given, Δ H vap. = 30 kJ mol − 1 Δ G ∘ = 0 at equilibrium, Δ S vap = T Δ H vap = 300 K 30 × 1 0 3 J mol − 1 = 100 J mol − 1 K − 1

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Chemistry - Thermochemistry Question with Solution | TestHub

ChemistryThermochemistryEnthalpy of ReactionHard2 minPYQ_2011

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If the enthalpy change for the transition of liquid water to steam is $30 kJ mol^{- 1}$ at $2 7^{\circ} C$ , the entropy change for the process would beNEET 2011 (Screening)

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Solution:

$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} Given, Δ H_{vap.} = 30 kJ mol^{- 1} Δ G^{\circ} = 0 at equilibrium, Δ S_{vap} = \frac{Δ H _{vap}}{T} = \frac{30 \times 1 0 ^{3} J mol ^{- 1}}{300 K} = 100 J mol^{- 1} K^{- 1}$

Stream:NEETSubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction

⏱ 2mℹ️ Source: PYQ_2011

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