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Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIGibb's Free EnergyMedium2 minQB

ChemistryMediuminteger

Passage / Comprehension

Gibbs energy of formation for cis- and trans-but-2-ene at $300 K$ are $66 k J m o l^{- 1}$ and $62 k J m o l^{- 1}$ respectively. The magnitude of Gibbs energy of the transformation is

$x J m o l^{- 1}$ , and the mole fraction of trans-but-2-ene in the equilibrium mixture is $y$ . [Given: $R = 8.3 J K^{- 1} m o l^{- 1}$ , $lo g 5 \approx 0.7$ , $ln z = 2.3 lo g z$ ]

The value of $y$ is $____$ .

Answer:

4096

Solution:

$Δ_{r} G^{\circ} Δ_{r} G^{\circ} - 4000 \Rightarrow lo g K \Rightarrow K K \Rightarrow m o l e f r a c t i o n o f t r an s = 62 - 66 = - 4 k J m o l^{- 1} = - 2.3 RT lo g K = - 2.3 \times 8.3 \times 300 lo g K = 0.698 \approx 0.7 = 1 0^{0.7} \approx 5 = \frac{[ t r an s ]}{[ c i s ]} = \frac{5}{1} = \frac{5}{6} = 0.83$

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Gibb's Free Energy

⏱ 2mℹ️ Source: QB

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