Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIGibb's Free EnergyMedium2 minQB

ChemistryMediumsingle choice

$PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)$

$Δ G_{f}^{\circ} [PC l_{5} (g)] = - 74 k c a l / m o l$

$Δ G_{f}^{\circ} [PC l_{3} (g)] = - 60 k c a l / m o l$

then calculate value of equilibrium constant for dissociation of PCl₅ (g) at 727ºC temperature ?

(ln2 = 0.7)

Answer:

Solution:

$Δ G^{o} = - RT ln k_{eq}^{o}$

$14 - 10 ln 2 k_{eq}^{o} = - (\frac{2}{1000}) (1000) ln k_{eq}^{o} = ln k_{eq}^{o} = 2^{- 10}$

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

⏱ 2mℹ️ Source: QB

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