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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

ΔfGo at 500...

Question

$Δ_{f} G^{o}$ at $500 K$ for substance 'S' in liquid state and gaseous state are $+ 100.7 k c a l m o l^{- 1}$ and $+ 103 k c a l m o l^{- 1}$ , respectively.
Vapour pressure of liquid 'S' at $500 K$ is approximately equal to: $(R = 2 c a l K^{- 1} m o l^{- 1})$ .

100

atm

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1

atm

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10

atm

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0.1

atm

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Solution

The correct option is B $10$ atm
$Δ G_{r x n}^{o} = Δ_{f} G^{o} (vapour) - Δ_{f} G^{o} (liquid)$

$Δ G_{r x n}^{o} = 103 - 100.7 = 2.3$ kcal/mol

$Δ G_{r x n}^{o} = - R T l n K$

$2.3 k c a l / m o l \times 1000 c a l / k c a l = - 2 c a l / m o l / K \times 500 K \times l n K$

$l n K = 2.3$

$K = 10 a t m =$ Vapour pressure of liquid 'S'
Vapour pressure of liquid 'S' at $500$ K is approximately equal to 10 atm.

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ΔfGo at 500 K for substance 'S' in liquid state and gaseous state are +100.7 kcal mol−1 and +103 kcal mol−1, respectively. Vapour pressure of liquid 'S' at 500 K is approximately equal to: (R=2 cal K−1mol−1).

The correct option is B 10 atm ΔGorxn=ΔfGo(vapour)−ΔfGo(liquid) ΔGorxn=103−100.7=2.3 kcal/mol ΔGorxn=−RTlnK 2.3kcal/mol×1000cal/kcal=−2cal/mol/K×500K×lnK lnK=2.3 K=10atm= Vapour pressure of liquid 'S'Vapour pressure of liquid 'S' at 500 K is approximately equal to 10 atm.

$Δ_{f} G^{o}$ at $500 K$ for substance 'S' in liquid state and gaseous state are $+ 100.7 k c a l m o l^{- 1}$ and $+ 103 k c a l m o l^{- 1}$ , respectively.
Vapour pressure of liquid 'S' at $500 K$ is approximately equal to: $(R = 2 c a l K^{- 1} m o l^{- 1})$ .