$Δ_{f} G^{0}$ at 400 K for substance ‘ S ’ in liquid state and gaseous state are $+ 200.7 k c a l m o l^{- 1}$ and $+ 203 k c a l m o l^{- 1}$ respectively. Vapour pressure of liquid ‘ S ’ at 400K is approximately equal to $(R = 2 c a l K^{- 1} m o l^{- 1})$

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$6 \times 10^{- 2} a t m$

1 atm

$6 \times 10^{- 3}$

0.1 atm

answer is C.

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Detailed Solution

$S_{(l)} \to S_{(g)} K_{p} = P_{v a p} Δ G_{Re a c t i o n}^{0} = Δ_{G}^{F^{0}} (V a p o u r) - Δ F {G^{0}}_{L i q} Δ G_{r e a c t i o n}^{0} = 203 - 200.7 = 2.3 K c a l / m o l e = 2300 C a l / m o l Δ G_{r e a c t i o n}^{0} = - R T \ln K_{p} 2300 = - 2.303 \times 2 \times 400 \log K_{p} \log K_{p} = - 1.24 \Rightarrow K_{p} = 10^{- 124} = 10^{- 2} \times 10^{0.76} = 6 \times 10^{- 2} a t m \ln K_{p} = - 2.875, \log_{10} K_{p} = - 1 = - 1.24 K_{p} = a n t i \log (- 1) = 0.1 a t m = 10^{- 2} \times 10^{0.76} = 6 \times 10^{- 12}$