Δr G° at 500 K for substance 'S' in liquid state and gascous 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 -1 mol -1)

Question

Δr G° at 500 K for substance 'S' in liquid state and gascous 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 -1 mol -1) (A) 0.1 atm (B) 1 atm (C) 10 atm (D) 100 atm

Tardigrade Admin · Accepted Answer

We have S text(liquid) leftharpoons S text (gas) Δ G text reaction ° =Δr G°( text vapor )-Δf G° text (liquid) =103-100.7 =2.3 kcal mol -1 We know Δ G text Extion °=-R T ln K 2.3 × 103=-2 × 500 × ln Kp ln Kp=-2.3 ⇒ Kp=0.1 ⇒ text vapour pressure =0.1 atm

Q. Δr​G∘ at 500K for substance 'S' in liquid state and gascous state are +100.7kcalmol−1 and +103kcalmol−1, respectively. Vapour pressure of liquid ' S ' at 500K is approximately equal to (R=2calK−1mol−1)

0.1 atm

1 atm

10 atm

100 atm

We have S(liquid)⇌S (gas) ΔGreaction ∘​=Δr​G∘( vapor )−Δf​G∘ (liquid) =103−100.7 =2.3kcalmol−1 We know ΔGExtion ∘​=−RTlnK 2.3×103=−2×500×lnKp​ lnKp​=−2.3 ⇒Kp​=0.1 ⇒ vapour pressure =0.1atm

Q. $Δ_{r} G^{\circ}$ at $500 K$ for substance 'S' in liquid state and gascous 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 $(R = 2 c a l K^{- 1} m o l^{- 1})$