Question
Chemistry Question on Thermodynamics
ΔrG∘ 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)
A
0.1 atm
B
1 atm
C
10 atm
D
100 atm
Answer
0.1 atm
Explanation
Solution
We have S(liquid)⇌S (gas) ΔGreaction ∘=ΔrG∘( vapor )−ΔfG∘ (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