As we know that, at equilibrium, ∆G∘=−2.303 RT logKp..... (1) S(I)⇌S(g) From the above reaction, ∆G∘=∆Gf(product) −∆G(reactant) ∘ ⇒∆G∘=103−100.7=2.3kcal∕mol=2.3×103kcal∕mol Given:- T=500K R=2cal∕mol−K From eq n (1), we have 2.3×103=−2.303×2×500×logKp ⇒logKp=−1 ⇒Kp=10−1atm Now, from the above reaction, Kp=
1
PS1
⇒PS1=
1
10−1
=10atm Hence the vapou pressure of liquid S is approximately equal to 10atm .