Question

In: Chemistry

The state diagram for substance X is characterized by a triple point (195 K, 0.0600 atm)...

The state diagram for substance X is characterized by a triple point (195 K, 0.0600 atm) and a critical point (406 K, 111 atm). These two points represent the two extreme conditions where the equilibrium X(l) ⇌ X(g) takes place; the pressure at each point represents the thermodynamic constant characterizing the process of evaporation of liquid X.

a) Estimate the values of ∆H°vap and ∆S°vap for the process X(l) → X(g).

b) From the values obtained part (a), estimate the normal boiling point of X.

Solutions

Expert Solution

ln (P2/P1) = Hvap/ R [1/T1-1/T2]

or, ln (111/0.06) = Hvap/8.314 J/K/mol [1/195-1/406]

or, Hvap = 23467.99 J/mol

--------------------------------------------------

S = Rln (P1/P2)

        = 8.413 J/K/mol ln (0.06/111) = -62.55 J/K

G = Hvap - TSvap

At boiling point, the substance is in equilibrium and at equilibrium G = 0

T = Hvap / Svap = 23467.99 J/mol/ -62.55 J/K

                                 =-375.18 K


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