The normal boiling point of iodomethane, CH3I, is 42.43 8C, and its vapor pressure at 0.00 8C is 140. Torr. Calculate (a) the stan- dard enthalpy of vaporization of iodomethane; (b) the standard entropy of vaporization of iodomethane; (c) the vapor pressure of iodomethane at 25.0 8C.

a=28600J; b=90.6 J/K; c=402 torr

Explanation:

(a) considering the data given

 Vapour pressure P1 =0  at Temperature T1 = 42.43˚C,

Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)

Using the Clausius-Clapeyron Equation

ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)

In 760/140 = ΔH/8.314 J/mol/K  × (1/315.58K -- 1/273.15K)

ΔH vap= +28.6 kJ/mol or 28600J

(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.

Since ΔG at boiling point is zero,

ΔS =(ΔH°vap/Τb)

 ΔS = 28600 J/315.58 K

= 90.6 J/K

(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)

ln P298 K/1 atm =  28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)

P298 K = 0.529 atm

                = 402 torr

i just learnt this recently but i need to look through my notes again