Question

The vapor pressure of Y is 1.620 atm @ 200K and 4.698 atm @ 273K. What are the standard free energies of vaporization temperatures? Hint: vaporization reaction is Y(l) = Y(g)

(at 200K)

AG =?kJ/mol

(at 273K)

AG =?kJ/mol

Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values.

(Delta S degrees)

(Delta H degrees)

What is the normal boiling point of Y in degrees Celcius?

t= ? degrees Celsius

What is the vapor pressure of Y @ 303K?

P = ? atm

Thank you! I need help with the entire problem!

Answer 1

e AG 200 = - 8:31 solution cas As we know AGE = - RT enk (at equilibrium) For geven Meaction YW = Yg) At T= 200 K , Prep = 162 Datin, Pyce) = 1 atm 1G200 = -8.314 7 X 200 K x Inl molik = -802 2 Tmoel 0.8 kJ/mol At T= 273K , Pycay = 4,698 atm , P46) = 1 atm + 16 273 = -8:314 X 273 KX en (41498) - molok - 3512 J/mol = -3.5 kJ/mal (6) since here it is given that as & AH are independent of Temperature so Obor = -802:2 J/meel - AH'- 20061 sº 16273 = - 3512 J/mol = AH - 273Ka s' -- On applying eq" (i) -(1) , we will get 2709,8J/mol = 73KAS" + Asº = 270918 J/mol = 137 1/malok] 73K Now on putting as' value in een co we will go -802.2T/mol = AH - 200 K x 37 /malok → AH = 6597-8 5/mol = 606 kJ/mol le) from clausius clapeyron equation

= 178 К here 2 = 1 atm Р. - 1.20 афт T, - ? т, = 2CD K o en liten) = - $598 I/mall ook] 8:314 Т/mo! -К І Т. T, = 17BK - т, = (178 - 273) C = -44 °C | (0) Р. = ? Т. = 3D, к P, = 1:620 т= або к Дn Р. = - - 6 548 Т/ml г 1 - 1 8.314 т/me.к ! ЗОЗk vok 2 (132) P2 = 1831 atm/