Question

10. Calculate the heat of vaporization and the entropy of vaporization of CCI, based on thermody- namic data in the appendix of your textbook. Will CCI, spontaneously boil at 90°C? Estimate the boiling point of CCl

This wasnt included in the thermodynamic table nor any of the formulas used in TRO ch 18 but i googled it and it came up 44.22 J/molK

Answer 1

The vaporization reaction of CCl4 is CCl4 (l) -----------> CCl4 (g)

∆H^o ( heat of vaporization)

∆H^orxn = ∆H^o_f of products - ∆H^o_f of reactants

= ∆H^o_f [CCl4 (g)] - ∆H^o_f [CCl4 (l)]

= -95.98 kJ/mol - [ -128.4 kJ/mol ]

= +32.42 kJ/mol

∆S^orxn ( Entropy of vaporization)

ΔS^o= S^o (products) - S^o(reactants)

= S^o [CCl4 (g)]- [CCl4 (l)]

      = [ 309.65 J/K/mol ] - [ 214.39 J/K/mol ]

     = + 95.26  J/K/mol

∆G^orxn Given that T = 90oC = 90 + 273 K = 363 K

We know that ∆G^orxn =  ∆H^orxn - T∆S^orxn

= 32.42 kJ/mol - ( 363 K x 95.26  J/K/mol)

= 32.42 x 10³ J/mol - ( 363 K x 95.26  J/K/mol)

= -2159.38 J/mol

  ∆G^orxn = -2159.38 J/mol

Since   ∆G^orxn is -ve, the reaction is spontaneous.

Therefore, CCl4 spontaneously boils at 90oC.

Boiling point of CCl4:

T = ∆H^orxn/ ∆S^orxn

= 32.42 kJ/mol / 95.26  J/K/mol

= 32.42 x 10³ J/mol / 95.26  J/K/mol

= 340.3 K

= 340.3- 273 oC

= 67.3 oC

T =   67.3 oC

Therefore,

Boiling point of CCl4 = 67.3 oC