Question

The vapor pressure for chlorine dioxide is 0.204 atm at –22.75 oC and 0.638 atm at 0.00^oC.
Calculate:
a) ∆G_vap^o at the two temperatures.
b) The boiling point at pressure 1 atm.

The answers are a) ∆G vap (250K) = 3,28 kJ/mol, ∆G vap (273) = 0,991 kJ/mol
b) 283 K

Answer 1

a) The vapor pressure is related to Gibb's free energy as follows:

Gvapor=RTln(p/p0) J

Where T is temparature in Kelvin or absolute scale.

R=8.314 J /mol/K

p->vapor pressure at the mention temparature

p0->vapor pressure at standard temparature=0.987 atm.

Substituting values,

In first case,

Gvapor=R×250.25(K)ln(0.204/0.987)J/mol=3280.13Joules/mol=3.28013kJoules/mol

In second case,

Gvapor=R×273(K)ln(0.638/0.987)=990.31 Joules=0.99031kJ/mol.

b)Boiling point is the temparature at which vapor pressure of substancebecomes 1 atm.

According to the relaion:

ln(p2/p1)=H/R (1/T1-1/T2)

Whre, Enthalpy change is H

Putting values p1 at T1=25025K=0.204atm

and p2=0.638 atm at T2.=273K.

H=431.28Joules/mol.

Now, we take boiling point T =T2 and T1=273K and p2=1 atm.

to find the required boiling point.

Calculating T=283 K.approx.

Hence boiling point of ClO2 is 283 K.