The vapor pressure of chlorine dioxide, ClO2, is 155 Torr of at -22.75 °C and 485 Torr at 0.00 °C.
Calculate(a) the standard enthalpy of vaporization; (b) the standard entropy of vaporization; (c) the standard Gibbs free energy of vaporization; (d) the normal boiling point of ClO2.

(a)  2.85 x 10⁴ J = 28.5 kJ

(b) 100.7 J/K

(c) 0

(d) 283 K

Explanation:

The strategy here is to use the Clausius Clayperon equation

ln( p₁/p₂ ) = - ΔvapH/R [ 1/T₁ - 1/T₂ ]

where p₁ and p₂ are the partial pressures at T₁ and T₂,  ΔvapH is the change in enthlpy and R the gas constant.

Then we can solve for  ΔvapH

ln ( 155/ 485 ) = - ΔvapHº/8.314 J/molK x [ 1/(- 22.75 + 273)K -  1/( 0.00 + 273)K]

- 1.141 =  - ΔvapHº/8.314 J/molK x (1/250.3 - 1/273)K

ΔvapHº = 2.85 x 10⁴ J = 28.5 kJ (a)

(b) The change in entropy is given by the expression

ΔvapSº = ΔvapHº / T

Where ΔvapS is the standard entropy of vaporization and T is tne normal boiling point which we do not know but can calculate from the Clausius Clayperon equation at p = 760 torr and any of the pressures given in this question:

ln( p₁/p₂ ) = - ΔvapHº/R [ 1/T₁ - 1/T₂ ]

ln(760/485) = -2.85 x 10⁴ J/ 8.314JK⁻¹  x [1/T₁ - 1/273 ] K

After doing some algebra we get  

T₂ = 283 K =  normal boiling point

Now we can compute

ΔvapSº = ΔvapHº / T =  2.85 x 10⁴ J / 283 K = 100.7 J/K (b)

(c) The standatd Gibbs free energy of reaction  for a phase change is zero since we are at equilibrium and the vaporization  change occurs at constant pressure and temperature

the answer is a and have a great day