Answer:
(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
