Question

The temperature of hte triple point: 195.41 K

Consider the phase diagram of NH3.   

Triple point: 195.410 K and 0.06921 atm

Normal boiling point: -33.342 oC

Normal freezing point: -77.728 oC

Density of liquid: 0.618g/mL

Density of solid: 0.817 g/mL

(a) Determine the heat of fusion, the heat of vaporization the heat of sublimation at near the triple point.

(b) Determine the entropy change of fusion, vaporization and sublimation at near the triple point.

(c) Plot the phase diagram in the temperature range T: [-90 oC, 130 oC]. For pressure (in unit of atm), use ln (p) other than p.

Answer 1

Multiple parts with parts within , (a) and (b) are answered :

(a) Using given data :

Triple point, a point at which the three phase (solid, liquid, and vapour) boundaries meet,at this all simultaneously coexist in equilibrium.

We determine  ΔHvap ;  ΔH_fus and  ΔHsub near triple points where solid, liquid and vapour exist in equilibrium.(The temperature at which the vapour pressure of a liquid is equal to the external pressure is called the boiling temperature at that pressure : we take triple point(T₃) and Tb as boiling T at respective P).

Similarly we can use triple point(T₃) and Tf as fusion T at respective pressures.

Using claussius -clayperon equation :

- ln(P1 /P2) =ΔHvap / R ((1/T2)-(1/T1))

From T₃ and Tb :

Tb: -33.342 oC = 239.8 K. Triple point: 195.410 K and 0.06921 atm

ln(0.06921 atm /1 atm) = ΔHvap / (8.314 J/K-mol) * ((1/239.8 K. )-(1/195.410 K))

we get , ΔHvap (ammonia near T3) = 23.44 kJ/mol

- Using general claussius -clayperon equation for phase transition :

dT/dP = T (V_l-V_s) /L

Normal freezing point: -77.728 oC

Density of liquid: 0.618g/mL

we take : V_l=1.62 ml/g

Density of solid: 0.817 g/mL ; V_l=1.22 ml/g

so,  (V_l-V_s) = 0.394 ml/g = 0.000394 L/g

Tf: = 195.422 K. Triple point: 195.41 K dT = 0.012 K

dP = 1atm - 0.06921 atm = 0.9308 atm

dT/dP = 0.01289 K/atm = 0.0003954 L/g*195.422 K / L

L = 5.99 L-atm /g ( 1L-atm = 101.325 J)

or L (specific heat) = 607.3 J/g

or ΔHfus = 607.3 J/g * 17 g/mol = 10.32 kJ/mol.

ΔHsub = ΔHvap + ΔHfus = 33.76 kJ/mol

(b)  the entropy change of fusion, vaporization and sublimation at near the triple point :

ΔSvap = ΔHvap /T = 23440 J/mol / 239.8 K = 97.75 J/K-mol

ΔSfus = ΔHfus /T = 10320 J/mol / 195.422 K = 52.81J/K-mol

ΔSsub = ΔSvap + ΔSfus =150.56 J/K-mol