CHM 330 Take-home Quiz Name Due by 8:00 AM on December 04, 2017. No late submiss

Question

CHM 330 Take-home Quiz Name Due by 8:00 AM on December 04, 2017. No late submission will be accepted. To receive partial credits, show your detailed calculation and/or explanation. 1. By using the phase diagram of an unknown pure Phase diagram for mysterious compound x substance on this page, 100 90 80 70 60 50 (a) (4pt) estimate the triple point, including pressure and temperature. (b) (14pt) estimate the heat of vaporization and the heat of sublimation. (c) (6pt) estimate the heat of fusion from your results in (b). (d) (6pt) Now consider the change of chemical potential of solid and liquid. By increasing the pressure from 60 atm to 80 atm, the change of chemical potential of solid should be (greater than, equal to, less than) the change of chemical potential 40 30 20 10 gas -100 0 100 200 300 400 500 600 700 800 of liquid. Temperature (degrees Celsius)

Explanation / Answer

1. From the above shown phase diagram

Using,

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

a. Triple point is the point where all three solid, liquid and gas phases exists in equilibrium with each other. In the given phase diagram triple point is at 350 oC and 50 atm.

b. dHvap,

From the vaporization curve (liquid-gas equilibrium)

P1 = 60 atm at T1 = 550 oC + 273 = 823 K

P2 = 70 atm at T2 = 660 oC + 273 = 933 K

So,

ln(70/60) = dHvap/8.314[1/823 - 1/933]

dHvap = 8.95 kJ/mol

c. dHfus,

From the melting curve (solid-liquid equilibrium)

P1 = 70 atm at T1 = 420 oC + 273 = 693 K

P2 = 80 atm at T2 = 440 oC + 273 = 713 K

So,

ln(80/70) = dHvap/8.314[1/693 - 1/713]

dHfus = 27.43 kJ/mol

d. The change in chemical potential for solid is greater than the change in potential for the liquid.

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