HW 11-3: p.480 # 59, 66, 67, 78, 80, 81, 82
Ch 11.8: Phase Changes (Boiling Pt, Liquid-Solid Equil and Solid-Vapor Equil)
59
What is a phase change? Name all possible changes that can occur among the vapor, liquid,
and solid phases of a substance.
A phase change is a transition from one phase to another.
Solid to liquid is melting. Liquid to vapor is vaporizing Solid to vapor is sublimation
Vapor to liquid is condensation Liquid to solid is freezing Vapor to solid is deposition.
66
Define boiling point. How does the boiling point of a liquid depend on external pressure?
Referring to Table 5.3, what is the boiling point of water when the external pressure is
187.5 mmHg?
Boiling point is the temperature at which the vapor pressure of the liquid is equal to the external
pressure. This is the temperature at which bubbles will form and rise to the surface of a liquid.
The boiling point is 65°C when Pext = 187.5 mmHg.
67
As a liquid is heated at constant pressure, its temperature rises. This trend continues until
the boiling point of the liquid is reached. No further rise in temperature of the liquid can be
induced by heating. Explain.
When a liquid is heated at constant pressure, the heat is used to increase the kinetic energy of
molecules. Thus, the temperature increases. However, once the boiling point is reached,
intermolecular forces must be broken to change from a liquid to a gas. Thus, the heat is absorbed
to break bonds instead of increasing the kinetic energy of molecules. Thus, the temperature stays
constant.
78
How much heat is needed to convert 866 g of ice at –10oC to steam at 126o C?
Step 1:
Warming ice to the melting point.
q1 = msΔt = (866 g H2O)(2.03 J/g°C)[0 − (−10)°C] = 17.6 kJ
Step 2:
Converting ice at the melting point to liquid water at 0°C. (See Table 11.8 of the
text for the heat of fusion of water.)
q2 = 866 g H 2O ×
Step 3:
1 mol
6.01 kJ
×
= 289 kJ
18.02 g H 2O 1 mol
Heating water from 0°C to 100°C.
q3 = msΔt = (866 g H2O)(4.184 J/g°C)[(100 − 0)°C] = 362 kJ
Step 4:
Converting water at 100°C to steam at 100°C. (See Table 11.6 of the text for the
heat of vaporization of water.)
q4 = 866 g H 2O ×
1 mol
40.79 kJ
×
= 1.96 × 103 kJ
18.02 g H 2O
1 mol
Step 5: Heating steam from 100°C to 126°C.
q5 = msΔt = (866 g H2O)(1.99 J/g°C)[(126 − 100)°C] = 44.8 kJ
qtotal = q1 + q2 + q3 + q4 + q5 = 2.67 × 103 kJ
2
Homework #11-3 Answer key
80
The molar heats of fusion and sublimation of molecular iodine are 15.27 kJ/mol and 62.30
kJ/mol, respectively. Estimate the molar heat of vaporization of liquid iodine.
ΔHvap = ΔHsub − ΔHfus = 62.30 kJ/mol − 15.27 kJ/mol = 47.03 kJ/mol
81
The following compounds, listed with their boiling points, are liquid at –10°C: butane,
–0.5°C; ethanol, 78.3°C; toluene, 110.6°C. At –10°C, which of these liquids would you
expect to have the highest vapor pressure? Which the lowest? Explain.
The substance with the lowest boiling point will have the highest vapor pressure at some
particular temperature. Thus, butane (BP = –0.5°C) will have the highest vapor pressure at
−10°C and toluene (BP = 110.6°C) the lowest.
82
Freeze-dried coffee is prepared by freezing brewed coffee and then removing the ice
component with a vacuum pump. Describe the phase changes taking place during the
process.
Two phase changes occur in this process. First, the liquid is turned to solid (freezing), then the
solid ice is turned to gas (sublimation).