Ch 11.8: Phase Changes: Boiling Point, Energy & Heating Curves
Input & Modeling
• Boiling
– Liquid evaporating inside a liquid
– Vapor pushing out, atmosphere pushing in
§ Bubbles collapse if Pvap < Patm but form when Pvap ≥ Patm
• Bubble rise due to buoyancy, form rolling boil
§ Boiling point is the temperature at which Pvap = Patm
Normal boiling point is boiling point at 1 atm (above)
Higher BP indicates higher IMFs: more energy needed to separate molecules
At lower pressure, boiling point is lower since need less energy to reach external pressure
♦ High altitude
• At higher pressure, boiling point is higher since need more energy to reach external pressure
♦ Pressure cooker, autoclave
Solid-Liquid Equilibrium
– Melting & Freezing point is the temperature at which a material exists
as a liquid and a solid in equilibrium (e.g. ice water)
§ Rate of melting = rate of solidification
§ Molar heat of fusion: ∆Hfus = energy needed to melt one mole of a
substance (endothermic)
§ Molar heat of solidification: ∆Hsol = –∆Hfus = energy released by
solidification of one mole of a substance (exothermic)
Phase Changes & Energy [energy diagram of phase changes]
§ All upward phase changes are endothermic; downward are
exothermic
§ ∆Hcond = –∆Hvap is amount of energy released when one mole of a
substance condenses
• ∆Hvap > ∆Hfus because the change in potential energy is much
greater for vaporization than for melting
§ ∆Hsub = molar heat of sublimation; ∆Hdep = –∆Hsub = molar heat of
deposition
• ∆Hsub = ∆Hfus + ∆Hvap
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Phase Changes: Boiling Point, Energy & Heating Curves
•
Heating Curve
– Graph shows effect of adding energy (over time)
on the temperature of a substance
– Solid (A-B), liquid (C-D) & gas (E-F) regions
show increasing T as E is added (time increases)
§ Energy q = ms∆T, where s is the specific heat
of the solid, liquid or gas
§ Kinetic energy is increasing
– Melting (B-C) and Vaporization (D-E) regions
have constant T
§ Potential energy of sample is increasing as
particles are separated (bonds broken)
§ Energy q = n∆H, where n is the number of
moles and ∆H is the molar enthalpy of the
process
§ To determine qtotal, determine q for all necessary regions then total
• Watch units: s is usually in J/g⋅°C, ∆H is usually in kJ/mol
• The melting point of gallium is 29.8°C. ∆Hfus = 5.59 kJ/mol, ssolid = 0.372 J/g·°C, and sliquid =
0.409 J/g·°C. How much energy is absorbed when 215.8 g of gallium at 21.5°C is heated to a
final temperature of 35.3°C?
q solid = ms ∆ T = (215.8 g)(0.372 J/g ⋅°C)(29.8°C – 21.5°C)
•
= 666 J ⇒ 0.666 kJ
⎛
1 mol ⎞
q melt = n ∆ H fus = ⎜ 215.8 g ×
⎟ × 5.59 kJ/mol
•
69.72 g ⎠
⎝
•
•
= 17.3 kJ
q liquid = ms ∆ T = (215.8 g)(0.409 J/g ⋅°C)(35.3°C – 29.8°C)
= 485 J ⇒ 0.485 kJ
• qtotal = 0.666 kJ + 17.3 kJ + 0.485 kJ = 18.5 kJ
Cooling curve
– Would be removing energy over time
– Reverse of heating curve
– Supercooling (diagram at right): liquid cooled
below its freezing point–heat removed so quickly
that the molecules have no time to become orderly
arranged (kinetic effect)
§ No solid forms until bottom of curve
p.2