### Lecture 35 (Slides) November 7

```Boiling and Boiling Point
•
•
FIGURE 12-21
Boiling water in a paper cup
Liquid boils at low
pressure
FIGURE12-21
Boiling water in a paper cup
General Chemistry: Chapter 12
Slide 2 of 61
Class Example – Clapeyron Equation:
• 1. The vapor pressure of acetone, (CH3)2C=O,
is 10.0 mm Hg at -31.1 oC. What is the vapor
pressure of acetone at 15.0 oC if the enthalpy
of vaporization of acetone is 31.3 kJ mol-1?
(Great care with significant figures is
required!) Does the answer make sense in
terms of the kinetic theory of matter? If so,
why? Use the data given to estimate the
normal boiling point of acetone.
Clapeyron Equation
• 2. How could you employ the ClausiusClapeyron equation to obtain thermodynamic
data without using calorimetry? What types of
thermodynamic data can be obtained? What
experiments would you conduct to determine a
∆Hvaporization value for liquid water using the
Clapeyron equation?
Coexistence Curves
• Under a range of temperature and pressure
conditions one can have liquid and vapor (gas)
in dynamic equilibrium. Gas phase molecules
are entering the liquid at the same rate as
liquid phase molecules are evaporating. (No
heat needs to be supplied for this process to
occur. Why?) The simplest cases involve a
pure liquid and a pure gas in equilibrium.
•
FIGURE 12-18
Vapor pressure curves of several liquids
General Chemistry: Chapter 12
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Coexistence Curve Calculations
• 1.We used the Clausius-Clapeyron and a vapor
pressure at one temperature to predict the
vapor pressure at a second temperature (given
ΔHVAP). 2.We also used a vapor pressure at one
temperature (and ΔHVAP) to calculate the
normal boiling point of a liquid (same as 1.?)
3. We can use the vapor pressure of a liquid at
two temperatures to find ΔHVAP. Mention
ΔHSUBLIMATION ?
Calculations Related to Coexistence
Curves
• One can imagine putting a volatile liquid in an
empty container and seeing one of two results:
• 1. All of the volatile liquid evaporates. In this
case PGas will be less than the equilibrium
vapor pressure of the substance/liquid.
• 2. Only some of the liquid eventually
evaporates. Here Pgas equals the equilibrium
vapor pressure.
Class Example:
• 1. A total of 3.00 g of liquid water was placed
in an evacuated (initially) 12.0L container
maintained at a temperature of 80.0 0C. Will a
liquid/gas equilibrium be established? What
piece of data is required to solve this problem?
If no liquid/gas equilibrium is established,
determine how much additional liquid water is
required to have some liquid in equilibrium
with H2O(g) in the container.
Critical Points
• The normal boiling point of water is 100 oC. At
this temperature the vapor pressure of water is
exactly 760mm Hg (the normal average
atmospheric pressure at sea level). If we heat
water in a sealed container all of the steam that
is formed is trapped above the liquid water.
The additional steam formed increases the
pressure of the gas above the liquid and
“naturally” the boiling point rises.
Critical Points
• There is a limit to how high we can raise the
temperature of a liquid/gas mixture and still
have a well-defined two phase equilibrium.
One eventually reaches a temperature/pressure
point beyond which there is only one fluid
phase. The specific T and P where this occurs
is referred to as the critical point of the
substance.
The Critical Point
•
•
FIGURE 12-22
Attainment of the critical point for benzene
General Chemistry: Chapter 12
Slide 12 of 61
Critical Points
• The next slide shows critical point data for a
number of substances. Do the “permanent”
and “non-permanent” gases differ appreciably
if one considers intermolecular forces?
General Chemistry: Chapter 12
Slide 14 of 61
Other Coexistence Curves
• For substances such as molecular iodine which
sublimes readily at ambient temperatures we
can have another type of equilibrium
• I2(s) ↔ I2(g)
• At a particular temperature and pressure the
vapor pressure of solid iodine is well defined.
We can construct a coexistence curve
illustrating the T’s and P’s where the solid and
gas are in equilibrium.
Some Properties of Solids
Melting, Melting Point, and Heat of Fusion
Figure 12-23
Figure 12-24
Heating curve for water
Cooling curve for water
H2O(s)
General Chemistry: Chapter 12
H2O(l)
ΔHfus(H2O) = +6.01 kJ/mol
Slide 16 of 61
General Chemistry: Chapter 12
Slide 17 of 61
Sublimation
ΔHsub = ΔHfus + ΔHvap
= -ΔHdeposition
•
FIGURE 12-25
• Sublimation of iodine
General Chemistry: Chapter 12
Slide 18 of 61
Phases Changes – Signs of ΔH’s:
• The left to right processes below are
endothermic (remember intermolecular
forces!).
• C2H5OH(s) → C2H5OH(l) → C2H5OH(g)
• The opposite processes are necessarily
exothermic. We will soon see processes where
ΔH for ionic deposition process is extremely
exothermic (Eg. Na+(g) + Cl-(g) → NaCl(s)).
More Coexistence Curves:
• We can usually represent (s) ↔(g) and , as
well, (l) ↔ (g) coexistence curves on the same
graph. We can also look at (s) ↔ (l)
coexistence curves – in this case we see
melting point values that change very slowly
as the pressure changes. Plotting all three
coexistence curves (and sometimes more!) on
a single graph gives us a phase diagram.
Iodine
• Phase diagram for iodine
General Chemistry: Chapter 12
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Carbon Dioxide
• Phase diagram for carbon dioxide
General Chemistry: Chapter 12
Slide 22 of 61