Heat of Vaporization
(loosely based on Chap. 12 Sec 8
of Jespersen 6th Ed)
Dr. C. Yau
Fall 2014
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Molar Heat of Vaporization =
amt of heat needed to vaporize 1 mole of liquid
p.552
2
For water, Hvap = +43.9 kJ mol-1
What exactly does this mean?
H2O (l)  H2O (g)
H = +43.9 kJ
Or
H2O (l) +43.9 kJ  H2O (g)
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How is Hvap Determined?
Clausius, a German physicist, and Clapeyron,
a French engineer, determined that
The relationship between ln P and T
where P is the vp and T is the temp in K.
Mathematically, as an equation, it is…
ln P = k (1/T)
where k is the proportionality constant
and it turns out to be a negative value.
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ln P is proportional to 1/T
ln P = k (1/T)
For P = vapor pressure of liquid
T = temp in deg Kelvin
k=
-
 H vap
where R = 8.314 J mol-1 K-1
R
  H vap   1 
ln P =    + C
R

 T 
This is an eqn for a straight line: y = mx + b
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Graphical Determination of Hvap
  H vap   1 
ln P =    + C
R

 T 
y =
m
x
+ b
Plot ln P vs. 1/T
  H vap 
Slope = 
R


Rearrange eqn to calculate Hvap = - (slope)(R)
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ln P vs. 1/T
Are the slopes positive or negative?
Ans. Negative. Slope for water is more
negative than that for acetone.
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slope
  H vap 
= 
R


Since the slope is negative,
Hvap must be positive, as expected
for vaporization (endothermic)
Slope of water being more negative
than acetone means
|Hvap| (water) > |Hvap | (acetone)
(Why is this so? What does this mean?)
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Graphical Determination of Hvap
1) Measure vp at various temp.
2) Plot ln P versus 1/T (kelvin)
3) Calculate the slope from the graph
slope =
y1 - y 2
x1 - x 2
4) From the slope calculate Hvap
Hvap = - (slope)(R)
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Two-Point Method to Determine Hvap
Instead of measuring a set of P and T
values, Hvap can also be determined by
measuring only two sets of P and T.
You need only vp at two different temp:
(P1, T1) and (P2, T2)
See lecture notes for derivation of
Clausius-Clapeyron Equation:
ln
P1
P2
 H vap  1
1 
=



R  T 2 T1 
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Example 12.3: Methanol, CH3OH,
experiences hydrogen bonding, dipoledipole interactions, and London forces. At
64.6 oC, it has a vp of 1.00 atm, and at
12.0oC, it has a vp of 0.0992 atm. What is
the heat of vaporization for methanol?
Given: R = 8.314 J mol-1 K-1
Write out the eqn & see what is known &
what is unknown.
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ln
P1
P2
 H vap  1
1 
=



R  T 2 T1 
T = 64.6 oC, P =1.00 atm,
T =12.0oC, P = 0.0992 atm
Which is the unknown?
R = 8.314 J mol-1K-1
T1 = 64.6+273.15 = 337.8 K
T2 = 12.0+273.15 = 285.2 K
P1 = 1.00 atm
P2 = 0.0992 atm
See page 554 for calculation. Ans 35.2 kJ/mol
Do Practice Exercises 12.10 & 12.11
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