This preview shows pages 1–2. Sign up to view the full content.
8.8
(a) The quantities
vap
vap
and
H
S
Δ
°
Δ °
can be calculated using the
relationship
vap
vap
1
ln
H
S
P
R
T
R
Δ
°
Δ °
= −
⋅
+
Because we have two temperatures with corresponding vapor pressures
(we know that the vapor pressure = 1 atm at the boiling point), we can set
up two equations with two unknowns and solve for
vap
vap
and
H
S
Δ
°
Δ °
. If
the equation is used as is,
P
must be expressed in atm, which is the
standard reference state. Remember that the value used for
P
is really
activity that, for pressure, is
P
divided by the reference state of 1 atm so
that the quantity inside the ln term is dimensionless.
vap
1
1
vap
vap
1
1
vap
8.314 J K
mol
ln 1
311.6 K
13 Torr
8.314 J K
mol
ln
760 Torr
227.94 K
H
S
H
S
−
−
−
−
Δ
°
⋅
⋅
×
= −
+ Δ °
Δ
°
⋅
⋅
×
= −
+ Δ °
which give, upon combining terms,
1
1
1
vap
vap
1
1
1
vap
vap
0J K
mol
0.003209K
33.9 J K
mol
0.004 3871K
H
S
H
S
−
−
−
−
−
−
⋅
⋅
= −
× Δ
°
+ Δ °
−
⋅
⋅
= −
× Δ
°
+ Δ °
Subtracting one equation from the other will eliminate the
vap
S
Δ °
term and
allow us to solve for
vap
:
H
Δ
°
1
1
1
vap
1
vap
33.9 J K
mol
0.001178 K
28.8 kJ mol
H
H
−
−
−
−
+
⋅
⋅
= +
× Δ
°
Δ
°
= +
⋅
(b) We can then use
vap
H
Δ
°
to calculate
vap
S
Δ °
using either of the two
equations:
1
1
vap
1
1
vap
1
1
1
1
vap
1
1
vap
0
0.003 209 K
( 28 800 J mol )
92.4 J K
mol
33.9 J K
mol
0.004 3871K
( 28 800 J mol )
92.4 J K
mol
S
S
S
S
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Crowell

Click to edit the document details