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Chem 10181, Fall 2007
Final Review Problems
SOLUTIONS
1
Chem 10181 Fall 2007
Final Review Problems
SOLUTIONS
(1) (a) The vapor pressure of liquid ammonia is the equilibrium constant for its vaporization:
NH
3
(l)
→
←
NH
3
(g)
K
eq
= p(NH
3
)
So, since
∆
Gº = –RT ln(K
eq
) =
∆
Hº – T
∆
Sº, then ln(P) =
–
∆
Hº
vap
R
(
1
T
)
+
∆
Sº
vap
R
.
In other
words, this is a van’t Hoff plot and from the slope and intercept we can conclude that
∆
Hº
vap
= +23.2 kJ/mol,
∆
Sº
vap
= +96.7 J/mol•K.
The sublimation of the solid can be treated similarly, with
NH
3
(s)
→
←
NH
3
(g)
K
eq
= p(NH
3
)
Using the data from the dashed line gives
∆
Hº
subl
= +32.7 kJ/mol,
∆
Sº
subl
= +144.1
J/mol•K.
Each van’t Hoff plot is linear because
∆
Hº and
∆
Sº do not vary significantly with
temperature.
But since the slopes and intercepts of these lines are determined by their
thermodynamic parameters, and the thermodynamic stabilities of the solid and liquid are
different, of course they are not going to lie on the same line!
(b)
At the melting point, the solid and liquid have equal stability, so the vapor pressure of the
solid and liquid must be equal.
One could use the
∆
Hº and
∆
Sº values calculated above, or
one can just find the intersection of the two lines:
ln(P
liquid
)
= ln(P
solid
)
11.63 – 2791(
1
T
) = 17.33 – 3931(
1
T
)
1140(
1
T
) = 5.70
T = 200 K
At that temperature, ln(P) = –2.33
P = 0.098 atm (technically this is the triple point, rather than the melting point)
(c)
From part (a):
NH
3
(s)
→
←
NH
3
(g)
∆
Hº = +32.7 kJ/mol
∆
Sº = +144.1 J/mol•K
NH
3
(g)
→
←
NH
3
(l)
∆
Hº = –23.2 kJ/mol
∆
Sº =
–96.7 J/mol•K
NH
3
(s)
→
←
NH
3
(l)
∆
Hº
fus
= +9.5 kJ/mol
∆
Sº
fus
= +47.4 J/mol•K
(d)
H
N
N
N
H
N
N
N
H
N
N
N
2–
major
major
very minor
(e)
Two factors are probably most important:
(1)
Inductive effects.
The two additional
electronegative nitrogen atoms in HN
3
withdraw electron density from the NH group,
relative to NH
3
.
(2)
Hybridization.
The N in NH
3
is sp
3
hybridized, while the N in HN
3
is sp hybridized; the increased s character in the latter stabilizes the conjugate base.
(Resonance may also play a role, as N
3
–
is resonancestabilized and NH
2
–
is not, but it is
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View Full DocumentChem 10181, Fall 2007
Final Review Problems
SOLUTIONS
2
unclear how important this is because it is unclear how much more resonancestabilized N
3
–
is relative to its conjugate acid, which is what counts.)
(2)(a)
N
O
O
N
O
O
(b)
The NO bond order in nitrite is 1.5 (two equally important resonance structures, so each
bond is average between single and double).
The molecular orbital picture for NO indicates
that it has the electron configuration (
σ
2s
)
2
(
σ
*
2s(2p)
)
2
(
π
2p
)
4
(
σ
2p(2s)
)
2
(
π
*
2p
)
1
, so the NO
bond order in NO is 2.5.
Thus nitrite would be expected to have a
longer
NO bond than
nitric oxide (and this is indeed observed).
(c)
Each atom is sp
2
hybridized (don't forget to distinguish between the shared and the
unshared lone pairs!).
sp
2
hybridization of the central N, which has one lone pair, implies a
bent geometry with a bond angle ~ 120º.
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 Fall '08
 Brown
 Equilibrium, Vaporization

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