Final Exam
CHEM 181: Introduction to Chemical Principles
December 14, 2015
Answer Key
1. The compound C4 H5 NOS has the connectivity shown below
H
H
N
C
C
O
C
H
C
H
S
H
On the next page:
• Draw all reasonable resonance structures and label them as major or
minor.
• Label all non-zero formal charges.
• Rank the resonance structures: 1=best, 2=next best, etc. If two resonant
structures are equivalent or very close to equivalent, you can assign them
the same number in the ranking.
If you prefer, you can skip writing the carbons and the hydrogens attached to
HN
the carbons, like so:
HO
S
1
There is only one structure with zero formal charges; everything else is minor.
Ranking of minor structures is based on electronegativity of the atom getting
the positive charge.
HN
HN
1, major
3
HO
HO
HN
HN
2, minor
HO
HO
S
HN
3
S
HN
2
HO
S
S
4
HO
S
S
2
2. A compound containing only C, H, and O with a molecular weight under
110 amu has the following 13 C NMR spectrum (and note that peak heights
in this 13 C NMR spectrum are not meaningful):
IR spectrum:
3
CDCl3
240
1
QE-300
220
200
180
160
140
120
100
80
60
40
20
0
8
7
6
5
4
3
2
1
0
and H NMR spectrum:
12
11
10
9
expanded to show detail:
Draw a Lewis structure for the compound. Also mark on each spectrum:
(a) which protons correspond to which peaks in the 1 H NMR spectrum, and
(b) labels for peaks in the IR spectrum that can be assigned unambiguously.
4
3. The electron configurations for the ground state and first three excited states of
N2 are shown. Use the following information about energies and bond lengths
to figure out the electron configurations for the ground and excited states of
CN. (All answers that are logical and self-consistent will be marked as correct.
You can use the MO diagram for N2 on the next page—the MO diagram for
CN will have states with different energies, but the order and names will be the
same.)
N2
σ2s
∗
σ2s
π2px , π2py
σ2pz
∗
∗
, π2p
π2p
y
x
∗
σ2p
z
1.10
2
2
4
2
0
0
1.00
1.29
2
2
4
1
1
0
1.18
1.21
2
2
3
2
1
0
1.77
1.15
2
1
4
2
1
0
energy (aJ)
bond length (nm)
0.0
CN
σ2s
∗
σ2s
π2px , π2py
σ2pz
∗
∗
, π2p
π2p
y
x
∗
σ2p
z
1.17
2
2
4
1
0
0
0.18
1.23
2
2
3
2
0
0
0.51
1.15
2
1
4
2
0
0
1.08
1.50
2
2
4
0
1
0
1.17
1.32
2
2
3
1
1
0
energy (aJ)
bond length (nm)
0.0
5
There are a few ways to answer this in a self-consistent manner. Here is one
line of reasoning:
• The 0.18 aJ state must come from a transition not available to N2 . Promoting a π2px (or y) to a σ2pz meets this requirement. Both are bonding,
so the overall bond order is the same, and the bond length changes only
slightly.
• The 1.08 aJ and 1.17 aJ states involve significant increases in bond length,
and should involve moving a bonding electron to an antibonding orbital.
Match the 1.08 aJ state for CN to the 1.00 aJ state for N2 , and the 1.17 aJ
to the 1.18 aJ state.
• The 0.51 aJ also has no clear counterpart in N2 . Since bond length decreases slightly, this means either bond order must stay the same or in∗
crease. For that to be the case, an electron must move out of the σ2s
orbital, and the relatively low energy of this state would place it into the
σ2pz . The 2s orbital for C will be significantly higher in energy than the
∗
N 2s, which will bring the σ2s
much closer in energy to the states derived
from the 2p orbitals.
6
4. Use the following enthalpies of formation
∆Hf◦
H
C2 H6 (g) −83.7
N2 H4 (g)
50.63
and the following bond enthalpies
◦
∆Hbond
C–C
C–N
N–N
H
368.0
331.0
301.0
Calculate the enthalpy of formation of CH3 NH2 (g).
H
H
H
C
C
H
H
H
H
N
N
H
H
C
N
H
H
H
H
The reaction
C2 H6 (g) + N2 H4 (g) −→ 2CH3 NH2 (g)
involves breaking a C–C and an N–N bond, and making two C–N bonds; viz.,
C2 H6 (g) + N2 H4 (g) −→ CH3 (g) + CH3 (g) + NH2 (g) + NH2 (g) −→ 2CH3 NH2 (g)
So
◦
∆Hrxn
=
=
=
=
◦
◦
∆Hbond
(reactants) − ∆Hbond
(products)
◦
◦
◦
∆Hbond (C–C) + ∆Hbond (N–N) − 2∆Hbond
(C–N)
368 + 301 − 2 · 331
7 kJ mol−1
◦
But we can also express ∆Hrxn
in terms of formation enthalpies:
◦
∆Hrxn
= ∆Hf◦ (products) − ∆Hf◦ (reactants)
7 kJ mol−1 = 2∆Hf◦ (CH3 NH2 (g)) − ∆Hf◦ (C2 H6 (g)) + ∆Hf◦ (N2 H4 (g))
7 kJ mol−1 = 2∆Hf◦ (CH3 NH2 (g)) − −83.7 kJ mol−1 + 50.63 kJ mol−1
2∆Hf◦ (CH3 NH2 (g)) = 7 kJ mol−1 − 83.7 kJ mol−1 + 50.63 kJ mol−1
= −26 kJ mol−1
∆Hf◦ (CH3 NH2 (g)) = −13 kJ mol−1
7
5. Molecular iodine (I2 ) is slightly soluble in water and somewhat more soluble in
cyclohexane (C6 H12 ):
Ksp @ 20 ◦ C
Ksp @ 50 ◦ C
I2 (s) −→ I2 (aq)
H2 O(`)
1.14 × 10−3 M 3.08 × 10−3 M
C6 H12 (`)
8.85 × 10−3 M 2.53 × 10−2 M
I2 (s) −→ I2 (C6 H12 sol.)
Water and cyclohexane do not mix, and iodine will distribute between the two
solvents according to the reaction:
I2 (aq) I2 (C6 H12 sol.)
For this reaction, what is K at 20 ◦ C, and what are ∆H and ∆S? (Assume
that ∆H and ∆S are constant over this temperature range.)
Our two reactions have very simple equilibrium constant expressions:
Ksp (aq) = [I2 (aq)]
Ksp (C6 H12 ) = [I2 (C6 H12 sol.)]
and
For
I2 (aq) I2 (C6 H12 sol.)
we have
[I2 (C6 H12 sol.)]
[I2 (aq)]
8.85 × 10−3
=
1.14 × 10−3
= 7.76
K =
This is at 20 ◦ C; at 50 ◦ C, K = 8.21. We can use the 20 ◦ C value to find ∆G◦ :
∆G◦ = −RT ln K
= −(8.314 J mol−1 K−1 ) · (293 K) · ln 7.76
= −4.99 kJ mol−1
The van’t Hoff equation gives us ∆H using the values of K at the two different
8
temperatures:
K2
ln
K1
8.21
ln
7.76
−1
−2
5.64 × 10 · (8.314 J mol K−1 )
∆H
1
1
−
T2 T1
1
∆H
1
= −
−
8.314 J mol−1 K−1 323 K 293 K
= −∆H · (−3.17 × 10−4 K−1 )
= 1.48 kJ mol−1
∆H
= −
R
Since the equilibrium constant increases (albeit slightly) with temperature,
LeChâtelier’s principle says ∆H should be positive (albeit slightly): add heat
to raise the temperature, and equilibrium shifts towards products in order to
take up some of that heat.
Then,
∆G = ∆H − T ∆S
−4.99 kJ mol−1 = 1.48 kJ mol−1 − (293 K) · ∆S
6.47 kJ mol−1
∆S =
293 K
= 22.1 J mol−1 K−1
9
6. Here are the chemical reactions to consider in the reaction of baking soda
(NaHCO3 ) with vinegar (CH3 COOH(aq)):
CO2 (g)
CO2 (aq) + H2 O(`)
H2 CO3 (aq)
HCO−
3 (aq)
CO2 (aq)
H2 CO3 (aq)
+
HCO−
3 (aq) + H (aq)
+
CO2−
3 (aq) + H (aq)
CH3 COOH(aq) CH3 COO− (aq) + H+ (aq)
H2 O(`) H+ (aq) + OH− (aq)
KH
K
Ka1
Ka2
=
=
=
=
3.4 × 10−2 M atm−1
1.3 × 10−3
2 × 10−4 M
4.7 × 10−11 M
Ka = 1.8 × 10−5 M
Kw = 1 × 10−14 M2
(a) 1 L of a 0.8 M aqueous solution of CH3 COOH has NaHCO3 added to it until
it reaches pH 7, at which point it is at equilibrium with respect to all of the
above reactions, and with atmospheric CO2 (g) with PCO2 =3.94×10−4 atm.
How many moles of NaHCO3 were added to reach this equilibrium? (Hint:
one way to do this is to figure out [Na+ ] at equilibrium.)
Equilibrium constant expressions for all of the above:
[CO2 (aq)]
PCO2 (g)
[H2 CO3 (aq)]
[CO2 (aq)]
+
[H ][HCO−
3 (aq)]
[H2 CO3 (aq)]
[H+ ][CO2−
3 (aq)]
−
[HCO3 (aq)]
+
[H ][CH3 COO− (aq)]
[CH3 COOH(aq)]
[H+ (aq)][OH− (aq)]
= 3.4 × 10−2 M atm−1
= 1.3 × 10−3
= 2 × 10−4 M
= 4.7 × 10−11 M
= 1.8 × 10−5 M
= 10−14 M2
We have PCO2 fixed at an equilibrium value of 3.94 × 10−4 atm. Plugging
in:
[CO2 (aq)]
PCO2 (g)
[CO2 (aq)]
3.94 × 10−4 atm
[CO2 (aq)]
[H2 CO3 (aq)]
10
= 3.4 × 10−2 M atm−1
= 3.4 × 10−2 M atm−1
= 1.34 × 10−5 M
= 1.74 × 10−8 M
At pH 7, [H+ ] must be 10−7 at equilibrium, so
[H+ ][HCO−
3 (aq)]
[H2 CO3 (aq)]
10−7 [HCO−
3 (aq)]
1.74 × 10−8
[HCO−
3 (aq)]
2−
[CO3 (aq)]
= 2 × 10−4 M
= 2 × 10−4 M
= 3.48 × 10−5 M
= 1.64 × 10−8 M
On the other hand, at pH 7 almost all of the acetic acid is in its conjugate
base form:
[H+ ][CH3 COO− (aq)]
= 1.8 × 10−5 M
[CH3 COOH(aq)]
10−7 [CH3 COO− (aq)]
= 1.8 × 10−5 M
[CH3 COOH(aq)]
[CH3 COO− (aq)]
= 180
[CH3 COOH(aq)]
So at equilibrium, the negative ions and their concentrations in solution
are
•
•
•
•
just under 0.8 M of CH3 COO− ,
3.48 × 10−5 M of HCO−
3,
−7
−
10 M of OH , and
1.64 × 10−8 M of CO2−
3 .
There is only 10−7 M of H+ , so a neutral solution requires
[Na+ (aq)] ≈ 0.8 M
0.8 moles of NaHCO3 were added.
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(b) 1 L of a 0.8 M aqueous solution of CH3 COOH has NaHCO3 added to it
until it reaches pH 7, at which point it is at equilibrium with respect to all
of the above reactions. In this case, the reaction takes place in a sealed 4 L
container. Compared to part (a), how much NaHCO3 was added: more,
slightly more, exactly the same amount, slightly less, or less? Explain your
answer.
In a sealed container, PCO2 is going to be much higher, resulting in significant concentrations of all of the carbonate species, and in particular
HCO−
3 (aq). The extra negative charge must be balanced by extra sodium,
so we will need more NaHCO3 .
Alternately, keeping higher pressures of CO2 (g) will act to acidify the solution, so you must add more base (NaHCO3 ) to counteract this.
(c) Calculate how many moles of NaHCO3 were added in part (b). (Use T =
298 K. Hint to simplify computations: the 3 L space left in the container
will hold 0.123 × P /atm moles of gas at this temperature.)
With a CO2 (g) pressure of P at equilibrium, the following come from our
equilibrium constant expressions:
[CO2 (aq)]
[H2 CO3 (aq)]
[HCO−
3 (aq)]
2−
[CO3 (aq)]
=
=
=
=
3.4 × 10−2 (P/atm) M
4.42 × 10−5 (P/atm) M
8.84 × 10−2 (P/atm) M
4.15 × 10−5 (P/atm) M
Each HCO−
3 that is protonated ends up either as an H2 CO3 (aq), a CO2 (aq),
or a CO2 (g). Since [CH3 COO− (aq)]≈0.8 M, there are 0.8 moles of protons
to be accounted for. The number of protons making H2 CO3 (aq) is more
or less cancelled out by the number released forming CO2−
3 (aq), so
0.8 mol
≈
=
=
P =
nCO2 (aq) + nCO2 (g)
3.4 × 10−2 (P/atm) M · 1 L + 0.123 (P/atm) mol
0.157 (P/atm) mol
5.1 atm
The HCO−
3 concentration must be
−2
[HCO−
(P/atm) M = 0.45 M
3 (aq)] = 8.84 × 10
In contrast to the situation in part (a), our negative-ion species in solution
are
12
• just under 0.8 M of CH3 COO− ,
• 0.45 M of HCO−
3,
• negligible amounts of OH− and CO2−
3 .
We must have added 1.25 M of NaHCO3 .
13