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Final Exam
CHEM 181: Introduction to Chemical Principles
December 12, 2012
Key
1. (a) The following diagrams show connectivity, but they are missing lone pairs,
double (or triple) bonds, and formal charges. Do the following:
•
• Fill in all of this missing information—lone pairs, double (or triple)
bonds, and formal charges.
• If there are resonance structures, draw them and indicate their relative
importance.
• Indicate the hybridization of every atom except H.
i.
H
H
O
C
C
sp3
H
sp2
sp2
H
N
C
sp3
H
H
H
O
C
C
H
N
C
H
H
sp2
H
H
H
minor
1
H
ii.
sp2
best
sp
N
N
N
sp
sp2
N
O
sp2
major
N
N
N
N
O
N
N
N
N
O
N
N
N
N
O
N
N
N
N
O
very minor
(can omit)
iii.
H
C
H
C
H
H
C
N
H
all sp2
C
C
N
O
best
O
H
H
C
C
C
N
H
major
H
H
C
H
H
O
C
C
C
N
O
minor
2
H
H
C
C
C
N
O
H
(b) Order the following molecules according to their boiling point:
CH3 Cl, CH3 CH3 , CH3 OH, and CH3 SH
Write your answer in the form:
lowest boiling = A < B < C < D = highest boiling
and give a brief explanation that explains this ordering.
lowest boiling = CH3 CH3 < CH3 Cl < CH3 SH < CH3 OH = highest boiling
−89 ◦ C < −23.8 ◦ C < 5.95 ◦ C < 64.7 ◦ C
C2 H6 has no dipole and the smallest London forces; it must be the lowest
boiling. Strong hydrogen bonding in CH3 OH will make it the highest
boiling. For CH3 Cl and CH3 SH, London forces are comparable. You could
guess that because CH3 Cl has a larger dipole moment, it will boil higher
than CH3 SH. As it happens, weak hydrogen bonding makes CH3 SH boil
higher. Either ordering of CH3 Cl and CH3 SH will be marked as correct, if
reasonably explained.
(c) Ethanol (C2 H5 OH) boils at 78.4 ◦ C, and chloroform (CHCl3 ) boils at
61.1◦ C. However, a 93/7% solution (by weight) of chloroform and ethanol
boils at 59.4 ◦ C. Give an explanation for this in terms of intermolecular
interaction strength.
Strong hydrogen bonding gives ethanol a relatively high boiling point,
while strong London forces and a large molecular dipole moment give
CHCl3 a high boiling point. However, ethanol–chloroform intermolecular interactions are weaker, as ethanol cannot hydrogen bond to chloroform, and both dipole-dipole and London forces are weaker for ethanolchloroform than for chloroform-chloroform.
3
(d) The following are illustrations of the four π molecular orbitals in the NO−
3
anion:
A
B
C
D
i. Indicate whether each one is bonding, non-bonding, or anti-bonding.
B is bonding, C and D are nonbonding, and A is antibonding.
ii. Arrange them from lowest in energy to highest. (Two are very close
in energy; as long as they are next to each other, don’t worry which
to put first.) How many electrons are in this π system, and which
orbitals are filled?
lowest → B < C = D < A ← highest
There are 6 electrons, which fill orbitals B, C, and D.
iii. What would you predict for N–O bond order, and how does this compare to the answer you get from Lewis electron structures?
There is a σ bond between the N and each O, and one π bonding pair
(in orbital A) distributed among all three N–O bonds. The bond order
is 4/3, which is also what you get from resonance structures.
iv. If you use the same MO diagram for the molecule SO3 , what S–O bond
order do you get? How does this compare to Lewis electron diagrams
for SO3 ?
This is still a 24-electron system, and with the same MOs, you still
get bond order 4/3. If you draw octet-rule-following Lewis structures,
they agree with this result. An expanded-octet Lewis structure gives a
bond order of 2 instead (which is yet another reason to think critically
about expanding octets.)
4
2. (a) The 1 H NMR spectra on this page are all of compounds with the molecular
formula C5 H10 O3 :
5
Choose from the compounds below (H atoms not drawn) and assign one
structure to each spectrum. Draw the structure (without H atoms is fine)
on the matching spectrum.
6
(b) A compound containing only C, H, and O with a molecular weight under
110 amu has the following 13 C NMR spectrum:
IR spectrum:
7
and 1 H NMR spectrum:
8
Draw a Lewis structure for the compound. Also mark on each spectrum:
i. which protons correspond to which peaks in the 1 H NMR spectrum,
and
ii. types of motion (type of bond, stretch vs. bend) for peaks in the IR
spectrum that can be assigned unambiguously.
The compound is ethyl glycolate:
H
O
H
O
C
C
O
H
H
C
C
H
H
H
H
Naming each set of protons:
B
C
H
O
D
A
H
O
C
C
H
O
H
H
C
C
H
H
H
The A protons give a quartet; they are shifted so far because of the
C(=O)O nearby. The B protons are tricky; they give a singlet, shifted
by both the O and the C=O. The alcohol C proton is unsplit and wide,
while the D protons give a triplet with small chemical shift.
9
3. (a) The Ka of nitrous acid (HNO2 ) is 4.0 × 10−4 . What is the pH of a 0.1 M
solution of sodium nitrite, NaNO2 ?
This is a 0.1 M solution of NO−
2 , which is the conjugate base of HNO2 . We
divide
Kb NO−
2 =
Kw
10−14
=
= 2.5 × 10−11
Ka (HNO2 )
4.0 × 10−4
because we’ll have the reaction
NO−
2 (aq)
−
+ H2 O(`) HNO2 (aq) + OH (aq)
Kb
NO−
2
[HNO2 ][OH− ]
=
[NO−
2]
use an ICE table:
[NO−
2]
[HNO2 ]
[OH− ]
0
x
x
≈0
x
x
I
0.1
−x
C
E 0.1 − x
So
Kb NO−
2
=
2.5 × 10−11 =
2.5 × 10−11 ≈
x2 ≈
x ≈
[HNO2 ][OH− ]
[NO−
2]
2
x
0.1 − x
x2
0.1
2.5 × 10−12
1.58 × 10−6
Given the size of x, our assumption that 0.1 − x ≈ 0.1 is really very good.
Autoionization will make the OH− concentration slightly higher, but you
can ignore that here. To calculate pH:
Kw
10−14
[H ] =
=
= 6.3 × 10−9
−
−6
[OH ]
1.58 × 10
+
10
pH = 8.2
(b) Citric acid (C6 H8 O7 ) is a triprotic acid:
+
H3 C6 H5 O7 (aq) H2 C6 H5 O−
7 (aq) + H (aq)
2−
+
H2 C 6 H5 O−
7 (aq) HC6 H5 O7 (aq) + H (aq)
2−
3−
HC6 H5 O7 (aq) C6 H5 O7 (aq) + H+ (aq)
Ka1 = 8.4 × 10−4
Ka2 = 1.8 × 10−5
Ka3 = 4.0 × 10−6
What is the pH of a 0.05 M solution of citric acid? Give an answer that is
accurate to 0.1 units of pH; if you are neglecting any of the above equilibria,
explain why you can do so.
This is going to need a few ICE tables. Start with the first equilibrium:
Ka1
I
C
E
[H+ ][H2 C6 H5 O−
7]
=
[H3 C6 H5 O7 ]
[H3 C6 H5 O7 ]
[H2 C6 H5 O−
7]
[H+ ]
0.05
−x
0.05 − x
0
x
x
≈0
x
x
x2
0.05 − x
x2
x2 + 8.4 × 10−4 x − 4.2 × 10−5
x
Move to the next:
Ka2 =
= 8.4 × 10−4
= 4.2 × 10−5 − 8.4 × 10−4 x
= 0
= 6.1 × 10−3
[H+ ][HC6 H5 O2−
7 ]
−
[H2 C6 H5 O7 ]
[H3 C6 H5 O7 ]
[H2 C6 H5 O−
7]
[H+ ]
0
x
x
6.1 × 10−3
x
x + 6.1 × 10−3
I
6.1 × 10−3
C
−x
E 6.1 × 10−3 − x
x(x + 6.1 × 10−3 )
6.1 × 10−3 − x
x2 + 6.1 × 10−3 x
x2 + 6.12 × 10−3 x − 1.1 × 10−7
x
11
= 1.8 × 10−5
= 1.1 × 10−7 − 1.8 × 10−5 x
= 0
= 1.8 × 10−5
So just considering the first ionization, we have
pH = − log10 (6.1 × 10−3 = 2.2
and with the second dissociation, we get
pH = − log10 (6.12 × 10−3 = 2.2
We will not need to worry about the third dissociation.
(c) What is the pH of a 0.1 M solution of disodium citrate, Na2 HC6 H5 O7 ?
You will need to consider the equilibrium:
2−
−
3−
HC6 H5 O2−
7 (aq) + HC6 H5 O7 (aq) H2 C6 H5 O7 (aq) + C6 H5 O7 (aq)
The equilibrium constant for this reaction is
3−
[H2 C6 H5 O−
7 ][C6 H5 O7 ]
2−
[HC6 H5 O2−
7 ][HC6 H5 O7 ]
[H2 C6 H5 O−
[H+ ][C6 H5 O3−
7]
7 ]
·
2−
2−
[HC6 H5 O7 ] [HC6 H5 O7 ][H+ ]
1
Ka3 ·
Ka2
4.0 × 10−6
1.8 × 10−5
0.22
K =
=
=
=
=
I
C
E
[HC6 H5 O2−
7 ]
[H2 C6 H5 O−
7]
[C6 H5 O3−
7 ]
0.1
−2x
0.1 − 2x
0
x
x
0
x
x
Don’t forget the −2x!
x2
(0.1 − 2x)2
x
0.1 − 2x
x
1.94x
x
12
= 0.22
= 0.47
= 0.047 − 0.94x
= 0.047
= 0.0242
So at equilibrium,
[H2 C6 H5 O−
7 ] = 0.024 M
[HC6 H5 O2−
7 ] = 0.052 M
[C6 H5 O3−
7 ] = 0.024 M
We get pH from the Henderson-Hasselbalch equation:
pH =
=
=
=
[A− ]
pKa + log10
[HA]
[HC6 H5 O2−
7 ]
pKa2 + log10
[H2 C6 H5 O−
7]
0.052
4.74 + log10
0.024
5.1
(you can use Ka3 and 0.024/0.052, with the same answer.)
(d) The amino acid histidine is shown below:
There are three protons with measurable acidities, with pKa1 =1.77, pKa2 =6.10,
and pKa3 =9.18. Label each acidic proton with its pKa value. When there
are resonance structures stabilizing a conjugate base, draw those resonance
structures.
H
H
H
H
N
O
C
C
N
H
N
O
C
C
C
H
H
N
N
H
H
H
H
C
C
C
H
H
C
O
H
C
C
N
C
H
H
13
O
H
H
H
H
N
O
C
C
N
H
N
O
C
C
C
H
H
N
N
H
H
C
C
C
H
H
C
O
H
C
C
N
C
O
H
H
(plus 3 minor structures with lone pairs on a carbon)
4. (a) Use the following enthalpies of combustion:
H2 (g)
−241.3
C(s)
−393.5
C2 H2 (g) −1256.2
C6 H6 (`) −3135.5
Calculate
i. ∆H◦f for C2 H2
We are combusting carbon and hydrogen, and uncombusting CO2 and
H2 O, in order to write the formation reaction:
+ 5 O2 (g)
− 5 O2 (g)
2
2
2C(s) + H2 (g) −→
2CO2 (g) + H2 O(g) −→
C2 H2 (g)
This is why we use reactant enthalpy changes minus product enthalpy
changes for enthalpies of combustion:
∆H◦f (C2 H2 (g)) = ∆H◦comb (reactants) − ∆H◦comb (products)
= (2(−393.5) + (−241.3)) − (−1256.2)
= 227.9 kJ mol−1
ii. ∆H◦f for C2 H2
A typo. ∆H◦f for C6 H6 comes from the reaction
6C(s) + 3H2 (g) −→ C6 H6 (`)
∆H◦f (C6 H6 (`)) = ∆H◦comb (reactants) − ∆H◦comb (products)
= (6(−393.5) + 3(−241.3)) − (−3135.5)
= 50.6 kJ mol−1
14
iii. ∆H◦rxn for the reaction
3C2 H2 (g) −→ C6 H6 (`)
We can use either our heats of formation or our heats of combustion
here.
∆H◦rxn = ∆H◦comb (reactants) − ∆H◦comb (products)
= (3(−1256.2)) − (−3135.5)
= −633.1 kJ mol−1
(b) At a temperature of 400 ◦ C, 0.80 atm of N2 (g), 0.89 atm of H2 (g), and
9.5 × 10−3 atm of NH3 (g) are at equilibrium with respect to the reaction
N2 (g) + 3H2 (g) 2NH3 (g)
The volume is held constant while the temperature is lowered to 200 ◦ C.
(Remember that this will decrease all gas pressures according to P V =
nRT .) When equilibrium is eventually reestablished, the total pressure
inside the container is 0.99 atm. What is ∆H◦f of NH3 ? (Assume ∆Hf
does not change with temperature.)
The first thing we can do is find KP at 400 ◦ C:
P2NH3
(9.5 × 10−3 )2
=
= 1.6 × 10−4
KP =
3
3
PN2 PH2
0.80 · 0.89
When the temperature drops, all pressures will come down by a multiplicative factor of
T1
473.15 K
=
= 0.703
T2
673.15 K
So our non-equilibrium pressures at 200 ◦ C are
PNH3 = 6.7 × 10−3 atm
PN2 = 0.56 atm
PH2 = 0.63 atm
PN2 = 0.56 − x
PH2 = 0.63 − 3x
Our equilibrium pressures are
PNH3 = 6.7 × 10−3 + 2x
15
Using the total pressure:
−3
(6.7 × 10
PNH3 + PN2 + PH2
+ 2x) + (0.56 − x) + (0.63 − 3x)
1.1967 − 2x
x
=
=
=
=
0.99 atm
0.99
0.99
0.103
The equilibrium pressures are
PNH3 = 0.21 atm
PN2 = 0.45 atm
PH2 = 0.32 atm
And KP at 200 ◦ C is
KP =
P2NH3
(0.21)2
=
= 3.0
PN2 P3H2
0.45 · 0.323
With T1 = 673 K and T2 = 473 K, the van’t Hoff equation lets us find
∆Hrxn :
∆H 1
1
K2
= −
−
ln
K1
R
T2 T1
∆H
1
3.0
1
= −
−
ln
1.6 × 10−4
8.314 J mol−1 K−1 473 K 673 K
∆H
−4
K−1 )
9.84 = −
−1
−1 (6.28 × 10
8.314 J mol K
∆H = −130.2 kJ mol−1
Since the formation reaction for NH3 is
1
3
N2 (g) + H2 (g) NH3 (g)
2
2
then
−130.2 kJ mol−1
= −65.1 kJ mol−1
2
This number is a bit off the actual value (−45.9).
∆Hf◦ (NH3 (g)) =
16