Name: 1: /25Grade: /100 2: /25

3: /254: /25

Final ExamCHEM 181: Introduction to Chemical Principles

December 12, 2012Key

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 C

H

H

C

O

N

H

C

H

H

H

H C

H

H

C

O

N

H

C

H

H

H

minor

sp3

sp3

sp2 sp2

sp2

1

ii.

N N N N O

N N N N O N N N N O

N N N N O N N N N O

best

major

very minor

sp2

sp2

sp2

sp

sp

(can omit)

iii.

N O

CC

C

H

H H

N O

CC

C

H

H H

N O

CC

C

H

H H

N O

CC

C

H

H H

N O

CC

C

H

H H

best major

minor

all sp2

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(b) Order the following molecules according to their boiling point:

CH3Cl, CH3CH3, CH3OH, and CH3SH

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 = CH3CH3 < CH3Cl < CH3SH < CH3OH = highest boiling−89 ◦C < −23.8 ◦C < 5.95 ◦C < 64.7 ◦C

C2H6 has no dipole and the smallest London forces; it must be the lowestboiling. Strong hydrogen bonding in CH3OH will make it the highestboiling. For CH3Cl and CH3SH, London forces are comparable. You couldguess that because CH3Cl has a larger dipole moment, it will boil higherthan CH3SH. As it happens, weak hydrogen bonding makes CH3SH boilhigher. Either ordering of CH3Cl and CH3SH will be marked as correct, ifreasonably explained.

(c) Ethanol (C2H5OH) boils at 78.4 ◦C, and chloroform (CHCl3) boils at61.1◦C. However, a 93/7% solution (by weight) of chloroform and ethanolboils at 59.4 ◦C. Give an explanation for this in terms of intermolecularinteraction strength.

Strong hydrogen bonding gives ethanol a relatively high boiling point,while strong London forces and a large molecular dipole moment giveCHCl3 a high boiling point. However, ethanol–chloroform intermolecu-lar interactions are weaker, as ethanol cannot hydrogen bond to chloro-form, and both dipole-dipole and London forces are weaker for ethanol-chloroform than for chloroform-chloroform.

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(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 closein energy; as long as they are next to each other, don’t worry whichto put first.) How many electrons are in this π system, and whichorbitals 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 com-pare 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 orderis 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 bondorder do you get? How does this compare to Lewis electron diagramsfor SO3?

This is still a 24-electron system, and with the same MOs, you stillget bond order 4/3. If you draw octet-rule-following Lewis structures,they agree with this result. An expanded-octet Lewis structure gives abond order of 2 instead (which is yet another reason to think criticallyabout expanding octets.)

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2. (a) The 1H NMR spectra on this page are all of compounds with the molecularformula C5H10O3:

5

Choose from the compounds below (H atoms not drawn) and assign onestructure 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 under110 amu has the following 13C NMR spectrum:

IR spectrum:

7

and 1H NMR spectrum:

8

Draw a Lewis structure for the compound. Also mark on each spectrum:

i. which protons correspond to which peaks in the 1H NMR spectrum,and

ii. types of motion (type of bond, stretch vs. bend) for peaks in the IRspectrum that can be assigned unambiguously.

The compound is ethyl glycolate:

H O C

H

H

C

O

O C

H

H

C

H

H

H

Naming each set of protons:

H O C

H

H

C

O

O C

H

H

C

H

H

H

AB

C

D

The A protons give a quartet; they are shifted so far because of theC(=O)O nearby. The B protons are tricky; they give a singlet, shiftedby 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.

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3. (a) The Ka of nitrous acid (HNO2) is 4.0× 10−4. What is the pH of a 0.1 Msolution 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

Ka (HNO2)=

10−14

4.0× 10−4= 2.5× 10−11

because we’ll have the reaction

NO−2 (aq) + H2O(`) � HNO2(aq) + OH−(aq) Kb

(NO−

2

)=

[HNO2][OH−]

[NO−2 ]

use an ICE table:[NO−

2 ] [HNO2] [OH−]I 0.1 0 ≈ 0C −x x xE 0.1− x x x

So

Kb

(NO−

2

)=

[HNO2][OH−]

[NO−2 ]

2.5× 10−11 =x2

0.1− x

2.5× 10−11 ≈ x2

0.1x2 ≈ 2.5× 10−12

x ≈ 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 youcan ignore that here. To calculate pH:

[H+] =Kw

[OH−]=

10−14

1.58× 10−6= 6.3× 10−9 pH = 8.2

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(b) Citric acid (C6H8O7) is a triprotic acid:

H3C6H5O7(aq) � H2C6H5O−7 (aq) + H+(aq) Ka1 = 8.4× 10−4

H2C6H5O−7 (aq) � HC6H5O2−

7 (aq) + H+(aq) Ka2 = 1.8× 10−5

HC6H5O2−7 (aq) � C6H5O3−

7 (aq) + H+(aq) Ka3 = 4.0× 10−6

What is the pH of a 0.05 M solution of citric acid? Give an answer that isaccurate 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 =[H+][H2C6H5O

−7 ]

[H3C6H5O7]

[H3C6H5O7] [H2C6H5O−7 ] [H+]

I 0.05 0 ≈ 0C −x x xE 0.05− x x x

x2

0.05− x= 8.4× 10−4

x2 = 4.2× 10−5 − 8.4× 10−4x

x2 + 8.4× 10−4x− 4.2× 10−5 = 0

x = 6.1× 10−3

Move to the next:

Ka2 =[H+][HC6H5O

2−7 ]

[H2C6H5O−7 ]

[H3C6H5O7] [H2C6H5O−7 ] [H+]

I 6.1× 10−3 0 6.1× 10−3

C −x x xE 6.1× 10−3 − x x x+ 6.1× 10−3

x(x+ 6.1× 10−3)

6.1× 10−3 − x= 1.8× 10−5

x2 + 6.1× 10−3x = 1.1× 10−7 − 1.8× 10−5x

x2 + 6.12× 10−3x− 1.1× 10−7 = 0

x = 1.8× 10−5

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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, Na2HC6H5O7?You will need to consider the equilibrium:

HC6H5O2−7 (aq) + HC6H5O

2−7 (aq) � H2C6H5O

−7 (aq) + C6H5O

3−7 (aq)

The equilibrium constant for this reaction is

K =[H2C6H5O

−7 ][C6H5O

3−7 ]

[HC6H5O2−7 ][HC6H5O

2−7 ]

=[H+][C6H5O

3−7 ]

[HC6H5O2−7 ]

· [H2C6H5O−7 ]

[HC6H5O2−7 ][H+]

= Ka3 ·1

Ka2

=4.0× 10−6

1.8× 10−5

= 0.22

[HC6H5O2−7 ] [H2C6H5O−

7 ] [C6H5O3−7 ]

I 0.1 0 0C −2x x xE 0.1− 2x x x

Don’t forget the −2x!

x2

(0.1− 2x)2= 0.22

x

0.1− 2x= 0.47

x = 0.047− 0.94x

1.94x = 0.047

x = 0.0242

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So at equilibrium,

[HC6H5O2−7 ] = 0.052 M [H2C6H5O

−7 ] = 0.024 M [C6H5O

3−7 ] = 0.024 M

We get pH from the Henderson-Hasselbalch equation:

pH = pKa + log10

[A−]

[HA]

= pKa2 + log10

[HC6H5O2−7 ]

[H2C6H5O−7 ]

= 4.74 + log10

0.052

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 thereare resonance structures stabilizing a conjugate base, draw those resonancestructures.

C

O

OC

H

N

C

H

H

CN

C

N C

H

H

H

H H

C

O

OC

H

N

C

H

H

CN

C

N C

H

H

H

H H

13

C

O

OC

H

N

C

H

H

CN

C

N C

H

H

H H

C

O

OC

H

N

C

H

H

CN

C

N C

H

H

H H

(plus 3 minor structures with lone pairs on a carbon)

4. (a) Use the following enthalpies of combustion:

H2(g) −241.3C(s) −393.5C2H2(g) −1256.2C6H6(`) −3135.5

Calculate

i. ∆H◦f for C2H2

We are combusting carbon and hydrogen, and uncombusting CO2 andH2O, in order to write the formation reaction:

2C(s) + H2(g)+ 5

2O2(g)−→ 2CO2(g) + H2O(g)

− 52O2(g)−→ C2H2(g)

This is why we use reactant enthalpy changes minus product enthalpychanges for enthalpies of combustion:

∆H◦f (C2H2(g)) = ∆H◦

comb(reactants)−∆H◦comb(products)

= (2(−393.5) + (−241.3))− (−1256.2)

= 227.9 kJ mol−1

ii. ∆H◦f for C2H2

A typo. ∆H◦f for C6H6 comes from the reaction

6C(s) + 3H2(g) −→ C6H6(`)

∆H◦f (C6H6(`)) = ∆H◦

comb(reactants)−∆H◦comb(products)

= (6(−393.5) + 3(−241.3))− (−3135.5)

= 50.6 kJ mol−1

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iii. ∆H◦rxn for the reaction

3C2H2(g) −→ C6H6(`)

We can use either our heats of formation or our heats of combustionhere.

∆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), and9.5× 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 PV =nRT .) When equilibrium is eventually reestablished, the total pressureinside 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:

KP =P2NH3

PN2P3H2

=(9.5× 10−3)2

0.80 · 0.893= 1.6× 10−4

When the temperature drops, all pressures will come down by a multi-plicative factor of

T1T2

=473.15 K673.15 K

= 0.703

So our non-equilibrium pressures at 200 ◦C are

PNH3 = 6.7× 10−3 atm PN2 = 0.56 atm PH2 = 0.63 atm

Our equilibrium pressures are

PNH3 = 6.7× 10−3 + 2x PN2 = 0.56− x PH2 = 0.63− 3x

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Using the total pressure:

PNH3 + PN2 + PH2 = 0.99 atm(6.7× 10−3 + 2x) + (0.56− x) + (0.63− 3x) = 0.99

1.1967− 2x = 0.99

x = 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

PN2P3H2

=(0.21)2

0.45 · 0.323= 3.0

With T1 = 673 K and T2 = 473 K, the van’t Hoff equation lets us find∆Hrxn:

lnK2

K1

= −∆H

R

(1

T2− 1

T1

)ln

3.0

1.6× 10−4= − ∆H

8.314 J mol−1 K−1

(1

473 K− 1

673 K

)9.84 = − ∆H

8.314 J mol−1 K−1 (6.28× 10−4 K−1)

∆H = −130.2 kJ mol−1

Since the formation reaction for NH3 is

1

2N2(g) +

3

2H2(g) � NH3(g)

then

∆H◦f (NH3(g)) =

−130.2 kJ mol−1

2= −65.1 kJ mol−1

This number is a bit off the actual value (−45.9).

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