Final ExamCHEM 181: Introduction to Chemical Principles

December 14, 2015Answer Key

1. The compound C4H5NOS has the connectivity shown below

N

CS

C

C

H

H

H

CO

H

H

On the next page:

• Draw all reasonable resonance structures and label them as major orminor.

• Label all non-zero formal charges.

• Rank the resonance structures: 1=best, 2=next best, etc. If two resonantstructures are equivalent or very close to equivalent, you can assign themthe same number in the ranking.

If you prefer, you can skip writing the carbons and the hydrogens attached to

the carbons, like so:

HN

SHO

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There is only one structure with zero formal charges; everything else is minor.Ranking of minor structures is based on electronegativity of the atom gettingthe positive charge.

HN

SHO

HN

SHO

HN

SHO

HN

SHO

HN

SHO

HN

SHO

1, major

2, minor

2

3

3

4

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2. A compound containing only C, H, and O with a molecular weight under110 amu has the following 13C NMR spectrum (and note that peak heightsin this 13C NMR spectrum are not meaningful):

IR spectrum:

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and 1H NMR spectrum:

12 11 10 9 8 7 6 5 4 3 2 1 0

240 220 200 180 160 140 120 100 80 60 40 20 0CDCl3 QE-300

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 1H NMR spectrum, and

(b) labels for peaks in the IR spectrum that can be assigned unambiguously.

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3. The electron configurations for the ground state and first three excited states ofN2 are shown. Use the following information about energies and bond lengthsto figure out the electron configurations for the ground and excited states ofCN. (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 forCN will have states with different energies, but the order and names will be thesame.)

N2

energy (aJ) bond length (nm) σ2s σ∗2s π2px , π2py σ2pz π∗

2px , π∗2py σ∗

2pz

0.0 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

CNenergy (aJ) bond length (nm) σ2s σ∗

2s π2px , π2py σ2pz π∗2px , π

∗2py σ∗

2pz

0.0 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

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There are a few ways to answer this in a self-consistent manner. Here is oneline of reasoning:

• The 0.18 aJ state must come from a transition not available to N2. Pro-moting 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 onlyslightly.

• 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 aJto the 1.18 aJ state.

• The 0.51 aJ also has no clear counterpart in N2. Since bond length de-creases 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 σ∗

2s

orbital, and the relatively low energy of this state would place it into theσ2pz . The 2s orbital for C will be significantly higher in energy than theN 2s, which will bring the σ∗

2s much closer in energy to the states derivedfrom the 2p orbitals.

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4. Use the following enthalpies of formation

∆H◦f

C2H6(g) −83.7N2H4(g) 50.63

and the following bond enthalpies

∆H◦bond

C–C 368.0C–N 331.0N–N 301.0

Calculate the enthalpy of formation of CH3NH2(g).

C C

H

H

H

H

H

H

C N

H

H

H

H

H

N N

H

H

H

H

The reactionC2H6(g) + N2H4(g) −→ 2CH3NH2(g)

involves breaking a C–C and an N–N bond, and making two C–N bonds; viz.,

C2H6(g) + N2H4(g) −→ CH3(g) + CH3(g) + NH2(g) + NH2(g) −→ 2CH3NH2(g)

So

∆H◦rxn = ∆H◦

bond(reactants)−∆H◦bond(products)

= ∆H◦bond(C–C) + ∆H◦

bond(N–N)− 2∆H◦bond(C–N)

= 368 + 301− 2 · 331

= 7 kJ mol−1

But we can also express ∆H◦rxn in terms of formation enthalpies:

∆H◦rxn = ∆H◦

f (products)−∆H◦f (reactants)

7 kJ mol−1 = 2∆H◦f (CH3NH2(g))−

[∆H◦

f (C2H6(g)) + ∆H◦f (N2H4(g))

]7 kJ mol−1 = 2∆H◦

f (CH3NH2(g))−[−83.7 kJ mol−1 + 50.63 kJ mol−1

]2∆H◦

f (CH3NH2(g)) = 7 kJ mol−1 − 83.7 kJ mol−1 + 50.63 kJ mol−1

= −26 kJ mol−1

∆H◦f (CH3NH2(g)) = −13 kJ mol−1

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5. Molecular iodine (I2) is slightly soluble in water and somewhat more soluble incyclohexane (C6H12):

Ksp @ 20 ◦C Ksp @ 50 ◦C

I2(s)H2O(`)−→ I2(aq) 1.14× 10−3 M 3.08× 10−3 M

I2(s)C6H12(`)−→ I2(C6H12 sol.) 8.85× 10−3 M 2.53× 10−2 M

Water and cyclohexane do not mix, and iodine will distribute between the twosolvents according to the reaction:

I2(aq) � I2(C6H12 sol.)

For this reaction, what is K at 20 ◦C, and what are ∆H and ∆S? (Assumethat ∆H and ∆S are constant over this temperature range.)

Our two reactions have very simple equilibrium constant expressions:

Ksp(aq) = [I2(aq)] and Ksp(C6H12) = [I2(C6H12 sol.)]

ForI2(aq) � I2(C6H12 sol.)

we have

K =[I2(C6H12 sol.)]

[I2(aq)]

=8.85× 10−3

1.14× 10−3

= 7.76

This is at 20 ◦C; at 50 ◦C, K = 8.21. We can use the 20 ◦C value to find ∆G◦:

∆G◦ = −RT lnK

= −(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

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temperatures:

lnK2

K1

= −∆H

R

(1

T2− 1

T1

)ln

8.21

7.76= − ∆H

8.314 J mol−1 K−1

(1

323 K− 1

293 K

)5.64× 10−2 · (8.314 J mol−1 K−1) = −∆H · (−3.17× 10−4 K−1)

∆H = 1.48 kJ mol−1

Since the equilibrium constant increases (albeit slightly) with temperature,LeChâtelier’s principle says ∆H should be positive (albeit slightly): add heatto raise the temperature, and equilibrium shifts towards products in order totake up some of that heat.Then,

∆G = ∆H − T∆S

−4.99 kJ mol−1 = 1.48 kJ mol−1 − (293 K) ·∆S

∆S =6.47 kJ mol−1

293 K= 22.1 J mol−1 K−1

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6. Here are the chemical reactions to consider in the reaction of baking soda(NaHCO3) with vinegar (CH3COOH(aq)):

CO2(g) � CO2(aq) KH = 3.4× 10−2 M atm−1

CO2(aq) + H2O(`) � H2CO3(aq) K = 1.3× 10−3

H2CO3(aq) � HCO−3 (aq) + H+(aq) Ka1 = 2× 10−4 M

HCO−3 (aq) � CO2−

3 (aq) + H+(aq) Ka2 = 4.7× 10−11 M

CH3COOH(aq) � CH3COO−(aq) + H+(aq) Ka = 1.8× 10−5 M

H2O(`) � H+(aq) + OH−(aq) Kw = 1× 10−14 M2

(a) 1 L of a 0.8 M aqueous solution of CH3COOH has NaHCO3 added to it untilit reaches pH 7, at which point it is at equilibrium with respect to all of theabove 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)

= 3.4× 10−2 M atm−1

[H2CO3(aq)]

[CO2(aq)]= 1.3× 10−3

[H+][HCO−3 (aq)]

[H2CO3(aq)]= 2× 10−4 M

[H+][CO2−3 (aq)]

[HCO−3 (aq)]

= 4.7× 10−11 M

[H+][CH3COO−(aq)]

[CH3COOH(aq)]= 1.8× 10−5 M

[H+(aq)][OH−(aq)] = 10−14 M2

We have PCO2 fixed at an equilibrium value of 3.94× 10−4 atm. Pluggingin:

[CO2(aq)]

PCO2(g)

= 3.4× 10−2 M atm−1

[CO2(aq)]

3.94× 10−4 atm= 3.4× 10−2 M atm−1

[CO2(aq)] = 1.34× 10−5 M[H2CO3(aq)] = 1.74× 10−8 M

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At pH 7, [H+] must be 10−7 at equilibrium, so

[H+][HCO−3 (aq)]

[H2CO3(aq)]= 2× 10−4 M

10−7[HCO−3 (aq)]

1.74× 10−8= 2× 10−4 M

[HCO−3 (aq)] = 3.48× 10−5 M

[CO2−3 (aq)] = 1.64× 10−8 M

On the other hand, at pH 7 almost all of the acetic acid is in its conjugatebase form:

[H+][CH3COO−(aq)]

[CH3COOH(aq)]= 1.8× 10−5 M

10−7[CH3COO−(aq)]

[CH3COOH(aq)]= 1.8× 10−5 M

[CH3COO−(aq)]

[CH3COOH(aq)]= 180

So at equilibrium, the negative ions and their concentrations in solutionare

• just under 0.8 M of CH3COO−,• 3.48× 10−5 M of HCO−

3 ,• 10−7 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 CH3COOH has NaHCO3 added to ituntil it reaches pH 7, at which point it is at equilibrium with respect to allof the above reactions. In this case, the reaction takes place in a sealed 4 Lcontainer. Compared to part (a), how much NaHCO3 was added: more,slightly more, exactly the same amount, slightly less, or less? Explain youranswer.

In a sealed container, PCO2 is going to be much higher, resulting in sig-nificant concentrations of all of the carbonate species, and in particularHCO−

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 so-lution, 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 containerwill hold 0.123× P/atm moles of gas at this temperature.)

With a CO2(g) pressure of P at equilibrium, the following come from ourequilibrium constant expressions:

[CO2(aq)] = 3.4× 10−2 (P/atm) M[H2CO3(aq)] = 4.42× 10−5 (P/atm) M[HCO−

3 (aq)] = 8.84× 10−2 (P/atm) M[CO2−

3 (aq)] = 4.15× 10−5 (P/atm) M

Each HCO−3 that is protonated ends up either as an H2CO3(aq), a CO2(aq),

or a CO2(g). Since [CH3COO−(aq)]≈0.8 M, there are 0.8 moles of protonsto be accounted for. The number of protons making H2CO3(aq) is moreor less cancelled out by the number released forming CO2−

3 (aq), so

0.8 mol≈ nCO2(aq) + nCO2(g)

= 3.4× 10−2 (P/atm) M · 1 L + 0.123 (P/atm) mol= 0.157 (P/atm) mol

P = 5.1 atm

The HCO−3 concentration must be

[HCO−3 (aq)] = 8.84× 10−2 (P/atm) M = 0.45 M

In contrast to the situation in part (a), our negative-ion species in solutionare

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• just under 0.8 M of CH3COO−,• 0.45 M of HCO−

3 ,• negligible amounts of OH− and CO2−

3 .

We must have added 1.25 M of NaHCO3.

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