Advanced Placement Chemistry

1980 Free Response Answers

Notes

• [delta] and [sigma] are used to indicate the capital Greek letters.
• [square root] applies to the numbers enclosed in parenthesis immediately following
• All simplifying assumptions are justified within 5%.
• One point deduction for a significant figure or math error, applied only once per problem.
• No credit earned for numerical answer without justification.
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1)

a) 7 points

CH₃NH₂ + H₂O <===> CH₃NH₃⁺ + OH¯

(0.160) (0.047) = moles / CH₃NH₂ ionizing = 7.5 x 10¯³ M

(0.160 - 7.5 x 10¯³) = 0.152 M = [CH₃NH₂] at equilibrium

[CH₃NH₃⁺] = [OH¯] = 7.5 x 10¯³ M

[H₃O⁺] = (1.0 x 10¯¹⁴ / 7.5 x 10¯³) = 1.3 x 10¯¹² M

pH = - log[H₃O⁺] = 11.89

b) 3 points

K_b = ([CH₃NH₃⁺][OH¯]) / [CH₃NH₂]

= (7.5 x 10¯³)² / 0.152 = 3.7 x 10¯⁴

c) 5 points

K_b = 3.7 x 10¯⁴ = [(0.20 + x) (x)] / (0.20 - x) = x = [OH¯]

Q = [La³⁺][OH¯]³ = (0.050) (3.7 x 10¯⁴)³ = 2.5 x 10¯¹²

Q > K_sp, therefore La(OH)₃ precipitation occurs.

2)

a) 5 points

M ---> M²⁺ + 2e¯, for this reaction E° = x

Cu²⁺ + 2e¯ ---> Cu, for this reaction E° = 0.34 volt

0.740 v = x + 0.34 v (summation E°_cell)

x = 0.40 volt (oxidation potential)

- x = - 0.40 volt (for reduction)

b) 7 points

M + Cu²⁺ ---> M²⁺ + Cu

Standard state, original concentrations = 1.00 M

At point of discharge [M²⁺] = 1.80 M (correct stoichiometry)

E = E° - (0.0592 / n) log Q: (recognition of Q)

E = 0.740 - (0.0592 / 2) log (1.80 / 0.20) (substitution) = 0.712 volt

c) 3 points

E = 0 at equilibrium

E° = (0.0592 / 2) log ( [M²⁺] / [Cu²⁺] )

log [M²⁺] / [Cu²⁺] = [2(0.740) / 0.0592] = 25.0 = log K

K = 1 x 10²⁵

3)

a) 5 points

[delta] T_f= K_fm

0.97 = 1.86 m

m = 0.52 moles solute / kg solvent

3.23 g solute / 100 g solvent = 32.3 g solute / kg solvent

Mol. Wt. = (32.3 g solute/kg solvent) / 0.52 mole solute/kg solvent = 62

b) 5 points

Per 100 g of compound:

9.74 mole H / 1.0 = 9.74

38.7 mole C / 12.0 = 3.22

51.56 mole O / 16.0 = 3.22

Emperical formula is CH₃O (fwt = 31)

Molecular formula is (CH₃O)₂ or C₂H₆O₂

c) 5 points

Combustion produces 5 moles gaseous products per 1 mole C₂H₆O₂

moles C₂H₆O₂ = 1.05 g / 62.0 g/mol = 0.0169 mole

moles gases = 5 (0.0169) = 0.0846

PV = nRT

P = [(0.0846) (0.0821) (400)] / 3.00 = 0.962 or 0.93 atmosphere

For the problems in Parts A and B, a maximum of 1 point per question was deducted for errors of the following kinds: arithmetic, significant figures, unit labels. Numerical answers without supporting work were valued at one-third credit.

4)

The 5 choices were scored at 3 points each: 1 point for reactants and 2 points for products. Totally spurious products were scored at -1/2 point per equation, as were inappropriate molecular (instead of ionic) formulations. No credit was given to species with incorrect formulas or charges. Alternate formulas or equation portions are included in parentheses.

(a) F¯ + H⁺ (H₃O⁺) ---> HF (+ H₂O)

(b) Ba²⁺ +OH¯ + Fe³⁺ + SO₄²¯ ---> BaSO₄ + Fe(OH)₃

(c) NH₄⁺ + OH¯ ---> NH₃ + H₂O

(d) CO₂ + OH¯ ---> CO₃²¯ + H₂O

(e) Cu + NO₃¯ + H⁺ ---> Cu²⁺ + NO+H₂O (NO₂, or other valid reduction product)

(f) Cl₂ + OH¯ ---> Cl¯ + ClO¯ (ClO₃¯) + H₂O

(g) MnO₄¯ + SO₃²¯ + OH¯ ---> MnO₂ (MnO₄²¯) + SO₄²¯

(h) CH₃I + OH¯ ---> CH₃OH+ I¯

5)

a) 4 points

At 1/2 point each: Curve shape, labeled axes, reactant energy, product energy, energy of activated complex, enthalpy change, activation energy (forward), and activation energy (reverse).

b) 2 points

Catalyzed path (dotted line on diagram)

Catalyst role: changes mechanism, or increases fraction of molecules with sufficient energy to react.

c) 2 points

The ratio, k_f / k_r increases with temperature, because...

k_f / k_r = K and LeChatelier shift

or

Boltzmann distribution graph

or

T[delta]S changes for net reaction influencing [delta]G

or

k_f increased more than k_r for endothermic reaction.

6)

5 effects and 5 explanations, valued at 0.8 point each. Proper prediction of equilibrium shift was necessary in order to receive credit for an explanation.

7)

a) 2 points

Entropy (S) is a measure of randomness or disorder in a system.

b) 6 points

(4 selections at 1/2 point each and 4 accompanying explanations at 1 point each.)

In part (a), partial credit of 1 point was allowed for confusing S with [delta]S, but writing correctly in terms of "change in randomness" or "tendency toward greater randomness." In the explanations of part (b), 1/2 point was credited for correct empirical relationships involving S as a function of P or T, or mentioning relative melting-points in parts (1), (3), and (4).

8)

a) 3 points

Recognition that dissolution is favored by shift to right of equilibrium,
BaX(s) <===> Ba²⁺ + X²¯

Recognition that conversions, CO₃²¯ ---> HCO₃¯ and SO₃²¯ ---> HSO₃¯ caused by H⁺, result in greater solubility of the carbonate and sulfite

Recognition that H⁺ does not react appreciably with SO₄²¯, so as to increase the solubility of the sulfate, since SO₄²¯ is a poor base (or equivalent)

b) 1 point

The sulfide ion is oxidized by warm nitric acid, shifting CuS(s)<===>Cu²⁺ + S²¯ to the right.

c) 2 points

AgCl(s) + 2 NH₃ ---> Ag(NH₃)₂⁺ + Cl¯

Recognition that NH₃ complexes Ag

Recognition of Ag(NH₃)₂⁺ complex and that Hg₂²⁺ and Pb²⁺ do not form soluble ammine complexes

d) 2 points

Al(OH)₃(s) + OH¯ ---> Al(OH)₄¯ (or reasonable equivalent);

Fe(OH)₃ does not react in this way.

Recognition of amphoterism of Al(OH)₃

9) This answer is missing from the 1980 scoring standards available to the ChemTeam. Here are solutions without the point divisions.

a) 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d¹⁰4p³

b) for the two electrons in the 4s: 4, 0, 0, +1/2 and 4, 0, 0, - 1/2

for the three electrons in the 4p: 4, 1, -1, +1/2; 4, 1, 0, +1/2 and 4, 1, +1, +1/2

c) Paramagnetic. It has three unpaired electrons.

d)

Na₃As - each Na gives up one electron to the As, the As has a complete octet and the sodium atoms are ionically bonded to the arsenic

AsCl₃ - the three chlorines each have one half-filled orbital and the aresenic has three. So three covalent bonds are created and the As has one non-bonding pair to make a pyramidal structure.

AsF₅ - fluorine is so eletronegative that it draws the two electrons of the non-bonding pair of AsCl₃ into bonding. A 4d orbital is involved in the sp³d hydridization, yielding a trigonal bipyramidal shape.

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