Advanced Placement Chemistry

1986 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) Average score = 3.14

a) one point

K_a = ([H⁺] [N₃¯]) ÷ [HN₃]

b) three points

2.8 x 10¯⁵ = x² ÷ 0.050

x = [H⁺] = [square root] (1.4 x 10¯⁶) = 1.2 x 10¯³ M

pH = - log[H⁺] = 2.93

c) two points

[N₃¯] = (0.80 grams x (1 mole/65 grams)) / 0.150 liters = 0.082 M

2.8 x 10¯⁵ = ([H⁺] (0.082)) / (0.050)

[H⁺] = 1.7 x 10¯⁵ M

pH= 4.77

d) three points

(0.075 liter) (0.100 M) = 0.0075 mole NaOH

(0.150 liter) (0.050 M) = 0.0075 mole HN₃

OH¯ + HN₃ ---> H₂O + N₃¯

neutralization is complete

N₃¯ + H₂O ---> HN₃ + OH¯

K_b = K_w / K_a

1 x 10¯¹⁴ ÷ 2.8 x 10¯⁵ = ([HN₃] [OH¯]) ÷ [N₃¯]

= x² ÷ (0.0075 / 0.225)

x = [OH¯] = 3.4 x 10¯⁶ M

Notes related to (d)

1) Neutralization of only H⁺ in solution to get only excess OH¯, and division of moles of OH¯ by total volume receives only one point
2) Neutralization of HN₃ and [OH¯] = 10¯⁷ received one point

2) Average score = 3.28

a) three points

1.100 hr x (3600 sec / hr) = 3960 sec

3960 sec x (0.125 coulomb / sec) = 495 coulombs

495 coulombs x (1 faraday / 96500 coulombs) = 5.13 x 10¯³ F

b) one point

Fe³⁺ + e¯ ---> Fe²⁺

c) two points

MnO₄¯ + 8 H⁺ + 5 e¯ ---> Mn²⁺ + 4 H₂O

5(Fe²⁺ ---> Fe³⁺ + e¯)

Sum of the above two half reactions:

MnO₄¯ + 8 H⁺ + 5 Fe²⁺ ---> Mn²⁺ + 4 H₂O + 5 Fe³⁺

d) three points

5.13 x 10¯³ faraday x (1 mole Fe²⁺ / 1 faraday)

= 5.13 x 10¯³ mole Fe²⁺

5.13 x 10¯³ mole Fe x (1 mole MnO₄¯ / 5 mole Fe2+)

= 1.03 x 10¯³ mole MnO₄¯

1.03 x 10¯³ mole MnO₄¯ ÷ 0.02465 liter = 0.0416 M MnO₄¯ (or KMnO₄)

Notes related to (a) and (d)

1) In (a) and (d), full credit for calculations on one line
2) in (d), stoichiometric ratio from the equation in (c) received one point
3) In (d), molarity of MnO₄¯ based on a candidate's solution received one point

3) Average score = 5.18

a) three points

PV= (grams / molec. wt) x RT

molec. wt = (3.53 grams / liter) (0.0821 liter atm/mole K) x (300 K) (1/[750/760])

= 88.1 grams/mole

OR

(3.53 grams / liter) (760/750) (300/273) x (22.4 L/mol) = 88.1 g/mol

b) one correct = 1 point; all correct = one additional point

gram Q/mole X = 0.648 x 88.1 = 57.1

gram Q/mole Y = 0.730 x 104 = 75.9

gram Q/mole Z = 0.593 x 64.0 = 38.0

c) one point

Masses in (b) must be integral multiples of atomic weight. Largest common denominator is 19.

Note: credit given for incorrect at. wt. if consistent with values in (b).

d) three points

1.37 grams CO₂ (1 mole/44.0 grams CO₂) = 0.0311 mole CO₂ = 0.0311 mole C

0.281 grams H₂O (1 mole/18.0 grams H₂O) = 0.0156 mole H₂O = 0.0312 mole H

1.00 gram Z = (1 mole/64 grams) = 0.0156 mole Z

Each mole Z contains 2 moles of CH, or 25 grams, which leaves (64 - 26) = 38 grams, corresponding to 2 moles of Element Q.

Mol. formula is C₂H₂Q₂

4) Average score = 6.95

a) Li + N₂ ---> Li₃N

b) H⁺ + SO₃²¯ ---> HSO₃¯ (or H₂SO₃ or SO₂ + H₂O)

c) Na₂O + H₂O ---> Na⁺ + OH¯

d) Zn²⁺ + S²¯ ---> ZnS

(or Zn²⁺ + HS¯ ---> ZnS + H⁺)

e) NH₃ + HC₂H₃O₂ ---> NH₄⁺ + C₂H₃O₂¯

f) Fe³⁺ + Fe ---> Fe²⁺

g) C₂H₄ + Br₂ ---> C₂H₄Br₂

h) Cl₂ + I¯ ---> I₂ + Cl¯

Notes:

1) 3 points; one for reactants and 2 for poducts
2) 2 of 3 points for correct responses in inappropriate form, e.g. molecular when it should be ionic
3) 1 point penalty for spurious product, e.g. redox products in acid-base reaction
4) 1 point penalty for inclusion of spectator ions (only once per paper)

5) Average score = 1.63

a) five points

c) three points

Solvent expansion is endothermic

OR

Increased exothermicity is associated with increased solubility

6) Average score = 3.24

a) two points

order of reaction determined by the slowest step in the mechanism

OR

order of reaction determined by exponents in the rate law

OR

providing a counterexample where the coefficients in equation and exponents in rate law are different

b) six points

1. Rate = k[XY] or equivalent

2. k = 1/time or units consistent with student's rate equation

3. Mechanism proposed should show:

a) steps adding up to the overall reaction
b) one step starting with XY
c) rate-determining step involving XY

7) Average score = 2.90

a) four points

1. As effective nuclear charge on central atom increases, acid strength increases.

OR

As number of lone oxygen atoms (oxygen atoms not bonded to hydrogen) increases, acid strength increases.

OR

As electronegativity of central atom increases, acid strength increases.

2. Loss H⁺ by neutral acid molecules reduces acid strength.

OR

K_a of H₂SO3 > K_a of HSO₃¯

b) four points

H₃BO₃ < HSO₃¯ < H₂SO₃ < HClO₃ < HClO₄

H₃BO₃ or HSO₃¯ weakest (must be together)

HClO₃ weaker than HClO₄

HSO₃¯ weaker than H₂SO₃

Both S acid weaker than both Cl acids

Note: if acids in exactly in exactly reverse, one point

8) Average score = 3.11

a) two points

Distinction or correctly implied distinction between the structures of graphite and diamond.

Freedom of movement of electrons in graphite restulting from the structure

Note: a good description of the structure of graphite or diamond and of the conductivity received one point.

b) two points

The rock salt forms a concentrated solution with a very little water from the ice.

The solution now has a freezing point lower than the temperature of the ice; therefore, the ice melts.

c) two points

Hot air is more dense then helium.

OR

hot air has much less lifting power per unit volume than helium has.

A scientific explanation of the volume/lift relation

d) two points

Carbon dioxide is more dense than air and so pushes the air away from the fire.

Water is more dense than the oil and so ends up below the oil, leaving the oil still in contact with the air; or spatters the oil into air.

9)

a) two points

1. Bubbling or dissolving of Zn

2. No reaction

b) two points

Zn + 2 H⁺ ---> Zn²⁺ + H₂

OR

Zn + 2 HCl ---> ZnCl₂ + H₂

Note: these last two answers are not off the official AP scoring standards, however the ChemTeam believes they are very representative of how the standards might look.

c) two points

2 H⁺ + 2e¯ --->H₂ E° = 0.00 V

Zn ---> Zn²⁺ + 2e¯ E° = 0.76 V

The E° of the overall reaction is positive, demonstrating that the forward reaction is spontaneous.

d) two points

1) The Cu dissolves with the evolution of a brown gas.

2) The Cu was non-reactive toward the HCl and reacted spontansously with the HNO₃

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