Advanced Placement Chemistry

1984 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.
• Other correct ways of solving the problems received full credit. Because of limitations of space, such other solutions are not shown.
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1) Average score 4.36

a) two points

pH = 8.60

[H⁺] = 1 x 10¯^8.60 M = 2.5 x 10¯⁹ M

[OH¯] = (1 x 10¯¹⁴) / (2.5 x 10¯⁹) M

= 4.0 x 10¯⁶ M

b) three points

C₆H₅CO₂¯ + HOH ---> C₆H₅CO₂H + OH¯

K = ([C₆H₅CO₂H][OH¯]) ÷ [C₆H₅CO₂¯]

= ((4.0 x 10¯⁸) (4.0 x 10¯⁶)) ÷ (0.10 - 4.0 x 10¯⁶)

= 1.6 x 10¯¹⁰

c) one point

C₆H₅CO₂H <===> H⁺ + C₆H₅CO₂¯

K_a = ([H⁺][C₆H₅CO₂¯]) ÷ [C₆H₅CO₂H]

= ((2.5 x 10¯⁹) (0.10)) ÷ (4.0 x 10¯⁶)

= 6.3 x 10¯⁵

d) two points

C₆H₅CO₂H <===> C₆H₅CO₂¯ + H¯

pH = 2.88

[H⁺ ] = 1 x 10¯^2.88M = 1.3 x 10¯³

K_a = ([H⁺ ][C₆H₅CO₂¯]) ÷ [C₆H₅CO₂H]

6.3 x 10¯⁸ = ((1.3 x 10¯³) (1.3 x 10¯³)) ÷ x

x = 2.8 x 10¯² M

Total C₆H₅CO₂H in solution =

(2.8 x 10¯² + 1.3 x 10¯³) M =

2.9 x 10¯² M

2) Average score 4.81

a) three points; one point for correct form of law and two points for correct methodology without an error; one point for correct methodology with an error

Rate = k[Y]

b) two points

7.0 x 10¯⁴ mole / L sec = k (0.10 mole / L)

k = 7.0 x 10¯³ sec¯¹

c) two points

2.3 log c_o / c = k t

2.3 log 0.60 / 0.40 = (7.0 x 10¯³) (t)

t = 58 s

d) two point

Mechanism 3 is correct.

The rate law shows that the slow reaction must involve one Y, consistent with mechanism 3.

Mechanisms 1 and 2 would involve both [X] and [Y] in the rate law, not consistent with the rate law.

3) Average score 5.52

a) three points

From Hess's law:

[delta]H°_f = [4 (393.5) + 4 (205.85)] - 2183.5 kJ = - 533.8 kJ

b) one point

4 C(s) + 4 H₂(g) + O₂(g) ---> C₃H₇COOH

c) two points

[delta]S°_f butyric acid = S° butyric acid - 4 S° carbon - 4 S° H₂ - S° O₂

[delta]S°_f butyric acid = [226.3 - 4(5.69) - 4(130.6) - 205] joules/K = - 523.9 joules/K

Note: If part c was based on the equation in part b, and was done correctly, credit was given for part c even if part b was wrong.

d) three points

[delta]G°_f butyric acid = [delta]H°_f - T [delta]S°_f

[delta]G°_f = [- 533.9 - (298) (- 0.524)] kJ

[delta]G°_f= - 377.7 kJ

Note: For each of the problems, a maximum of one point was subtracted for gross misuse of significant figures and. or for a mathematical error if correct principles were used.

4) Average score 3.52

a) two points

Water boils at a lower temperature in Denver than in New York City because the atmospheric pressure is less at high altitudes.

At a lower temperature, the cooking process is slower, so the time to prepare a hard-boiled egg is longer.

b) two points

S + O₂ ---> SO₂ (as coal is burned)

SO₂ + H₂O ---> H₂SO₃ (in the atmosphere)

SO₂ + 1/2 O₂ + H₂O ---> H₂SO₄ (in the atmosphere)

c) two points

Vaporization or evaporation of sweat from the skin.

These processes are endothermic and so cool the skin.

d) two points

Colligative properties, which depend on the number of particles present, are involved.

Solute (the antifreeze) causes the lowering of the vapor pressure of the solvent. When the vapor pressure of the solvent is lowered, the freezing point is lowered and the boiling point is raised.

5) Average score 3.05

Points awarded are indicated at the end of each statement.

An indicator signals the endpoint of a titration by changing color. (2)

An indicator is a weak acid or a weak base where the acid form and the basic form of the indicator are different colors. (2)

An indicator usually changes color when the concentrations of the acid form and the basic form are about equal; that is, when the pH is near the pK for the indicator. (1)

When an indicator is selected, the pH at which the indicator changes color should bracket the pH of the titration solution at the equivalence point. (2)

For example, when a strong acid is titrated with a strong base, the pH at the equivalence point is 7, so an indicator that changes color at a pH of 7 should be used. (1)

6) Average score 2.60

Two points (to a maximum of eight) were awarded for each correct test with the expected results. Possibilities include the following:

1. Flame test: K₊ lavender; (NH₄)₂CO₃, no lavender

2. Add Cl₂and CH₂Cl₂: KI, pink color in organic layer; (NH₄)₂CO₃, no change

3. Add Pb²⁺: KI, yellow precipitate of PbI₂; (NH₄)₂CO₃, white precipitate of PbCO₃

4. Add Ag⁺: KI, pale yellow precipitate of AgI; (NH₄)₂CO₃, white precipitate of Ag₂CO₃

5. Add I₂: KI, brown color of I₃¯; (NH₄)₂CO₃, no change

6. Add good oxidizing agent: KI, brown color of I₃¯; (NH₄)₂CO₃, no change

7. Add strong base: KI, no change: (NH₄)₂CO₃, odor of NH₃ or color change of red litmus

8. Add Ba₂₊ or Ca₂₊ or Mg₂₊: KI, no change; (NH₄)₂CO₃, with precipitate of a carbonate

9. Dissolve salts and use litmus: KI, neutral; (NH₄)₂CO₃, basic solution

10. Add nonoxidizing acid: KI, no change; (NH₄)₂CO₃, bubbles of CO₂

11. Test melting points: KI, high; (NH₄)₂CO₃, decomposes before melting

7) Average score 4.28

Points awarded for each statement follow that statement.

Constant a is related to the attractive forces that exist between real molecules. (1 point)

Constant b is related to the fact that real molecules do have volumes. (1 point)

H₂S has a larger value for a. (1 point)

H₂S is a polar molecule and, therefore, has stronger intermolecular forces. (2 points)

H₂S has a larger value for b because an H₂S molecule has a larger volume than H₂ has. (1 point)

The constant a correlates with the boiling point since a is related to the intermolecular forces, which must be overcome in the process of boiling a liquid. (2 points)

8) Average score 4.16

Electron configurations (1 point each statement): Metals: Valence electrons in s or d sublevels of their atoms. (A few heavy elements have atoms with one or two electrons in p sublevels.) Nonmetals: Valence electrons in the s and p sublevels of their atoms.

There are more metals than nonmetals because filling d orbitals in a given energy level involves the atoms of ten element and filling the f orbitals involves the atoms of 14 elements. In the same energy levels, the maximum number of elements with atoms receiving p electrons is six.

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