AP Free Response Answers

Advanced Placement Chemistry

1988 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 = 4.95

a) four points

(1) 89.7 g SbCl₅ / 299.0 g mole = 0.300 mole SbCl₅

[SbCl₅] initial = 0.300 mole / 15.0 liter = 0.200 M

(2) T = 182 °C + 273 = 455 K

P = nRT / V = [(0.300 mole) (0.0821 L atm) (455 K)] / [(15.0 L) (mole K)] = 0.747 atm

(15.0 L / 0.300 mol) x (273 K / 455 K) = 30.0 L / mol (at std. temperature)

1 atm x [(22.4 L / mol) ÷ (30.0 L / mol)] = 0.747 atm

b) three points

Equilibrium concentrations:

[SbCl₃] = [Cl₂] = (0.0200 mol / L) x 0.292 = 5.84 x 10¯³ M

[SbCl₅] = (0.0200 mol L) x 0.708 L = 1.42 x 10¯² M

K_c = ([SbCl₃][Cl₂]) ÷ [SbCl₅] = (5.84 x 10¯³)² ÷ (1.42 x 10¯²) = 2.41 x 10¯³

Equilibrium pressures:

P_SbCl₃ = P_Cl₂

= 0.747 atm x 0.292 = 0.218 atm

P_SbCl₅ = 0.747 atm x 0.708 = 0.529 atm

K_p = (P_SbCl₃ x P_Cl₂) ÷ P_SbCl₅ = (0.218)² ÷ 0.529 = 8.98 x 10¯²

c) two points

K = ([SbCl₃][Cl₂]) ¯ [SbCl₅] = 0.117

Equilibrium concentrations:

[SbCl₅] = (1.00 - 0.70) mol / 2.00 L = 0.15 M

[SbCl₃] = 0.700 mole / 2.00L = 0.350 M

[Cl₂] = x

K_c = [(0.350) (x)] ÷ (0.15) = 0.117

x = [Cl₂] = 0.50 M

Moles Cl₂ at equilibrium = 0.050 mol L x 2.00 L = 0.10 mol

Moles Cl₂ needed to make 0.300 mol SbCl₃ into SbCl₅ = 0.30 mol

Moles Cl₂ that must be added = 0.40 mol

2) Average score = 3.25

a) two points

C + O₂ --> CO₂

2 H₂ + O₂ --> 2 H₂O

C₂H₅OH --> 2CO₂ + 3 H₂O

(Half credit for recognition that combustion is reaction with O₂)

b) three points

2 C + 2 O₂ ---> 2 CO₂	[delta]H° = 2 (- 393.5) = - 787.0 kJ
2 H₂ + O₂ ---> 2 H₂O	[delta]H° = 2(- 285.8) = - 571.6 kJ
2 CO₂ + 3 O₂ ---> C₂H₅OH + 3 O₂	[delta]H° = - (- 1366.7) = 1.366.7 kJ

Sum of three equations above

2 C + 2 H₂ + H₂O ---> C₂H₅OH

[delta]H° = - 8.1 kJ

(1 point for correct [delta]H° for each of the first 3 reations)

[delta]H° combustion C(s) = [delta]H_f° CO₂(g)

[delta]H° combustion H₂(g) = [delta]H_f° H₂O(l)

C₂H₅OH + 3 O₂ ---> 2 CO₂ + 3 H₂O [delta]H° = -1,366.7kJ

[delta]H° reaction = [sigma] [delta]H_f° (products) - [sigma] [delta]H_f° (reactants)

= [2( -393.5) + 3(- 258.8) - [delta]H_f°_C₂H₅OH - 0] kJ = - 277.7 kJ

2 C + 2 H₂ + H₂O ---> C₂H₅OH [delta]H° reaction = [sigma] [delta]H° (products) - [sigma] [delta]H° (reactants)

= [- 277.7 - 0 - 0 - (- 285.8)] kJ = + 8.1 kJ

c) one point

[delta]S° = [sigma] S° (products) - [sigma] S° (reactants)

= (160.7 - 11.5 - 261.2 - 69.9) J / mol K = - 181.9 J / mol K

d) three points

(1) [delta]G° = [delta]H° - T [delta]S°

= 8,100 J - (298) (- 181.9) J = 8,100 J + 54,200 J = 62,300 J

(2) [delta]G° = - RTlnK

lnK = - [delta]G / RT

= - 62.300J ÷ [(8.31J/mol-K) (298K)] = - 25.2

K = 1.1 x 10¯¹¹

[delta]G° = - 2.303 RT logK

logK = - [delta]G / 2.303 RT

= - 62.000J ÷ [(2.303) (8.31) (298)] = - 10.9

K = 1.3 x 10¯¹¹

If the terms are not rounded, K = 1.2 x 10¯¹¹
correct substitutions in (1) and (2) earns one point

3) Average score = 5.02

a) two points

Sn ---> Sn²⁺ + 2e¯

Ag⁺ + e¯ ---> Ag

b) two points

2 Ag⁺ + Sn ---> 2 Ag + Sn²⁺

E° = [0.80 - (- 0.14)] V = 0.94 V

c) two points

E = (0.0591 ÷ n) log K (or - nFE = - RT ln K)

log K = (0.94 x 2) ÷ 0.0591 = 31.8

K = 6 x 10³¹

d) two points

E = E° - (0.0591 ÷ n) log [Sn²⁺] / [Ag⁺]²

E = E° - (RT / nF) ln Q

Q = [Sn²⁺] / [Ag⁺]²

E = 0.94 = (0.0591 ÷ 2) log (1 ÷ (0.02²))

E = 0.94 - 0.10 = 0.84 V

The final point is for completion of calculations in (c) and (d).

4) Average score = 5.66

3 points for each reaction; 1 pt. for reactants and 2 for products. Two points (of the three) for correct response in inappropriate form, i.e. molecular when ionic is correct. One point penalty for spurious product, e.g. redox product in an acid-base reaction. One point penalty for inclusion of spectator ions.

a) H⁺ (or H₃O⁺) + I¯ + Cr₂O₇²¯ ---> I₂ + Cr³⁺ H₂O

(Any reasonable I¯ oxidation product accepted.)

b) OH¯ + NH₄⁺ ---> H₂O + NH₃ (or NH₄OH)

c) Mg + Ag⁺ ---> Mg²⁺ + Ag

d) KClO₃ ---> KCl + O₂

(MnO₂ is also shown over the reaction arrow in the original.)

e) H⁺ + CO₃²¯ ---> CO₂ + H₂O (or H₂CO₃ or HCO₃¯)

f) SO₃ + H₂O ---> H⁺ + HSO₄¯ (or H⁺ + SO₄²¯)

g) Ba²⁺ + SO₄²¯ ---> BaSO₄

(or Ba²⁺ + HSO₄¯ ---> BaSO₄ + H⁺)

h) Cu²⁺ + NH₃ ---> Cu(NH₃)₄²⁺

(Partial credit for other logical Cu(II) ammonia complexes and for Cu(OH)₂)

5) Average score = 2.99

a) two points

Xe and Ne are monatomic elements held together by London dispersion (van der Waals) forces. The magnitude of such forces is determined by the number of electrons in the atom. A Xe atom has more electrons than a neon atom has. (Size of the atom was accepted but mass was not.)

b) two points

The electrical conductivity of copper metal is based on mobile valence electrons (partially filled bands). Copper chloride is a rigid ionic solid with the valence electrons of copper localized in individual copper(II) ions.

c) two points

SiO₂ is a covalent network solid. There are strong bonds many of which must be broken simultaneously to volatize SiO₂. CO₂ is composed of discrete, nonpolar CO₂ molecules so that the only forces holding the molecules together are the weak London dispersion (van der Waals) forces.

d) two points

A lone pair of electrons on the central atom results in a pyramidal shape. The dipoles don't cancel, thus the molecule is polar.

There is no lone pair on the central atom so the molecule has a trigonal planar shape in which the dipoles cancel, thus the molecule is nonpolar.

6) Average score = 4.31

a) two points

The equilibrium pressure of NH₃ gas would be unaffected K_p = (P_NH₃) (P_H₂S). Thus the amount of solid NH₄HS present does not affect the equilibrium.

b) two points

The equilibrium pressure of NH₃ gas would decrease. In order for the pressure equilibrium constant, K_p, to remain constant, the equilibrium pressure of NH₃ must decrease when the pressure of H₂S is increased.

K_p = (P_NH₃) (P_H₂S)

(A complete explanation based on Le Chatelier's principle is also acceptable.)

c) two points

The mass of NH₄HS increases. A decrease in volume causes the pressure of each gas to increase. To maintain the value of the pressure equilibrium constant, K_p, the pressure of each of the gases must decrease. That decrease realized by the formation of more solid NH₄HS.

K_p = (P_NH₃) (P_H₂S)

(A complete explanation based on Le Chatelier's principle is also acceptable.)

d) two points

The mass of NH₄HS decreases because the endothermic reaction absorbs heat and goes nearer to completion (to the right) as the temperature increases. (One point was assigned for each correct prediction and one point for each correct explanation.)

7) Average score = 1.76

a) two points

The sharp vertical rise in the pH on the pH-volume curve appears at the equivalence point (about 23 mL). Because the acid is monoprotic, the number of moles of acid equals the number of moles of NaOH. That number is the product of the exact volume and the molarity of the NaOH. The molarity of the acid is the number of moles of the acid divided by 0.030 L, the volume of the acid.

b) two points

At the half-equivalence point (where the volume of the base added is exactly half its volume at the equivalence point), the concentration [HX] of the weak acid equals the concentration [X¯] of its anion. Thus, in the equilibrium expression, [H⁺] [X¯] / [HX] = K_a. Therefore, pH at the half-equivalence point equals pK_a

c) one point

Cresol red is the best indicator because its pK_a (about 8) appears midway in the steep equivalence region. This insures that at the equivalence point the maximum color change for the minimal change in the volume of NaOH added is observed.

d) three points

8) Average score = 3.82

a) four points

One point each for:

correct identification of axes
diagram of correct shape
correct labeling of regions
correct position of normal boiling point

b) one point

The argon sublimes.

c) one point

The argon vaporizes.

d) two points

The liquid phase is less dense than the solid phase. Since the freezing point of argon is higher than the triple point temperature, the solid-liquid equilibrium line slopes to the right with increasing pressure. Thus, if a sample of liquid argon is compressed (pressure increased) at constant temperature, the liquid becomes a solid. Because increasing pressure favors the denser phase, solid argon must be the denser phase.

9) Average score = 3.31

a) two points

Equipment needed includes a thermometer, and a container for the reaction, preferably a container that serves as a calorimeter, and volumetric glassware (graduated cylinder, pipet....)

b) two points

Measurements include the difference in temperatures between just before the start of the reaction and the completion of the reaction, and amounts (volume, moles) of the acid and the base.

c) two points

Determination of heat (evolved or absorbed): The sum of the volumes (or masses) of the two solutions, the change in temperature and the specific heat of water are multiplied together to determine the heat of solution for the sample used, (q = m x C_p x [delta]T). For full credit, two of the three factors must be identified.

Division of the calculated heat of neutralization by moles of water produced, or moles of H⁺, or moles of OH¯, or moles of limiting reagent.

d) two points

Experimental errors: heat loss to the calorimeter wall, to air, to the thermometer; incomplete transfer of acid base from graduated cylinder; spattering of some of the acid or base so that incomplete mixing occurred....Experimenter errors: dirty glassware, spilled solution, misread volume or temperature....(One point for each of two experimental errors: one point for an experimenter error if only one experimental error is given.)

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