Advanced Placement Chemistry s3

Advanced Placement Chemistry

1986 Free Response Questions

1) In water, hydrazoic acid, HN3, is a weak acid that has an equilibrium constant, Ka, equal to 2.8 x 10¯3 at 25 °C. A 0.300-liter sample of a 0.050-molar solution of the acid is prepared.

(a) Write the expression for the equilibrium constant, Ka, for hydrazoic acid.

(b) Calculate the pH of this solution at 25 °C.

(c) To 0.150 liter of this solution, 0.80 gram of sodium azide, NaN3, is added. The salt dissolves completely. Calculate the pH of the resulting solution at 25 °C if the volume of the solution remains unchanged.

(d) To the remaining 0.150 liter of the original solution, 0.075 liter of 0.100-molar NaOH solution is added. Calculate the [OH¯] for the resulting solution at 25 °C.

2) A direct current of 0.125 ampere was passed through 200 milliliters of a 0.25-molar solution of Fe2(SO4)3 between platinum electrodes for a period of 1.100 hours. Oxygen gas was produced at the anode. The only change at the cathodes was a slight change in the color of the solution. At the end of the electrolysis, the electrolyte was acidified with sulfuric acid and was titrated with an aqueous solution of potassium permanganate. The volume of the KMnO4 solution required to reach the end point was 24.65 milliliters.

(a) How many faradays were passed through the solution?

(b) Write a balanced half-reaction for the process that occurred at the cathode during the electrolysis.

(d) Calculate the molarity of the KMnO4 solution.

3) Three volatile compounds X, Y, and Z each contain element Q. The percent by weight of element Q in each compound was determined. Some of the data obtained are given below.

Compound / Percent by Weight
of Element Q / Molecular Weight
X / 64.8% / ?
Y / 73.0% / 104.
Z / 59.3% / 64.0

(a) The vapor density of compound X at 27 °C and 750. mm Hg was determined to be 3.53 grams per liter. Calculate the molecular weight of compound X.

(b) Determine the mass of element Q contained in 1.00 mole of each of the three compounds.

(d) Compound Z contains carbon, hydrogen, and element Q. When 1.00 gram of compound Z is oxidized and all of the carbon and hydrogen are converted to oxides, 1.37 grams of CO2 and 0.281 gram of water are produced. Determine the most probable molecular formula of compound Z.

4) Use appropriate ionic and molecular formulas to show the reactants and the products for the following, each of which occurs in aqueous solution except as indicated. Omit formulas for any ionic or molecular species that do not take part in the reaction. You need not balance. In all cases a reaction occurs.

(a) A piece of lithium metal is dropped into a container of nitrogen gas.

(b) Dilute hydrochloric acid is added to a solution of potassium sulfite.

(d) A solution of sodium sulfide is added to a solution of zinc nitrate.

(e) A solution of ammonia is added to a dilute solution of acetic acid.

(f) A piece of iron is added to a solution of iron(III) sulfate.

(g) Ethene (ethylene) gas is bubbled through a solution of bromine.

(h) Chlorine gas is bubbled into a solution of potassium iodide.

5) The first ionization energy of sodium is +496 kilojoules per mole, yet the standard heat of formation of sodium chloride from its elements in their standard state is -411 kilojoules per mole.

(a) Name the factors that determine the magnitude of the standard heat of formation of solid sodium chloride. Indicate whether each factor makes the reaction for the formation of sodium chloride from its elements more or less exothermic.

(b) Name the factors that determine whether the reaction that occurs when such a salt dissolves in water is exothermic or endothermic and discuss the effect of each factor on the solubility.

6) The overall order of a reaction may not be predictable from the stoichiometry of the reaction.

(a) Explain how this statement can be true.

(b) 2 XY ---> X2 + Y2

1. For the hypothetical reaction above, give a rate law that shows that the reaction is first order in the reactant XY.
2. Give the units for the specific rate constant for this rate law.
3. Propose a mechanism that is consistent with both the rate law and the stoichiometry.

H2SO3 / HSO3¯ / HClO4 / HClO3 / H3BO3

Oxyacids, such as those above, contain an atom bonded to one or more oxygen atoms; one or more of these oxygen atoms may also be bonded to hydrogen.

(a) Discuss the factors that are often used to predict correctly the strengths of the oxyacids listed above.

(b) Arrange the examples above in the order of increasing acid strength.

8) Give a scientific explanation for each of the following observations. Use equations or diagrams if they seem relevant.

(a) Graphite is used to make electrodes, while diamond, another allotrope of carbon, is a very poor conductor of electricity.

(b) Putting rock salt on an icy driveway melts the ice even when the air temperature is -10 °C.

(d) Carbon dioxide, rather than a stream of water, should be used to extinguish an oil fire.

(a) Describe what you would see if you added

1. a piece of zinc metal to a test tube that contains 6-molar hydrochloric acid;
2. a piece of copper metal to another test tube that contains 6-molar hydrochloric acid.

(b) Write balanced equations for any reactions that occur.

(d) In a separate experiment, concentrated nitric acid is added to a test tube containing a piece of copper metal.

1. Describe what you would see.
2. Explain any differences between the results obtained in this experiment and those obtained with copper metal in part (a).

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.

1) Average score = 3.14

a) one point

Ka = ([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¯

Kb = Kw / Ka

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

(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 + H+

(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

Heat of sublimation of sodium / Endothermic
First ionization energy of sodium / Endothermic
Heat of dissociation of Cl2 / Endothermic
Electron affinity of chlorine / Exothermic
Lattice energy of NaCl / Exothermic

c) three points

Lattice energy of NaCl / Endothermic
to solution
Hydration energy of the ions / Exothermic

Solvent expansion is endothermic

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

order of reaction determined by exponents in the rate law

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.

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

As electronegativity of central atom increases, acid strength increases.

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

Ka of H2SO3 > Ka 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.

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.

a) two points

1. Bubbling or dissolving of Zn

2. No reaction

b) two points

Zn + 2 H+ ---> Zn2+ + H2

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