Advanced Placement Chemistry s2

Advanced Placement Chemistry

1991 Free Response Questions

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1) The acid ionization constant, Ka, for propanoic acid, C2H5COOH, is 1.3 x 10¯5.

(a) Calculate the hydrogen ion concentration, [H+], in a 0.20-molar solution of propanoic acid.

(b) Calculate the percentage of propanoic acid molecules that are ionized in the solution in (a).

(c) What is the ratio of the concentration of propanoate ion, C2H5COO¯, to that of propanoic acid in a buffer solution with a pH of 5.20 ?

(d) In a 100-milliliter sample of a different buffer solution, the propanoic acid concentration is 0.50-molar and the sodium propanoate concentration is 0.50-molar. To this buffer solution, 0.0040 mole of solid NaOH is added. Calculate the pH of the resulting solution.

2) The molecular formula of a hydrocarbon is to be determined by analyzing its combustion products and investigating its colligative properties.

(a) The hydrocarbon burns completely, producing 7.2 grams of water and 7.2 liters of CO2 at standard conditions.

(b) Calculate the mass in grams of O2 required for the complete combustion of the sample of the hydrocarbon described in (a).

(c) The hydrocarbon dissolves readily in CHCl3. The freezing point of a solution prepared by mixing 100. grams of CHCl3 and 0.600 gram of the hydrocarbon is -64.0 °C. The molal freezing-point depression constant of CHCl3 is 4.68 °C / molal and its normal freezing point is -63.5 °C. Calculate the molecular weight of the hydrocarbon.

(d) What is the molecular formula of the hydrocarbon?

2 ClO2(g) + F2(g) ---> 2 ClO2F(g)

The following results were obtained when the reaction represented above was studied at 25 °C.

(a) Write the rate law expression for the reaction above.

(b) Calculate the numerical value of the rate constant and specify the units.

(d) Which of the following reaction mechanisms is consistent with the rate law developed in (a)? Justify your choice.

ClO2 + F2 <---> ClO2F2 (fast)
ClO2F2 ---> ClO2F + F (slow)
ClO2 + F ---> ClO2F (fast)

II.

F2 ---> 2 F (slow)
2 (ClO2 + F ---> ClO2F) (fast)

4) Give the formulas to show the reactants and the products for FIVE of the following chemical reactions. Each of the reactions occurs in aqueous solution unless otherwise indicated. Represent substances in solution as ions if the substance is extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. In all cases a reaction occurs. You need not balance.

Example: A strip of magnesium is added to a solution of silver nitrate.

Mg + Ag+ ---> Mg2+ + Ag

(a) Solid aluminum oxide is added to a solution of sodium hydroxide.

(b) Solid calcium oxide is heated in the presence of sulfur trioxide gas.

(d) Calcium metal is heated strongly in nitrogen gas.

(e) Solid copper(II) sulfide is heated strongly in oxygen gas.

(f) A concentrated solution of hydrochloric acid is added to powdered manganese dioxide and gently heated.

(g) A concentrated solution of ammonia is added to a solution of zinc iodide.

(h) A solution of copper(II) sulfate is added to a solution of barium hydroxide.

BCl3(g) + NH3(g) <---> Cl3BNH3(s)

The reaction represented above is a reversible reaction.

(a) Predict the sign of the entropy change, DS, as the reaction proceeds to the right. Explain your prediction.

(b) If the reaction spontaneously proceeds to the right, predict the sign of the enthalpy change, DH. Explain your prediction.

(c) The direction in which the reaction spontaneously proceeds changes as the temperature is increased above a specific temperature. Explain.

(d) What is the value of the equilibrium constant at the temperature referred to in (c); that is, the specific temperature at which the direction of the spontaneous reaction changes? Explain.

An experiment is to be performed to determine the molecular mass of a volatile liquid by the vapor density method. The equipment shown above is to be used for the experiment. A barometer is also available.

(a) What data are needed to calculate the molecular mass of the liquid?

(b) What procedures are needed to obtain these data?

(d) If the volatile liquid contains nonvolatile impurities, how would the calculated value of the molecular mass be affected? Explain your reasoning.

7) Explain each of the following.

(a) When an aqueous solution of NaCl is electrolyzed, Cl2(g) is produced at the anode, but no Na(s) is produced at the cathode.

(b) The mass of Fe(s) produced when 1 faraday is used to reduce a solution of FeSO4 is 1.5 times the mass of Fe(s) produced when 1 faraday is used to reduce a solution of FeCl3.

Zn + Pb2+ (1-molar) ---> Zn2+ (1-molar) + Pb

(c) The cell that utilized the reaction above has a higher potential when [Zn2+] is decreased and [Pb2+] held constant, but a lower potential when [Pb2+] is decreased and [Zn2+] is held constant.

(d) The cell that utilizes the reaction given in (c) has the same cell potential as another cell in which [Zn2+] and [Pb2+] are each 0.1-molar.

8) Experimental data provide the basis for interpreting differences in properties of substances.

Account for the differences in properties given in Tables 1 and 2 above in terms of the differences in structure and bonding in each of the following pairs.

(a) MgCl2 and SiCl4

(b) MgCl2 and MgF2

(d) F2 and N2

9) Explain each of the following in terms of nuclear models.

(a) The mass of an atom of 4He is less than the sum of the masses of 2 protons, 2 neutrons, and 2 electrons.

(b) Alpha radiation penetrates a much shorter distance into a piece of material than does beta radiation of the same energy.

(c) Products from a nuclear fission of a uranium atom such as 90Sr and 137Ce are highly radioactive and decay by emission of beta particles.

(d) Nuclear fusion requires large amounts of energy to get started, whereas nuclear fission can occur spontaneously, although both processes release energy.

Advanced Placement Chemistry

1991 Free Response Answers

Notes

· [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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a) three points

Ka = ( [H+] [C3H5O2¯] ) ÷ [HC3H5O2]

1.3 x 10¯5 = x2 ÷ 0.20

x = [H+] = 1.6 x 10¯3

b) one point

% dissoc. = [H+] ÷ [HC3H5O2]

= 1.6 x 10¯3 ÷ 0.20 = 0.80%

c) two points

[H+] = antilog (- 5.20) = 6.3 x 10¯6

1.3 x 10¯5 = (6.3 x 10¯6) x ([C3H5O2¯] ÷ [HC3H5O2]

[C3H5O2¯] ÷ [HC3H5O2] = 1.3 x 10¯5 ÷ 6.3 x 10¯6 = 2.1

An alternate solution for (c) based on the Henderson-Hasselbalch equation.

pH = pKa + log ([base] ÷ [acid])

5.20 = 4.89 + log ([C3H5O2¯] ÷ [HC3H5O2])

log ([C3H5O2¯] ÷ [HC3H5O2]) = 0.31

[C3H5O2¯] ÷ [HC3H5O2] = 2.0

d) six points

0.10 L x 0.35 mol/L = 0.035 mol HC3H5O2

0.10 L x 0.50 mol/L = 0.050 mol C3H5O2¯

0.035 mol - 0.004 mol = 0.031 mol HC3H5O2

0.050 mol + 0.004 mol = 0.054 mol C3H5O2¯

1.3 x 10¯5 = [H+] x [(0.054 mol/0.1 L) ÷ (0.031 mol/0.1 L)]

Can use 0.54 and 0.31 instead.

[H+] = 7.5 x 10¯6

pH = 5.13

An alternate solution for (d) based on the Henderson-Hasselbalch equation.

use [ ]s or moles of HC3H5O2 and C3H5O2¯

pH = pKa + log (0.054 / 0.031)

= 4.89 + 0.24 = 5.13

a) three points

7.2 g H2O ÷ 18.0 g/mol = 0.40 mol H2O

0.40 mol H2O x (2 mol H / 1 mol H2O) = 0.80 mol H

7.2 L CO2 ÷ 22.4 L/mol = 0.32 mol CO2

0.32 mol CO2 x (1 mol C / 1 mol CO2) = 0.32 mol C

n = PV ÷ RT = [(1 atm) (7.2 L)] ÷ [(0.0821 L atm mol¯1 K1) (273 K)] = 0.32 mol CO2

0.80 mol H ÷ 0.32 = 2.5

0.32 mol C ÷ 0.32 = 1

2.5 x 2 = 5 mol H

1 x 2 = 2 mol C

empirical formula = C2H5

b) two points

mol O2 for combustion = mol CO2 + 1/2 mol H2O = 0.32 + 0.20 = 0.52 mol O2

0.52 mol O2 x 32 g/mol = 17 g O2

alternate approach for mol O2 from balanced equation

C2H5 + 13/4 O2 ---> 2 CO2 + 5/2 H2O

other ratio examples:
1, 6.5 ---> 4, 5
0.25, 1.625 ---> 1, 1.25

mol O2 = 0.40 mol H2O x (13/4 mol O2 / 5/2 mol H2O) = 0.52 mol O2

Note: starting moles of C2H5 = 0.16 mol C2H5

c) three points

MM stands for molar mass.

DT = (Kf (g/MM)) / kg of solvent

0.5 °:C = ((4.68 °:C kg mol¯:1) x (0.60 g / MM)) / 0.1 kg

MM = (4.68 x 0.60) / (0.5 x 0.1) = 56 or 6 x 101

an alternate solution for (c)

molality = 0.5 °:C / (4.68 0.5 °:C/m) = 0.107 m

mol solute =( 0.107 mol / kg solvent) x 0.100 kg solvent = 0.0107 mol

MM = 0.60 g / 0.0107 mol = 56 or 6 x 101

d) one point

(56 g/mol of cmpd) / (29 g/mol of empirical formula) = 1.9 empirical formula per mol

6 x 101 / 29 = 2.1

empirical formula times 2 equals molecular formula = C4H10

a) four points

rate = k [ClO2] [F2]

one point - rate equation form, k
one point - F2 order
two points - ClO2 order

b) two points

k = rate / ([ClO2] [F2])

= 2.4 x 10¯3 mol L¯1 sec¯1 / ((0.010 mol/L) (0.10 mol/L))

= 2.4 L mol¯1 sec¯1

one point - value consistent with equation in (a)
one point - units consistent with equation in (a)

c) one point

2 ClO2 + F2 ---> 2 ClO2F

- d[F2] / dt = 1/2 (d[ClO2F] / dt)

= 1/2 (9.6 x 10¯3)

= 4.8 x 10¯3 mol L¯1 sec¯1

d) two points

mechanism I

defense:

slow step is first order
three equations add to proper stoichiometry

Note: if ClO2 order in rate equation of part (a) is zero, mechanism II must be chosen to obtain credit.

a) Al2O3 + OH¯ ---> Al(OH)4¯

Al2O3 + H2O ---> Al(OH)3

(Personal note by John Park: I think the H2O in the second equation above comes from the fact that the NaOH concentration was not specified in the original problem. For example, suppose [OH¯] were 10¯12 M. Then the second equation becomes the predominate one.)

b) CaO + SO3 ---> CaSO4

c) H+ + OH¯ ---> H2O

d) Ca + N2 ---> Ca3N2

e) CuS + O2 ---> Cu + SO2 (also CuO, Cu2O)

f) H+ + Cl¯ + MnO2 ---> Mn2+ + Cl2 + H2O (one pt. for either redox product, two pts. for all three products)

g) Zn2+ + NH3 ---> Zn(NH3)42+

Zn2+ + NH3 + H2O ---> Zn(OH)2 + NH4+

h) Cu2+ + SO42¯ + Ba2+ + OH¯ ---> Cu(OH)2 + BaSO4

A rare double precipitation.
Partial credit was allowed for some alternate solutions, e.g.

Cu2+ + OH¯ ---> Cu(OH)2

Ba2+ + SO42¯ ---> BaSO4

a) two points

DS will be negative. The system becomes more ordered as two gases form a solid.

b) two points

DH must be negative. For the reaction to be spontaneous, DG must be negative, so DH must be more negative than -TDS is positive.

c) two points

As T increases, -TDS increases. Since DS is negative, the positive -TDS term will eventually exceed DH (which is negative), making DG positive. (In the absence of this, DG = DH - TDS and general discussion of the effect of T and DS gets 1 point.)

d) two points

The equilibrium constant is 1. The system is at equilibrium at this temperature with an equal tendancy to go in either direction.

DG = 0 at equilibrium so K = 1 in DG = -RT ln K

(In the absence of these, DG = -RT ln K gets 1 point).

The above concludes the AP scoring standards published in 1991. The following is simply alternate ways of answering which the AP readers may or may not have given full credit to.

a) The amount of entropy goes down, DS is negative.

b) DG = DH - TDS. If DS is negative, then DH must also be negative to get a negative DG.

c) Let us say DG is positive when DH is positive and DS is positive. As T goes up - TDS becomes more negative until it makes DG (which equals DH - TDS) become negative.

d) At the temperature when the direction changes, the rate forward = the rate reverse. Since K = kf / kr, this equals 1.

a) two points

Mass of vaporized liquid (or liquid or substance)

two of three in parts a or b

atmosperic pressure
volume of flask
temperature of vapor (water)