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 freezingpoint 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?
3)
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.
(c) In experiment 2, what is the initial rate of decrease of [F2]?
(d) Which of the following reaction mechanisms is consistent with the rate law developed in (a)?
Justify your choice.
I.
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.
(c) Equal volumes of 0.1-molar sulfuric acid and 0.1-molar potassium hydroxide are mixed.
(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.
5)
BCl3(g) + NH3(g) <---> Cl3BNH3(s)
The reaction represented above is a reversible reaction.
(a) Predict the sign of the entropy change, S, 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,
H. 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.
6)
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?
(c) List the calculations necessary to determine the molecular mass.
(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
(c) F2 and Br2
(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
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[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.
Return to Questions
Return to Additional Materials Menu
1)
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
2)
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
OR
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.
T = (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
OR
6 x 101 / 29 = 2.1
empirical formula times 2 equals molecular formula = C4H10
3)
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.
4)
a) Al2O3 + OH¯ ---> Al(OH)4¯
OR
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+
OR
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
5)
a) two points
S will be negative. The system becomes more ordered as two gases form a solid.
b) two points
H must be negative. For the reaction to be spontaneous, G must be negative, so H must be
more negative than -TS is positive.
c) two points
As T increases, -TS increases. Since S is negative, the positive -TS term will eventually
exceed H (which is negative), making G positive. (In the absence of this, G = H - TS and
general discussion of the effect of T and S 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.
OR
G = 0 at equilibrium so K = 1 in G = -RT ln K
(In the absence of these, G = -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, S is negative.
b) G = H - TS. If S is negative, then H must also be negative to get a negative G.
c) Let us say G is positive when H is positive and S is positive. As T goes up - TS becomes
more negative until it makes G (which equals H - TS) become negative.
d) At the temperature when the direction changes, the rate forward = the rate reverse. Since K =
kf / kr, this equals 1.
6)
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)
b) three points
Procedure for:
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mass - the mass difference between flask + air and flask + vaporized liquid
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volume - volume of flask by filling with H2O and then using graduated cylinder for
measuring.
Flask containing liquid is heated until liquid disappears
c) one point
mass ÷ mole where mole is determined from PV = nRT
d) two points
Molar mass is too high
because the non-volatile inpurity contribute additional mass (but insignificant volume).
7)
a) two points
Cl¯ is more easily oxidized than H2O
H2O is more easily oxidized than Na+
no 2nd pt is awarded for H+ ---> H2 unless H2O is implied.
no 2nd pt for Na(s) + H2O ---> Na + OH¯ unless H2 is indicated.
b) two points
Fe2+ requires 2 Faraday / mol Fe (s) or 1Faraday ---> 1/ 2 mol Fe(s)
Fe3+ require 3 Faraday / mol Fe (s) or 1 Faraday ---> 1/3 mol Fe (s)
for equal numbers of Faraday (1/2 : 1/3 as 1.5 : 1) (Or inverse relationship is clear)
no 2nd point unless flow of e¯ to mass (moles) is clear and logically correct
c) two points
Le Châtlier's argument
if [Zn2+ ] goes down; reaction shifts right, i.e. cell potential goes up
if [Pb2+ ] goes down; reaction shifts left, i.e. cell potential goes down
OR
Nernst Equation argument
E = E° - RT ln Q with Q = [Zn2+ ] / [Pb2+ ]
if [Zn2+ ] goes down Q < 1, therefore E > E°
if [Pb2+ ] goes down Q > 1, therefore E < E°
reasoning must indicate correct usage of equation
d) two points
[Zn2+ ] / [Pb2+ ] does not change; regardless of values; i.e. E=E°
OR
[Zn2+ ] / [Pb2+ ] = 1 so ln Q = 0; i.e. E=E°
no pt is awarded for just stating concentrations are equal
ratio or proportion concept is required for 2nd point
8)
a) three points
MgCl2 is ionic and SiCl4 is covalent. The elctrostatic, interionic forces in MgCl2 are much
stronger than the intermolecular (dispersion) forces in SiCl4 and lead to a higher melting point.
Molten MgCl2 contains mobile ions that conduct electricity whereas molten SiCl4 is molecular,
not ionic, and has no conductivity.
b) two points
MgF2 has a higher melting point than MgCl2 because the smaller F¯ ions and smaller interionic
distances in MgF2 cause stronger forces and higher melting point.
c) one point
The bond length in Br2 is larger than in F2 because the Br atom is larger than the F atom.
d) two points
The bond length in N2 is less than in F2 because the N-N bond is triple and the F-F is single.
Triple bonds are stonger and therefore shorter than single bonds.
9)
a) two points
When nucleons are combined in nuclei, some of their mass is converted to energy (binding
energy) which is released and stabilizes the nucleus. (Key concepts: mass defect; binding energy)
b) two points
Alpha particles have a greater mass than beta particles. Thus their speed (penetrating potential) is
less. (Alternate explanation could be based on charge.)
c) two points
The neutron/proton ratio in Sr-90 and Cs-137 is too large and they emit beta particles (converting
neutrons to protons) to lower this ratio.
d) two points
Large amounts of energy are neded to initiate fusion reactions in order to overcome the repulsive
forces between the positively charged nuclei. Large amounts of energy are not required to cause
large nuclei to split.