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AP Chemistry
End of Year Review
Name:
Period:
Unit 2:
1971
Permanganate ion, MnO4-, oxidizes sulfite ions to sulfate ion. The manganese product depends upon the
pH of the reaction mixture. The mole ratio of oxidizing to reducing agent is two to five at pH 1, and is
two to one at pH 13. For each of these cases, write a balanced equation for the reaction, and indicate the
oxidation state of the manganese in the product containing manganese.
1981 B
A 1.2516 gram sample of a mixture of CaCO3 and Na2SO4 was analyzed by dissolving the sample and
completely precipitating the Ca2+ as CaC2O4. The CaC2O4 was dissolved in sulfuric acid and the resulting
H2C2O4 was titrated with a standard KMnO4 solution.
(a)
Write the balanced equation for the titration reaction, shown unbalanced below.
Indicate which substance is the oxidizing agent and which substance is the reducing agent.
(b)
The titration of the H2C2O4 obtained required 35.62 milliliters of 0.1092 molar MnO4- solution.
Calculate the number of moles of H2C2O4 that reacted with the MnO4-
(c)
Calculate the number of moles of CaCO3 in the original sample.
(d)
Calculate the percentage by weight of CaCO3 in the original sample.
Unit 3
1996 D (Required)
atmosphere pressure with the pure gases indicated.
(a) Which balloon contains the greatest mass of gas? Explain.
(b) Compare the average kinetic energies of the gas molecules in the balloons. Explain.
(c) Which balloon contains the gas that would be expected to deviate most from the behavior of an ideal
gas? Explain.
(d) Twelve hours after being filled, all the balloons have decreased in size. Predict which balloon will be
the smallest. Explain your reasoning.
AP Chemistry
End of Year Review
Name:
Period:
2003 B
A rigid 5.00 L cylinder contains 24.5 g of N2(g) and 28.0 g of O2(g)
(a) Calculate the total pressure, in atm, of the gas mixture in the cylinder at 298 K.
(b) The temperature of the gas mixture in the cylinder is decreased to 280 K. Calculate each of the
following.
(i) The mole fraction of N2(g) in the cylinder.
(ii) The partial pressure, in atm, of N2(g) in the cylinder.
(c) If the cylinder develops a pinhole-sized leak and some of the gaseous mixture escapes, would the
ratio
N2 (g )
in the cylinder increase, decrease, or remain the same? Justify your answer.
O2 ( g )
A different rigid 5.00 L cylinder contains 0.176 mol of NO(g) at 298 K. A 0.176 mol sample of O2(g) is
added to the cylinder, where a reaction occurs to produce NO2(g).
(d) Write the balanced equation for the reaction.
(e) Calculate the total pressure, in atm, in the cylinder at 298 K after the reaction is complete.
Unit 4
1988 D
An experiment is to be performed to determine the standard molar enthalpy of neutralization of a strong
acid by a strong base. Standard school laboratory equipment and a supply of standardized 1.00 molar HCl
and standardized 1.00 molar NaOH are available.
(a) What equipment would be needed?
(b) What measurements should be taken?
(c) Without performing calculations, describe how the resulting data should be used to obtain the
standard molar enthalpy of neutralization.
(d) When a class of students performed this experiment, the average of the results was -55.0 kilojoules
per mole. The accepted value for the standard molar enthalpy of neutralization of a strong acid by a
strong base is -57.7 kilojoules per mole. Propose two likely sources of experimental error that could
account for the result obtained by the class.
1995 B
Propane, C3H8, is a hydrocarbon that is commonly used as fuel for cooking.
(a) Write a balanced equation for the complete combustion of propane gas, which yields CO2(g) and
H2O(l).
AP Chemistry
End of Year Review
Name:
Period:
(b) Calculate the volume of air at 30C and 1.00 atmosphere that is needed to burn completely 10.0
grams of propane. Assume that air is 21.0 percent O2 by volume.
(c) The heat of combustion of propane is -2,220.1 kJ/mol. Calculate the heat of formation, Hf, of
propane given that Hf of H2O(l) = -285.3 kJ/mol and Hf of CO2(g) = -393.5 kJ/mol.
(d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00
kilograms of water (specific heat = 4.18 J/g.K), calculate the increase in temperature of water.
Unit 5
2007 Part A
1. Answer the following problems about gases.
(a) The average atomic mass of naturally occurring neon is 20.18 amu. There are two common isotopes
of naturally occurring neon as indicated in the table below.
Isotope
Mass (amu)
Ne-20
19.99
Ne-22
21.99
(i)
Using the information above, calculate the percent abundance of each isotope.
(ii)
Calculate the number of Ne-22 atoms in a 12.55 g sample of naturally occurring neon.
(b) A major line in the emission spectrum of neon corresponds to a frequency of 4.341014 s-1. Calculate
the wavelength, in nanometers, of light that corresponds to this line.
(c) In the upper atmosphere, ozone molecules decompose as they absorb ultraviolet (UV) radiation, as
shown by the equation below. Ozone serves to block harmful ultraviolet radiation that comes from the
Sun.
O3(g) UV
 O2(g) + O(g)
A molecule of O3(g) absorbs a photon with a frequency of 1.001015 s-1.
(i)
How much energy, in joules, does the O3(g) molecule absorb per photon?
(ii)
The minimum energy needed to break an oxygen-oxygen bond in ozone is 387 kJ mol-1.
Does a photon with a frequency of 1.001015 s-1 have enough energy to break this bond?
Support your answer with a calculation.
AP Chemistry
End of Year Review
Name:
Period:
2006 D
2. Suppose that a stable element with atomic number 119, symbol Q, has been discovered.
(a) Write the ground-state electron configuration for Q, showing only the valence-shell electrons.
(b) Would Q be a metal or a nonmetal? Explain in terms of electron configuration.
(c) On the basis of periodic trends, would Q have the largest atomic radius in its group or would it
have the smallest? Explain in terms of electronic structure.
(d) What would be the most likely charge of the Q ion in stable ionic compounds?
(e) Write a balanced equation that would represent the reaction of Q with water.
(f) Assume that Q reacts to form a carbonate compound.
(i)Write the formula for the compound formed between Q and the carbonate ion, CO32–.
(ii)Predict whether or not the compound would be soluble in water. Explain your reasoning.
1997 D
3. Answer each of the following questions regarding radioactivity.
(a) Write the nuclear equation for decay of 234
94 Pu by alpha emission.
(b) Account for the fact that the total mass of the products of the reaction in part (a) is slightly less
than that of the original 234
94 Pu .
(c) Describe, or trace, how , , and  rays each behave when they pass through an electric field.
Use the diagram below to illustrate your answer.
(d) Why is it not possible to eliminate the hazard of nuclear waste by the process of incineration?
AP Chemistry
End of Year Review
Name:
Period:
Unit 6
2000 D
1. Answer the following questions about the element selenium, Se (atomic number 34).
a) Samples of natural selenium contain six stable isotopes. In terms of atomic structure,
explain what these isotopes have in common, and how they differ.
b) Write the complete electron configuration (e.g., 1s2 2s2... etc.) for a selenium atom in the
ground state. Indicate the number of unpaired electrons in the ground-state atom, and
explain your reasoning.
c) In terms of atomic structure, explain why the first ionization energy of selenium is
i
less than that of bromine (atomic number 35), and
ii
greater than that of tellurium (atomic number 52).
Selenium reacts with fluorine to form SeF4. Draw the complete Lewis electron-dot structure for SeF4 and
sketch the 1997 D (Required)
2. Consider the molecules PF3 and PF5.
a) Draw the Lewis electron-dot structures for PF3 and PF5 and predict the molecular
geometry of each.
b) Is the PF3 molecule polar, or is it nonpolar? Explain.
c) On the basis of bonding principles, predict whether each of the following compounds
exists. In each case, explain your prediction.
(i)
NF5
(ii)
AsF5
d) molecular structure. Indicate whether the molecule is polar or nonpolar, and justify your
answer.
Unit 8
2005 B
1. Answer the following questions related to the kinetics of chemical reactions.
–

 IO–(aq) + Cl–(aq)
I–(aq) + ClO–(aq) OH
Iodide ion, I–, is oxidized to hypoiodite ion, IO–, by hypochlorite, ClO–, in basic solution according
to the equation above. Three initial-rate experiments were conducted; the results shown in the
following table.
Experiment
[I–]
(mol L–1)
[ClO–]
(mol L–1)
Initial Rate of
Formation of IO–
(mol L–1 s–1)
1
0.017
0.015
0.156
2
0.052
0.015
0.476
3
0.016
0.061
0.596
(a) Determine the order of the reaction with respect to each reactant listed below. Show your work.
(i) I–(aq)
(ii)
ClO–(aq)
AP Chemistry
End of Year Review
Name:
Period:
(b) For the reaction,
(i) write the rate law that is consistent with the calculations in part (a);
(ii) calculate the value of the specific rate constant, k, and specify units.
The catalyzed decomposition of hydrogen peroxide, H2O2(aq), is represented by the following
equation.
2 H2O2(aq) catalyst
 2 H2O(l) + O2(g)
The kinetics of the decomposition reaction were studied and the analysis of the results show that it is
a first-order reaction. Some of the experimental data are shown in the table below.
[H2O2]
(mol L–1)
Time
(minutes)
1.00
0.0
0.78
5.0
0.61
10.0
(c) During the analysis of the data, the graph below was produced.
(i) Label the vertical axis of the graph
(ii) What are the units of the rate constant, k, for
the decomposition of H2O2(aq) ?
(iii) On the graph, draw the line that represents
the plot of the uncatalyzed first-order
decomposition of 1.00 M H2O2(aq).
2007 part B, question #6
2.
The reaction between SO2(g) and O2(g) to form SO3(g) is represented below.
2 SO2(g) + O2(g)  2 SO3(g)
The reaction is exothermic. The reaction is slow at 25˚C; however, a catalyst will cause the reaction to
proceed faster.
(a) Using the axes provided, draw the complete potential-energy diagram for both the catalyzed and
uncatalyzed reactions. Clearly label the curve that represents the catalyzed reaction.
AP Chemistry
End of Year Review
(b) Predict how the ratio of the equilibrium pressures,
Name:
Period:
PSO2
PSO3
, would change when the temperature of the
uncatalyzed reaction mixture is increased. Justify your prediction.
(c) How would the presence of a catalyst affect the change in the ratio described in part (b)? Explain.
Unit 9
2000 A
1. 2 H2S(g)  2 H2(g) + S2(g)
When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of
H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K,
and 3.7210–2 mol of S2(g) is present at equilibrium.
(a) Write the expression for the equilibrium constant, Kc, for the decomposition reaction represented
above.
(b) Calculate the equilibrium concentration, in molL-1, of the following gases in the container at 483 K.
(i)
H2(g)
(ii) H2S(g)
(c) Calculate the value of the equilibrium constant, Kc, for the decomposition reaction at 483 K.
(d) Calculate the partial pressure of S2(g) in the container at equilibrium at 483 K.
1
(e) For the reaction H2(g) + 2 S2(g)  H2S(g) at 483 K, calculate the value of the equilibrium constant,
Kc.
1998 D
C(s) + H2O(g)  CO(g) + H2(g)
Hº = +131kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(g) at equilibrium.
State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same
after each of the following disturbances is applied to the original mixture. For each case, assume that all
other variables remain constant except for the given disturbance. Explain each answer with a short
statement.
(a) Additional H2(g) is added to the equilibrium mixture at constant volume.
(b) The temperature of the equilibrium mixture is increased at constant volume.
(c) The volume of the container is decreased at constant temperature.
(d) The graphite pellets are pulverized.
Unit 10
1986 D
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.
AP Chemistry
End of Year Review
Name:
Period:
(b) Arrange the examples above in the order of increasing acid strength.
2005 A
HC3H5O2(aq) ↔ C3H5O2–(aq) + H+(aq)
Ka = 1.34 10–5
Propanoic acid, HC3H5O2, ionizes in water according to the equation above.
(a) Write the equilibrium constant expression for the reaction.
(b) Calculate the pH of a 0.265 M solution of propanoic acid.
The methanoate ion, HCO2–(aq) reacts with water to form methanoic acid and hydroxide ion, as
shown in the following equation.
HCO2–(aq) + H2O (l) ↔ H2CO2(aq) + OH–(aq)
(d) Given that [OH–] is 4.18 10–6 M in a 0.309 M solution of sodium methanoate, calculate
each of the following.
(i) The value of Kb for the methanoate ion, HCO2–(aq)
(ii) The value of Ka for methanoic acid, HCO2H
(e) Which acid is stronger, propanoic acid or methanoic acid? Justify your answer.
Unit 11
1994 A
MgF2(s)  Mg2+(aq) + 2 F-(aq)
In a saturated solution of MgF2 at 18ºC, the concentration of Mg2+ is 1.2110-3 molar. The
equilibrium is represented by the equation above.
(a) Write the expression for the solubility-product constant, Ksp, and calculate its value at 18ºC.
(b) Calculate the equilibrium concentration of Mg2+ in 1.000 liter of saturated MgF2 solution at
18ºC to which 0.100 mole of solid KF has been added. The KF dissolves completely.
Assume the volume change is negligible.
(c) Predict whether a precipitate of MgF2 will form when 100.0 milliliters of a 3.0010-3-molar
Mg(NO3)2 solution is mixed with 200.0 milliliters of a 2.00l0-3-molar NaF solution at 18ºC.
Calculations to support your prediction must be shown.
(d) At 27ºC the concentration of Mg2+ in a saturated solution of MgF2 is 1.1710-3 molar. Is the
dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation
to support your conclusion.
1985 A
At 25ºC the solubility product constant, Ksp, for strontium sulfate, SrSO4, is 7.610-7. The
solubility product constant for strontium fluoride, SrF2, is 7.910-10.
(a) What is the molar solubility of SrSO4 in pure water at 25ºC?
(b) What is the molar solubility of SrF2 in pure water at 25ºC?
(c) An aqueous solution of Sr(NO3)2 is added slowly to 1.0 litre of a well-stirred solution
containing 0.020 mole F- and 0.10 mole SO42- at 25ºC. (You may assume that the added
Sr(NO3)2 solution does not materially affect the total volume of the system.)
AP Chemistry
End of Year Review
Name:
Period:
1. Which salt precipitates first?
2. What is the concentration of strontium ion, Sr2+, in the solution when the first precipitate
begins to form?
(d)
As more Sr(NO3)2 is added to the mixture in (c) a second precipitate begins to form. At
that stage, what percent of the anion of the first precipitate remains in solution?
Unit 12
2006 B
CO(g) +
1
2
O2(g)  CO2(g)
The combustion of carbon monoxide is represented by the equation above.
(a) Determine the value of the standard enthalpy change, ∆H˚rxn for the combustion of CO(g) at 298 K
using the following information.
C(s) +
1
O2(g)  CO(g)
2
∆H˚298 = –110.5 kJ mol-1
C(s) + O2(g)  CO2(g)
∆H˚298 = –393.5 kJ mol-1
(b) Determine the value of the standard entropy change, ∆S˚rxn, for the combustion of CO(g) at 298 K
using the information in the following table.
Substance
S˚298
(J mol-1 K-1)
CO(g)
197.7
CO2(g)
213.7
O2(g)
205.1
(c) Determine the standard free energy change, ∆G˚rxn, for the reaction at 298 K. Include units with your
answer.
(d) Is the reaction spontaneous under standard conditions at 298 K? Justify your answer.
(e) Calculate the value of the equilibrium constant, Keq, for the reaction at 298 K.
2007 part A, form B, question #1
A sample of solid U308 is placed in a rigid 1.500 L flask. Chlorine gas, Cl2(g), is added, and the
flask is heated to 862˚C. The equation for the reaction that takes place and the equilibriumconstant expression for the reaction are given below.
U308(s) + 3 Cl2(g),  3 UO2Cl2(g) + O2(g)
KP 
(PUO2 Cl2 )3 (PO2 )
(PCl2 )3
When the system is at equilibrium, the partial pressure of Cl2(g) is 1.007 atm and the partial
pressure of UO2Cl2(g) is 9.73410-4 atm
(a) Calculate the partial pressure of O2(g) at equilibrium at 862˚C.
(b) Calculate the value of the equilibrium constant, KP, for the system at 862˚C.
(c) Calculate the Gibbs free-energy change, ∆G˚, for the reaction at 862˚C.
AP Chemistry
End of Year Review
Name:
Period:
(d) State whether the entropy change, ∆S˚ for the reaction at 862˚C is positive, negative, or zero.
Justify your answer.
(e) State whether the enthalpy change, ∆H˚, for the reaction at 862˚C is positive, negative, or
zero. Justify your answer.
(f) After a certain period of time, 1.000 mol of O2(g) is added to the mixture in the flask. Does
the mass of U308(s) in the flask increase, decrease, or remain the same? Justify your answer.
Unit 13
1998 D
Answer the following questions regarding the electrochemical cell shown.
(a) Write the balanced net-ionic equation for the spontaneous reaction that occurs as the cell operates,
and determine the cell voltage.
(b) In which direction do anions flow in the salt bridge as the cell operates? Justify your answer.
(c) If 10.0 mL of 3.0-molar AgNO3 solution is added to the half-cell on the right, what will happen to
the cell voltage? Explain.
(d) If 1.0 gram of solid NaCl is added to each half-cell, what will happen to the cell voltage? Explain.
(e) If 20.0 mL of distilled water is added to both half-cells, the cell voltage decreases. Explain.
2004 D Required
AP Chemistry
End of Year Review
Name:
Period:
An electrochemical cell is constructed with an open switch, as shown in the diagram above. A strip of Sn
and a strip of unknown metal, X are used as electrodes. When the switch is closed, the mass of the Sn
electrode increases. The half-reactions are shown below.
Sn2+ (aq) + 2 e–  Sn(s)
E˚ = –0.14 V
X3+(aq) + 3 e–  X(s)
E˚ = ?
(a) In the diagram above, label the electrode that is the cathode. Justify your answer.
(b) In the diagram above, draw an arrow indicating the direction of electron flow in the external circuit
when the switch is closed.
(c) If the standard cell potential E˚cell is +0.60 V, what is the standard potential, in volts for the X3+/X
electrode?
(d) Identify metal X.
(e) Write balanced net-ionic equation for the overall chemical reaction occurring in the cell.
(f) In the cell, the concentration of Sn2+ is changed from 1.0 M to 0.50 M, and the concentration of X3+
is changed from 1.0 M to 0.10 M.
(i) Substitute all appropriate values for determining the cell potential, Ecell, into the Nernst
equation. (Do not do any calculations.)
(ii) On the basis of your response in (f) (i), will the cell potential be greater than, less than, or equal
to E˚cell? Justify your answer.
AP Chemistry
End of Year Review
Name:
Period:
Answers:
Unit 2:
1971
Answer:
2 MnO4- + 5 SO32- + 6 H+  2 Mn2+ + 3 H2O + 5 SO42- (oxidation state Mn = +2)
2 MnO4- + SO32- + 2 OH-  2 MnO42- + H2O + SO42- (oxidation state Mn = +6)
1981 B
Answer:
(a) 2 MnO4- + 5 H2C2O4 + 6 H+  2 Mn2+ + 10 CO2 + 8 H2O
oxidizing agent: MnO4- , reducing agent: H2C2O4
(b)
0.03562 L 
0.1092 mol 5 mol H 2C2O4
3

–  9.7210 mol H 2C2O 4
L
2 mol MnO 4
(c) moles of H2C2O4 = moles CaCO3, therefore, 9.7210-3 mol H2C2O4 = 9.7210-3 mol CaCO3
100.1 g CaCO3
9.72103 mol CaCO3 
1 mol CaCO3
100 77.7% CaCO3
1.2516 g sample
(d)
Unit 3:
1996 D (Required)
Answer:
(a) CO2; according to Avogadro’s Hypothesis, they all contain the same number of particles, therefore,
the heaviest molecule, CO2 (molar mass = 44), will have the greatest mass.
(b) all the same; at the same temperature all gases have the same kinetic energy.
(c) CO2; since they are all essentially non-polar, the largest intermolecular (London) force would be
greatest in the molecule/atom with the largest number of electrons.
(d) He; it has the smallest size and has the greatest particulate speed and, therefore, it’s the easiest to
penetrate the wall and effuse.
2003 B
Answer:
1mol
= 0.875 mol N2
28.0 g
1mol
28.0 g O2 x
= 0.875 mol O2
32.0 g
(a) 24.5 g N2 x
P=
L•atm
1.75mol   0.0821 mol•K
  298K 
nRT
=
5.00L
V
= 8.56 atm
0.875 mol N 2
= 0.500 mole fraction N2
1.75 mol mix
P P
P T (8.56atm)(280K)
(ii) 1  2 ; P2  1 2 
T1 T2
T1
298K
(b) (i)
AP Chemistry
End of Year Review
Name:
Period:
= 8.05 atm x mole fraction = 8.05 atm x 0.500 = 4.02 atm N2
(c) decrease; since N2 molecules are lighter than O2 they have a higher velocity and will escape more
frequently (Graham’s Law), decreasing the amount of N2 relative to O2
(d) 2 NO + O2 → 2 NO2
(e) all 0.176 mol of NO will react to produce 0.176 mol of NO2, only 1/2 of that amount of O2 will react,
leaving 0.088 mol of O2, therefore, 0.176 + 0.088 = 0.264 mol of gas is in the container.
L•atm
 0.264 mol   0.0821 mol•K
  298 K  = 1.29 atm
nRT
=
5.00L
V
P=
Unit 4:
1988 D
Answer:
(a) 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,
etc.).
(b) 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) Determination of head (evolved or absorbed): The sum of the volumes (or masses) of the
two solutions, and change in temperature and the specific heat of water are multiplied
together to determine the heat of solution for the sample used. q = (m)(cp)(T).
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) Experimental errors: heat loss to the calorimeter wall, to air, to the thermometer; incomplete
transfer of acid or 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, …
1995 B
Answer:
(a) C3H8 + 5 O2  3 CO2 + 4 H2O
(b) 10.0 g C3H8  1 mol C3H8/44.0 g  5 mol O2/1 mol C3H8) = 1.14 mol O2
VO
2

o
(c)
nR T
P

1.14 mol 0 . 0821
L a t m
m ol K
303 K 
1 . 00 atm
o
o
o
= 28.3 L O2 ; f(28.3 L,21.0%) = 135 L of air
o
H comb  H f (CO )  H f (H O ) H f (C H )  H f (O ) 
2
2
3 8
2
-2220.1 = [3(-393.5) + 4(-285.3)] - [X+ 0]
X = Hcomb) = -101.6 kJ/mol
(d) q = 30.0 g C3H8  1 mol/44.0 g  2220.1 kJ/1 mol = 1514 kJ
q = (m)(Cp)(T)
1514 kJ = (8.00 kg)(4.184 J/g.K)(T)
T = 45.2
AP Chemistry
End of Year Review
Name:
Period:
Unit 5:
2007
Answer:
3.(a) (i) let X = decimal percentage of Ne-20, then (1-X) = decimal percentage of Ne-22
19.99X + 21.99(1-X) = 20.18; X = 0.905 or 90.50% Ne-20 and 9.500% Ne-22
(ii) (12.55 g)(9.5%) 
(b) c = ;  =

c

1 mol Ne 6.02  1023 atoms
= 3.5571022 atoms

20.18 g
1 mol
3.0  10 8 m s-1 10 9 nm
= 690 nm

4.34  1014 s-1
1m
(c) (i) E = h = (6.6310-34 J s)(1.001015 s-1) = 6.6310-19 J
(ii) 6.6310-19 J 
6.02  10 23 1 kJ
 3 = 399 kJ; this is more than 387 kJ, so there is enough energy to
1 mol
10 J
break the bond.
2006 D
Answer:
(a) 8 s1
(b) metal. It has one valence electron that it easily looses, consistent with the other alkali metals in its
family.
(c) Q would have the largest atomic radius in its group. Every time another s-orbital is added to an
element, it creates a bigger electronic shell.
(d) +1
(e) 2 Q + 2 H2O → 2 Q+ + H2 + 2 OH–
(f) (i) Q2CO3
(ii) soluble in water. With very few exceptions, all alkali metal salts are soluble in water.
1997
Answer:
(a)
234
94
Pu  24 +
230
92
U
Due to a printing error, the student’s answer booklet had the Pu-239 isotope. Therefore, the
following is a valid response.
239
94
Pu  24 +
235
92
U
(b) This mass defect has been converted into energy. E = mc2
(c) An alpha particle,  or He nuclei, has a 2+ charge and would be attracted to the (-) side of the
electric field. A beta particle, , or electron, has a single negative charge and is attracted to the
positive side of the electric field, but since it is much lighter and faster than an alpha it would not be
as strongly deflected. Gamma, , rays are not charged and, therefore, not deflected by the electric
field.
AP Chemistry
End of Year Review
Name:
Period:
(d) The half-life of a radionuclide is independent of its environment. Incineration will neither accelerate
its decay nor render it non-radioactive. Half-life is a function of its nucleus, incineration is a function
of its electrons.
Unit 6:
2000 D
1. Answer:
a) all the isotopes have 34 protons but a different number of neutrons in the nucleus.
b) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4
2. 2 unpaired electrons. 4s [] 4p [][ ][ ] Hund’s rule indicates that each of the orbitals will
be filled with a single electron before it gets paired.
(c) (i) in Se, the single paired 4p electrons has 1 electron easily removed to create the 3 unpaired 4p
orbitals which is energetically favorable; in bromine, the removal of 1 electron still leaves a
paired 4p orbital.
3. (ii) the shielding effect is stronger in Te and makes it easier to remove a electron (lower
ionization energy).
F
··
·F·
· ·
· Se
·
·· F ··
··
··
·F ·
· ·
F
··
Se
F
·· F ··
··
F
4. (d)
see-saw shape
5. Because F is very electronegative and the molecule is asymmetric with respect to the fluorines,
this molecule is polar.
1997 D (Required)
1. Answer:
c) PF3 = tripod (pyramid); PF5 = trigonal bipyramid
2.
(b) polar; net dipole moment toward the non-symmetrical position of the fluorines.
(c) (i) NF5 - doesn’t exist, nitrogen can’t hybridize to form the dsp3 orbitals and is also too small to
accommodate 5 fluorine atoms around it.
(i) AsF5 - does exist, arsenic can hybridize to form the dsp3 orbitals and is large enough to
accommodate 5 fluorine atoms around it.
Unit 8:
2005 B
Answer:
(a) (i) comparing expt. 1 to expt. 2, while the hypochlorite concentration remains constant, the
0.052
3.06
iodide concentration is essentially tripled { 0.017 =
} and the initial rate is
1
0.476
3.05
essentially tripled { 0.156 =
}. This indicates a first order with respect to the
1
iodide ion.
AP Chemistry
End of Year Review
Name:
Period:
(ii) comparing expt. 1 to expt. 3, while the iodide concentration remains essentially constant
0.061
4.07
(a 2.7% drop), the hypochlorite concentration is essentially quadrupled { 0.015 = 1
}
0.596
3.82
and the initial rate is essentially quadrupled { 0.156 = 1
}. This indicates a first order
with respect to the hypochlorite ion.
OR
(i) from experiments 1 & 2
rate2
k[I–]2m[ClO–]2n
=
rate1
k[I–]1m[ClO–]1n
0.476
k(0.052)m(0.015)n
=
0.156
k(0.017)m(0.015)n
0.052m
3.05 =
= 3.1m, where m = 1
0.017m
(ii) from experiments 1 & 3
rate3
k[I–]3m[ClO–]3n
=
rate1
k[I–]1m[ClO–]1n
0.596
k(0.016)m(0.061)n
=
0.156
k(0.017)m(0.015)n
0.061n
3.82 = (0.94)
0.015n
4.06 = 4.1n, where n = 1
(b) (i) rate = k[I–] [ClO–]
rate
0.156mol L–1 s–1
(ii) k = –
= (0.017mol L–1)(0.015mol L–1) = 610 L mol –1 s –1
–
[I ] [ClO ]
(c) (i) vertical axis is “ln of [H2O2]”
(ii) units for k are min–1
uncatalyzed
(iii) ln [conc]
Time(minutes)
2007 part B, question #6 (repeated in bonding section)
Answer:
(a)
(b) T-shaped
AP Chemistry
End of Year Review
(c)
Name:
Period:
SO2 is a resonance structure that switches between the two
forms and “evens out” the bond length
(d) sp2
(e)
(f) Since the reaction is exothermic, an increase in temperature would cause a LeChâtelier shift
towards the endothermic process, i.e., the reverse direction. SO3 would decrease and SO2
would increase, so the ratio would get larger.
(g) the presence of a catalyst would increase the speed of the forward reaction as much as the
reverse reaction and there would be no change in the equilibrium concentrations, therefore,
no change in the ratio.
Unit 9:
2000 A Required
Answer:
[H2]2[S2]
(a) Kc =
[H2S]2
(b) (i)
(ii)
3.7210–2 mol S2 2 mol H2
 1 mol S = 5.9510–2 M H2
1.25 L
2
2 mol H 2S 
1 mol  

-2
 3.40 g H 2S  34.0 g  -  3.72  10 mol S2  1 mol S 


2
1.25 L
= 2.0510–2 M H2S
2
2  3.72  10 
5.952 102  

 1.25  = 0.251
(c) Kc =
2
0.02048
(d) PV=nRT = 1.18
(e) K’c =
1
Kc
= 2.00
1998 D
Answer
(a) CO will decrease. An increase of hydrogen gas molecule will increase the rate of the reverse reaction
which consumes CO. A LeChatelier Principle shift to the left.
AP Chemistry
End of Year Review
Name:
Period:
(b) CO will increase. Since the forward reaction is endothermic (a H > 0) an increase in temperature
will cause the forward reaction to increase its rate and produce more CO. A LeChatelier Principle
shift to the right.
(c) CO will decrease. A decrease in volume will result in an increase in pressure, the equilibrium will
shift to the side with fewer gas molecules to7104 decrease the pressure, , a shift to the left.
(d) CO will remain the same. Once at equilibrium, the size of the solid will affect neither the reaction
rates nor the equilibrium nor the concentrations of reactants or products.
Unit 10
2005 A Required
Answer:
[C3H 5O-2 ][H + ]
(a)
= Ka
[HC3H 5O2 ]
(b) let X be the amount of acid that ionizes, then
X = [C3H5O2–] = [H+]
0.265 – X = [HC3H5O2]
X2
= Ka = 1.34x10–5
0.265X
X= 0.00188 M = [H+]
[you can assume that 0.265 – X ≈ 0.265 in order to simplify your calculations]
pH = –log[H+] = 2.73
(d) (i) [HCO2] = [OH–] = 4.18x10–6 M
[HCO2–] = 0.309 – 4.18x10–6
Kb =
[H 2CO2 ][OH- ]
(4.18  10 6 )2
=
= 5.65x10–11
6
[HCO2 - ]
(0.309  4.18  10 )
Kw
1  1014
(ii) Ka =
=
= 1.77x10–4
Kb
5.65  1011
(e) methanoic acid is stronger;
the larger the Ka, the stronger the acid
OR
for monoprotic organic acids, the longer the carbon chain, the weaker the acid. Propanoic
has 3 carbons, whereas, methanoic has only 1.
1986 D
Answer:
(a) 1) As effective nuclear charge on central atom increases, acid strength increases. OR
As number of lone oxygen atoms (oxygen atoms not bonded to hydrogen) increases, acid
strength increases. OR
As electronegativity of central atom increases, acid strength increases.
2) Loss of H+ by a neutral acid molecule reduces acid strength. OR
AP Chemistry
End of Year Review
Name:
Period:
Ka of H2SO3 > Ka of HSO3–
(b) H3BO3 < HSO3– < H2SO3 < HClO3 < HClO4
H3BO3 or HSO3– weakest (must be together)
Unit 11
1994 A
Answer:
(a) Ksp = [Mg2+][F-]2 = (1.2110-3)(2.4210-3)2
= 7.0910-9
(b) X = concentration loss by Mg2+ ion
2X = concentration loss by F- ion
[Mg2+] = (1.2110-3 - X) M
[F-] = (0.100 + 2.4210-3 - 2X) M
since X is a small number then (0.100 + 2.4210-3 - 2X)  0.100
Ksp = 7.0910-9 = (1.2110-3 - X)(0.100)2
X = 1.209291410-3
[Mg2+]= 1.2110-3-1.2092910-3 = 7.0910-7M
(c) [Mg2+] = 3.0010-3M  100.0 mL/300.0 mL = 1.0010-3M
[F-] = 2.0010-3 M  200.0 mL/300.0 mL = 1.3310-3 M
trial Ksp = (1.0010-3)(1.3310-3)2 = 1.7810-9
trial Ksp < = 7.0910-9,  no ppt.
(d) @ 18ºC, 1.2110-3 M MgF2 dissolves
@ 27ºC, 1.1710-3 M MgF2 dissolves
MgF2  Mg2+ + 2 F- + heat
dissolving is exothermic; if heat is increased it forces the equilibrium to shift left (according
to LeChatelier’s Principle) and less MgF2 will dissolve.
1985 A
Answer:
(a) SrSO4(s)  Sr2+(aq) + SO42-(aq)
At equilibrium: [Sr2+] = X M = [SO42-]
X2 = Ksp = 7.610-7
X = 8.710-4 mol/L, solubility of SrSO4
(b) SrF2(s)  Sr2+(aq) + 2 F-(aq)
At equilibrium: [Sr2+] = X M = [F-] = 2X M
KSP = [Sr2+][F-]2 = (X)(2X)2 = 7.910-10
X = 5.810-4 mol/L, solubility of SrF2
(c) Solve for [Sr2+] required for precipitation of each salt.
0.020 mol
Ksp = [Sr2+][F-]2 = 7.910-10 = X 
; X = 2.010-6 M
1 mol
AP Chemistry
End of Year Review
Name:
Period:
0.010 mol
; y = 7.610-6 M
1.0 L
Since 2.010-6 M < 7.610-6 M, SrF2 must precipitate first.
When SrF2 precipitates, [Sr2+] = 2.010-6 M
(d) The second precipitate to form is SrSO4, which appears when [Sr2+] = 7.610-6 M (based on
calculations in Part c.)
When [Sr2+] = 7.610-6 M, [F-] is determined as follows:
Ksp = [Sr2+][F-]2 = 7.910-10 = (7.610-6)(z)2 = 7.910-10 ; z = 1.010-2 M
Ksp = [Sr2+][SO42-] = 7.610-7 = Y 
Unit 12
2006 B
Answer:
(a) ∆H˚rxn = H f ( prod )  H f (reactants) = (–393.5 kJ mol-1) – (–110.5 kJ mol-1 + 1/2(0)) = –283.0 kJ
(b) ∆S˚rxn = S prod  Sreactants = (213.7) – (197.7 + 1/2(205.1) J mol-1 K-1 = -86.6 J K-1
(c) ∆G˚ = ∆H˚ – T∆S˚ = (–283.0 kJ) – (298K)(-0.0866 J K-1) = –257.2 kJ
(d) spontaneous; any reaction in which the ∆G˚ < 0 is spontaneous
(e) Keq = e–(∆G/RT) = e–(–257208.1J/(8.31298) = 1.28  1045
2007 part A, form B, question #1
Answer:
(a) using the equation, the ratio is 3:1, UO2Cl2:O2; PO2 
(b)
KP 
4 3
PUO2 Cl2
3

9.734 104
= 3.24510-4 atm
3
4
(9.734 10 ) (3.245 10 )
= 2.93110-13
(1.007)3
(c) ∆G˚ = –RTlnK = -(8.31)(862 + 273) ln (2.93110-13) = 272 kJ
(d) positive; a solid (low entropy) and 3 gases are converting into a mixture of gases (high entropy)
(e) positive; ∆G˚ = ∆H˚ – T∆S˚; ∆H˚ = ∆G˚ + T∆S˚; from (c) and (d), ∆G˚ and ∆S˚ are both positive,
making ∆H˚ positive
(f) increase; adding oxygen gas will cause a LeChâtelier Principle shift to the left, producing more
reactants
Unit 13
1998 D
Answer
(a) 2 Ag+ + 2 e-  2 Ag
E = +0.80 v
2+
Cd – 2 e-  Cd
E = +0.40 v
2 Ag+ + Cd  2 Ag + Cd2+
E = +1.20v
(b) Anions flow into the cadmium half-cell. As the cell operates, Cd2+ cations increase in number and
need to be balanced by an equal number of anion charges from the salt bridge.
(c) Cell voltage will increase. An increase in silver ion concentration will result in faster forward
reaction and a higher cell potential.
AP Chemistry
End of Year Review
Name:
Period:
(d) Cell voltage will decrease. As the salt dissolves, the Cl – ion will cause the Ag+ ion to precipitate as
AgCl and decrease the [Ag+]. This will result in a slower forward reaction and a decrease in cell
potential. Since cadmium chloride is a soluble salt, it will not affect the cadmium half-cell.
0.05 92
(e) Ecell = 1.20v –
2
log
[Cd
2+
[Ag
+ 2
]
]
; while both concentrations are 1.0M, the cell potential is 1.20v.
0.05 92
But if each solution’s concentration is cut in half, then, Ecell = 1.20v –
2
log
[.5 ]
[.5 ]
2
= 1.19v
2004 D Required
Answer:
(a) tin electrode is the cathode; cathode is the site of reduction (gain in electrons) and will convert metal
ions into a metal.
(b) (see diagram)
(c) red: Sn2+ (aq) + 2 e–  Sn(s)
oxid: X(s) – 3 e–  X3+(aq)
red:
(d) Cr
X3+(aq) + 3 e–  X(s)
E˚ = –0.14 V
E˚ = +0.74 V
E˚cell = +0.60 V
E˚ = –0.74 V
(e) 3 Sn2+ + 2 Cr  3 Sn + 2 Cr3+
(f) (i) E˚cell = 0.60 –
(ii) greater; •
0.0592
(0.10)2
log
(
)
6
(0.50)3