1994
CHEMISTRY
Three hoursare allottedfor thisexamination:1 hourand30 minutesfor SectionI, whichconsistsof multiple-choice
questions,and 1 hourand30 minutesfor SectionII, whichconsistsof problemsandessayquestions.SectionI is printed
in thisexaminationbooklet;SectionII, in a separatebooklet.
Battery-operated
hand-heldcalculatorsmay be usedin bothsectionsof the examination.All calculatormemoriesmust
be clearedof bothprogramsanddata;no peripheraldevicessuchas magneticcardsor tapeswill be allowed.Calculators
may not be shared.
SECTION I
Time -
1 hourand30 minutes
Numberof questions
- 75
Percentof totalgrade- 45
This examinationcontains75 multiple-choicequestions.Therefore, pleasebe careful to fill in only the
ovalsthat are precededby numbers 1 through 75 on your answer sheet.
GeneralInstructions
DO NOT OPEN THIS BOOKLET UNTIL YOU ARE INSTRUCTED TO DO SO.
INDICATE ALL YOUR ANSWERS TO QUESTIONS IN SECTION I ON THE SEPARATE ANSWER SHEET.
No creditwill be givenfor anythingwrittenin thisexaminationbooklet,but you may usethebookletfor notesor
scratchwork.After you havedecidedwhichof the suggested
answersis best,COMPLETELY fill in the corresponding
oval on the answersheet.Give only oneanswerto eachquestion.If you changean answer,be surethattheprevious
markis erasedcompletely.
Example:
SampleAnswer
Chicagois a
(A)
(B)
(C)
(D)
(E)
state
city
country
continent
village
Many candidates
wonderwhetheror not to guesstheanswersto questions
aboutwhichthey are not certain.In this
sectionof theexamination,asa correctionfor haphazardguessing,
one-fourthof thenumberof questionsyou answer
incorrectlywill be subtracted
from the numberof questionsyou answercorrectly.It is improbable,therefore,thatmere
guessingwill improveyour scoresignificantly;it may evenlower your score,andit doestake time. If, however,you are
not sureof thecorrectanswerbut havesomeknowledgeof thequestionandare ableto eliminateoneor moreof the
answerchoicesas wrong,your chanceof gettingthe rightansweris improved,andit may be to your advantageto answer
sucha question.
Use your time effectively, workingasrapidlyas you canwithoutlosingaccuracy.Do not spendtoo muchtimeon
questionsthatare too difficult. Go on to otherquestions
andcomebackto the difficult oneslaterif you havetime. It is
not expectedthateveryonewill be ableto answerall the multiple-choicequestions.
Copyright 0 1994 by Educational Testing Service. All rights reserved.
Note: For all questionsinvolving solutionsand/or chemical equations,assumethat the systemis in pure water and at
G
temperatureunlessotherwisestated.
Part A
Directions: Each set of letteredchoicesbelow refers to the numberedquestionsor statementsimmediately following
it. Select the one letteredchoice that bestanswerseach questionor bestfits each statementand then fill in the
correspondingoval on the answersheet.A choice may be used once, more than once, or not at all in each set.
Questions5-7 refer to the phasediagrambelow of a
pure substance.
Questionsl-4
(4
Heisenberguncertaintyprinciple
(B) Pauli exclusionprinciple
K> Hund’s rule (principle of maximum multiplicity)
0) Shieldingeffect
03 Wave nature of matter
P
1. Can be usedto predict that a gaseouscarbon atom
in its groundstateis paramagnetic
(atm)
2. Explains the experimentalphenomenonof electron
diffraction
’ lb ’ io ’
’
do ’ iio’ i30-
T (“Q
3. Indicatesthat an atomic orbital can hold no more
than two electrons
(A)
(B)
(C)
(D)
(E)
4. Predictsthat it is impossibleto determinesimultaneouslythe exact positionand the exact velocity of
an electron
Sublimation
Condensation
Solvation
Fusion
Freezing
5. If the temperatureincreasesfrom 10” C to 60” C at a
constantpressureof 0.4 atmosphere,which of the
processesoccurs?
6. If the temperaturedecreasesfrom 110” C to 40” C at
a constantpressureof 1.1 atmospheres,which of
the processesoccurs?
7. If the pressureincreasesfrom 0.5 to 1.5 atmospheres
at a constanttemperatureof 50” C, which of the
processesoccurs?
q 11a
Questions8-10 refer to the following diatomic species.
(A)
(B)
(C)
0)
(E)
Liz
I%
N
02
FZ
Questions11-13
(4 Pb
(W Ca
cc>zn
0% As
(E) Na
8. Has the largestbond-dissociation
energy
11. Utilized as a coating to protect Fe from corrosion
9. Has a bond order of 2
12. Is addedto silicon to enhanceits propertiesas a
semiconductor
10. Contains1 sigma(CT)and 2 pi (7~)bonds
13. Utilized as a shieldfrom sourcesof radiation
Part B
Directions: Each of the questionsor incompletestatementsbelow is followed by five suggestedanswersor completions.
Select the one that is bestin each caseand then fill in the correspondingoval on the answersheet.
14. Which of the following is lower for a l.O-molar
aqueoussolutionof anysolutethan it is for pure
water?
(A) PH
(B)
(C)
(D)
(E)
Vapor pressure
Freezing point
Electrical conductivity
Absorptionof visible light
15 In a moleculein which the central atom exhibits
sp3d2 hybrid orbitals,the electronpairs are directed
16. Commercial vinegar was titrated with NaOH
solutionto determinethe contentof acetic acid,
HC2H302. For 20.0 milliliters of the vinegar,
26.7 milliliters of 0.600-molar NaOH solutionwas
required.What was the concentrationof acetic
acid in the vinegar if no other acid was present?
(A)
(B)
(C)
(D)
(E)
1.60 A4
0.800 M
0.600 M
0.450 A4
0.200 M
towardthe cornersof
(A)
(B)
(C)
(D)
(E)
a tetrahedron
a square-based
pyramid
a trigonal bipyramid
a square
an octahedron
17. Relatively slow ratesof chemical reactionare associated with which of the following?
(A)
(B)
(C)
(D)
(E)
The presenceof a catalyst
High temperature
High concentrationof reactants
Strong bondsin reactantmolecules
Low activationenergy
18.
2 HZ0 + 4 Mn04- + 3 C102- + 4 MnOz + 3 ClOh- + 4 OHWhich speciesactsas an oxidizing agentin the reactionrepresented
above?
(A)
(B)
(C)
(D)
(E)
Hz0
ClodClOzMn02
Mn04-
19. In which of the following compoundsis the massratio of chromium
to oxygen closestto 1.62 to 1.00 ?
(A)
(B)
(C)
(D)
(E)
20.
Cr-03
CrOr
CrO
Cr,O
CrzO,
. . . Ag’ + . . . ASH,(~) + . . . OH- + . . . Ag(s) + . . . H3As03(aq) + . . . Hz0
When the equationaboveis balancedwith lowestwhole-numbercoefficients,the
coefficient for OH- is
(A)
(B)
(C)
(D)
(E)
2
4
5
6
7
2 1. Correct statementsaboutalpha particlesinclude
which of the following?
I. They havea massnumberof 4 and a charge
of +2.
II. They are more penetratingthan beta particles.
III. They are helium nuclei.
(A)
(B)
(C)
(D)
(E)
I only
III only
I and II
I and III
II and III
HSO,- + HZ0 =+ H30+ + SO:-
22.
In the equilibrium representedabove,the species
that act as basesinclude which of the following?
I. HS04
II. HZ0
III. sod2(A)
(B)
(C)
(D)
(E)
II only
III only
I and II
I and III
II and III
Step 1: Ce4++ Mn2++ Ce3++ Mn3+
23.
Step 2: Ce4++ Mn3++ Ce3++ Mn4+
Step 3: Mn4++ Tl’ + T13++ Mn2+
The proposedstepsfor a catalyzedreaction
betweenCe4+and Tl’ are representedabove.The
productsof the overall catalyzedreactionare
(A)
(B)
(C)
(D)
(E)
Ce4+and Tl’
Ce3+and T13+
Ce3+and Mn3+
Ce3+and Mn4+
T13+and Mn2’
24. A sampleof 0.0100 mole of oxygen gasis confined
at 37” C and 0.216 atmosphere.What would be the
pressureof this sampleat 15” C and the same
volume?
(A)
(B)
(C)
(D)
(E)
0.0876 atm
0.175 atm
0.201 atm
0.233 atm
0.533 atm
25.
Hz(g) + $ Wg)
+ HzO@)
AH”
= -286 kJ
2 Na(s) + i O?(g) + NalO(s)
AH” = -414 kJ
Na(s)+$Ol(g)+~H~(g)+NaOH(s)
AH”=-425kJ
Basedon the informationabove,what is the standardenthalpy
changefor the following reaction?
NazO(s) + H,O(!?) + 2 NaOH(s)
(A) -1,125 kJ
(B) -978 kJ
(C) -722 kJ
(D) -150 kJ
(E)
+275 kJ
26. Which of the following actionswould be likely to
changethe boiling point of a sampleof a pure
liquid in an open container?
I. Placing it in a smallercontainer
II. Increasingthe numberof molesof the liquid in
the container
III. Moving the container and liquid to a higher
altitude
(A)
(B)
(C)
(D)
(E)
I only
II only
III only
II and III only
I, II, and III
27. Which of the following setsof quantumnumbers
(n, I, mp, ms) bestdescribesthe valenceelectron of
highestenergy in a ground-stategallium atom
(atomic number3 1) ?
(A) 4, 0, 0, ;
(B) 4, 0, 1, ;
(C) 4, 1, 1, ;
(D) 4, 1, 2,;
(E) 4, 2, 0, ;
28. Given that a solutionis 5 percentsucroseby mass,what
additionalinformation is necessaryto calculatethe
molarity of the solution?
I. The density of water
II. The density of the solution
III. The molar massof sucrose
(A)
(B)
(C)
(D)
(E)
I only
II only
III only
I andII1
II andIII
29. When an aqueoussolutionof NaOH is addedto
an aqueoussolutionof potassiumdichromate,
K&O,,
the dichromateion is convertedto
(A) CrO:(B)
Cfl2-
(C) Cr3+
(D) CrzO&)
(E) Cr(OH)&)
ReactionCoordinate
30. The energy diagram for the reactionX + Y + Z is
shownabove.The additionof a catalystto this reaction would causea changein which of the indicated energy differences?
(A)
(B)
(C)
(D)
(E)
1 only
II only
III only
I and II only
I, II, and III
q 17B
31.
H2C204+ 2 Hz0 ti 2 H30+ + C20$Oxalic acid, H2C204, is a diprotic acid with
K1 = 5.36 x lo-* and K2 = 5.3 x 10m5.For the
reactionabove,what is the equilibrium constant?
(A)
(B)
(C)
(D)
(E)
5.36 x lo-*
5.3 x 1O-5
2.8 x lOA
1.9 x 10-l’
1.9 x lo-l3
32. CH3CH20H boils at 78” C and CH30CH3 boils at
-24” C, althoughboth compoundshavethe same
composition.This difference in boiling pointsmay
be attributedto a difference in
(A)
(B)
(C)
(D)
(E)
molecularmass
density
specific heat
hydrogenbonding
heat of combustion
33. A hydrocarbongaswith an empirical formula CH2
has a density of 1.88 gramsper liter at 0” C and 1.00
atmosphere.A possibleformula for the hydrocarbonis
(B)
C2H4
CC>
C3He
CD)
Cd-b
(E)
Cd-&o
34.
CH3-CH2-CH2-CH2-CH3
X
CH3-CH2-CH2-CH2-OH
HO-CH2-CH2-CH2-OH
Y
z
Basedon conceptsof polarity and hydrogenbonding,which of the following sequences
correctly liststhe compoundsabovein the order of their increasingsolubility in water?
(A)
(B)
(C)
(D)
(E)
2
Y
Y
X
X
<
<
<
<
<
Y< X
2 <X
x < z
Z < Y
Y< Z
35. For which of the following processeswould OS
have a negativevalue?
I. 2 Fe203(s) -+ 4 Fe(s) + 3 O?(g)
II. Mg2++ 2 OH- + Mg(OH)&)
III.
(A)
(B)
(C)
(D)
(E)
H?(g) + GH&)
I. There must be enoughsamplein the tube
to cover the entire light path.
II. The instrumentmust be periodically reset
using a standard.
III. The solutionmustbe saturated.
+ GH&)
I only
I and II only
I and III only
II and III only
I, II, and III
Zn(s) + Cu2+4
36.
(A)
(B)
(C)
(D)
(E)
Zn*+ + Cu(s)
An electrolytic cell basedon the reactionrepresentedabovewas constructedfrom zinc and copper
half-cells. The observedvoltagewas foundto be
1.00 volt insteadof the standardcell potential,E”,
of 1.10 volts. Which of the following could correctly
accountfor this observation?
(A) The copperelectrodewas larger than the zinc
electrode.
(B) The Zn” electrolyte was Zn(N03)2, while the
Cu’+ electrolyte was CuS04.
(C) The Zn” solutionwas more concentratedthan
the Cu” solution.
(D) The solutionsin the half-cells had different
volumes.
(E) The salt bridge containedKC1 as the electrolyte.
37. A sampleof 3.30 gramsof an ideal gas at 15O.O”C
and 1.25 atmospherespressurehas a volume of
2.00 liters. What is the molar massof the gas?
The gasconstant,R, is 0.0821 (L atm)/(mol K).
l
(A)
(B)
(C)
(D)
(E)
0.0218 gram/mole
grams/mole
16.2
grams/mole
37.0
grams/mole
45.8
grams/mole
71.6
38. Concentrationsof colored substances
are commonly
measuredby meansof a spectrophotometer.
Which
of the following would ensurethat correct values
are obtainedfor the measuredabsorbance?
l
I only
II only
I and II only
II and III only
I, II, and III
39. Samplesof F2 gasand Xe gasare mixed in a
containerof fixed volume. The initial partial pressureof the F2 gasis 8.0 atmospheresand that of
the Xe gasis 1.7 atmospheres.When all of the Xe
gasreacted,forming a solid compound,the pressureof the unreactedF2 gas was 4.6 atmospheres.
The temperatureremainedconstant.What is the
formula of the compound?
(A)
(B)
(C)
(D)
(E)
XeF
XeF3
XeF4
XeF6
XeFs
IL
Vacuum
Closed-endManometer
42.
Mass of an empty container 3.0 grams
Mass of the containerplus
25.0 grams
the solid sample
Volume of the solid sample 11.0 cubic
centimeters
The data abovewere gatheredin order to determine
the density of an unknown solid. The density of the
sampleshouldbe reported as
(A) 0.5 g/cm3
(B) 0.50 g/cm3
40. The systemshownaboveis at equilibrium at 28” C.
At this temperature,the vapor pressureof water is
28 millimeters of mercury. The partial pressureof
O,(g) in the systemis
(A) 28 mm Hg
(B) 56 mm Hg
(C) 133 mm Hg
(D) 161 mm Hg
(E) 189 mm Hg
(C) 2.0 g/cm3
(D) 2.00 g/cm3
(E) 2.27 g/cm3
43. Which of the following pairs of compoundsare
isomers?
(A) CH3--Hz----
CHz--CH3 and CH3-CH-CH3
CH3
41. A strip of metallic scandium,SC, is placed in a
beaker containingconcentratednitric acid. A
brown gas rapidly forms, the scandiumdisappears,
and the resultingliquid is brown-yellow but
becomescolorlesswhen warmed. These observationsbest supportwhich of the following statements?
(B) CH3-FH-CH3
and CH3-y=CH2
CH3.
CH3
(C) CH3-0-CH3
::
and CH3-C-CH3
(A) Nitric acid is a strongacid.
m
In solutionscandiumnitrate is yellow and
scandiumchloride is colorless.
Nitric
acid reactswith metals to form
(0
hydrogen.
09 Scandiumreactswith nitric acid to form a
brown gas.
03 Scandiumand nitric acid react in mole proportionsof 1 to 3.
(D) CH3-OH
and CH3-CH2-OH
(E) CH4 and CH*=CH;!
44. Which of the following solutionshas the lowest
freezing point?
(A) 0.20 m CaH1206,glucose
(B) 0.20 m NI-&Br
(C) 0.20 m ZnS04
(D) 0.20 m KMn04
(E) 0.20 m MgCl2
45. A sampleof an ideal gasis cooled from 50.0’ C to
25.0” C in a sealedcontainerof constant
volume.
Which of the following valuesfor the gas will
decrease?
I. The averagemolecularmassof the gas
II. The averagedistancebetweenthe molecules
III. The averagespeedof the molecules
(A)
(B)
(C)
(D)
(E)
I only
II only
III only
I and III
II and III
49. The isomerizationof cyclopropaneto propyleneis
a first-order processwith a half-life of 19 minutes
at 500”C. The time it takesfor the partial pressure
of cyclopropaneto decreasefrom 1.0 atmosphere
to 0.125 atmosphereat 500”C is closestto
(A) 38 minutes
(B) 57 minutes
(C) 76 minutes
(D) 152 minutes
(E) 190 minutes
50, Which of the following acidscan be oxidized to
form a strongeracid?
46. Which of the following solidsdissolvesin water to
form a colorlesssolution?
(A)
(B)
(C)
(D)
(E)
Cc13
FeC13
coc12
CuC12
ZnCl?
47. Which of the following has the lowestconductivity?
(A)
(B)
(C)
(D)
(E)
@I
HNO3
(c>
H2co3
0%
H3BO3
09
H2so3
4 HCl(g) + 02(g) *
2 Cl,(g) + 2H&?)
Equal numbersof molesof HCl and O2 in a closed
systemare allowedto reach equilibrium as representedby the equationabove.Which of the
following mustbe true at equilibrium?
I. [HCI] mustbe lessthan [Cl,].
II.
PCls(g) + energy
III.
Some PC13and Cl2 are mixed in a containerat
200” C and the systemreachesequilibrium
accordingto the equationabove.Which of the
following causesan increasein the number of
molesof PCls presentat equilibrium?
(A)
(B)
(C)
(D)
(E)
I. Decreasingthe volume of the container
II. Raisingthe temperature
III. Adding a mole of He gas at constantvolume
(A)
(B)
(C)
(D)
(E)
H3PQ
51.
0.1 A4CuSO4
0.1 it4 KOH
0.1 M BaC&
0.1 MI-IF
0.1 M HNO3
PC&(g > + C12(g) @
48.
(A)
I only
II only
I and III only
II and III only
I, II, and III
121 II
[02] mustbe greater than [HCl].
[Cl,] must equal [H,O].
I only
II only
I and III only
II and III only
I, II, and III
59. When a 1.00~gramsampleof limestonewas
dissolvedin acid, 0.38 gram of CO* was generated.
If the rock containedno carbonateother than
CaC03, what was the percentof CaC03 by massin
the limestone?
63. What is the maximum massof copper that could be
plated out by electrolyzingaqueousCuC12for
16.0 hoursat a constantcurrentof 3.00 amperes?
(1 faraday = 96,500 coulombs)
(A) 28 grams
(B) 57 grams
(C) 64 grams
(D) 114 grams
(E) 128 grams
(A) 17%
(B) 51%
(C) 64%
(D) 86%
(E) 100%
60.
12(g)
+
3 Cl2(g)
+
2
ICl,(g)
According to the data in the table below, what is
the value of AH0 for the reactionrepresented
above?
Bond
AverageBond Energy
(kilojoules/mole)
I-I
Cl-Cl
I-Cl
149
239
208
64. At 25” C, a sampleof NH3 (molar mass 17 grams)
effusesat the rate of 0.050 mole per minute. Under
the sameconditions,which of the following gases
effusesat approximatelyone-half that rate?
(A)
(B)
(C)
(D)
(E)
O2 (molar mass32 grams)
He (molar mass4.0 grams)
CO2 (molar mass44 grams)
Cl, (molar mass71 grams)
C& (molar mass 16 grams)
65. Barium sulfate is LEAST solublein a O.Ol-molar
solutionof which of the following?
(A) -860 kJ
(B) -382 kJ
(C) +180 kJ
(D) +450 kJ
(E) +1,248 kJ
(A)
A12(S04)3
(W
(Nb)2S04
(C>
Na2S04
0)
NH3
(E) BaC12
61. A l-molar solutionof which of the following salts
has the highestpH ?
(A)
NflO3
@I
Na2C03
(C>
66. What is the pH of a 1.0 x 10-2-molarsolutionof
HCN ? (For HCN, Ka = 4.0 x lo-“.)
(A) 10
(B) Between 7 and 10
WC1
(C) 7
(D) Between 4 and 7
(E) 4
(D) NaHSO‘,
(E) Na2SQ
62. The electron-dotstructure(Lewis structure)for
which of the following moleculeswould havetwo
unsharedpairs of electronson the central atom?
(A)
H2S
(B)
NH3
Cl
cH4
(D) HCN
(E) COz
n 23 n
67. SubstancesX and Y that were in a solutionwere
separatedin the laboratory using the techniqueof
fractionalcrystallization.This fractionalcrystallization is possiblebecausesubstances
X and Y have
different
(A)
(B)
(C)
(D)
(E)
70. To determinethe molar massof a solid monoprotic acid, a studenttitrated a weighedsampleof
the acid with standardizedaqueousNaOH. Which
of the following could explain why the student
obtaineda molar massthat was too large?
boiling points
melting points
densities
crystalcolors
solubilities
I. Failure to rinse all acid from the weighingpaper
into the titration vessel
II. Addition of more water than was neededto
dissolvethe acid
III. Addition of somebasebeyondthe equivalence
point
68. Which of the following moleculeshas a dipole
momentof zero?
(A)
(B)
(C)
(D)
(E)
C6H6(benzene)
NO
SO1
NH3
HS
(A)
(B)
(C)
(D)
(E)
71. . . . Fe(OH)z + . . . O2 + . . . Hz0 + . . . Fe(OH)3
If 1 mole of O2 oxidizesFe(OH)2 accordingto the
reactionrepresentedabove,how many molesof
Fe(OH)j can be formed?
69. Correct proceduresfor a titration include which of
the following?
I. Draining a pipet by touchingthe tip to the side
of the containerusedfor the titration
II. Rinsingthe buret with distilledwaterjustbefore
filling it with the liquid to be titrated
III. Swirling the solutionfrequentlyduringthe titration
(A)
(B)
(C)
(D)
(E)
I only
III only
I and II only
II and III only
I, II, and III
(A)
(B)
(C)
(D)
(E)
I only
II only
I and III only
II and III only
I, II, and III
n 24 n
2
3
4
5
6
72. The nuclide 249
,,Cm is radioactiveand decaysby the
loss of one beta (p-) particle. The productnuclide is
74. A solutionof calcium hypochlorite,a common
additiveto swimming-poolwater, is
(A) basicbecauseof the hydrolysisof the OCl- ion
(B) basic becauseCa(OH)2 is a weak and insoluble
base
(C) neutral if the concentrationis kept below
0.1 molar
(D) acidic becauseof the hydrolysisof the Ca”
ions
(E) acidic becausethe acid HOC1 is formed
(A) ‘;;Pu
(B) “z:Arn
(C) ‘ZiCrn
(D) ‘;;&-I
73.
2 sol(g) + 02(g) +
2
SO3(g)
When 0.40 mole of SO?and 0.60 mole of 02 are
placed in an evacuatedl.OO-liter flask, the reaction
representedaboveoccurs.After the reactants
and the productreach equilibrium and the initial
temperatureis restored,the flask is found to contain
0.30 mole of S03. Basedon theseresults,the equilibrium constant,K,, for the reactionis
75. A direct-currentpower supplyof low voltage(less
than 10 volts) has lost the markingsthat indicate
which outputterminal is positiveand which is
negative.A chemistsuggeststhat the power supply
terminalsbe connectedto a pair of platinumelectrodes that dip into O.l-molar KI solution.Which
of the following correctly identifies the polaritiesof
the power supplyterminals?
(A) 20.
(B) 10.
(C) 6.7
(D) 2.0
(E) 1.2
(A) A gas will be evolvedonly at the positiveelectrode.
(B) A gas will be evolvedonly at the negativeelectrode.
(C) A brown color will appearin the solutionnear
the negativeelectrode.
(D) A metal will be depositedon the positiveelectrode.
(E) None of the methodsabovewill identify the
polaritiesof the power supplyterminals.
c
STOP
END OF SECTION I
IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS SECTION.
DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.
n 25 n
m REVISED QUANTITATIVE
ITEMS FROM THE 1994
CHEMISTRY EXAM
There were a total of 20 quantitative questionsin the
multiple-choice section of the 1994 AP Chemistry
Exam. Below are 10 quantitative questionsfrom this
group rewritten to conform to the new format being
introduced in 1996, in which calculators will not be
allowed for the multiple-choice questions.(The capital “R” following the questionnumber indicatesthat it
is a Revised question.) The remaining quantitative
questions from the 1994 exam are those for which
students would not typically need a calculator, and
therefore represent quantitative questionsthat would
appear exactly as they are now on the new format of
the exam.
24R. A sample of 0.010 mole of oxygen gas is confined
at 127°C and 0.80 atmosphere.What would be the
pressureof this sample at 27°C and the same
volume?
(A)
(B)
(C)
(D)
(E)
0.10 atm
0.20 atm
0.60 atm
0.80 atm
1.1 atm
25~ H,(g) + l/2 O,(g) -
H,O@)
AHO=_X
2 Na(s) + l/2 O,(g) - N$O(s)
AP=y
Na(s) + l/2 O,(g) + l/2 H,(g) - NaOH(s)
AH0 = z
Based on the information above, what is the
enthalpy change for the following reaction?
N%O(s) + H,O@) - 2 Na0I-W
(A)
(B)
(C)
(D)
(E)
16R. Commercial vinegar was titrated with NaOH solution
to determine the content of acetic acid, HC,H,O,.
For 20.0 milliliters of the vinegar, 32.0 milliliters of
0.500-molar NaOH solution was required. What was
the concentration of acetic acid in the vinegar if no
other acid was present?
(A)
(B)
(C)
(D)
(E)
1.60 M
0.800 M
0640M
0.600 M
0.400 M
19R. In which of the following compounds is the mass
ratio of chromium to oxygen closest to 1.6 to l.O?
(A)
(B)
(C)
(D)
(E)
CrO,
CrO,
CrO
Cr,O
Cr,O,
31R.
x+y+z
x+y-z
x+y-22
22-x-y
z-x-y
H,C,O, + 2 H,O = 2 H,O+ + C,O,2Oxalic acid, H&O,, is a diprotic acid with
X10-5.Whichofthe
K, =5 X lo-*andK,=5
following is equal to the equilibrium constant for the
reaction representedabove?
(A)
(B)
(C)
(D)
(E)
5 X lo-*
5 X10-’
2.5 x 10”
5 x1O-7
2.5 X10-*
37R. A sample of 3.0 grams of an ideal gas at 127°C and
1.O atmospherepressurehas a volume of 1.5 liters.
Which of the following expressionsis correct for the
molar mass of the gas?The ideal gas constant, R, is
0.08 (L= atm)/(moleeK).
(A)
(0.08)(400)
(3.0)( 1.O)(1.5)
(B)
(1.0)(1.5)
(3.0)(0.08)(400)
(C)
(0.08)( 1.O)(1.5)
(3.0)(400)
(D)
(3.0)(0.08)(400)
(1.0)(1.5)
(E)
(3.0)(0.08)( 1.5)
(1.0)(400)
59R. When a 1.25-gram sample of limestone was
dissolved in acid, 0.44 gram of CO, was generated.
If the rock contained no carbonate other than
CaCO,, what was the percent of CaCO, by mass in
the limestone?
(A)
(B)
(C)
(D)
(E)
35%
44%
67%
80%
100%
63R. Which of the following expressionsis correct for
the maximum mass of copper, in grams, that could
be plated out by electrolyzing aqueousCuCl, for
16 hours at a constant current of 3.0 amperes?
(1 faraday = 96,500 coulombs)
(A)
(16)(3,600)(3.0)(63.55)(2)
(96,500)
(B)
(16)(3,600)(3.0)(63.55)
(96,500)(2)
(C)
(16)(3,600)(3.0)(63.55)
(96,500)
(D)
(16)(60)(3.0)(96,500)(2)
(63.55)
(E)
(16)(60)(3.0)(96,500)
. (63.55)(2)
73R.
2 SO,(g) + Q,(g) = 2 SO,(g)
When 0.40 mole of SO, and 0.60 mole of 0, are
placed in an evacuated l.OO-liter flask, the reaction
representedabove occurs.After the reactants and the
product reach equilibrium and the initial temperature
is restored, the flask is found to contain 0.30 mole of
SO,. Based on these results, the expression for the
equilibrium constant, K,, of the reaction is
(A)
(0.45)(0. 1o)2
(B)
60R.
I,(g) + 3 Cl,(g) -
(C)
(D)
Average Bond Energy
(kilojoules/mole)
I-I
150
Cl - Cl
240
I - Cl
210
(o.30)2
(0.60)(0.40)2
2 ICl,(g)
According to the data in the table below, what is the
value of AHo for the reaction representedabove?
Bond
(o.30)2
(2 x 0.30)
(0.45)(2 X 0.10)
(0.30)
(0.45)(0.10)
(E)
(0.30)
(0.60)(0.40)
I
Answers to
Revised Questions
I
(A)
(B)
(C)
(D)
(E)
- 870kJ
-390kJ
+ 180 kJ
+45OkJ
+1,260 kJ
I
I
16R.
19R.
24R.
25R.
31R.
37R.
59R.
60R.
63R.
73R.
B
B
C
D
C
D
D
B
B
A
1994
CHEMISTRY
SECTION II
Time-l
hour and 30 minutes
Percentof total grade-55
PartsA, B, and C: Suggestedtime-50 minutes
Part D: Suggestedtime--40 minutes
General Instructions
The suggestedtimeswill not be announced,and you may proceedfreely from one questionto the next. Do not
spendtoo long on any one problem.
Pagescontaininga periodic table and the electrochemicalseriesare printedon the greeninsertand in the pink
essaybooklet for your use.
You may write your answerswith either a pen or a pencil. Be sureto write CLEARLY andLEGIBLY. If you
make an error, you may savetime by crossingit out rather thantrying to eraseit.
When you are told to begin, open your booklet,carefully tear out the greeninsert, and start work. The questions
are also printed in your essaybooklet,but it may be easierto work from the insert when answeringquestions.
Writeall v0ur answers in the Dink essav booklet.Number vour answersas the auestionsare numberedin the
examinationbook.
n 29 n
INFORMATION
THIS SECTION
IN THE FOLLOWING
TABLES MAY BE USEFUL
OF THE EXAMINATION.
IN ANSWERING
THE QUESTIONS
R = 8.31 joules (mole - K) = 0.0821 liter-atm/(mole - K)
= 62.4 liter-mm Hg/(mole - K) = 1.99 calories/(mole - K)
= 8.3 1 (volt)(coulombs)/(mole - K)
1 faraday ( 3 ) = 96,500 coulombs = 23.060 calories/volt = 96,500 joules/volt
1 caloric = 4.184 joules
1 electron volt/atom = 23.1 kilocalories/mole = 96.5 kilojoules/mole
Speed of light in vacuum = 2.998 x lo8 m/set
Universal gas constant:
In, = 2.303 log,,
Planck’s constant It = 6.63 x 10 -34joule * set
Boltzmann’s constant k = 1.38 x 10 “joule/K
Avogadro’s number = 6.022 x 1013molecules/mole
0.059 1
In Q = ~
log Q
At 25” C, RT
n
n3
STANDARD
REDUCTION
Li.
POTENTIALS,
E”, IN WATER SOLUTION
+ e
AT 25” C (in V)
Li(s)
- 3.05
Cs’ + e
K++e
Cs(s)
K(s)
- 2.92
- 2.92
Rb‘
Rb(s)
- 2.92
Ba(s)
Ws)
- 2.90
Q(s)
- 2.87
-2.71
fe
Ba2+ + 2e
Sr“
+ 2e
Ca”
Na’
+ 2e
+ em
Mg”
+ 2e
Na(s)
Mg(s)
Be’ ’ + 2 e
Be(s)
-
Al’
Al(s)
- 1.66
+ 3e
- 2.89
- 2.37
1.I0
+ 2e
Mn(s)
- 1.18
Zn” + 2 e
Cr’ ’ + ?e
Zn(s)
-076
- 0.74
Mn”
Fe”
Cr-”
Cr(s)
Fe(s)
Cr’ ’
+ 2e
+ e
Cd”
+ 2e
Tl’
+ e
- 0.44
-0.41
Cd(s)
- 0.40
Tl(s)
- 0.34
- 0.28
Co”
+ 2e
Co(s)
Ni’.
Sn’
+ 2e
+ 2e
Ni(s)
- 0.25
Sri(s)
Pb(s)
-0.14
’
Pb”
2H’
+ 2e
+2e
Hz(g)
H$
S(s) + 2 H ’ + 2 e
Sn4 ’ + 2e
Cu”
+ e
Cu’+
+ 2e
Cu.
+e
-0.13
0.00
0.14
Sn’ +
Cu’
0.15
Cu(s)
Cu(s)
0.34
0.52
0.15
I:(s) + 2 e
Fe’+ + e
21
0.53
Fe”
0.77
Hg,’ ’ + 2e
Ag’ +e
Hg” + 2em
2 Hg(Q)
0.79
Ag(s)
0.80
Hg(Q)
Hgz’ +
2 Br-
0.85
2 H,O
I .23
1.36
2Hg”
+ 2e
Br,(Q) + 2e
O,(g) + 4H’
+ 4e
2 Cl
Au(s)
0.92
1.07
Cl,(g) + 2 e
Au’* + 3 e
Co’+ + e
co:+
I .82
FJg) + 2 e
2F
2.87
n 31 n
1.50
IN
CHEMISTRY
SECTION II
Time-l
hour and 30 minutes
The percentagesgiven for the partsrepresentthe scoreweightingsfor this sectionof the examination.Spend about
50 minuteson Parts A, B, and C combinedand about40 minuteson Part D.
THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS MUST BE SHOWN
CLEARLY. It is to your advantageto do this, sinceyou may obtain partial credit if you do and you will receive
little or no credit if you do not. Attention shouldbe paid to significantfigures.
Be sureto write your answersin the spaceprovided following each question.
Data necessaryfor the solutionof the problemsmay be found in the tableson the precedingpages.
Part A
(20 percent)
Solve the following problem.
1.
MgW)
+
Mg*+(aq) + 2 F-(aq)
In a saturatedsolutionof MgF7 at 18” C, the concentrationof Mg*+ is 1.21 x 10m3
molar. The equilibrium is
representedby the equationabove.
(4 Write the expressionfor the solubility-productconstant,Ksp, and calculateits value at 18” C.
(b) Calculatethe equilibrium concentrationof Mg” in 1.OOOliter of saturatedMgF2 solutionat 18” C to
which 0.100 mole of solid KF has been added.The KF dissolvescompletely.Assumethe volume change
is negligible.
x 10-3-molarMg(N03)2
solutionis mixed with 200.0 milliliters of a 2.00 x 10-3-molarNaF solutionat 18” C. Calculationsto support
your predictionmust be shown.
(4 Predict whether a precipitateof MgF2 will form when 100.0 milliliters of a 3.00
w
At 27” C the concentrationof Mg” in a saturatedsolutionof MgF2 is 1.17 x 10T3molar. Is the dissolvingof
MgF2 in water an endothermicor an exothermicprocess?Give an explanationto supportyour conclusion.
N 32 n
Part B
(20 percent)
Solve EITHER problem 2 OR problem 3 in this part. (A secondproblem will not be scored.)
2
2 NO(g) + 2 Hz(g) + N(g) + 2 I-W(g)
Experimentswere conductedto studythe rate of the reactionrepresentedby the equationabove.Initial concentrations and ratesof reactionare given in the table below.
Initial Concentration
(mom)
Experiment
[NOI
1
0.0060
0.0060
0.0010
0.0020
2
3
4
B321
Initial Rate of Formation
of N2
(mol/L*min)
0.0010
0.0020
0.0060
0.0060
1.8 x
3.6 x
0.30 x
1.2 x
1O-4
1O-4
1o-4
1o-4
(a> (i) Determine the order for each of the reactants,NO and HZ, from the data given and show your reasoning.
(ii) Write the overall rate law for the reaction.
(b) Calculatethe value of the rate constant, k, for the reaction.Include units.
w
For experiment2, calculatethe concentrationof NO remainingwhen exactly one-half of the original amount
of H2 had been consumed.
Cd)The following sequenceof elementarystepsis a proposedmechanismfor the reaction.
I.
II.
III.
NO + NO+=N202
N202 + H2 + H20 + N20
N20 + H2 + N2 + H20
Basedon the data presented,which of the aboveis the rate-determiningstep?Show that the mechanismis
consistentwith
(i) the observedrate law for the reaction,and
(ii) the overall stoichiometryof the reaction.
q 33 n
rr
Gas Sample
Water
3. A studentcollecteda sampleof hydrogengasby the displacementof water as shownby the diagram above.The
relevantdata are given in the following table.
GAS SAMPLE DATA
I
Volume of sample
90.0 mL
Temperature
25” c
Atmospheric
Pressure
Equilibrium
Vapor Pressure
of Hz0 (25” C)
745 mm Hg I
23.8 mm Hg
(a) Calculatethe numberof molesof hydrogengascollected.
(b) Calculatethe numberof moleculesof water vapor in the sampleof gas.
(c) Calculatethe ratio of the averagespeedof the hydrogenmoleculesto the averagespeedof the water vapor
moleculesin the sample.
(d) Which of the two gases,H2 or H20, deviatesmore from ideal behavior?Explain your answer.
n 34 n
Part C
( 15 percent)
4. Answer FIVE of the eight optionsin this part. (Answersto more than five optionswill not be scored.)
Give the formulas to showthe reactantsand the productsfor FIVE of the following chemical reactions.Each of
the reactionsoccursin aqueoussolutionunlessotherwiseindicated.Representsubstances
in solutionas ions if the
substanceis extensivelyionized. Omit formulas for any ions or moleculesthat are unchangedby the reaction.In
all casesa reactionoccurs.You need not balance.
Example: A strip of magnesiumis addedto a solutionof silver nitrate.
Mg + Ag’ -+ Mg*+ + Ag
(4 Excesssodiumcyanidesolutionis addedto a solutionof silver nitrate.
0-9 Solutionsof manganese(I1)sulfate and ammonium sulfide are mixed.
(4 Phosphorus(V)oxide powder is sprinkledover distilledwater.
Cd)Solid ammoniumcarbonateis heated.
Cd Carbon dioxide gasis bubbledthrougha concentratedsolutionof potassiumhydroxide.
(0 A concentratedsolutionof hydrochloricacid is addedto solid potassiumpermanganate.
(g) A small piece of sodiummetal is addedto distilledwater.
(h) A solutionof potassiumdichromateis addedto an acidified solutionof iron(I1) chloride.
Part D
(45 percent)
Spendabout40 minuteson this part of the examination.Answeringthesequestionsprovidesan opportunity to
demonstrateyour ability to presentyour material in logical, coherent,and convincingEnglish.Your responseswill be
judged on the basisof accuracyand importanceof the detail cited and on the appropriateness
of the descriptive
material used.Specific answersare preferableto broad, diffuse responses.Illustrativeexamplesand equationsmay be
helpful.
ANSWER THE FOLLOWING ESSAY QUESTION.
5 Discussthe following phenomenain terms of the chemical and physicalpropertiesof the substances
involvedand
generalprinciplesof chemical and physicalchange.
(a) As the systemshownon the right
approachesequilibrium, what change
occursto the volumeof water in
beaker A ? What happensto the
concentrationof the sugarsolution
in beaker B ? Explain why these
changesoccur.
(b) A bell jar connectedto a vacuumpump
is shownon the right. As the air
pressureunder the bell jar decreases,
what behaviorof water in the beaker
will be observed?Explain why this occurs.
Pure
H2O
To
Vacuum
Pump
(c) What will be observedon the surfaces
of zinc and silver stripsshortly after
they are placed in separatesolutionsof
CuS04, as shownon the right? Account for
theseobservations.
I, in Water
(d) A water solutionof I2 is shakenwith an
equal volume of a nonpolarsolventsuch
as TTE (trichlorotrifluoroethane).Describe
the appearanceof this systemafter shaking.
(A diagrammay be helpful.) Account for this
observation.
n 36 n
SELECT TWO OF THE FOUR ESSAY QUESTIONS, NUMBERED 6 THROUGH 9.
(Additional essayswill not be scored.)
6.
2 HIS(g) + SO*(g) +
At 298 K, the standardenthalpychange,AH’,
3 S(s) + 2 HzO(g)
for the reactionrepresentedaboveis -145 kilojoules.
64 Predict the sign of the standardentropy change,ASo, for the reaction.Explain the basisfor your prediction.
What change,if any, would occur in
(b) At 298 K, the forward reaction(i.e., toward the right) is spontaneous.
the value of AG” for this reactionas the temperatureis increased?Explain your reasoningusing thermodynamic principles.
(4 What change,if any, would occur in the value of the equilibrium constant,Keq, for the situationdescribed
in (b)? Explain your reasoning.
can be predicted.Write the
(4 The absolutetemperatureat which the forward reactionbecomesnonspontaneous
equationthat is usedto make the prediction.Why doesthis equationpredict only an approximatevalue for
the temperature?
7. A chemical reactionoccurswhen 100. milliliters of 0.200-molar HCl is addeddropwiseto 100. milliliters of
O.lOO-molarNa3P04 solution.
Write the two net ionic equationsfor the formation of the major products.
Identify the speciesthat acts as both a Bronstedacid and as a Bronstedbasein the equationsin (a). Draw the
Lewis electron-dotdiagram for this species.
Sketcha graph using the axes provided, showingthe shapeof the titration curve that resultswhen 100. milliliters
of the WC1solutionis addedslowly from a buret to the Na3P04 solution.Account for the shapeof the curve.
0
mL HCl
Write the equationfor the reactionthat occursif a few additionalmilliliters of the HCl solutionare addedto
the solutionresultingfrom the titration in (c).
n 38 n
8. For each of the following, use appropriatechemical principlesto explain the observation.
(a) Sodium chloride may be spreadon an icy sidewalkin order to melt the ice; equimolaramountsof calcium
chloride are even more effective.
(b) At room temperature,NH3 is a gas and HZ0 is a liquid, even thoughNH3 has a molar massof 17 gramsand
Hz0 has a molar massof 18 grams.
(c) C (graphite) is usedas a lubricant,whereasC (diamond) is usedas an abrasive.
Cd)Pouring vinegar onto the white residueinside a kettle usedfor boiling water resultsin a fizzing/bubbling
phenomenon.
n 39 n
9. Use principlesof atomic structureand/or chemicalbondingto answereach of the following.
(a) The radiusof the Ca atom is 0.197 nanometer;the radiusof the Ca2+ion is 0.099 nanometer.Accountfor
this difference.
(b) The lattice energy of CaO(s) is -3,460 kilojoulesper mole; the lattice energy for M20(s) is -2,240 kilojoules
per mole. Account for this difference.
Ionization Energy
(kJ/mol)
First
Second
K
419
3,050
Ca
590
1,140
(c) Explain the difference betweenCa and K in regardto
(i) their first ionizationenergies,
(ii) their secondionizationenergies.
(d) The first ionizationenergy of Mg is 738 kilojoulesper mole and that of Al is 578 kilojoulesper mole.
Accountfor this difference.
END OF EXAMINATION
n 40 n
Chapter III
Answers to the
1994 AP Chemistry Examination
n SECTION I: MULTIPLE-CHOICE
answered an individual question in this section also
achieved a higher mean score on the test as a whole
than candidates who did not answer that question
correctly. An answer sheet gridded with the correct
responsesappearson the next page.
Listed below are the correct answersto the multiplechoice questionsand the percentageof AP candidates
who attempted each question and answered it correctly. As a general rule, candidates who correctly
Section I Answer Key and Percent Answering Correctly
c
Item
No.
1
2
3
Correct
Answer
Percent
Correct
Item
No.
Correct
Answer
Percent
Correct
Item
No.
Correct
Answer
Percent
Correct
C
E
32%
38%
44%
82%
73%
74%
66%
21%
47%
57%
52%
21%
78%
46%
50%
69%
82%
62%
83%
58%
58%
62%
71%
76%
63%
26
C
C
E
D
B
29
A
30
32
D
C
D
33
C
34
E
35
D
61%
48%
58%
36%
55%
39%
77%
52%
39%
54%
46%
81%
41%
38%
67%
58%
39%
55%
31%
44%
46%
38%
35%
49%
58%
51
27
28
29%
29%
55%
43%
35%
35%
46%
29%
29%
47%
21%
64%
24%
23%
21%
64%
25%
41%
36%
15%
33%
53%
24%
34%
16%
B
4
A
5
A
6
7
B
B
8
C
9
D
10
11
C
12
D
13
A
14
C
15
E
16
B
D
17
18
19
20
21
22
C
E
B
D
D
E
23
24
B
25
D
C
:
31
36
C
37
38
D
39
C
40
C
41
D
D
42
C
43
44
A
45
C
46
47
E
D
48
A
49
B
50
E
E
52
53
E
54
D
55
C
56
B
57
A
58
B
59
60
65
D
B
B
A
B
D
A
66
D
61
62
63
64
67
E
68
A
69
70
C
A
71
72
C
73
74
A
A
75
B
E
H SECTION II: FREE-RESPONSE
Report of the Chief Faculty Consultant
Robert W. Gleason
Middlebury College
Grading the Examination
The free-response section of the AP Chemistry
Examination is read and scoredby faculty consultants
- AP Chemistry teachersand college chemistry professors- who are under the direction of a chemistry
teacherdesignatedas the chief faculty consultant.The
faculty consultantsare from secondary schools and
colleges throughout the United States, and also from
Canada.The faculty consultantsdo not have accessto,
and therefore are not influenced by, the multiplechoice section of the examination, which is scored
separatelyby machine. Studentscoreson both partsof
the examination are combined and used by the chief
faculty consultant to determine the levels of student
performance on the AP 1 to 5 grading scale.
The Reading
In June the faculty consultantsmeet for six days on a
college campusto score the free-responsesectionsof
the AP Chemistry Examination. The chief faculty
consultantdivides the faculty consultantsinto groups,
each under the direction of a designatedfaculty consultantcalled a table leader. One or more table leaders
with their group of faculty consultantsis assignedto
score each free-responsequestion, depending on the
number of studentswho chose to answer it. The table
leaders train their groups to score their designated
question, and scoring of student papers commences
according to standardsdeveloped as describedbelow.
Each answer booklet is circulated among the various
groups until all the studentresponsesin that booklet
have been scored. The finished booklets are then
removed from the Reading site and the scores are
enteredinto computersand matchedwith the students’
multiple-choice section scores.Composite scoresare
calculated and theseand other data are provided to the
chief faculty consultantfor the grade-settingsession,
which occurs shortly after the Reading is over.
Developing Free-Response
Scoring Standards
Scoring standardsfor the free-responsequestionsare
a consideration throughout the development of the
examination. Members of the AP Chemistry Development Committee submit suggestedscoring standards
with each question that they write. Scoring standards
are discussedfurther when questionsare revised and
chosenby the committee to be included in an examination.At this stage,considerationis given to potential
difficulties that might interfere with the reliable scoring
of a question,and the scoringstandardmay be revised
accordingly. Prior to the Reading, the chief faculty
consultantgeneratesa draft of the scoringstandardfor
each of the nine questionsin the free-responsesection,
taking into considerationissuesraisedduring previous
reviews.The generalscoringguidefor the free-response
questionsis as follows:
Problems
9 pointseach
Chemical Reactions 3 pointseach (15 pointstotal)
Essays
8 points each
Two days before the Reading begins, the chief faculty consultantmeets with the table leaders to review
the draft standards,and the group reachesa consensus
on a possible standardfor each question. The table
leadersbreak into groupsto test the standardsagainst
a number of studentresponses.During this phase of
the process,the standardsmay be modified somewhat.
Meeting again as a whole group, the chief faculty
consultantand the table leadersreach anotherconsensuson the standards,after which each table leader is
assignedto a particular questionand is also assigneda
list of faculty consultantswith whom he or she will
score the questionduring the Reading.
On the first day of the Reading, the table leaders
train the faculty consultantsin applying the scoring
standardsto a setof samplestudentresponsesselected
for that purpose.During this process,the standardsare
refined and may be modified slightly once again.
After the group is proficient in applying the scoring
standards,the actual scoring of the student papers
begins. The final results of the rigorous standardsetting procedure described above are standardsthat
canbe appliedreliably not only to the commonmethods
of solution seen in the studentresponses,but also to
commonerrorsandto alternativeor unusualapproaches.
Becausevarious studentresponsesto a given question are scoredover a six-day period by more than one
faculty consultant,it is important to monitor the application of the standardsduring the Reading. This is
done by having a certain number of student papers
independently graded more than once, either by different faculty consultants,or by the sameconsultantat
a later time. The original scores are concealed from
subsequentfaculty consultants, and the two sets of
scoresare then compared. The checking of one consultant againstanother quickly identifies any remaining ambiguities that may exist in the standardsand
allows their further refinement, to help assurethat a
student’s score is independent of the person scoring
the paper. Checking a consultant against his or her
own work also helps assurethat a student’s score is
independentof what day or time the paper is scored.
Rarely is there a discrepancyof more than 1 point on
the scale. Other procedureshelp maintain consistent
scoring standards.After the Reading is underway, the
table leaders select and score another series of questions for yet another kind of consistencycheck. This
one does not compare the faculty consultants with
themselves,but rather it comparesthem with the other
faculty consultants in their group and individually
with the table leaders.
The philosophyof the faculty consultantsin scoring
the free-responsequestionsis to award credit for correct work. When questionsinvolve calculations,most
of the points awarded are given for setting up the
solution correctly rather than actually carrying out the
computation. Partial credit is awarded within each
part of a question,so studentsshouldbe encouragedto
show their work. Faculty consultantstry to determine
whether an incorrect answer to a previous part has
been correctly usedin a subsequentpart of a question.
Full credit for the latter part may be awarded if the
consultant can successfully trace the student’s work
to make that determination. Students should also be
encouragedto continue on to later parts of a question
if they get stuck at some point. Parts of a questionare
often independentof each other; even when they are
dependent, credit can be earned on later parts when
earlier answersare missing.Also, a student’s explanation of what he or shewould do, if possible,could earn
some credit. Finally, when final answersare numerical, studentsshouldpay attentionto significantfigures
since 1 point is deducted (once per problem) if the
number of significant figures in a student’s answer
differs by more than one from the appropriatenumber.
n FREE-RESPONSE QUESTIONS,
SCORING GUIDES, AND SAMPLE
STUDENT ANSWERS
On the pages that follow are a selection of student
responsesto each of the questionsthat made up the
free-response section of the 1994 Advanced Placement Examination in Chemistry. Also included are
the standardsthat were applied in the scoringprocess,
and an explanation of why each responsereceived the
score it did. For each question, two studentresponses
have been selected to illustrate a superior answer and
one of somewhatlower quality.
From our experience in reading Advanced Placement Examinations we know that no set of scoring
standardscan possiblyanticipatethe creativity of high
school studentsin developing solutions to problems.
Therefore, you should understandthat readers make
every possibleeffort to give credit for every response
that reflects an understandingof basic principles of
chemistry regardless of how far the approach used
deviates from what might be a more conventional
route developed in the standards.
In developing the standardsfor the 1994 AP Chemistry Exam, the chief faculty consultanthad the assistance of 11 faculty consultantswho served as table
leaders, and two test development specialists from
ETS. The group met for two days prior to the Reading.
After a draft set of standardswas established, the
application of the standard for each question was
testedwith about 100 papers.The standardswere then
reviewed in the light of that experience and either
revised accordingly or adopted.
If a studentmade an error in part (a) of a four-part
problem and the answer to part (a) was essential to
working the rest of the problem, the reader of the
paper was obliged to work through the solutionsto the
subsequentparts of the question with the erroneous
answer to part a. Thus, every effort was made to
reward the students with points for the appropriate
applicationof chemicalprinciples.Studentswere, however, penalized for mathematical errors and errors in
significant figures (exceeding one too many or one too
few) to the maximum extent of 1 point for an error of
each kind on any one problem.
n 44 n
Question 1
This required problem provided studentswith an opportunity to demonstratetheir understandingof an ionic
equilibrium and the common ion effect, to make a prediction basedon their calculations,and to relate solubility
data to thermodynamics.
ScoringStandards
Ksp -
(4
[Mg*+J[F-]*
(1.21
7.09
-
x 1.21
x 10’3)2
U pt.1
if number of significant
figures
in final
answer differs
more than one from the appropriate
number, 1 point is
deducted ONCE PER PROBLEM.
Note:
(W
KsP
7.09 x 10-9
IMg*+l
OK if
Note:
Cc>
(1 pt.>
x 10’3)(2
x 10-9
IQ*+] .-
-
[M$+](2x
-
[Mg*+](o
010)2
(7.09 x io+)/(lo-2)
7.09 x 10-7 M
0.102
is used for
100.0
x 3.00
I%*+1
[F”] :
200.0
x 2.00
WI
Q -
Ion Product
**
Correct
[F-j,
then Ksp - 6.76
300 . 0 x [Mg2+]
-
-
x 1O'3 M
1.00
x 1O-3
-
-
x loo3 M
1.33
300.0
-
[Mg*+][F']*
(1.00 x 10’3)(1.33
-
1.77
will
must be consistent
substitution
concentration
the first.
x [F-l
x lo-S)*
x 10'9
no precipitate
(1 pt.) for
substitution
2x << 0.100
x 103
-
Since Q < Ksp ,
Note: conclusion
+ O.lOO)*
by
x 10”
I**
form
with
and calculation
(1 pt.)
if both
concentrations
are
correct
(1 pt.>
(1 ptJ
Q value.
of the wrong
values earns the second point,
proper
but not
(d)
Solubility
of MgF, decreases
thus dissolution
temperature,
MgFz(s)
+
Mg2+
+
with increasing
process is exothermic
2 F-
+
Q (cr
(1 pt.>
H)
Reason:
EITHER
i)
Increased
temperature
puts a stress on the system
The system will
reduce the stress
(Le Chatelier).
by shifting
the equilibrium
in the endothermic
(left)
direction
(1 pt*)
0%
ii)
a data supported argument such as comparing ion
concentrations,
calculating
second Ksp and giving
proper interpretations.
n 46m
i
Comment:This paper earned all of the 9 possiblepoints. Although the studentdid not explicitly indicate that the
simplifying approximationwas being made in part b, it clearly was; and the substitutionand calculationwere made
correctly. The calculationsin part c were clear and correct and the conclusionbased on the relationship between
Q, and K,, was unambiguous.The analysisand conclusion in part d were straightforward and to the point.
Sample Student Response 2
Comment:In part a of this question,the studentlost a point for failing to squarethe concentrationof the fluoride
ion in the calculation. This was a common flaw in part a. Although the studenthad an incorrect value for K,, in
part a, he or sheearnedfull credit in part b for usingit correctly in the calculations.The studentlost a point in part
c for the mathematicserror made in calculating Q,. Although the analysisin part d is somewhatunusual,it is clear
the studentunderstandsthe implications of raising the temperatureon a systemin equilibrium in which the forward
reaction is exothermic, thus full credit was awarded for this part. The total score for this paper was 7.
Question 2
Studentsfrequently did parts a and b correctly (although many omitted units in part b). Credit was given in (ii) of
part a if the rate law given was consistentwith the kinetic orders of (i). The fact that the initial concentrationsof
the reactantsin part c were not equal led many studentsto miss this part. Many studentswho lost points in part c
simply failed to take advantage of the simple stoichiometry but tried to plug concentrationsinto the rate law.
Part d was difficult for many studentswho tried to show that the reaction stoichiometry was consistentwith the
rate-determining step. Many studentswho correctly chose Step II of the mechanismas the rate-determining step
simply statedthat because2N0 = N202, then [N202] = [NO]2 and lost a point as a result.
ScoringStandards
(a)
Doubling [HZ] while keeping
From exps. 1 and 2:
constant
doubles the rate,
therefore
the reaction
is first
order in [HZ].
(i)
Doubling
From exps. 3 and 4:
constant
quadruples
the rate,
is second order in [NO].
(ii)
Note:
Rate
-
k[Hz
From exp.
(c)
(1 pt.)
for order and
jus tification
(1 pt.)
for order and
jus tification
(1 pt.>
for (ii)
as long as rate
orders in (i).
expression
[Hz;:O]L
1:
k
-
Note:
[NO] while keeping
[Hz ]
therefore
the reaction
I WI2
full
credit
earned
is consistent
with
(b) k -
[NO]
same result
Stoichiometry:
(1.0
5.0
1.8 x 10s4 M/min
x 1O-3 M)(6.0 x 1O-3 M)2
X lo3 !I'*
from initial
NO:H2
is
rate
min'l
data
for
value
(2 pt.)
for
units
4 experiments
1:l
When 0.0010 mole of Hz had reacted,
it
reacted with 0.0010 mole of NO; thus
[NO] remaining
from all
(1 pt.)
- 0.0060
- 0.0010
must have
- 0.0050
M.
(1 pt.>
1
IN202
(d)
(i)
For I:
Keq
-
Rate
For II:
=
IN2021 Rate
Note:
w,
I [NZOZ
1
KeqlNOl'
(1 pt.1
WH21[W2
there must be some clear algebraic
showing that [N202] is proportional
Step II
is
(ii)
the
I:
II:
III:
I
-
INW
+ II
+ III:
rate-determining
NO
+
NO
step.
to
[NO]*.
(1 pt.1
N2O2
N2O2 +
H2
-
H,O
N,O
H,
-
N2
2 NO + 2 H2
-
N2+
+
manipulation
(NOT equal)
+
+
N20
H,O
2H20
(1 pt.>
SampleStudent Response1
Comment:In part a, the kinetic ordersderived are well justified and the rate law is consistentwith them. In part
b, the studentderives an expressionfor the rate constantfrom the rate law and even specifies the number of the
experiment from which he or she takesthe experimental data in evaluating k. In part c, the studentsystematically
tabulatesthe concentrationsof the reactantsin a fashion similar to that used in equilibrium problems and clearly
recognizesthe stoichiometricratio of hydrogen to nitric oxide. The studentearnsall 3 of the possiblepoints in part
d by choosing the secondstep of the mechanism as the rate-determining step, showing that its rate is consistent
with the rate law, and demonstratingthat the sumof the three stepsin the mechanismis equal to the overall reaction
stoichiometry. This was an excellent answer that received a perfect score of 9.
Sample Student Response 2
Comment:Although the studentexpressesthe rate law in part (ii) of a as a proportionality, the equation is written
correctly in part b and full credit is given in part a. The solutionto part b is clear and straightforward.The student
runs into trouble in part c by substitutingconcentrationsinto the rate expression(a common error) and failing to
note the 1:1 stoichiometry in the reaction. In part d, a point was deductedbecausethe studentdid not showthat the
mechanismwas consistentwith the reaction stoichiometry. The total score for this answer was 7.
Question3
Although the difficulty of this question was judged by the readersto be similar to that of question 2, significantly
lessthan half of the examineeschoseto work on this problem. Points were frequently lost in part a when students
failed to consider the water vapor in the collected gas. Failure to read part b carefully led many studentsto
calculate the number of molecules of hydrogen rather than water vapor. In part c credit was awarded for answers
derived from equating the average kinetic energies of the gas molecules or by calculating the root-mean-square
speedsof the molecules of the two gases.In part d many studentscited the difference in the massesof the two
molecules and lost credit for the explanation part of the question.
Scoring Standards
(a)
z2 -
-$
(721)(0.090)
(62.4)(298)
-
-
3.49
x low3 mol Hz
25°C 298 K
745 - 24 721 mm Hg
calculation
of moles of Hz
(23*8)(o*ogo)
(62.4) (298)
tb)
(1.15
(c)
x
_
(+
(1 pt.>
(1 pt.>
mol H 0
x
1023)
kinetic
)H,O
-
energies
-
(+&H2
(1 pt.1
2
’
10B4)(6.03
The average
1 15 x 10-4
(1 pt.>
6.92
x
1019 molecules
are equal,
H20
(1 pt.>
so
.
(1 pt.)
(1 pt.)
Note:
(d)
credit
also
H20 deviates
given
for
correct
more from ideal
use of
vrms
-
-
3RT
M
behavior.
(1 pt.>
Explanation:
EITHER
i)
The volume of the Hz0 molecule
than that of the Hz molecule
is
larger
W
ii)
The intermolecular
forces among Hz0 molecules
stronger
than those among Hz molecules
are
I
(1 pt.>
for
for
formula
calculatiorz
Sample Student Response 1
Comment:In part a, the conversionof Celsiustemperatureto Kelvins, the application of Dalton’s Law, and the use
of the ideal gasequation were all done correctly. In part b, the studentclearly recognized the relationshipbetween
the pressureof a gas and the number of moles presentand made the correct calculation usingNA. The solution to
part c is clear and straightforward.The analysisin part d wanderssomewhat,but credit was awarded becausethe
studentwas aware that the hydrogen bonding in water is an intermolecular force and that ideal gasesexperience no
intermolecular forces. This responsereceived a score of 9 points.
n 55m
Sample Student Response 2
Comment:The correct answerswere obtained in both parts a and b in routine fashion. Although the answer in
part b was expressedin too many significant figures, no points were deductedbecausethe number of significant
figures given exceeded the appropriate number by only one. The student earned no credit in part c since the
responsereflected no recognition of the relationshipbetween molecular speedand the squareroot of the molecular
mass nor the inverse relationship between molecular mass and speed. In part d, the studentcorrectly identifies
water as the gasthat deviatesmore from ideal behavior, but, like many students,attributedthat greater deviation to
water’s larger molar mass.The total score was 6 points.
Question4
The ability to write the formulas of chemical speciesfrom their namesand knowing the reactionsthat commonly
Historically,the
encounteredchemical systemsundergoapparentlycontinuesto confoundAP Chemistrystudents.
performanceof studentson the required“equation”questionhas reflectedtheir lack of exposureto enough
descriptivechemistry.The performancesof studentson this year’s requiredequationquestion,however,were
significantlyimprovedover1993(meanscoresof5.3 versus4.4). Only0.7% of thestudents
takingtheexamination
hadperfectscoresof 15 on thisquestion.
ScoringStandards
Guiding
principles
:
Each reaction
is worth a total
of 3 points
+l point;
products +2 points
3) Ignore balancing
and states
4) Inappropriate
ionization
- maximum 1 point
1)
2) Reactants
(a)
CN’
Note:
w
any complex ion of Ag with cyanide with consistent
AgCN given as product earns 1 product point
If
3’
-
+
of Mn, maximum possible
Hz0
-
score
CO,
Note:
charge
earns
is 2 points
Acidic
species (H+ or oxyacid of phosphorus) earns 1 product
P in +5 oxidation
state in oxyanion earns 1 product point;
anions of oxyacids of phosphorus require
H+ for full credit
for products
-
NH3 +
Any one product
+
OH- -
C+2’
+ H20
Caj2’
alone
HCOj - + Hz0
3 points;
H3 PO4
Hz0
earns
+
all
1 point;
three
earns
2 product
points
HC%’
as products
as product
earns
earns
earns
1 product
point;
CO2
NH4OH+ co* earns 1 product point
NH3 + HtCO3 earns 1 product point
(e)
equation
Mns
Mg used instead
lNH4)ZcC3
Note:
per
Ag(CN)z’
PbOjO (or PZOS)
Note:
W
Ag+ -
Mn*+ +
Note:
(4
+
penalty
1 product
point
point
2 product
points
(f)
Ii+
Note:
W
Na
Note:
W
+
+ KMrlO& -
K+
+
Mn*+
+
Cl2
+
Hz0
as reactants
HCl and MnO4' acceptable
Any valid
redox product earns 1 point
All four products earns 2 points
K+ and/or Hz0 only as products earns no credit
If both H+ and Hz0 omitted,
then maximum of 2 points
+
Hz0
-
Hz0
+
Na+
+
possible
OH'
All three products earns 2 product points
Any vaid redox product earns 1 product point
Crz+2'
Note:
Cl'
All
Any
Hz0
If
of
+
Fe2+
+
H+' -
Cr3+
+
Fe3+
+
Hz0
three products earns 2 product points
valid
redox product earns 1 product point
only earns no credit
instead of
Fe2+ Fe3+, then maximum
Cl' Cl,
2 points
possible
n
s8m
Sample Student Response 1
Comment:This responsefailed to earn a perfect scorefor omitting the water in equationh. This was a common
error in the redox reactions.The total score for this answer was 14 points.
Sample Student Response2
Comment:A point was lost in equation g for failure to represent sodium hydroxide as an ionized species. In
equation a, a point was deducted for the failure to recognize that a complex ion was formed (see the Scoring
Standards).In equationd, a “product”point was deductedfor the hydroxide ion (ratherthan water). In equatione, the
c?udentlost a point for the incorrectchargeon the carbonateion product.The scorefor this responsewas 11 points.
n 59 n
Question5
For this requiredessayquestioninvolving simulatedlaboratoryexercises,studentswere asked to explain the result
of a change in conditions on an illustrated experimental situation. The committee hoped it would encourage a
significant increasein the laboratory componentin AP Chemistry courses.
ScoringStandards
(a)
Volume decreases
in beaker A; the concentration
decreases
in beaker B (either
observation
earns
provided
other one is not wrong)
The vapor pressure
of pure Hz0 is greater
pressure
of Hz0 in solution,
than
1 point
(1 pt.>
1
the vapor
7
(1 pt.>
7
(1 pt.>
OR,
the rate of evaporation
of Hz0 molecules from pure Hz0 is
greater
than that from the sugar solution,
while the
condensation
rates are the same.
(b)
The water
will
The external
to the vapor
begin
pressure
pressure
to boil
(or
J
evaporate).
on the water will become equal
of the water, causing it to boil,
OR,
the drop in external
pressure causes the boiling
to drop to the temperature
of the water.
(c)
(d)
Solid
strip
copper is
goes into
deposited
solution.
on the zinc
No reaction
point
/
strip;
the zinc
occurs with silver.
(1 pt.>
Zinc is a better
reducing agent or a more active metal
than copper and will
be oxidized.
Silver
is a less
reactive
metal than copper is.
(1 pt.>
Two layers
(1 pt.>
will
form,
one of which is colored.
Iodine is nonpolar and will
dissolve
Water-is
polar and will
not dissolve
Note:
placement
of 12 must be correctly
in TTE.
in TTE.
indicated
n 60 H
(1 pt.>
for
2nd point.
Sample StudentResponse1
n 61 n
Comment:The responsein part a presentsan unusualcombination of a reference to an increasein entropy and the
differential rates of evaporation of pure water and an aqueoussolution of sugar.In part b, the responsedoes not
match the standardsbut clearly indicates that the student understandswhat is required for this system to reach
equilibrium and why this one never can. The student’s responseto part c essentially embodies the ideas in the
standards.The responseto part d quite simply covers all of the bases.One would have to look hard for a place to
deduct points from this response,which received the maximum score of 8 points.
SampleStudent Response2
Comment: This paper earned a score of 6. The responsein part a was seen in many papers.The awarding of the
“observation point” was routine, but simply stating that there is a tendency for the concentrationsof the two
systemsto become equal was insufficient to earn the “explanation point.” Both the “observation” point and the
“explanation” point were clearly articulated in the responseto part b. The student’s responsein part c also clearly
earnscredit for both the “observation” and “explanation” points. In part d a point was lost since there was a clear
statementthat the solutionswould separateand two layers would form. Failure to specify that the iodine will be in
the TTE cost this studenta point.
Question6
The importance of thermodynamicsin even an introductory chemistry course is reflected by the fact that it is
weregiventhe
almosta standardessaytopic on theAP ChemistryExamination.In part a of thisquestion,students
opportunityto relatetheentropychangefor a reactionto thestoichiometry
andto thephasechangethattakesplace
as reactantsare convertedto products.Parts b and c gave the student an opportunity to demonstrate an
understandingof the implicationsof a change in the temperatureon the spontaneityof a reaction and the K,, of a
systemat equilibrium. In each of the parts a, b, andc, the writingof an equationalonewasnot suffkientto earn
the “explanation”point;the studentwasrequiredto connecttheequationto the issueto beexplained.In part d, a
knowledgeof the Gibbs-Helmholtzequationand the dependenceof the enthalpy and entropy changeson
temperaturewere required.Pointswere not deductedfor sloppinessor confusionrelativeto the differences
betweenthe standardandnonstandard
valuesof changesin freeenergy,entropy,or enthalpy.
ScoringStandards
(a)
Statement
that
ASo is negative
(1 pt.>
3 moles of gas 2 moles of gas plus solid,
(3 moles 2 moles earns no points)
0%
2 gases -
1 gas + solid,
(1 pt.>
0%
use of
Note:
(b)
A@ -A@
Ati -
0
If
statement
is that AS0 is positive,
then explanation
3 moles gas 5 moles of gas earns I point
(3 moles 5 moles earns no points)
If correct
explanation
for ASo being negative
is given
but wrong sign is stated,
1 point is earned.
A@ is
less
Explanation
Note:
- TAS" with
goes to 0, goes +, gets larger
negative,
using
A8
- A@ - TAS”
if answer to (a) is that ASo is positive,
then full
credit can be earned here for correct reasoning based
on that assumption.
An explanation that uses
Le Ch&teli.er's
principle
based on sign of @ is NOT
valid
here since system not at equilibrium.
of
(1 pt.>
(1 PU
(4
decreases
&q
(exponent -
more negative)
as 2' increases
CR.
(1 pt.>
g oes
Keq
Correct
A#
-
from > 1, to 1, to < 1, as T increases
explanation using the equation
In Keq (or In&/K;!)
- AH%l/T2
-RT
- WI)
OR.
(1 pt.>
higher 2' favors the reverse reaction (Le Chatelier)
because the forward reaction is exothermic
Note:
(d)
if answer for (a) is that AS0 is positive then statement
that Ke will decrease or increase depending on the relative
magnitu 3e of T and A8 change earns 2 points.
Recognition that
BOTH A@ and T are changing in A8 - -RT In Keq is necessary.
Or, ignoring part (a), use of AHO< 0 explanation to correctly
predict that Qq will decrease earns 2 points.
Since A@ - 0 at this
W
-
&
Prediction
General
Note:
-
TAS”
is
point, the equation is T - f@/AS”.
NOT sufficient
without Afl - 0.)
is not exact because &
and/or AS0 vary with T
(1 pt.>
(1 pt.)
For parts (a), (b), and (c), just writing an equation is not
sufficient
for the "explanation" point.
To earn credit,
the
student must connect the equation to issue to be explained.
Sample Student Response 1
Comment:Full credit is given for part a sincethe studentcorrectly evaluatesthe changein disorderasthe reactants
are converted to products. In part b, the student appropriately discussesthe Gibbs-Helmholtz equation and
correctly notes how the two state functions will change as temperature is increased.The answer in part c could
have been clearer; nonetheless,the answerdoes reflect an understandingof the relationshipbetween temperature,
K,,, and free energy change,so both points were awarded.In part d, the studentcorrectly notesthat at the “changeover” point the free energy change becomes zero and that both changes in entropy and enthalpy are slightly
temperature dependent.The total score for this answer was 8
Sample Student Response2
Comment:This paper earned4 of the possible8 points. No points were given in part a, becausethe studentfailed
to recognize that changes in phase are more important in determining changesin entropy than changes in the
number of moles of products versusthe number of moles of reactants.Full credit was given in part b, since the
analysisusingthe Gibbs-Helmholtz equation is correct usingthe incorrect sign of entropy changefrom part a. No
points were awarded in part c, althoughthere is somemerit in the analysisof the exothermic nature of the reaction
and the effect of increasingthe temperature on the position of the equilibrium. In fact, the studentdid not answer
the question,“What is the changein the value of the equilibrium constant?” Two points were awarded in part d for
the correct use of the Gibbs-Helmholtz equation and the recognition that (at least) one of the state functions is
dependenton temperature.
Question7
The student’s performanceon this question is reflected in the liberal nature of the scoring standards.It seemsthat
most studentsare not exposedappropriately to titration curvesor taught the propertiesof polyprotic acids.
Scoring Standards
POb3'
(a)
+
H+
*
HPObt'
HP04*- +
H+
_
H2P0,-
transfer
to any
only
Note:
any proton
:;:
H : 'b' 0.;:b':
**
53':
.#
Note:
l
*
3
HPOb*- (formula
correct
(c)
diagram
explicit
32 e’
explicit
2 - charge (somewhere)
not more than 1 double P-O bond
only)
earns
Graph goes from upper left
Two protons
transferred
Note:
H2P04-
+
or other
1 point.
to lower right
H+
at least
(2 pts*>
PxOr species with
Two "buffers"
Two "equivalence"
c
Explain/correctly
label
"equivalence" region
w
PxOY species earns 1 point.
2-
L
w
(2 p-4
(1
in either
direction
(1 pt.>
7
one "buffer"
*
H3f04
other proton transfer earns 1 point if
consistent with product in part (a)
pm
(pH decreases)
or
(1 PC*>
(1 pt.1
Sample Student Response 1
(c) Sketch a graphusing the axes provided, showingthe shapeof the titrationkurve that resultswhen 100. milliliters
of the HCl solutionis added slowly from a buret to the Na3P04 soiution.Account for the shapeof the curve.
t
0
nL HCI
(d) Write the equationfor the reaction that occursif a few additionalmilliliters of the HCl solutionare addedto
the solutionresultingfrom the titration in (c).
$ H-,?OY * Ofl’
‘Y I- t I@
!-Jr3
Comment:This was one of a very few perfect responses(a score of 8 points) to this question. Both points were
clearly earned in part a and in part b. The nature of this student’s titration curve (part c) reflects a real
understandingof the nature of this system,the somewhatdiffuse equivalencepoints, and the buffering regions.The
responseto part d is exactly right.
.
SampleStudent Response2
0
rnL HCI
1s
(d) Write the equationfor the reaction that occursif a few additional milliliters of the HCl solution are addedto
the solution resulting from the titration in (c).
Comment:The studentclearlyearnsthetwo pointsin part a, buthe or shefailsto identifytheamphotericspecies,
andthe Lewisstructurehaslittle validityin referenceto the octetrule.This studentearnsa pointin part c for the
shapeof the titrationcurveandanotherpointfor identifyingtheequivalence
pointsas“drops”occurringwith the
formationof eachsuccessive
moleof acid.The studentwasalsoawardeda pointin part d for correctlywritingthe
appropriateequationfor a total scoreof 5.
Question8
This questiongave the studentsan opportunity to relatechemicalprinciplestoeverydayexperiences.Theattempt
tomakechemistry“relevant"hasbecomemoreandmoreapparentinrecentyearsandthattrendisreflectedineach
ofthe four parts to this question.
ScoringStandards
(a)
w
The addition
of water.
of a solute
lowers the freezing
point
(1 pt.1
A mole of NaCl contains (dissociates
into) 2 moles of
ions/particles,
whereas a mole of CaC12 contains (dissociates
into) 3 moles of ions. Therefore CaC12 is more effective.
(1 pt.)
Hydrogen bonding is the most important intermolecular
attractive
force between molecules of Hz0 and between
molecules of NHs.
(1 pt*)
Water is a liquid because the hydrogen-bonding forces are
stronger between adjacent Hz0 molecules than between adjacent
NH3 molecules.
(1 l-4
Further explanations for the stronger hydrogen bonding in Hz0
include the larger dipole moment (or more polar character)
of Hz0 compared to NH, and the fact that 0 is more electronegative
than N is.
(c)
Graphite's structure consists of 2-dimensional sheets
forces
of covalently bonded carbon atoms. The attractive
between sheets (layers)
are weak London (dispersion)
forces,
which allow the sheets to slide easily over one another.
Note:
(d)
must indicate
layers
and sliding
(1 pt.1
to earn point.
Diamond consists of an extended 3-dimensional covalent network
of carbon atoms. This makes diamond a very hard substance.
(1 pt.)
Vinegar, a dilute solution of acteic acid, reacts
white solid, which contains metal carbonates,
(1 pt.1
in a neutralization
reaction
to form gaseous CO2.
n 7i n
with the
(1 PW
Comment:Since studentswere not required to relate freezing-point depressionto vapor-pressurelowering, the
statementthat the added salt lowers the freezing point of water earned the first point in part a. The secondpoint in
this part was awarded for the clear recognition of the difference in the i factor for the two solutes.Mentioning
hydrogen bonding and relating the strength of the hydrogen bonds in the two molecules to electronegativity in
oxygen and nitrogen earnedboth points in part b. In part c, the responseclearly earnedthe first point in explaining
the lubricating properties of graphite as a function of sliding layers and, even though the three-dimensionalnature
of the covalent network was not mentioned, the secondpoint was also awarded.The responseto part d is right on
the mark. The total score was 8 points.
n 72 n
Sample Student Response 2
Comment: Although the responseis rather cryptic, since the student notes in part a that the freezing point is
reduced by dissolved solutesand that calcium chloride producesmore ions in solution than sodium chloride, the
responseearned both points. In part b, the studentlosesa point for mistakenly identifying hydrogen bonding as an
intramolecular force and in part c, a point was deductedbecausethe studentfailed to note the covalent network
structure of diamond. In part d, the studentgets back on track by concisely explaining the effervescence when
vinegar is added to the white residue in a tea kettle.
Question9
Atomic structureandchemicalbonding,frequentlythe topicsof essayquestionson theAP Chemistry Examination, were the foci in this question. In addressingthe four parts, studentswere required to relate differences in
atomic size, crystal-lattice energy, and ionization energy to chemical principles.The fact that very few students
scoredwell on part b is indicative,perhaps,thatthistopicis oftenglossedoverin AP coursesand shouldreceive
more attention.
ScoringStandards
(a)
Cat+ has fewer
The outermost
the outermost
Note:
The first
electrons,
electrons,
electron
electron
thus
it
is smaller
Ca
than
in Ca is in a 4s orbital,
in Ca*+ in in a 3p orbital
(1 pt.)
whereas
(1 pt.)
point is earned for indicating
the loss of
the second point for indicating
the outermost
electrons are in different
shells -- must account for
the magnitude of the size difference between Ca and Ca2+.
(b)
ZI for CaO is more negative than U for K20, so it is more
difficult
to break up the CaO lattice
(stronger bonds in CaO).
Ca*+ is
smaller
cations
and 02-)are
than @,
so internuclear
separations
(between
less,
OR,
(1 pt.>
Ca*+ is more highly
are
(1 pt.)
charged
than K+, thus cation--02’
bonds
stronger
Note: understanding what "lattice
energy" is earns 1 point; size
or charge explanation needed for the second point.
Responses
that use Lewis structures or otherwise indicate molecules rather
than ionic lattice
earn no points.
n 74 n
(c)
(i)
(ii)
Note:
(d)
Ca has more protons e
is smaller.
The outermost
electrons are more strongly held by the nuclear charge
of Ca.
The outermost electrons
a higher energy orbital
electron in K.
in Ca are in the 4s, which is
(more shielded) than the 3p
for (i),
the idea of attraction
between nucleus
and electrons must be present; for (ii),
a
"noble-gas configuration"
argument must be tied
to an energy argument in order to earn credit.
The highest energy (outermost) electron in Al is in
a 3p orbital,
whereas that electron in Mg is in a 3s orbital.
(1 pt*)
The 3p electron in Al is of higher energy (is more shielded)
than is the 3s electron in Mg.
(1 pt*)
Note: noting
point;
that different
orbitals
are involved earns the first
a correct energy argument earns the second point.
Responses that attribute
the greater stability
of Ca over K
(or K+ over Ca+,or Mg over Al) to the stability
of a completely
filled
(vs. half or partially
filled)
orbital
earn NO credit.
SampleStudent Response2
Comment:Thisresponseearned 4 points. Both points in part a were given for noting that electronsare lost in the
ionization while implying that the nuclear chargeremainsthe same,increasingthe force of attractionof the nucleus
per electron. In part b, the studentlost both points sinceit is clear he or shedoesnot understandthe nature of lattice
energy. In part c, the studentearned only 1 of the 2 points, for invoking a common error in responsesto this
question(that filled subshellsare especially stable) in the first part. The point for part c (ii) was awarded since the
studentclearly knew that the secondelectron lost from potassiumcomes from a 3p orbital. In part d, a point was
earned by noting the electron configurationsof the speciesinvolved, but the explanationpoint was lost by using the
same fallacious argument as in part c.
Chapter IV
Statistical Information
n SECTION II SCORES
Table 4.1 shows the score distribution for the freeresponsesectionof the 1994 AP Chemistry Examination. Studentswere required to complete Questions 1,
4, and 5, and then had to choosea fourth questionfrom
Questions2 and 3, and two more from Questions6-9.
Questions l-3 were scored on a 9-point scale, Question 4 on a 15-point scale, and Questions5-9 on an 8point scale. For each question,the number of students
Table 4.1 Score
Question 1
(9)*
Question 2
(9)
T
Question 3
(9)
at each score point is listed, along with the total number of candidatesattempting that essay.
Question 2 had the highest mean as percent of
maximum possible score, indicating that the score
earned by this question’s typical studentwas closer to
the question’s highest score than for any other question. Question 1 had the greatest standarddeviation,
indicating that the scorestended to be spreadout more
than for the other questions,
Section II Scores
Question 4
(15)
Question 5
(8)
Question 6
(8)
Question 7
(8)
Question 8
(8)
Question 9
(8)
16
15
14
215
13
12
710
892
11
1,226
527
10
1,532
357
800
8
2,116
2,618
2,674
1,155
2,210
735
7
9
1,473
1,740
2,165
2,344
486
337
35
938
1,787
57
165
412
41
175
3,398
963
1,096
2,593
1,722
2,300
4,390
1,401
2,880
2,601
2,320
101
152
712
1,242
293
2,525
2,518
2,863
3,102
1,543
3,183
3,840
2,573
268
1,673
1,839
1,459
1,578
3,325
5,197
2,470
825
2,165
1,137
798
1,356
1,018
3,406
2,107
6,131
5,642
2,992
2,072
0
NR**
3,983
1,989
1,111
624
173
1,750
1,094
3,076
1,997
1,598
2,305
2,162
2,99 1
887
3,446
5,104
1
3,87 1
4,94 1
1 11,114
31,630
31,630
18,449
6
5
4
3
2
1,226
I
Number of
Candidates
3 1,630
19,117
x,t
x(
1
2,396
2,78 1
1,840
3,191
283
7,478
14,737
4,760
1,422
242
I
t(
I *
3.45
846
4.71
1
4.32
2.61
1 17,926
:(219/
3.60
1
1.73
2.41
1
2.24
* Numbers
in brackets
indicate the maximum possiblescore.
__
*y No response.Studentsgave either no responseor a responsenot on the topic. Responsesthat fall into this category were not included in the calculation
of number of candidates, mean, standard deviation, or means as a percentageof maximum.
18060-12162*S35M20*255118