Advanced Placement Chemistry
1992 Free Response Questions
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1)
2 NaHCO3(s) <===> Na2CO3(s) + H2O(g) + CO2(g)
Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation
above.
(a) A sample of 100. grams of solid NaHCO3 was placed in a previously evacuated rigid 5.00liter container and heated to 160. °C. Some of the original solid remained and the total pressure
in the container was 7.76 atmospheres when equilibrium was reached. Calculate the number of
moles of H2O(g) present at equilibrium.
(b) How many grams of the original solid remained in the container under the conditions
described in (a)?
(c) Write the equilibrium expression for the equilibrium constant, Kp, and calculate its value for
the reaction under the conditions in (a)
(d) If 110. grams of solid NaHCO3 had been placed in the 5.00-liter container and heated to 160
°C, what would the total pressure have been at equilibrium? Explain.
2) An unknown metal M forms a soluble compound M(NO3)2.
(a) A solution of M(NO3)2 is electrolyzed. When a constant current of 2.50 amperes is applied
for 35.0 minutes, 3.06 grams of the metal M is deposited. Calculate the molar mass of M and
identify the metal.
(b) The metal identified in (a) is used with zinc to construct a galvanic cell, as shown below.
Write the net ionic equation for the cell reaction and calculate the cell potential, E°.
(c) Calculate the value of the standard free energy change, G°, at 25 °C for the reaction in (b)
(d) Calculate the potential, E, for the cell shown in (b) if the initial concentration of ZnSO4 is
0.10-molar, but the concentration of the M(NO3)2 solution remains unchanged.
3)
Cl2(g) + 3 F2(g) ---> 2 ClF3(g)
ClF3 can be prepared by the reaction represented by the equation above. For ClF3 the standard
enthalpy of formation, H°f, is - 163.2 kilojoules/mole and the standard free energy of
formation, G°f, is - 123.0 kilojoules/mole.
(a) Calculate the value of the equilibrium constant for the reaction at 298 K.
(b) Calculate the standard entropy change, S°, for the reaction at 298 K.
(c) If ClF3 were produced as a liquid rather than as a gas, how would the sign and magnitude of
S for the reaction be affected? Explain.
(d) at 298 K the absolute entropies of Cl2(g) and ClF3(g) are 222.96 joules per mole-Kelvin and
281.50 joules per mole-Kelvin, respectively.
(i) Account for the larger entropy of ClF3(g) relative to that of Cl2(g).
(ii) Calculate the value of the absolute entropy of F2(g) at 298 K.
4) Give the formulas to show the reactants and the products for FIVE of the following chemical
reactions. Each of the reactions occurs in aqueous solution unless otherwise indicated. Represent
substances in solution as ions if the substance is extensively ionized. Omit formulas for any ions
or molecules that are unchanged by the reaction. In all cases a reaction occurs. You need not
balance.
Example: A strip of magnesium is added to a solution of silver nitrate.
Mg + Ag+ ---> Mg2+ + Ag
(a) An excess of sodium hydroxide solution is added to a solution of magnesium nitrate.
(b) Solid lithium hydride is added to water.
(c) Solutions of ammonia and hydrofluoric acid are mixed.
(d) A piece of aluminum metal is added to a solution of silver nitrate.
(e) A solution of potassium iodide is electrolyzed.
(f) Solid potassium oxide is added to water.
(g) An excess of nitric acid solution is added to a solution of tetraaminecopper(II) sulfate.
(h) Carbon dioxide gas is bubbled through water containing a suspension of calcium carbonate.
5)
H2(g) + I2(g) ---> 2 HI(g)
For the exothermic reaction represented above, carried out at 298 K, the rate law is as follows.
Rate = k [H2] [I2]
Predict the effect of each of the following changes on the initial rate of the reaction and explain
your prediction.
(a) Addition of hydrogen gas at constant temperature and volume.
(b) Increase in volume of the reaction vessel at constant temperature.
(c) Addition of a catalyst. In your explanation, include a diagram of potential energy versus
reaction coordinate.
(d) Increase in temperature. In your explanation, include a diagram showing the number of
molecules as a function of energy.
6) The equations and constants for the dissociation of three different acids are given below.
HCO3¯ <===> H+ + CO32¯
Ka = 4.2 x 10¯7
H2PO4¯ <===> H+ + HPO42¯
Ka = 6.2 x 10¯8
HSO4¯ <===> H+ + SO42¯
Ka = 1.3 x 10¯2
(a) From the systems above, identify the conjugate pair that is best for preparing a buffer with a
pH of 7.2.
(b) Explain briefly how you would prepare the buffer solution described in (a) with the conjugate
pair you have chosen.
(c) If the concentrations of both the acid and the conjugate base you have chosen were doubled,
how would the pH be affected? Explain how the capacity of the buffer is affected by this change
in concentrations of acid and base.
(d) Explain briefly how you would prepare the buffer solution in (a) if you had available the
solid salt of only one member of the conjugate pair and solutions of a strong acid and a strong
base.
7) Four bottles, each containing about 5 grams of finely powdered white substance, are found in
a laboratory. Near the bottles are four labels specifying high purity and indicating that the
substances are glucose (C6H12O6), sodium chloride (NaCl), aluminum oxide (Al2O3), and zinc
sulfate (ZnSO4).
Assume that these labels belong to the bottles and that each bottle contains a single substance.
Describe the tests you would conduct to determine which label belongs to which bottle. Give the
results you would expect for each test.
8) Explain each of the following in terms of atomic and molecular structures and/or
intermolecular forces.
(a) Solid K conducts an electric current, whereas solid KNO3 does not.
(b) SbCl3 has a measurable dipole moment, whereas SbCl5 does not.
(c) The normal boiling point of CCl4 is 77 °C, whereas that of CBr4 is 190 °C.
(d) NaI(s) is very soluble in water whereas I2(s) has a solubility of only 0.03 gram per 100 grams
of water.
9)
NO2 NO2¯ NO2+
Nitrogen is the central atom in each of the species given above.
(a) Draw the Lewis electron-dot structure for each of the three species.
(b) List the species in order of increasing bond angle. Justify your answer.
(c) Select one of the species and give the hybridization of the nitrogen atom in it.
(d) Identify the only one of the species that dimerizes and explain what causes it to do so.
Advanced Placement Chemistry
1992 Free Response Answers
Notes
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[delta] and [sigma] are used to indicate the capital Greek letters.
[square root] applies to the numbers enclosed in parenthesis immediately following
All simplifying assumptions are justified within 5%.
One point deduction for a significant figure or math error, applied only once per problem.
No credit earned for numerical answer without justification.
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1) average points = 3.9
a) three points
mole fraction is 50:50 from equation
therefore 7.76 atm ÷ 2 = 3.88 atm (contribution by H2O)
PV = nRT; rearrange to get n = (PV) ÷ (RT)
n = [(3.88 atm) (5.00 L)] ÷ [(0.0821 L atm mol¯1 K¯1) (433 K)]
= 0.545 mol
Personal note by John Park: the actual AP standard uses 7.76 atm in its PV = nRT calculation
and then applies the 50:50 criterion to the value for n (1.09 mol) calculated. I prefer the above
explanation.
b) two points
from equation, the ratio of NaHCO3 to H2O is 2:1
x / 0.545 = 2 / 1
x = 1.09 mol of NaHCO3 decomposed
1.09 mole x 84.01 g/mol = 91.6 g of NaHCO3 decomposed
100. g - 91.6 g = 8.4 g of NaHCO3 remaining
c) two points
Kp = (PH2O) (PCO2)
= (3.88 atm) (3.88 atm) = 15.1 atm2
d) two points
The total pressure would remain at 7.76 atm. Since some solid remained when 100. g was used
(and there had been no temperature change), then using 110 g would not affect the equilibrium.
2) average points = 4.0
a) three points
2.50 amps = 2.50 C/sec; 35.0 min = 2.10 x 103 sec
(2.50 C/sec) (2.10 x 103 sec) = 5.25 x 103 Coulombs delivered
(5.25 x 103 C) ÷ (1 Faraday / 96,500 C) = 0.0544 mol e¯ delivered
since M2+ + 2e¯ ---> M, therefore 0.0544 ÷ 2 = 0.0272 mol M2+ deposited
3.06 g ÷ 0.0272 mol = 112.5 g/mol
From an examination of the periodic table, the metal is seen to be cadmium.
b) two points
Cd2+ + 2e¯ ---> Cd
Zn ---> Zn2+ + 2e¯
Cd2+ + Zn ---> Cd + Zn2+
E° = - 0.40 V
E°= - 0.76 V
E° = 0.36 V
c) two points
[delta]G° = - FE°
= - (2 mol) (96,500 C/mol) (0.36 J/C) [or do the units as () (J/V) (V)]
= - 69,480 J = - 69 kJ (after appropriate rounding off)
d) two points
E = E° - (0.0591 ÷ n) log ([ Zn2+] / [Cd2+])
E = 0.36 - (0.0591 / 2) log (0.01 / 1)
E = 0.36 - (- 0.029) = 0.39 V
3) average = 3.5 (Only 30 scores of nine; kids did not see stoichiometry in (b), had problems on
which gas constant to use, and a hard time in (c) in relating a more negative value.)
a) two points [delta]G° = - RT ln K; rearranging gives ln K = [delta]G° ÷ - RT ln K = - 123,000 J
÷ - ((8.31 J/mol K) (298 K)) = 49.7 K = 3.72 x 1021
b) two points
[delta]G° = [delta]H° - T[delta]S°
- 246,000 J = - 326,400 J - (298) (x)
x = - 270 J / K
c) two points
[delta]S is a larger negative number
ClF3 (liquid) is more ordered (less disordered) than ClF3 (gas)
This was my answer before I saw the standard given above: A liquid is more ordered than a gas.
There would be a greater entropy change in gas + gas ---> liquid than in gas + gas ---> gas.
Therefore, sign is the same, but absolute magnitude is greater.
d) three points
i) ClF3 is a more complex molecule (i.e. more atoms) with more vibrational and rotational
degrees of freedom than Cl2
ii) Cl2 + 3 F2 ---> 2 ClF3; use Hess's Law
[sigma]Srxn = [sigma]S products - [sigma]S reactants
- 270 = [2 (281.50)] - [222.96 + 3x]
x = 203 J mol¯1 K¯1
4) average = six points. Be careful not to over process equation; do not cancel. (I think this
means to write the final equation answer without strikeouts.)
Count positive credit; single concept mistake = minus one point; products = first allowable is one
point; all correct are two points. Spurious product if all other right is minus one point.
a) Mg2+ + 2 OH¯ ---> Mg(OH)2
b) LiH + H2O ---> Li+ + OH¯ + H2
OH¯ or H2 earns one point, all three for two points.
c) NH3 + HF ---> NH4+ + F¯
NH3 + H+---> NH4+ earns two points
d) Al + Ag+ ---> Al3+ + Ag
e) I¯ + H2O ---> I2 + H2 + OH¯
I2 or H2 (one point); all three products for two points
f) K2O + H2O ---> K+ + OH¯
KOH product alone is one point.
g) H+ + Cu(NH3)42+ ---> Cu2+ + NH4+
h) CaCO3 + CO2 + H2O ---> Ca2+ + HCO3¯
5) average = 4.1
a) two points
EFFECT: addition of H2 would increase the rate.
EXPLANATION: Since the rection is first-order in H2, doubling the concentration would double
the rate. The inclusion of H2 in the rate law indicates it participates in the rate-determining step.
OR
Relate the increase in concentration of hydrogen to an increase in collision rate.
b) one point
EFFECT: The initial rate would decrease.
EXPLANATION: Increasing the volume would decrease the concentration of both H2 and I2. At
the lower concentration there would be a lesser number of overall collisions (due to greater
distance between individual molecules), leading to a lesser number of effective collisions.
c) three points
EFFECT: addition of a catalyst will increase the rate of both forward and reverse reactions.
EXPLANATION:
d) two points
EFFECT: The initial rate of reaction will increase.
EXPLANATION:
If Ea is missing from explanation, this gets one point.
6) average = 3.2
a) two points
best conjugate pair = H2PO4¯ and HPO42¯ pair with pKa = 7.2.
7.2 = pH when [HPO42¯] = [H2PO4¯]
b) one point
dissolve equal number of moles of a H2PO4¯ salt and a HPO42¯ salt.
If answer in (a) was HCO3¯ and CO32¯, students would receive one point in (b) if they stated that
the CO32¯/HCO3¯ mole ratio was 6.7 : 1
c) three points
the pH would not change.
The ratio [salt]/[acid] in the Henderson-Hasselbalch equation would remain the same if the
concentrations were doubled.
There is now more of each ion to react when an acid or a base is added.
d) two points
Add one mole of conjugate acid to 0.5 mole of strong base or one mole of conjugate base to 0.5
mole strong acid
OR
use pH meter to monitor addition of strong base (acid) to conjugate acid (base)
7)
General Principles
A series of chemical and/or physical tests must be performed which lead to distinct and
unambiguous identification of these unknowns relative to each other. By the structure of the
question, if 3 out of 4 substances are unambiguously identified, the fourth may be identified by
exclusion. If attempted, this must be stated to earn final 2 points.
The series may take to form of a flow chart, excluding samples from further tests as they are
identified. Or, the series may be a series of tests, each applied to each substance, the total results
of all tests being used to identify the substance.
Scoring
1) Each of the 4 substances correctly and unambiguously identified earns 2 points.
2) Tests which are usable and are misapplied, misinterpreted, etc. can earn at most 1 point.
3) Full credit for a substance cannot be earned unless the identification is both correct and
unambiguous.
4) The use of physical tests only may be acceptible if:
i) Unambiguous results can be stated based upon well-known properties, (and easily separated).
ii) Potentially ambiguous results can be resolved by reference to established values from sources
(for example, solubility from CRC).
iii) Unusual physical behaviors, different between compounds, but characteristics of the
compound are explained. (For example; when using a colligative property to find FW, the
dissociation factor must be cited.)
iv) Properties must be distinguishable to within errors of reasonably performed experiments.
Note: Tasting is explicitly excluded as a test technique.
8) average score = 3.0
a) two points
K conducts because of its metallic bonding or "sea" of mobile e's (or "free" e's)
KNO3 does not conduct because it is ionically bonded and has immobile ions (or imm. e's)
b) two points
SbCl3 has a measurable dipole moment because it has a lone pair of e's which causes a dipole
or its dipoles do not cancel
or it has a trigonal pyramidal structure
or clear diagram illustrating any of the above
SbCl5 has no dipole moment because its dipoles cancel
or it has a trigonal bipyramidal structure
or clear diagram illustrating either of the above
c) two points
CBr4 boils at a higher T than CCl4 because it has stronger intermolecular forces (or van der
Waals or London dispersion). These stronger forces occur because CBr4 is larger and/or has
more electrons than CCl4. (Note added to scoring standard: student misconception of inter-, with
"inter-" double underlined.)
d) two points
NaI has greater aqueous solubility than I2 because NaI is ionic (or polar) whereas I2 is nonpolar
(or covalent). H2O, being polar, interacts with the ions of NaI but not with I2. (Like dissolves like
accepted if polarity of H2O clearly indicated.)
9) average score = 2.5
a) three points; one point per structure; minus one point for one or more missing charges.
NO2
A correct structure with one electron on oxygen is OK.
Note added to standards: Although not required by the wording of the question, both resonance
forms are shown.
NO2¯
Note added to standards: Although not required by the wording of the question, both resonance
forms are shown.
NO2+
b) three points
NO2¯ < NO2 < NO2+
[This order, whether or not structures are drawn in (a) are correct, receives one point. If
structures in (a) are incorrect, an order that is consistent with the structures actually given in (a)
receives one point.]
Whether structures in (a) are correct or incorrect, one point is awarded for any one correct angle
or ranking consistent with a particular structure, with appropriate justification; two points are
awarded for two correct comparisons with justifications; for the correct order:
NO2¯
<
NO2
<
NO2+
3 charge centers on
3 charge centers on
2 charge centers on
N;
N;
N
lone pair of e¯ on N
single e¯ on N
c) one point
Hybridization:
NO2+ is sp
NO2 is sp2
NO2¯ is sp2 also.
d) one point
NO2 will dimerize because it contains an odd electron that will pair readily with another, giving
N2O4.