Advanced Placement Chemistry
1984 Free Response Questions
1) Sodium benzoate, C6H5COONa, is a salt of the weak acid, benzoic acid, C6H5COOH. A 0.10molar solution of sodium benzoate has a pH of 8.60 at room temperature.
(a) Calculate the [OH¯] in the sodium benzoate solution described above.
(b) Calculate the value for the equilibrium constant for the reaction
C6H5COO¯ + H2O <===> C6H5COOH + OH¯
(c) Calculate the value of Ka, the acid dissociation constant for benzoic acid.
(d) A saturated solution of benzoic acid is prepared by adding excess solid benzoic acid to pure
water at room temperature. Since this saturated solution has a pH of 2.88, calculate the molar
solubility of benzoic acid at room temperature.
2) For a hypothetical chemical reaction that has the stoichiometry 2 X + Y ---> Z, the following
initial rate data were obtained. All the measurements were made at the same temperature.
Initial Rate of
Initial [X]o, Initial [Y]o,
Formation of Z,
mol L¯1
mol L¯1
mol L¯1 sec¯1
7.0 x 10¯4
0.20
0.10
3
0.40
0.20
3
2.8 x 10¯
0.40
0.40
4.2 x 10¯3
0.60
0.60
1.4 x 10¯
(a) Give the rate law for this reaction from the data above
(b) Calculate the specific rate constant for this reaction and specify its units.
(c) How long must the reaction proceed to produce a concentration of Z equal to 0.20 molar, if
the initial reaction concentrations are [X]o = 0.80 molar, [Y]o = 0.60 molar, and [Z]o = 0 molar?
(d) Select from the mechanisms below the one most consistent with the observed data, and
explain your choice. In these mechanisms M and N are reaction intermediates. (NB - the words
fast and slow are placed above the reaction arrow on the original test.)
(1)
(2)
(3)
X + Y ----> M (slow) X + X <===> M (fast)
Y ---> M (slow)
X + M ----> Z (fast) Y + M----> Z (slow) M + X ---> N (fast)
N + X ----> Z (fast)
3)
Substance
Standard Heat of
Formation, Hf°,
in kJ mol¯1
Absolute Entropy
S°, in J mol¯1 K¯1
C(s)
0.00
5.69
CO2(g)
-393.5
213.6
H2(g)
0.00
130.6
H2O(l)
-285.85
69.96
O2(g)
0.00
205.0
C3H7COOH(l)
?
226.3
The enthalpy change for the combustion of butyric acid at 25 °C, H°comb, is -2,183.5 kilojoules
per mole. The combustion reaction is
C3H7COOH(l) + 5 O2(g) <===> 4 CO2(g) + 4 H2O(l)
(a) From the data above, calculate the standard heat of formation, Hf°, for butyric acid at 25 °C.
(b) Write a correctly balanced equation for the formation of butyric acid from its elements.
(c) Calculate the standard entropy change, Sf°, for the formation of butyric acid at 25 °C. The
entropy change, S°, for the combustion reaction above is -117.1 J K¯1 at 25 °C.
(d) Calculate the standard free energy of formation, Gf°, for butyric acid at 25 °C.
4) Give a scientific explanation for the following observations. Use equations of diagrams if they
are relevant.
(a) It takes longer to cook an egg until it is hard-boiled in Denver (altitude 1 mile above sea
level) than it does in New York City (near sea level).
(b) Burning coal containing a significant amount of sulfur leads to "acid rain."
(c) Perspiring is a mechanism for cooling the body.
(d) The addition of antifreeze to water in a radiator decreases the likelihood that the liquid in the
radiator will either freeze or boil.
5) Discuss the roles of indicators in the titration of acids and bases. Explain the basis of their
operation and the factors to be considered in selecting an appropriate indicator for a particular
titration.
6) Given solid samples of KI and of (NH4)2CO3. Briefly describe four simple laboratory tests by
which these two compounds can be distinguished. For each test, report the expected result for
each compound.
7) The van der Waals equation of state for one mole of a real gas is as follows:
(P + a/V2) (V - b) = RT
For any given gas, the values of the constants a and b can be determined experimentally. Indicate
which physical properties of a molecule determine the magnitudes of the constants a and b .
Which of the two molecules, H2 or H2S has the higher value for a and which has the higher value
for b ? Explain. One of the van der Waals constants can be correlated with the boiling point of a
substance. Specify which constant and how it is related to the boiling point.
8) Discuss some differences in physical and chemical properties of metals and nonmetals. What
characteristic of the electronic configurations of atoms distinguishes metals from nonmetals? On
the basis of this characteristic, explain why there are many more metals than nonmetals.
Advanced Placement Chemistry
1984 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.
Other correct ways of solving the problems received full credit. Because of limitations of
space, such other solutions are not shown.
1) Average score 4.36
a) two points
pH = 8.60
[H+] = 1 x 10¯8.60 M = 2.5 x 10¯9 M
[OH¯] = (1 x 10¯14) / (2.5 x 10¯9) M
= 4.0 x 10¯6 M
b) three points
C6H5CO2¯ + HOH ---> C6H5CO2H + OH¯
K = ([C6H5CO2H][OH¯]) ÷ [C6H5CO2¯]
= ((4.0 x 10¯8) (4.0 x 10¯6)) ÷ (0.10 - 4.0 x 10¯6)
= 1.6 x 10¯10
c) one point
C6H5CO2H <===> H+ + C6H5CO2¯
Ka = ([H+][C6H5CO2¯]) ÷ [C6H5CO2H]
= ((2.5 x 10¯9) (0.10)) ÷ (4.0 x 10¯6)
= 6.3 x 10¯5
d) two points
C6H5CO2H <===> C6H5CO2¯ + H¯
pH = 2.88
[H+ ] = 1 x 10¯2.88M = 1.3 x 10¯3
Ka = ([H+ ][C6H5CO2¯]) ÷ [C6H5CO2H]
6.3 x 10¯8 = ((1.3 x 10¯3) (1.3 x 10¯3)) ÷ x
x = 2.8 x 10¯2 M
Total C6H5CO2H in solution =
(2.8 x 10¯2 + 1.3 x 10¯3) M =
2.9 x 10¯2 M
2) Average score 4.81
a) three points; one point for correct form of law and two points for correct methodology without
an error; one point for correct methodology with an error
Rate = k[Y]
b) two points
7.0 x 10¯4 mole / L sec = k (0.10 mole / L)
k = 7.0 x 10¯3 sec¯1
c) two points
2.3 log co / c = k t
2.3 log 0.60 / 0.40 = (7.0 x 10¯3) (t)
t = 58 s
d) two point
Mechanism 3 is correct.
The rate law shows that the slow reaction must involve one Y, consistent with mechanism 3.
Mechanisms 1 and 2 would involve both [X] and [Y] in the rate law, not consistent with the rate
law.
3) Average score 5.52
a) three points
From Hess's law:
[delta]H°f = [4 (393.5) + 4 (205.85)] - 2183.5 kJ = - 533.8 kJ
b) one point
4 C(s) + 4 H2(g) + O2(g) ---> C3H7COOH
c) two points
[delta]S°f butyric acid = S° butyric acid - 4 S° carbon - 4 S° H2 - S° O2
[delta]S°f butyric acid = [226.3 - 4(5.69) - 4(130.6) - 205] joules/K = - 523.9 joules/K
Note: If part c was based on the equation in part b, and was done correctly, credit was given for
part c even if part b was wrong.
d) three points
[delta]G°f butyric acid = [delta]H°f - T [delta]S°f
[delta]G°f = [533.9 - (298) (- 0.524)] kJ
[delta]G°f= - 377.7 kJ
Note: For each of the problems, a maximum of one point was subtracted for gross misuse of
significant figures and. or for a mathematical error if correct principles were used.
4) Average score 3.52
a) two points
Water boils at a lower temperature in Denver than in New York City because the atmospheric
pressure is less at high altitudes.
At a lower temperature, the cooking process is slower, so the time to prepare a hard-boiled egg is
longer.
b) two points
S + O2 ---> SO2 (as coal is burned)
SO2 + H2O ---> H2SO3 (in the atmosphere)
SO2 + 1/2 O2 + H2O ---> H2SO4 (in the atmosphere)
c) two points
Vaporization or evaporation of sweat from the skin.
These processes are endothermic and so cool the skin.
d) two points
Colligative properties, which depend on the number of particles present, are involved.
Solute (the antifreeze) causes the lowering of the vapor pressure of the solvent. When the vapor
pressure of the solvent is lowered, the freezing point is lowered and the boiling point is raised.
5) Average score 3.05
Points awarded are indicated at the end of each statement.
An indicator signals the endpoint of a titration by changing color. (2)
An indicator is a weak acid or a weak base where the acid form and the basic form of the
indicator are different colors. (2)
An indicator usually changes color when the concentrations of the acid form and the basic form
are about equal; that is, when the pH is near the pK for the indicator. (1)
When an indicator is selected, the pH at which the indicator changes color should bracket the pH
of the titration solution at the equivalence point. (2)
For example, when a strong acid is titrated with a strong base, the pH at the equivalence point is
7, so an indicator that changes color at a pH of 7 should be used. (1)
6) Average score 2.60
Two points (to a maximum of eight) were awarded for each correct test with the expected results.
Possibilities include the following:
1. Flame test: K+ lavender; (NH4)2CO3, no lavender
2. Add Cl2and CH2Cl2: KI, pink color in organic layer; (NH4)2CO3, no change
3. Add Pb2+: KI, yellow precipitate of PbI2; (NH4)2CO3, white precipitate of PbCO3
4. Add Ag+: KI, pale yellow precipitate of AgI; (NH4)2CO3, white precipitate of Ag2CO3
5. Add I2: KI, brown color of I3¯; (NH4)2CO3, no change
6. Add good oxidizing agent: KI, brown color of I3¯; (NH4)2CO3, no change
7. Add strong base: KI, no change: (NH4)2CO3, odor of NH3 or color change of red litmus
8. Add Ba2+ or Ca2+ or Mg2+: KI, no change; (NH4)2CO3, with precipitate of a carbonate
9. Dissolve salts and use litmus: KI, neutral; (NH4)2CO3, basic solution
10. Add nonoxidizing acid: KI, no change; (NH4)2CO3, bubbles of CO2
11. Test melting points: KI, high; (NH4)2CO3, decomposes before melting
7) Average score 4.28
Points awarded for each statement follow that statement.
Constant a is related to the attractive forces that exist between real molecules. (1 point)
Constant b is related to the fact that real molecules do have volumes. (1 point)
H2S has a larger value for a. (1 point)
H2S is a polar molecule and, therefore, has stronger intermolecular forces. (2 points)
H2S has a larger value for b because an H2S molecule has a larger volume than H2 has. (1 point)
The constant a correlates with the boiling point since a is related to the intermolecular forces,
which must be overcome in the process of boiling a liquid. (2 points)
8) Average score 4.16
Physical properties:
Metals
Nonmetals
Melting points
Relatively high
Relatively low
Electrical conductivity
Good
Insulators
Luster
High
Little or none
Physical state
Most are solids
Gases, liquids, or solids
Chemical properties:
Metals
Nonmetals
redox
Reducing agents
Oxidizing
or reducing agents
attraction to electrons
Electropositive
Electronegative
acid/base character
Oxides basic
or amphoteric
Oxides acidic
reactivity
React with nonmetals
React with both metals
and nonmetals
Electron configurations (1 point each statement): Metals: Valence electrons in s or d sublevels of
their atoms. (A few heavy elements have atoms with one or two electrons in p sublevels.)
Nonmetals: Valence electrons in the s and p sublevels of their atoms.
There are more metals than nonmetals because filling d orbitals in a given energy level involves
the atoms of ten element and filling the f orbitals involves the atoms of 14 elements. In the same
energy levels, the maximum number of elements with atoms receiving p electrons is six.