AP Chem Exam - ‘94
MgF2(s) <===> Mg 2+(aq) + 2 F -(aq)
1.
In a saturated solution of MgF2 at 18° C, the concentration of Mg2+ is
1.21 x10-3 M. The equilibrium is represented by the equation above.
a)
Write the expression for the solubility-product constant, Ksp, and calculate its
value at 18° C.
Ksp, = [Mg 2+][F -]2 = (1.21 x10-3)(2 x 1.21 x10-3)2 = 7.09 x10-9
b)
Calculate the equilibrium concentration of Mg2+ in 1.000 L of saturated MgF2
solution at 18° C to which 0.100 mole of solid KF has been added. KF dissolves
completely. Assume the volume change is negligible.
Ksp, = [Mg 2+](2x + 0.100)2
[Mg 2+] = (7.09 x10-9)/(10-2) = 7.09 x10-7
c)
Predict whether a precipitate of MgF2 will form when 100.0 mL of a 3.00 x10-3 M
Mg(NO3)2 solution is mixed with 200.0 mL of 2.00 x10-3 M NaF solution at 18° C.
Calculations to support your prediction must be shown.
[Mg 2+] = (0.1)(3.00 x10-3)/0.3 = 1.00 x 10-3
[F -] = (0.2 L)(2.00 x10-3)/0.3 = 1.33 x10-3
Q = [Mg 2+][F -]2 = (1.00 x10-3)(1.33 x10-3)2 = 1.77 x10-9
Since Q < Ksp, no precipitate forms
d)
At 27° C the concentration of Mg2+ in a saturated solution of MgF2 is 1.17 x10-3
M. Is the dissolving of MgF2 in water endothermic or exothermic? Give an
explanation to support your conclusion.
MgF2(s) <===> Mg 2+ + 2 F Based on the concentration of Mg2+ at 27° C is 1.17 x10-3 M and at 18° C the
concentration is 1.21 x10-3 M it appears that the forward reaction is exothermic.
2.
2 NO(g) + 2 H2(g) ===> N2(g) + 2 H2O(g)
Experiments were conducted to study the rate of the reaction represented by the
equation above. Initial concentration and rates of reaction are given in the table
below
a)
(i)
Trial
[NO]
[H2]
1
2
3
4
0.0060
0.0060
0.0010
0.0020
0.0010
0.0020
0.0060
0.0060
Rate ==> N2
1.8 x10-4
3.6 x10-4
0.3 x10-4
1.2 x10-4
Determine the order for each of the reactants, NO and H2, from the data
given and show your reasoning
Doubling [H2] doubles the rate; doubling [NO] quadruples the rate [H2] is
first order [NO] is second order
(ii)
Write the overall rate law for the reaction
r = k[H2][NO]2
b)
Calculate the value of the rate constant, k, for the reaction. Include units.
k = 1.8 x10-4 M/min/(1.0 x10-3 M)(6.0 x10-3 M)2 = 5.0 x103 M-2-min-1
c)
For experiment 2, calculate the concentration of NO remaining when exactly onehalf of the original amount of H2 has been consumed.
NO : H2 is 1 : 1
When 0.0010 mol H2 has reacted, it must have reacted with 0.0010 mol NO
[NO] remaining = 0.0060 - 0.0010 = 0.0050 M
d)
The following sequence of elementary steps is a proposed mechanism for the
reaction.
I.
II.
III.
NO + NO <===> N2O2
N2O2 + H2 <===> H2O + N2O
N2O + H2 ===> N2 + H2O
Based on the data presented, which of the above is the rate-determining step?
Show that the mechanism is consistent with
i)
the observed rate law for the reaction, and
For I:
For II:
Keq = [N2O2]/[NO]2
rate = k [H2][N2O2]
[N2O2] = K eq [NO]2
rate = k' [H2][NO]2
ii)
the overall stoichiometry of the reaction.
Step II is the rate determining step
I
II
III
NO + NO <===> N2O2
N2O2 + H2 ===> H2O + N2O
N2O + H2 ===> N2 + H2O
I + II + III:
3.
2 NO + 2 H2 ===> N2 + 2 H2O
A student collected a sample of hydrogen gas by the displacement of water as shown
by the diagram above. The relevant data are given in the following table
Gas Sample Data
Volume of sample
Temperature
Atmospheric Pressure
PH2O (25° C)
a)
90.0 mL
25° C
745 mm Hg
23.8 mm Hg
Calculate the number of moles of hydrogen gas collected.
n = PV/RT = (0.95)(0.090)/0.0821)(298) = 3.49 x10-3 mol H2
b)
Calculate the number of molecules of water vapor in the sample of gas.
(0.0313)(0.09)/0.0821)(298) = 1.15 x10-4 mol H2O
(1.15 x10-4)(6.023 x1023) = 6.92 x1019 molecules H2O
c)
Calculate the ratio of the average speed of the hydrogen molecules to the average
speed of the water vapor molecules in the sample.
vH2
v H 2O
d)

MM H O
2
MM H 2

18
2
 3
Which of the two gases, H2 or H2O, deviates more from ideal behavior? Explain
your answer.
The intermolecular forces among H2O molecules are stronger than those among
H2 molecules
4.
Net Ionics
a) Excess sodium cyanide solution is added to a solution of silver nitrate.
CN - + Ag + ===> Ag(CN)2 -
b) Solution of manganese (II) sulfate and ammonium sulfide are mixed.
Mn 2+ + S 2- ===> MnS
c) Phosphorus(V) oxide powder is sprinkled over distilled water.
P4O10 (P2O5) + H2O ===> H3PO4
d) Solid ammonium carbonate is heated.
(NH4)2CO3 ===> NH3 + H2O + CO2
e) Carbon dioxide gas is bubbled through a concentrated solution of potassium
hydroxide.
CO2 + OH - ===> HCO3 -
or
CO3 2- + H2O
f) A concentrated solution of hydrochloric acid is added to solid potassium
permanganate.
H + + Cl - + KMnO4 ===> K + + Mn 2+ + Cl2 + H2O
g) A small piece of sodium metal is added to distilled water.
Na + H2O ===> H2 + Na + + OH h) A solution of potassium dichromate is added to an acidified solution of iron (II)
chloride.
Cr2O7 2- + Fe 2+ + H + ===> Cr 3+ + Fe 3+ + H2O
5. Discuss the following phenomena in terms of the chemical and physical properties of
the substances involved and general principles of chemical and physical change.
a) As the system shown on the right approaches equilibrium, what change occurs to
the volume of water in beaker A? What happens to the concentration of the
sugar solution in beaker B? Explain why these changes occur.
Volume decreases in beaker A; the concentration decreases in beaker B
The rate of evaporation of H2O molecules from pure H2O is greater than that
from the sugar solution, while the condensation rates are the same.
b) A bell jar connected to a vacuum pump is shown on the right. As the air pressure
under the bell jar decreases, what behavior of water in the beaker will be
observed? Explain why this occurs.
The external pressure on the water will become equal to the vapor pressure of the
water, causing it to boil
c) What will be observed on the surfaces of zinc and silver strips shortly after they
are placed in separate solution of CuSO4, as shown on the right? Account for
these observations.
Solid copper is deposited on the zinc strip; the zinc strip goes into solution. No
reaction occurs with silver.
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.
d) A water solution of I2 is shaken with an equal volume of a non-polar solvent such
as TTE. Describe the appearance of this system after shaking. (A diagram may
be helpful.) Account for this observation.
Two layers will form, one of which is colored. Iodine is nonpolar and will
dissolve in TTE.
6.
2 H2S(g) + SO2(g) <===> 3 S(s) + 2 H2O(g)
At 298 K, the standard enthalpy change, ∆H , for the reaction represented above is
- 145 kJ.
a) Predict the sign of the standard entropy change, ∆S , for the reaction. Explain the
basis for your prediction.
∆S° is negative because:
3 mol gas ===> 2 mol gas + solid
2 gases ===> 1 gas + solid
b) At 298K, the forward reaction is spontaneous. What change, if any, would occur
in the value of ∆G for this reaction as the temperature is increased? Explain your
reasoning using thermodynamic principles.
∆G° begins negative and then becomes positive.
c) What change, if any, would occur in the value of the equilibrium constant, K eq,
for the situation described in (b)? Explain your reasoning.
Keq decreases as T increases because the forward reaction is exothermic
∆G° = - RTln Keq (or ln(K1/K2) = ∆H°/R (1/T2 = 1/T1)
d) The absolute temperature at which the forward reaction becomes
nonspontaneous can be predicted. Write the equation that is used to make the
prediction. Why does this equation predict only an approximate value for the
temperature?
Since ∆G° = 0 at this point, the equation is T = ∆H°/∆S°.
7.
A chemical reaction occurs when 100. mL of 0.200 M HCl is added dropwise to
100. mL of 0.100 M Na3PO4 solution.
a) Write the two net ionic equations for the formation of the major products.
PO4 3- + H + <===> HPO4 2HPO4 2- + H + <===> H2PO4 b) Identify the species that acts as both a Brønsted acid and as a Brønsted base in the
equation is (a). Draw the Lewis electron-dot diagram for this species.
Base: PO4 3- /HPO4 2-
Acid: HPO4 2-/H2PO4 -
c) Sketch a graph using the axes provided, showing the shape of the titration curve
that results when 100. mL of the HCl solution is added slowly from a burette to
the Na3PO4 solution. Account for the shape of the curve.
Two protons are transferred, two buffers are produced, and two equivalence points
are found.
d) Write the equation for the reaction that occurs if a few additional mL of the HCl
solution are added to the solution resulting from the titration in (c).
H2PO4 - + H + <===> H3PO4
8.
For each of the following, use appropriate chemical principles to explain the
observation.
a) Sodium chloride may be spread on an icy sidewalk in order to melt the ice;
equimolar amounts of calcium chloride are even more effective.
The addition of a solute lowers the freezing point of water. NaCl dissociates into
2 moles of ions, whereas CaCl2 dissociates into 3 moles of ions.
b) At room temperature, NH3, is a gas and H2O is a liquid, even though NH3 has a
molar mass of 17 g and H2O has a mass of 18 g
Water is a liquid because the hydrogen-bonding forces are stronger between
adjacent H2O molecules than between adjacent NH3 molecules.
c) C(graphite) is used as a lubricant, whereas C(diamond) is used as an abrasive.
Graphite consists of 2-dimensional sheets of covalently bonded carbon atoms.
The attractive forces between sheets are weak London forces, which allow the
sheets to slide easily over one another.
Diamond consists of an extended 3-dimensional covalent network of carbon
atoms. This makes diamond a very hard substance.
d) Pouring vinegar onto the white residue inside a kettle used for boiling water
results in a fizzing/bubbling phenomenon.
Vinegar, a dilute solution of acetic acid, reacts with white solid, which contains
metal carbonates, in a neutralization reaction to form gaseous CO2.
9. Use principles of atomic structure and/or chemical bonding to answer each of the
following.
a) The radius of the Ca atom is 0.197 nm; the radius of the Ca2+ ion is 0.099 nm.
Account for this difference.
The outermost electron in Ca is in a 4s orbital, whereas the outermost electron in
Ca 2+ is in a 3p orbital.
b) The lattice energy of CaO(s) is - 3460 kJ/mole; the lattice energy for K2O(s) is 2240 kJ/mole. Account for this difference.
K
Ca
Ionization Energy (kJ/mol)
1st
2nd
419
3,050
590
1,140
Ca 2+ is smaller than K +, so internuclear separations are less, and Ca 2+ is more
highly charges than K +, thus cation O 2- bonds are stronger
c) Explain the difference between Ca and K in regard to
(i)
the first ionization energies,
Ca has more protons and is smaller. The outermost electrons are more
strongly held by the nuclear charge of Ca.
(ii)
their second ionization energies.
The outermost electrons in Ca are in the 4s, which is a higher energy orbital
(more shielded) than the 3p electron in K.
d) The first ionization energy of Mg is 738 kJ/mole and that of Al is 578 kJ/mole.
Account for this difference.
The highest energy (outermost) electron in Al is in a 3p orbital, whereas that
electron in Mg is in 3s orbital.
The 3p electron in Al is of higher energy (is more shielded) than is the 3s electron
in Mg.