AP Chem Exam - ‘99
NH3(aq) + H2O(l) <==> NH4 +(aq) + OH –(aq)
1.
In aqueous solution, ammonia reacts as represented above. In 0.0180 M NH3(aq) at
–4 M. In answering the
25 °C, the hydroxide ion concentration, [OH –
following, assume that temperature is constant at 25 °C and that volumes are
additive.
(a) Write the equilibrium-constant expression for the reaction represented above.
[NH4+][OH–]
Kb =
[NH3]
(b) Determine the pH of 0.0180 M NH3(aq).
pOH = - log(5.60 x10–4) = 3.252
pH = 10.748
(c) Determine the value of the base ionization constant, Kb, of NH3(aq).
Kb =
(5.60¥10–4)(5.60¥10–4)
= 1.80
(0.0180 – 5.60¥10–4)
–5
(d) Determine the percent ionization of NH3 in 0.0180 M NH3(aq).
5.60¥10–4
0.0180
(e) In an experiment, a 20.0 mL sample of 0.0180 M NH3(aq) was placed in a flask
and titrated to the equivalence point and beyond using 0.0120 M HCl(aq).
(i)
Determine the volume of 0.0120 M HCl(aq) that was added to reach
the equivalence point.
(0.0120 N)(VA) = (0.0180 N)(20.0 mL) = 30.0 mL
(ii) Determine the pH of the solution in the flask after a total of
15.0
mL of 0.0120 M HCl(aq) was added.
at 15.0 mL it is half-titrated and [NH4+] = [NH3],
then the Kb = [OH–] = 1.80
pOH = - log(1.80 x10–5) = 4.745
–5.
pH = 9.255
(iii) Determine the pH of the solution in the flask after a total of 40.0 mL of
0.0120 M HCl(aq) was added.
at 40.0 mL, there is an excess of 10.0 mL of HCl past equivalence point,
(0.010 L)(0.0120 M) = 0.00012 mol H + ions
1.2 x10-4 mol /0.06 L= 0.00200M = [H +]
pH = 2.70
2. Answer the following questions regarding light and its interactions with molecules,
atoms, and ions.
(a) The longest wavelength of light with enough energy to break the Cl–Cl boned in
Cl2(g) is 495 nm.
(i)
Calculate the frequency, in s–1, of the light.
c =
1014 waves/sec
(ii) Calculate the energy, in J, of a photon of the light.
10–34 joule.s)(6.06 1014 s–1) = 4.02 10–19 J
(iii) Calculate the minimum energy, in kJ mol–1, of the Cl–Cl bond.
(4.02 x10–19 J)(6.02 x1023 mol–1) = = 241,870 J = 242 kJ
(b) A certain line in the spectrum of atomic hydrogen is associated with the
electronic transition of the H atom from the sixth energy level (n = 6) to the second energy level (n = 2).
(i)
Indicate whether the H atom emits energy or whether it absorbs
energy during the transition. Justify your answer.
emits energy. The n = 6 state is at a higher energy than n = 2, the electron
must release energy go to the lower state.
(ii) Calculate the wavelength, in nm, of the radiation associated with the
spectral line.
1
1
∆E = – 2.178 10–18 (22 – 62 ) J = 4.84 10–19 J
(6.626¥10–34 J sec)(3.00¥1017 nm/sec)
 = hc/E =
= 411 nm
4.84¥10–19 J
(iii) Account for the observation that the amount of energy associated
with the same electronic transition (n = 6 to n = 2) in the He + ion is
greater than that associated with the corresponding transition in the
H atom.
The He + has a nuclear charge of 2 + vs H with a 1+, therefore, it has a
stronger hold on the electron. This requires more energy for the
electron to move to higher energy levels and when the electron
moves from higher to lower energies, it releases more energy.
3.
2 NO(g) + Br2(g)
(g)
A rate study of the reaction represented above was conducted at 25 °C. The data
that were obtained are shown in the table below.
Initial [NO]
Trial
(mol L–1)
1
2
3
0.0160
0.0160
0.0320
Initial [Br2] (mol
Initial Rate of Appearance of NOBr (mol
L -1 )
0.0120
0.0240
0.0060
L–1 s–1)
–4
–4
–4
(a) Calculate the initial rate of disappearance of Br2(g) in experiment 1.
Since the disappearance of 1 Br2 produces 2 NOBr, then the rate would be half as
much or rate = - 1
–4.
(b) Determine the order of the reaction with respect to each reactant, Br 2(g)and
NO(g). In each case, explain your reasoning.
[NO]2 [Br2]1
(c) For the reaction,
(i)
write the rate law that is consistent with the data
rate = k[NO]2[Br2]
(ii) calculate the value of the specific rate constant, k, and specify units.
k=
rate
3.24¥10–4 mol L–1 s–1
=
= 105 L2mol–2s–1
[NO]2[Br2]
(0.0160 mol L–1)2(0.0120 mol L–1)
(d) The following mechanism was proposed for the reaction:
Br2(g) + NO(g)
OBr2(g)
NOBr2(g) + NO(g)
slow
(g)
fast
Is this mechanism consistent with the given experimental observations?
Justify your answer.
No; since the rate determining step is the slowest step (and in this case, the first
step), then the rate for this proposed mechanism depends only on the
cencentration of the reactants in the first step and would be:
rate = k[NO][Br2]
4.(a)
Calcium oxide powder is added to distilled water.
CaO + H2
2
(or Ca 2+ + OH –)
(b) Solid ammonium nitrate is heated to temperatures above 300 °C.
NH4NO3
2
+ O2 + H2O (or N2O + H2O)
(c) Liquid bromine is shaken with 0.5M sodium iodide solution.
Br2 + I –
2
(or I3 –) + Br –
(d) solid lead(II) carbonate is added to a 0.5M sulfuric acid solution.
PbCO3 + H + + SO4 2–
4
+ CO2 + H2O
(e) A mixture of powdered iron(III) oxide and powdered aluminum metal is heat
strongly.
Fe2O3
2O3
+ Fe
(f) Methylamine gas is bubbled into distilled water.
(g)
+
–
CH3NH2 + H2
3NH3 + OH
(h) Carbon dioxide gas is passed over hot, solid sodium oxide.
CO2 + Na2
2CO3
(i) A 0.2M barium nitrate solution is added to an alkaline 0.2M potassium chromate
solution.
Ba 2+ + CrO4 2–
4
D (Required)
5. A student performs an experiment to determine the molar mass of an unknown gas.
A small amount of the pure gas is released from a pressured container and collected
in a graduated tube over water at room temperature, as shown in the diagram
above. The collection tube containing the gas is allowed to stand for several
minutes, and its depth is adjusted until the water levels inside and outside the tube
are the same. Assume that:
•
•
•
the gas is not appreciably soluble in water
the gas collected in the graduated tube and the water are in thermal
equilibrium
a barometer, a thermometer, and analytical balance, and a table of the
equilibrium vapor pressure of water at various temperatures are also
available.
(a) Write the equation(s) needed to calculate the molar mass of the gas.
barometric pressure
– vapor pressure water
= pressure of dry gas
mass of gas bottle, start
– mass of gas bottle, end
= mass of gas in collection tube
PV
mol of gas = n = RT ;
mass of gas
= molar mass
n
(b) List the measurements that must be made in order to calculate the molar mass of
the gas.
mass of gas bottle, start
mass of gas bottle, end
barometric pressure
volume of gas in collection tube
temperature of gas
look-up in handbook or table: equilibrium vapor pressure of water at
specific temperature and ideal gas constant, R
(c) Explain the purpose of equalizing the water levels inside and outside the gas
collection tube.
equilizing water levels equilizes the internal collection tube pressure to the
external barometric pressure.
(d) The student determines the molar mass of the gas to be 64 g mol –1. Write the
expression (set-up) for calculating the percent error in the experimental value,
assuming that the unknown gas is butane (molar mass 58 g mol–1). Calculations
are not required.
(64 g – 58 g)
58 g
(e) If the student fails to use information from the table of the equilibrium vapor
pressures of water in the calculation, the calculated value for the molar mass of
the unknown gas will be smaller than the actual value. Explain.
if the water vapor pressure is not subtracted from the total pressure, the
apparent gas pressure is too large; this gives a larger value for the number of
moles; in turn, since the mass of the gas is divided by too large a number of
moles, the molar mass is too small.
D (Required)
6. Answer the following question in terms of thermodynamic principles and concepts
of kinetic molecular theory.
(a) Consider the reaction represented below, which is spontaneous at 298 K.
CO2(g) + 2 NH3(g)
2)2(s)
+ H2O(l)
∆H˚298 = – 134 kJ
(ii) For the reaction, indicate whether the standard entropy change, ∆S˚298,
is positive, negative, or zero. Justify your answer.
∆S˚ is negative because three molecules of reactant make two
molecules of product (a decrease in entropy) and two different gases
make a solid and a liquid (both with smaller entropies).
(ii) Which factor, the change in enthalpy, ∆H˚298, or the change in entropy,
∆S˚298, provides the principle driving force for the reaction at 298 K?
Explain.
natural tendency to maximize entropy and since this reaction decreases
entropy and is spontaneous (-∆G˚), then ∆H˚ must be negative to
overcome the entropy change and drive this reaction.
(iii) ∆G˚ = ∆H˚ –T∆S˚ ; a T increase the value of –T∆S° increases and the
value of –∆G˚ becomes a smaller negative number (i.e., moves toward
zero).
For the reaction, how is the value of the standard free energy change,
∆G˚, affected by an increase in temperature? Explain.
(b) Some reactions that are predicted by their sign of ∆G˚ to be spontaneous at room
temperature do not proceed at a measurable rate at room temperature.
(i) Account for this apparent contradiction.
the sign of ∆G˚ (thermodynamics) does not account for activation
energy (kinetics); a large activation energy would effectively prevent a
reaction even though there is a favorable free energy change.
(ii) A suitable catalyst increases the rate of such a reaction. What effect
does the catalyst have on ∆G˚ for the reaction? Explain.
a catalyst changes neither the ∆H˚ nor the ∆S˚ for a reaction, therefore,
it will have no effect on the ∆G˚.
0.10M
NaF
0.10M
MgCl2
0.10M
C2H5OH
0.10M
CH3COOH
7. Answer the following questions, which refer to the 100 mL samples of aqueous
solutions at 25˚C in the stoppered flasks shown above.
(a) Which solution has the lowest electrical conductivity? Explain.
0.10 M C2H5OH, flask 3. It is the only solute that is not partially or wholly
ionized (or dissociated) since it is a covalently bonded molecule. Ions are needed
to conduct an electric current.
(b) Which solution has the lowest freezing point? Explain.
0.10 M MgCl2, flask 2. The freezing point depression is a colligative property,
and the magnesium chloride dissolves to give 3 particles (ions) for each dissolved MgCl2. In effect, the solution acts more like a 0.3 M solution, the van’t
Hoff factor.
(c) Above which solution is the pressure of the water vapor greatest? Explain.
0.10 M C2H5OH, flask 3. The vapor pressure is also a colligative property. Since
the other solutions give an effective concentration greater than their measured
molarity, the non-ionized ethyl alcohol will allow the water to vaporize the
easiest.
(d) Which solution has the highest pH? Explain.
0.10 M NaF, flask 1. HF is a weak acid and the F – ion will react with water to
produce hydroxide ions and a pH greater than 7.
F – + H2O  HF + OH –
The MgCl2 is essentially neutral as well as the C2H5OH, the CH3COOH is a
weak acid and will give a pH less than 7.
8. Answer the following questions using principles of chemical bonding and
molecular structure.
(a) Consider the carbon dioxide molecule, CO2, and the carbonate ion, CO3 2–.
(i) Draw the complete Lewis electron-dot structure for each species.
there are two other similar resonance structures for the carbonate ion.
(ii) Account for the fact at the carbon-oxygen bond length in CO3 2– is
greater than the carbon-oxygen bond length in CO2.
the pi O=C double bond in CO2 is shorter than a single O-C resonance
sigma-bond (all are identical and are about 1-1/3 bond) found in a carbonate
ion.
(b) Consider the molecules CF4 and SF4.
(i) Draw the complete Lewis electron-dot structure for each molecule.
••
F
•• •• ••
F C F
•• •• ••
F
••
(ii) In terms of molecular geometry, account for the fact that the CF4
molecule is nonpolar, whereas the SF4 molecule is polar.
in the tetrahedral CF4, the polar C-F bonds are cancelled out by the
equiangular pull of the 4 bonds. With an expanded octet and distorted
tetrahedral structure, SF4 has a pair of unbonded electrons at the center
of the bipyramid, this gives a “seasaw” shape to the molecule and an
uneven pull to the polar S-F bonds.