SPOTLIGHT EDUCATION INSTITUTE ▪ AP CHEMISTRY
2009 AP CHEMISTRY FREE RESPONSE EXAM
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2009 AP Chemistry Free Response Exam (5/14/2009, 6:00 pm)
(This is my first draft; please let me know of any apparent errors; point distributions are entirely guesses.)
1. Answer the following questions that relate to the chemistry of halogen oxoacids.
(a)
Use the information in the table below to answer part (a)(i).
Acid
Ka at 298 K
HOCl
2.9 × 10–8
HOBr
2.4 × 10–9
(i) Which of the two acids is stronger, HOCl or HOBr ? Justify your answer in
terms of Ka.
HOCl is a stronger acid. A stronger acid dissociates to a greater degree.
Because the Ka for HOCl is larger than that of HOBr, HOCl dissociates more
than HOBr, and is a stronger acid.
(ii) Draw a complete Lewis electron-dot diagram for the acid that you
identified in part (a)(i).
H
O Cl
(iii) Hypoiodous acid has the formula HOI. Predict whether HOI is a stronger
acid or a weaker acid than the acid that you identified in part (a)(i). Justify your
prediction in terms of chemical bonding.
(b)
(c)
Cl is more electronegative than I, so it polarizes the electrons away from the
H–O bond, resulting in a weaker bond. It is easier to remove the H + from
HOCl than from HOI, and HOI is a weaker acid than HOCl.
Write the equation for the reaction that occurs between hypochlorous acid and
water.
+1
+1
+1
+1
HOCl + H2O  OCl– + H3O+
A 1.2 M NaOCl solution is prepared by dissolving solid NaOCl in distilled water at 298 K. The
hydrolysis reaction OCl− (aq) + H2O(l) HOCl (aq) + OH− (aq) occurs.
(i) Write the equilibrium-constant expression for the hydrolysis reaction that
occurs between OCl– (aq) and H2O (l) .
[OCl–][H3O+]
[HOCl]
(ii) Calculate the value of the equilibrium constant at 298 K for the hydrolysis
reaction.
+1
Kb =
+1
Kw 1.0 × 10–14
Kb = K = 2.9 × 10–8 = 3.4 × 10–7
a
(iii) Calculate the value of [OH–] in the 1.2 M NaOCl solution at 298 K .
M
I
C
E
(d)
OCl− + H2O
1.2
–x
1.2–x
HOCl + OH−
0
0
+x
+x
x
x
Kb =
[OCl–][H3O+]
= 3.4 × 10–7
[HOCl]
x2
= 1.2 – x = 3.4 × 10–7
x = 6.4 × 10–4 M = [OH–]
A buffer solution is prepared by dissolving some solid NaOCl in a solution of
HOCl at 298 K. The pH of the buffer solution is determined to be 6.48.
+2
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2009 AP CHEMISTRY FREE RESPONSE EXAM
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(i) Calculate the value of [H3O+] in the buffer solution.
+1
[H3O+] = 10–pH = 10–6.48 = 3.3 × 10–7 M
(ii) Indicate which of HOCl (aq) or OCl− (aq) is present at the higher
concentration in the buffer solution. Support your answer with a calculation.
+1
[A–]
For a buffer, pH = pKa + log [HA]
6.48 = – log (2.9 × 10–8) + log
[OCl–]
[OCl–]
= 7.54 + log
[HOCl]
[HOCl]
[OCl–]
log [HOCl] = 6.48 – 7.54 = –1.06.
[OCl–]
[OCl–]
Because log [HOCl] < 0, [HOCl] < 1, so [HOCl] > [OCl–]
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2009 AP CHEMISTRY FREE RESPONSE EXAM
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2. A student was assigned the task of determining the molar mass of an unknown gas. The student
measured the mass of a sealed 843 mL rigid flask that contained dry air. The student then flushed the flask
with the unknown gas, resealed it, and measured the mass again. Both the air and the unknown gas were at
23.0°C and 750. torr. The data for the experiment are shown in the table below.
Volume of sealed flask 843 mL
Mass of sealed flask and dry air 157.70 g
Mass of sealed flask and unknown gas 158.08 g
(a)
Calculate the mass, in grams, of the dry air that was in the sealed flask. (The density
+1
of dry air is 1.18 g L−1 at 23.0°C and 750. torr.)
(b)
1 L  1.18 g
843 mL 
= 0.995 g
1000
mL  1 L 

Calculate the mass, in grams, of the sealed flask itself (i.e., if it had no air in it).
+1
(c)
157.70 g – 0.995 g = 156.71 g
Calculate the mass, in grams, of the unknown gas that was added to the sealed
flask.
+1
(d)
158.08 g – 156.71 g = 1.37 g
Using the information above, calculate the value of the molar mass of the unknown
gas.
+2
1 atm
P = 750. torr 760 torr = 0.987 atm


T = 23.0 + 273.15 = 296.2 K
PV
(0.987 atm)(0.843 L)
n = RT = (0.0821 atm L mol–1 K–1)(296.2 K) = 0.0342 mol
m
1.37 g
MM = n = 0.0342 mol = 40.0 g mol–1
After the experiment was completed, the instructor informed the student that the unknown gas was carbon
dioxide (44.0 g mol−1).
(e)
Calculate the percent error in the value of the molar mass calculated in part (d).
|theoretical – experimental|
× 100%
theoretical
|44.0 – 40.0|
=
× 100% = 9.09%
44.0
% error =
+1
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2009 AP CHEMISTRY FREE RESPONSE EXAM
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(f)
Date:
For each of the following two possible occurrences, indicate whether it by itself
could have been responsible for the error in the student’s experimental result. You
need not include any calculations with your answer. For each of the possible
occurrences, justify your answer.
+2
Occurrence 1: The flask was incompletely flushed with CO2 (g) , resulting in some
dry air remaining in the flask.
If the flask was incompletely flushed, then some of the mass is from the dry air.
Because the density of dry air is less than that of CO 2, the number of moles in the
flask is too high, so the calculated molar mass would be too small. This would
account for the error. We can also notice that the density of a gas is directly
related to the molar mass (D = (P × MM)/(R × T)), so a lower density would
result in a lower molar mass.
Occurrence 2: The temperature of the air was 23.0°C, but the temperature of the
CO2 (g) was lower than the reported 23.0°C.
(g)
If the actual temperature was lower than 23.0°C, then the calculated moles of gas
would be too small, resulting in a molar mass that is too large. This would not
account for the error, because the calculated molar mass is too small.
Describe the steps of a laboratory method that the student could use to verify that
the volume of the rigid flask is 843 mL at 23.0°C. You need not include any
calculations with your answer.
Fill the flask completely with distilled water, and weigh. Because water has a
density of 1.0 g mL–1, it should weigh 843 g more than the mass of the empty
flask (calculated in (b)).
+1
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2009 AP CHEMISTRY FREE RESPONSE EXAM
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CH4 (g) + 2 Cl2 (g)  CH2Cl2 (g) + 2 HCl (g)
3. Methane gas reacts with chlorine gas to form dichloromethane and hydrogen chloride, as represented by
the equation above.
(a)
A 25.0 g sample of methane gas is placed in a reaction vessel containing 2.58 mol of Cl2 (g).
(i) Identify the limiting reactant when the methane and chlorine gases are
combined. Justify your answer with a calculation.
+3
1 mol CH4
25.0 g CH4 16.04 g CH  = 1.56 mol CH4


4
2 mol Cl2
If all of CH4 reacts, then we need 1.56 mol CH4 1 mol CH  = 3.12 Cl2.


4
Only 2.58 mol Cl2 are available, so Cl2 is the limiting reactant.
(ii) Calculate the total number of moles of CH2Cl2 (g) in the container after the
limiting reactant has been totally consumed.
+1
1 mol CH2Cl2
2.58 mol Cl2  2 mol Cl  = 1.29 mol CH2Cl2

2

Initiating most reactions involving chlorine gas involves breaking the Cl–Cl bond, which has a bond energy
of 242 kJ mol–1.
(b)
Calculate the amount of energy, in joules, needed to break a single Cl–Cl bond.
+1
(c)
242 kJ
 1 mol Cl–Cl bond  1000 J
–19
1 mol Cl–Cl bond 6.02 × 1023 Cl–Cl bond  1 kJ  = 4.02 × 10 J
Calculate the longest wavelength of light, in meters, that can supply the energy per
photon necessary to break the Cl–Cl bond.
+1
hc
hc (6.626 × 10–34 J s)(3.0 × 108 m s–1)
 = =
= 4.94 × 10–7 m
E
(4.02 × 10–19 J)

The following mechanism has been proposed for the reaction of methane gas with chlorine gas. All species
are in the gas phase.
Step 1 Cl2 2 Cl
fast equilibrium
Step 2 CH4 + Cl  CH3 + HCl
slow
Step 3 CH3 + Cl2  CH3Cl + Cl fast
Step 4 CH3Cl + Cl  CH2Cl2 + H fast
Step 5 H + Cl  HCl
fast
(e)
In the mechanism, is CH3Cl a catalyst, or is it an intermediate? Justify your answer.
+1
E = h =
(f)
CH3Cl is an intermediate because it is produced in step 3 and used up in step 4.
Identify the order of the reaction with respect to each of the following according to the mechanism.
In each case, justify your answer.
(i) CH4 (g)
+1
The reaction is first order with respect to CH4, because it is first-order in step 2,
the rate-determining step.
(ii) Cl2 (g)
The reaction is 1/2-order with respect to Cl2. The rate law of step 2, the ratedetermining step, is rate = k[Cl][CH4]. Because Cl is an intermediate, we can
obtain [Cl] from the first, fast-equilibrium step. In equilibrium, the rates of
forward and reverse reactions are equal, so [Cl] 2 = [Cl2], so [Cl] = [Cl2]1/2.
+1
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2009 AP CHEMISTRY FREE RESPONSE EXAM
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4. For each of the following three reactions, in part (i) write a balanced equation for the reaction and in part
(ii) answer the question about the reaction. In part (i), coefficients should be in terms of lowest whole
numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions
as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are
unchanged by the reaction. You may use the empty space at the bottom of the next page for scratch work,
but only equations that are written in the answer boxes provided will be graded.
(a)
A sample of solid iron(III) oxide is reduced completely with solid carbon.
5 points
(a) Balanced equation:
2 Fe2O3 (s) + 3 C (s)  3 CO2 (g) + 4 Fe (s)
(b) What is the oxidation number of carbon before the reaction, and what is the
oxidation number of carbon after the reaction is complete?
(b)
Carbon changes from an oxidation number of 0 before the reaction to +4 after
the reaction is complete.
Equal volumes of equimolar solutions of ammonia and hydrochloric acid are
combined.
(a) Balanced equation:
5 points
NH3 (aq) + H+ (aq)  NH4+ (aq)
(b) Indicate whether the resulting solution is acidic, basic, or neutral. Explain.
(c)
The resulting solution is acidic because NH4+ is the conjugate acid of NH3.
Solid mercury(II) oxide decomposes as it is heated in an open test tube in a fume
hood.
(a) Balanced equation:
2 HgO (s)  2 Hg (l) + O2 (g)
(b) After the reaction is complete, is the mass of the material in the test tube
greater than, less than, or equal to the mass of the original sample? Explain.
The mass of the materials in the test tube is less than the mass of the original
sample because the gaseous product is released from the open test tube.
5 points
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5. Answer the following questions using the information related to reactions X, Y, and Z in the table above.
Reaction
Equation
∆H°298
∆S°298
∆G°298
–1
–1
–1
C (s) + H2O (g) CO (g) + H2 (g)
X
+131 kJ mol
+134 J mol K
+91 kJ mol–1
Y
CO2 (g) + H2 (g)
Z
(a)
2 CO (g)
CO (g) + H2O (g)
C (s) + CO2 (g)
+41 kJ mol–1
+42 J mol–1 K–1
+29 kJ mol–1
?
?
?
For reaction X, write the expression for the equilibrium constant, Kp .
(PCO)(PH2)
(PH2O)
For reaction X, will the equilibrium constant, Kp, increase, decrease, or remain the
same if the temperature rises above 298 K? Justify your answer.
+1
Kp =
(b)
(c)
(d)
(e)
Reaction X is an endothermic reaction, so an increase in temperature will increase
Kp. (An increase in temperature will cause the reaction to shift in the forward
direction, because when Kp increases, Q < Kp).
For reaction Y at 298 K, is the value of Kp greater than 1, less than 1, or equal to 1?
Justify your answer.
For reaction Y, ∆G °> 0. Because ∆G° = – RT ln Kp, ln Kp must be negative, so Kp is
less than 1.
For reaction Y at 298 K, which is larger: the total bond energy of the reactants or the
total bond energy of the products? Explain.
In reaction Y, ∆H° > 0, so the energy required to break the bonds in the reactants
is greater than the energy released when the bonds are formed in the products.
The total bond energy of the reactants are greater than the total bond energy of
the products.
Is the following statement true or false? Justify your answer.
+1
+1
+1
+1
“On the basis of the data in the table, it can be predicted that reaction Y will
occur more rapidly than reaction X will occur.”
(f)
The statement is false. The table provides thermodynamic data and can be used to
determine the spontaneity of the reactions. However, it does not provide any
kinetic information to determine the speed of the reactions.
Consider reaction Z at 298 K.
(i) Is ∆S° for the reaction positive, negative, or zero? Justify your answer.
+1
∆S is positive, because the system is decreasing in entropy (less disordered) when
going from gaseous reactants to solid product.
(ii) Determine the value of ∆H° for the reaction.
+1
Reverse Reaction X: CO (g) + H2 (g)  C (s) + H2O (g) ∆H = –131 kJ mol–1
Reverse Reaction Y: CO (g) + H2O (g)  CO2 (g) + H2 (g)∆H = –41 kJ mol–1
Reaction Z:
2 CO (g)  C (s) + CO2 (g)
∆H = –172 kJ mol–1
SPOTLIGHT EDUCATION INSTITUTE ▪ AP CHEMISTRY
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(iii) A sealed glass reaction vessel contains only CO (g) and a small amount of C (s). If
a reaction occurs and the temperature is held constant at 298 K, will the pressure in
the reaction vessel increase, decrease, or remain the same over time? Explain.
Because there is no CO2, reaction Z will proceed forward, decreasing CO and
forming CO2. According to the balanced equation, for every 2 moles of CO used,
1 mole of CO2 will be produced; as the reaction proceeds forward, there will be
fewer moles of gas in the vessel, so the pressure will decrease.
+1
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6. Answer the following questions related to sulfur and one of its compounds.
(a)
(b)
(c)
(d)
Consider the two chemical species S and S2–.
(i) Write the electron configuration (e.g., 1s2 2s2 …) of each species.
+2
S: 1s2 2s2 2p6 3s2 3p4
S2–: 1s2 2s2 2p6 3s2 3p6
(ii) Explain why the radius of the S2− ion is larger than the radius of the S atom.
+1
Both S and S2– have 16 protons. S has 16 electrons and S2– has 18 electrons.
Because S2– has more electrons than S, there are more repulsions, so the radius is
larger.
(iii) Which of the two species would be attracted into a magnetic field? Explain.
+1
In S, there are two unpaired electrons in the 3p orbital, so S is paramagnetic, and
will be attracted into a magnetic field. InS2–, all the electrons are paired, so S2– is
diamagnetic, and will not be attracted in a magnetic field.
The S2− ion is isoelectronic with the Ar atom. From which species, S 2− or Ar, is it
easier to remove an electron? Explain.
S2– has 16 protons, Ar has 18 protons. S2– has fewer protons than Ar, so it has a
weaker effective nuclear charge, and the valence electrons are less attracted to the
nucleus. It will require less energy to remove an electron from S 2– than from Ar.
In the H2S molecule, the H–S–H bond angle is close to 90°. On the basis of this
information, which atomic orbitals of the S atom are involved in bonding with the H
atoms?
+1
+1
3p orbitals.
Two types of intermolecular forces present in liquid H2S are London (dispersion) forces and dipoledipole forces.
(i) Compare the strength of the London (dispersion) forces in liquid H2S to the
strength of the London (dispersion) forces in liquid H2O. Explain.
H2S has more electrons than H2O, so it is more polarizable and will have stronger
London (dispersion) forces than H2O.
(ii) Compare the strength of the dipole-dipole forces in liquid H2S to the strength
of the dipole-dipole forces in liquid H2O. Explain.
S is less electronegative than O, so the H–S bond is less polar than H–O bond.
H2S will have weaker dipole-dipole forces than H2O.
+1
+1
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2009 AP CHEMISTRY FREE RESPONSE EXAM
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Approximate Scoring
Question 1:
/10 × 1.5000
=
Question 2:
/9 × 1.6667
=
Question 3:
/9 × 1.6667
=
Question 4:
/15 × 0.5000
=
Question 5:
/8 × 1.40625
=
Question 6:
/8 × 1.40625
=
Total:
AP Score
5
4
3
2
1
/75
FR Total
55
45
35
25
0