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AP Chemistry
Quarter 2
PRACTICE Exam 2 – A
January 13, 2012
This exam consists of two sections. You will have 55 minutes to complete it.
When you finish the multiple-choice section and turn in your scantron sheet, you may take out your calculator.
Name: _______________________________________________ Block: ______
Multiple-Choice Section – Write “Q2-2A” on the Test Name line of the scantron sheet.
You may use your periodic table only. You may NOT use a calculator.
Please record all answers on the scantron sheet.
You may use a MAXIMUM of 25 minutes on this section.
Use the reactions below to answer the following 3 questions.
A. Cu (s) + Zn(NO3)2 (aq)  Cu(NO3)2 (aq) + Zn (s)
B. H+(aq) + Cl-(aq) + Na+ (aq) + OH- (aq) NaCl (aq) + H2O (l)
C. Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq)  AgCl (s) + Na+ (aq) + NO3- (aq)
D. H+ (aq) + OH- (aq)  H2O (l)
E. CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
1) Which of the reactions represents a net ionic equation? D (all others have spectators or don’t have ions)
2) Which of the above reactions is a redox reaction? A, E
3) Which of the above reactions represents a precipitation reaction? C
10 HI + 2 KMnO4 + 3 H2SO4  5 I2 + 2 MnSO4 + 2 K2SO4 + 8 H2O
4) A chemist is running the reaction above. She would like to produce 2.5 mol of I2. She has 4.0 mol of
KMnO4 and 3.0 mol of H2SO4. How many mol of HI should she use?
(A) It doesn’t matter. There is an insufficient quantity of another reactant.
(B) 20. mol
(C) 10. mol
(D) 8.0 mol
(E) 5.0 mol
Look at the mole ratio between HI and I2
5) The balanced net ionic equation for a reaction between solid manganese(II) hydroxide, Mn(OH)2, and
aqueous nitric acid, HNO3, is
(A) Mn(OH)2 + 2 HNO3  Mn(NO3)2 + 2 H2O
(B) Mn2+ + 2 NO3-  Mn(NO3)2
(C) Mn(OH)2 + 2 H+  Mn2+ + 2 H2O
(D) H+ + OH-  H2O
(E) OH- + HNO3  H2O + NO33 Ag(s) + 4 HNO3(aq)  3 AgNO3(aq) + NO(g) + 2 H2O
6) The reaction of silver metal and dilute nitric acid proceeds according to the equation above. If 0.10 mol of
powdered silver is added to 10. milliliters of 6.0-molar nitric acid, the number of moles of NO gas that can
be formed is
(A) 0.015 mol
(B) 0.020 mol
(C) 0.030 mol
(D) 0.045 mol
(E) 0.090 mol
7) Which of the following best describes the role of the spark from the spark plug in an automobile engine?
(A) The spark decreases the activation energy for the slow step.
(B) The spark increases the concentration of the volatile reactant.
(C) The spark provides a more favorable activated complex for the combustion reaction.
(D) The spark provides the heat of vaporization for the volatile hydrocarbon.
(E) The spark supplies some of the activation energy for the combustion reaction.
Experiment
Initial [NO]
(mol L-1)
Initial [O2]
(mol L-1)
Initial Rate of
Formation of NO2
(mol L-1s-1)
1
0.10
0.10
2.5 x 10-4
2
0.20
0.10
5.0 x 10-4
3
0.20
0.40
8.0 x 10-3
NO(g) + 1/2 O2(g)  NO2(g)
8) The initial-rate data in the table above were obtained for the reaction represented above. What is the
experimental rate law for the reaction?
(A) Rate = k[NO][O2]
(C) Rate = k[NO][O2]2
(E) Rate = k[NO]2[O2]
(B) Rate = k[NO]2[O2]2
(D) Rate = k[O2]2
9) When the concentration of B in the reaction below is doubled, all other factors being held constant, it is
found that the rate of the reaction remains unchanged.
2 A(g) + B(g)  2 C(g)
The most probable explanation for this observation is that
(A) The order of the reaction with respect to substance B is 1.
(B) Substance B is not involved in any of the steps of the mechanism of the reaction.
(C) Substance B is probably a catalyst, and as such, its effect on the rate of reaction does not depend on
concentration.
(D) Substance B is not involved in the rate-determining step of the mechanism, but is involved in subsequent
steps.
(E) The reactant with the smallest coefficient in the balanced equation generally has little or no effect on the
rate of the reaction.
10) Which of the following statements about a catalyst is NOT true?
(A) It increases the value of the rate constant, k.
(B) It makes the reaction proceed more quickly.
(C) It decreases the time it takes for the reaction to finish.
(D) It is used up in the reaction.
(E) It lowers the activation energy.
___C3H7OH + ___ O2  ___ CO2 + ___ H2O
One mole of C3H7OH underwent combustion as shown in the unbalanced reaction above. How many
moles of oxygen were consumed?
(A) 2 moles
(B) 3 moles
(C) 7/2 moles
(D) 9/2 moles
(E) 5 moles
11)
12)
2MnO4- + 5SO32- + 6H+  2Mn2+ + 5SO43- + 3H2O
Which of the following is true regarding the reaction above?
(A) MnO4- acts as the reducing agent.
(B) H+ acts as the oxidizing agent.
(C) SO32- acts as the reducing agent.
(D) Manganese is oxidized.
(E) Sulfur is reduced.
The four questions that follow refer to this reaction mechanism:
C2H4 + H+  C2H5+
C2H5+ + H2O  C2H7O+
C2H7O+  C2H6O + H+
13) Which of these gives the correct overall equation for this reaction?
(A) C2H4 + H+ + H2O  C2H6O
(B) C2H4 + H2O  C2H6O
+
+
(C) C2H7O  C2H4 + H3O
(D) C2H4 + H+  C2H6O
(E) C2H4 + H+ + C2H5+ + H2O  C2H6O + H+ + C2H5+
14) If the first step in this reaction mechanism is rate-limiting, what is the rate law for the overall reaction?
(A) Rate = k[C2H4][H+]
(B) Rate = k[C2H4]
(C) Rate = k[C2H4][H2O]
+
+
(D) Rate = k[C2H5 ][H2O]
(E) Rate = k[H ][C2H4][H2O]
15) Which of the following could represent a catalyst in this reaction?
I. H2O
II. H+
III. C2H5+
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
16) Which of the following represent(s) an intermediate in this reaction?
I. H+
II. C2H7O+
III. C2H5+
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
17)
S(s) + O2(g)  SO2(g)
S(s) + 3/2O2(g)  SO3(g)
Ho=x
Ho=y
Based on the information above, what is the standard enthalpy change for the following reaction?
2SO2(g) + O2(g)  2SO3(g)
(A) x-y
18)
(B) y-x
(C) 2x-y
(D) 2x-2y
(E) 2y-2x
H2 (g) + Cl2 (g)  2HCl (g)
Based on the information given below, what is the Ho of the reaction above?
Bond
H-H
Cl-Cl
H-Cl
(A) -860 kJ/mol
(B) -620 kJ/mol
Bond Energy (kJ/mol)
440
240
430
(C) -440 kJ/mol
(D) -180 kJ/mol
(E) 240 kJ/mol
19) A student added a solution of salt A to an aqueous solution of sodium iodide. After the salt was added, a
precipitate was observed. Which of the following could have been salt A. NO ANSWER Should be II+III
I. BaCl2
II. Pb(NO3)2
III. AgC2H3O2
(A) I only
(B) III only
(C) I + III only
(D) I + II only
(E) I, II, + III
20) Which of the following chemical equations represents the balanced chemical reaction that occurs when
calcium carbonate is placed into excess hydrochloric acid
(A) CaCO3(s) + HCl(aq)  CaCl2(aq) + H2CO3 (aq)
(B) Ca2C(s) + 4HCl(aq)  2CaCl2(aq) + H4C (aq)
(C) CaCO2(s) + HCl(aq)  CaCl2(aq) + HCO2 (aq)
(D) CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O (l) + CO2 (g)
(E) CaCO4(s) + 2HCl(aq)  CaCl2(aq) + H2CO4 (aq)
Open Response Section
You may use your periodic table, formula sheet, and calculator.
Please write all answers in the space provided.
1) Answer these questions about the laboratory situation described. Powdered calcium is added to a solution
of copper(II) chloride, CuCl2(aq). The powder disappears and a red-brown solid appears.
(a) Write a complete balanced equation for this reaction, including states of matter.
Ca(s) + CuCl2(aq)  CaCl2(aq) + Cu(s)
(b) Write a balanced net ionic equation for this reaction, including states of matter.
Ca(s) + Cu2+(aq)  Ca2+(aq) + Cu(s)
(c) Identify the oxidizing and reducing agents in this reaction.
Oxidizing agent = __Copper________
Reducing agent = _____Calcium________________
2) Several chemical equations are shown below. For each one, predict the products and write a balanced net
ionic equation, including states of matter. (4 pt each)
a) HC2H3O2(aq) + Ca(OH)2(aq)  H2O(l) + Ca(C2H3O2)2(aq)
H+(aq) + OH-(aq)  H2O(l)
b) Mg(OH)2(s) + HCl(aq)  H2O(l) +MgCl2(aq)
H+(aq) + OH-(aq)  H2O(l)
c) Al(s) + AgNO3(aq)  Al(NO3)3(aq) + Ag(s)
Al(s) + 3Ag+(aq)  Al3+(aq) + Ag(s)
d) Pb(NO3)2(aq) + CaBr2(aq)  PbBr2(s) + Ca(NO3)2(aq)
Pb2+(aq) + 2Br-(aq)  PbBr2(s)
3) Some metals can be used to trap hydrogen gas as a hydride. In one experiment, 0.780 g Nb(s) was sealed in a
28.0 mL glass tube at 25°C under 7.33 atm of hydrogen gas, H2. After reacting with the hydrogen for one week,
all of the niobium had been converted to niobium hydride, NbH.
2 Nb(s) + H2(g)  2 NbH(s)
a. Calculate the number of moles of hydrogen gas that are used up in the reaction.
Convert grams of Nb to moles : 0.780 g/92.90 g/mol = 0.00840 mol Nb
Use mole ratio to find moles of H2 used : 0.00840 mol Nb * (1 mol H2/2 mol Nb) = 0.00420 mol
b. Calculate the final pressure of hydrogen gas in the system at 25°C.
Calculate pressure of “used” H2 : PV=nRT: P(0.028L)=(0.00420mol)(0.082 atm*L/mol*K)(298K) :
P = 3.67 atm
Final P = Initial P – Used P = 7.33 – 3.67 = 3.66 atm
Note: You could also solve this by using initial moles and final moles, but I think this way is more efficient.
2 NO(g) + Br2(g)  2 NOBr(g)
4) 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 rate of
Initial [NO]
Initial [Br2]
appearance of
Experiment
(mol L-1)
(mol L-1)
NOBr
(mol L-1 s-1)
1
0.0163
0.0120
3.24 x 10-4
2
0.0159
0.0240
6.46 x 10-4
3
0.0321
0.0061
6.51 x 10-4
4
0.0160
0.0060
1.62 x 10-4
(a) Calculate the initial rate of disappearance of Br2(g) in experiment 1.
½*[NOBr]/t =[Br2]/t = (3.24 x 10-4 mol L-1 s-1)/2 = 1.62 x 10-4 mol L-1 s-1
(b) Determine the order of the reaction with respect to each reactant, Br2(g) and NO(g). In each case, explain
your reasoning or provide calculations to justify your answer.
Order of reaction for NO: look at experiments 3 + 4 (keep [Br2] constant). The initial [NO] in run 3 is 2x the
initial [NO] of run 4 and the initial rate of appearance is 4x. 2x = 4, x=2
Order of reaction for Br2: look at experiments 1+2 (or 1+4 or 2+4). This initial [Br2] in run 2 is 2x the initial
[Br2] in run 1. The initial rate in run 2 is 2x the initial rate in run 1. 2x=2, x=1
(c) For the reaction,
(i) write the rate law that is consistent with the data, and
(ii) calculate the value of the specific rate constant, k, and specify units.
(i) r=k[NO]2[Br2]
(ii) choose any experiment # and plug in the numbers: run 1:
3.24 x 10-4 mol L-1 s-1 = k[0.0163 mol/L]2[0.0120 mol/L]
k=102 L2 mol-2 s-1
(d) The following mechanism was proposed for this reaction:
Br2(g) + NO(g)  NOBr2(g)
fast (equilibrium)
NOBr2(g) + NO(g)  2 NOBr(g)
slow
Is this mechanism consistent with the given experimental observations? Justify your answer.
The above mechanism is consistent with experimental observations. Since the second step is the rate limiting
step, the rate law is: r=k2[NOBr2][NO]. However, we do not want an intermediate in our rate law, so we have
to substitute for [NOBr2] so we use reaction 1 (at equilibrium). k-1[NOBr2] = k1[Br2][NO]. Substituting this
into the original rate law equation we get: r=(k2*k1/k-1)[Br2][NO]2. The “k” values simplify to a single k (since
they are all constants), leaving us with our experimentally determined rate law. This means the mechanism is
consistent.
5) 12.2 g of acetylene (C2H2) is combusted in a calorimeter containing 2.00 L of water. The temperature of the
water rose from 25.0 oC to 39.2 oC.
a. Based on the experimental data, determine the amount of energy absorbed by the water.
Q=mcT : m = density * volume : m = 2000 g
Q = 2000 g * 4.184 J g-1 oC-1 * 14.2 oC = 119 kJ
b. Based on the experimental data, determine the molar heat of combustion, ΔHcomb, for acetylene.
-Q = ΔHcomb*mol (remember, heat absorbed by water was removed from acetylene) : 119 kJ = ΔHcomb * (12.2 g
/ 26 g mol-1) : ΔHcomb = 119 kJ/0.469 mol = -254 kJ/mol
The accepted value for the heat of combustion for acetylene is -227 kJ/mol.
c. What is the percent error in your experimental value?
% error = | actual – experimental / actual | * 100% = |(-227 - (-254))/-227| = 11.9% error
d. If the system allowed heat to escape into the surroundings, would that explain the error above?
If the system allowed heat to escape into the surroundings, some of the heat given off during combustion would
be lost. This would mean that the energy going into the water would be lower. We would therefore expect a
calculated ΔHcomb that is lower than the accepted value. Since we obtained a higher number, this cannot explain
our error.
e. If the sample of acetylene was not pure, but contained traces of ethane, C2H6, would that explain the error?
This problem can be answered in a couple of different ways, but we will focus on comparing the bonds made
and the bonds broken in the combustion of C2H6 and C2H2. In the combustion of 1 mol of C2H6 we break 1 CC  bond and 6 C-H bonds. We make 4 C=O bonds and 6 H-O bonds. In the combustion of 1 mol of C2H2 we
break 1 C-C  bond, 2 C-C  bonds, and 2 C-H bonds. We make 4 C=O bonds and 2 H-O bonds. If you make
a logical explanation from here, you would receive credit (i.e. there are more net bonds made with C2H2 so it
will release more energy or  bonds are less energetic than  bonds so it takes less energy to break them).
To completely solve the problem, we should use the heat of combustion per gram (instead of molar), which we
don’t have in this question. We were looking for qualitative explanations.