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Advanced 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.
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 is an acid base neutralization? B, D
2) Which of the above reactions is a combustion reaction? 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. How many mol of HI
should she use?
(A) 20 mol
(B) 10 mol
(C) 8.0 mol
(D) 5.0 mol
(E) 2.5 mol
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)
5) 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]
2
2
2
(B) Rate = k[NO] [O2]
(D) Rate = k[O2]
6) 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.
7) 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, so 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 other steps.
(E) The reactant with the smallest coefficient in the balanced equation generally has little or no effect on the rate
of the reaction.
2 C3H7OH + 9 O2  6 CO2 + 8 H2O
One mole of C3H7OH underwent combustion as shown in the 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
8)
2MnO4- + 5SO32- + 6H+  2Mn2+ + 5SO43- + 3H2O
Which of the following is true regarding the reaction above?
(A) MnO4- is oxidized.
(C) SO32- is oxidized.
(B) H+ is reduced.
(D) Manganese is oxidized.
9)
(E) Sulfur is reduced.
The four questions that follow refer to this reaction mechanism:
C2H4 + H+  C2H5+
C2H5+ + H2O  C2H7O+
C2H7O+  C2H6O + H+
10) 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
11) 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]
12) 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
13) 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
14)
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
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 _____________________
(d) If 50.0 mL of a 0.471 M copper(II) chloride solution is used,
i)
calculate the number of moles of CuCl2(aq) in this amount of solution.
50.0 ml x (1 L / 1000 mL) * (0.471 mol / 1 L) = 0.0236 mol
ii)
calculate the number of moles of Cu(s) that can be produced.
Multiply the answer to i) by the mole ratio. In this case the mole ratio is 1:1, so the answer is 0.0236 mol.
iii)
calculate the mass of Cu(s) that can be produced.
Multiply the number of moles by the molar mass of Cu: 0.0236 mol * 63.55 g / mol = 1.50 g
3) Some metals can be used to trap hydrogen gas as a hydride. In one experiment, 0.680 g Nb(s) was sealed in a
0.0260 L glass tube at 298 K under 6.33 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 Nb that are in the original 0.680 g sample.
Convert grams of Nb to moles : 0.780 g/92.90 g/mol = 0.00840 mol Nb
b. Calculate the number of moles of hydrogen gas (H2) that would be needed to react with the niobium
sample.
Use mole ratio to find moles of H2 used : 0.00840 mol Nb * (1 mol H2/2 mol Nb) = 0.00420 mol
c. Calculate the number of moles of hydrogen gas that were originally present in the 0.0260 L container at
298 K and 6.33 atm.
Use PV = nRT: (6.33 atm)(0.0260 L) = n (0.082 atm L mol-1 K-1)(298K) : n = 0.00674 mol
d. Calculate the number of moles of hydrogen gas that remain after the reaction is complete.
Subtract answer b from answer c : 0.00674 – 0.00420 = 0.00254 mol remain
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) 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
(b) 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
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, in kJ/mol.
-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.