Exam #2 F93

CHEM 1515
Exam II
John II. Gelder
October 14, 1993
Name ________________________
TA's Name ________________________
Lab Section ______________________
INSTRUCTIONS:
1. This examination consists of a total of 8 different pages.
The last two pages include a periodic table, a solubility
table and some useful equations. All work should be done
in this booklet.
2. PRINT your name, TA's name and your lab section
number now in the space at the top of this sheet. DO
NOT SEPARATE THESE PAGES.
3. Answer all questions that you can and whenever called
for show your work clearly. Your method of solving
problems should pattern the approach used in lecture.
You do not have to show your work for the multiple
choice or short answer questions.
4. No credit will be awarded if your work is not shown in
problems 2 – 5.
5. Point values are shown next to the problem number.
6. Budget your time for each of the questions. Some
problems may have a low point value yet be very
challenging. If you do not recognize the solution to a
question quickly, skip it, and return to the question after
completing the easier problems.
7. Look through the exam before beginning; plan your
work; then begin.
8. Relax and do well.
Page 2 Page 3 Page 4 Page 5 Page 6 TOTAL
SCORES _____ _____ _____ _____ _____ ______
(33) (20) (23) (12) (12) (100)

CHEM 1515 Exam II PAGE 2
(12) 1. Write the chemical formula(s) of the product(s) and balance the following reactions. Identify the phase
of each product as either (g)as, (l)iquid, (s)olid or (aq)ueous. Soluble ionic compounds should be
written in the form of their component ions.
a) MgO(s) + CO 2 (g) →
b) Al(s) + Fe 2 O 3 (s) →
c) Na 2 CO 3 (aq) + Ca(NO 3 ) 2 (aq) →
(21) 2. The rate constant for the second order decomposition of NOBr
is 0.80 M –1·s –1 at 10 ˚C.
2NOBr(g) → 2NO(g) + Br 2 (g)
a) If the concentration of NOBr is initially 0.0289 M, calculate the concentration of NOBr after 75.0
seconds.
b) Calculate the half-life for this reaction under the same set of conditions as described in part a).
c) How long would it take for 90.0% of the NOBr to react Assume the same initial conditions as
described in part a).

CHEM 1515 Exam II PAGE 3
(20) 3. The half-life for the first order decomposition of N 2 O
2N 2 O(g) → 2N 2 (g) + O 2 (g)
is 3.58 x 10 3 min at 730 ˚C. A sample of N 2 O is placed in a flask at this temperature and sealed.
a) What will be the pressure of N 2 O in the flask after 5.50 x 10 2 minutes, given that the initial pressure
of N 2 O is 2.30 atm
b) Assuming the initial pressure of N 2 is zero, what is the pressure of N 2 after 5.50 x 10 2 minutes
c) Calculate the total pressure exerted in the flask after one half-life (3.58 x 10 3 min).
d) Suggest a two step mechanism which is consistent with the rate law.

CHEM 1515 Exam II PAGE 4
(11) 4. At 200 ˚C, 0.500 mol of H 2 , 0.500 mol of N 2 and 0.500 mol of NH 3 are introduced into a 1.00 liter
container and allowed to react according to the equation,
At equilibrium the concentration of NH 3 is 0.384 M. Calculate K c for the reaction.
(12) 5. The magnitude of the equilibrium constant for the reaction,
is 54.7 at 700 K. If the initial concentrations of H 2 and I 2 are 0.0714 M and the initial concentration
of HI is 0.800 M at 700 K, calculate the concentrations of all species when the reaction reaches
equilibrium.

CHEM 1515 Exam II PAGE 5
Short Answer:
(12) 6. Consider the reaction
for which ∆H rxn = –233 kJ. Assume a 1.00 L vessel containing an equilibrium mixture, predict how the
[CH 4 ] will change when the equilibrium is disturbed by,
a) addition of H 2
b) removal of CS 2
c) increase in temperature
d) decrease in the volume of the reaction container

CHEM 1515 Exam II PAGE 6
Multiple Choice:
Print the letter (A, B, C, D, E) which corresponds to the answer selected.
7. ________ 8. ________ 9. ________ 10. ________
ONLY THE ANSWERS IN THE AREA ABOVE WILL BE GRADED. Select the most correct answer
for each question. Each question is worth 3 points.
7. How does a catalyst increase the rate of a reaction
A) by changing the mechanism of the reaction.
B) by increasing the activation energy of the reaction.
C) by increasing the concentration of one of the reactants.
D) by decreasing the difference in relative energy of the reactants and the products.
8. The activation energy for a given reaction is 83.1 kJ . By what factor will the rate constant
mol
increase for a 10.0 ˚ temperature change when the initial temperature is 50.0 ˚C
A) 0.395
B) 0.929
C) 2.53
D) 10
9. The bromination of acetone is acid-catalyzed as shown in the equation below:
CH 3 COCH 3 (aq) + Br 2 (aq) H+ → products
The following initial rate data was obtained for the several experiments;
Exp. # [CH 3 COCH 3 ] [Br 2 ] [H + ] initial rate ⎛
M
⎞
⎝ s ⎠
1 0.30 M 0.050 M 0.050 M 5.7 x 10 –5
2 0.30 M 0.100 M 0.050 M 5.7 x 10 –5
3 0.30 M 0.050 M 0.100 M 1.2 x 10 –4
4 0.40 M 0.050 M 0.200 M 3.1 x 10 –4
The experimental rate is;
A) rate = k[CH 3 COCH 3 ] 1 [Br 2 ] 1 [H + ] 1
B) rate = k[CH 3 COCH 3 ] 1 [Br 2 ] 0 [H + ] 1
C) rate = k[CH 3 COCH 3 ] 0 [Br 2 ] 1 [H + ] 1
D) rate = k[CH 3 COCH 3 ] 1 [Br 2 ] 1 [H + ] 0
10. What conditions of temperature and pressure favor the formation of products in the reaction,
A) high temperature and low pressure.
B) high temperature and high pressure
C) low temperature and low pressure
D) low temperature and high pressure

CHEM 1515 Exam II PAGE 7
1
2
3
4
5
6
7
IA
1
H
1.008
3
6.94
11
IIA
4
Li Be
9.01
12
Na Mg
22.99 24.30
19 20
K
Ca
39.10 40.08
37
Rb Sr
38
85.47 87.62
55 56
Cs
Ba
132.9 137.3
87 88
Fr
(223)
Ra
226.0
IIIA IVA VA VIA VIIA 4.00
5 6 7 8
B C N O F
9 10
Ne
10.81 12.01 14.01 16.00 19.00 20.18
13 14 15 16 17 18
IIIB IVB VB VIB VIIB VIII IB IIB 26.98 28.09 30.97 32.06 35.45 39.95
21 22 23 24 25 26 27 28 29
Sc Ti V Cr Mn Fe Co Ni Cu 30 31 32 33 34 35 36
Zn Ga Ge As Se Br Kr
44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.38 69.72 72.59 74.92 78.96 79.90 83.80
39
Y Zr
40 41 42 43 44 45 46 47
Nb Mo Tc Ru Rh Pd A g 48 49 50 51 52 53 54
Cd In Sn Sb Te I Xe
88.91 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
57 72 73 74 76 77 78 79 81 82 83 84 85 86
La
Periodic Table of the Elements
Hf
Al
Si
P
S
Cl
VIIIA
2
Ta W Re 75
Os Ir Pt Au Hg 80
Tl Pb Bi Po At Rn
138.9 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210)
89
Ac 104 105 106
227.0 (261) (262) (263)
He
Ar
(222)
Lanthanides
Actinides
58 59 60 61 62 63 64
Ce Pr Nd Pm Sm Eu Gd 65 66 67 68 69 70 71
Tb Dy Ho Er Tm Yb Lu
140.1 140.9 144.2 (145) 150.4 152.0 157.2 158.9 162.5 164.9 167.3 168.9 173.0 175.0
90 91 92 93 94 95 96
Th Pa U Np Pu AmCm 97 98 99 100 101 102 103
Bk Cf Es Fm Md No Lr
232.0 231.0 238.0 237.0 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)
Useful Information
PV = nRT
R = 0.0821 L·atm
mol·K
ln⎛
k 1 ⎞
⎝ k
= E a
2 ⎠ R ⎝ ⎛ 1
T
– 1 ⎠ ⎞
J
R = 8.314
2 T 1 K·mol
ln⎛
[A] t ⎞
1 1
⎝ [A]
= –kt o ⎠
[A]
– t [A]
= kt o
K p = K c (RT) ∆n
x 1,2 = –b±√⎺⎺⎺⎺⎺⎺ b 2 – 4ac
2a
for ax 2 + bx + c = 0
ln⎛
K 1 ⎞
⎝ K 2 ⎠
= ∆H rxn
R ⎝ ⎛ 1
T
– 1 ⎠ ⎞
2 T 1

CHEM 1515 Exam II PAGE 8
Solubility Table
Ion Solubility Exceptions
NO – 3 soluble none
ClO – 4 soluble none
Cl – soluble except Ag + , Hg 2+ 2 , *Pb 2+
I – soluble except Ag + , Hg 2+ 2 , Pb 2+
SO 2– 4 soluble except Ca 2+ , Ba 2+ , Sr 2+ , Hg 2+ , Pb 2+ , Ag +
CO 2– 3 insoluble
+
except Group IA and NH 4
PO 3–
+
4 insoluble except Group IA and NH 4
- OH insoluble except Group IA, *Ca 2+ , Ba 2+ , Sr 2+
S 2– insoluble except Group IA, IIA and NH + 4
Na + soluble none
NH + 4 soluble none
K + soluble none
*slightly soluble