CHM 142 Exam 1A F23 key-revised-12-11-23
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CUNY College of Staten Island *
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Course
142
Subject
Chemistry
Date
Feb 20, 2024
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6
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CHM 142 Exam 1 (6 pages) Name: _____________________________ Section 1.
Multiple choice questions:
51 points; 3
points each; no need to show your work 1. Which of the following molecules dose have the HIGHEST standard molar enthalpy of vaporization? a) Br
2
b) ICl c) Cl
2
d) FCl
e) same for all ICl and Br
2
have a similar molar mass, but ICl is polar and has higher b.p. ICl has higher molar mass than FCl. 2. Which of the following compounds does have the LEAST exothermic hydration
energy in water? a) CsBr b) Na
2
S c) Li
2
S d) LiBr e) same for all CsBr has monovalent cation and anion, and both Cs
+
and Br
- are larger than other ions in the question. 3. Consider the liquid of the compounds in Figure 1
. Which compound has the HIGEST boiling point? a) Compound 1 b) Compound 2 c) Compound 3 d) same boiling point for all A similar molar mass for all 3, but only Compound 2 has hydrogen bonding forces. 4. Which of the following compounds does have the LOWEST solubility in water? a) CH
3
CH
2
NH
2
b) CH
3
CH
2
CH
2
NH
2
c) CH
3
CH
2
CH
2
CH
2
NH
2 (longer hydrocarbon segment)
d) same solubility for all 5. A translucent solid has a melting point of 2100 o
C. It is highly soluble in water, and becomes a very good conductor of electricity in aqueous solution. The solid easily becomes fragments by grinding. The solid most likely is _____. a) molecular solid b) network solid c) ionic solid d) metallic solid e) none of the above Ionic solids are fragile, has a higher melting point, and is a good conductor of electricity in aqueous solution. 6. Which of the following compounds does have the MOST exothermic lattice energy? a) CaCl
2 b) MgF
2
c) CsCl d) LiF e) same for all MgF
2
has a divalent Mg
2+
ion. Mg
2+
is smaller than Ca
2+
, and F
-
is smaller than Cl
-
.
7. Which of the following compounds does have the HIGHEST melting point? a) HBr b) CH
3
Br c) Br
2
d) KBr e) same melting point for all KBr is an ionic compound and the others are molecular compounds.
8. Which of the following strong electrolyte aqueous solutions does have the HIGHEST freezing point? a) 1.2 m Ca(NO
3
)
2
(aq) b) 1.6 m MgSO
4
(aq)
c) 1.3 m KI (aq) d) 1.4 m Na
2
SO
4
(aq) e) same for
all
1.2 x 3 = 3.6 m 1.6 m x 2 = 3.2 m 1.3 x 2 = 2.6 m 1.4 x 3 = 4.2 m Lowest solute particle concentratin. 9. The normal freezing point of Substance A is -82 o
C and the normal freezing point of Substance B is -33 o
C. At -60 o
C, which of the two substances is solid? a) Substance A b) Substance B c) Both substances d) None of the above -60 o
C is below its freezing point 10. Which of the following partial pressure and temperature will H
2
gas have the GREATEST solubility in water? a) 5 atm and 80 o
C b) 5 atm and 20 o
C c) 1 atm and 80 o
C d)
1 atm and 20 o
C e) same for all Henry’s law: higher pressure, higher gas solubility; low temperature (lower gas kinetic energy), higher gas solubility. 11. Which of the following expressions of K is TRUE for the reaction: 4 NO (g) + 2 O
2
(g) 4 NO
2
(g)? Figure 1 Compound 1 Compound 2 Compound 3
𝑎𝑎
) 𝐾𝐾
=
[
𝑁𝑁𝑂𝑂
2
]
2
[
𝑁𝑁𝑂𝑂
]
2
[
𝑂𝑂
2
]
𝑏𝑏
) 𝐾𝐾
=
[
𝑁𝑁𝑂𝑂
2
]
4
[
𝑁𝑁𝑂𝑂
]
4
[
𝑂𝑂
2
]
2
𝑐𝑐
) 𝐾𝐾
=
[
𝑁𝑁𝑂𝑂
]
4
[
𝑂𝑂
2
]
2
[
𝑁𝑁𝑂𝑂
2
]
4
𝑑𝑑
) 𝐾𝐾
=
[
𝑁𝑁𝑂𝑂
]
2
[
𝑂𝑂
2
]
[
𝑁𝑁𝑂𝑂
2
]
2
𝑒𝑒
)
None of the above By definition of equilibrium constant expressions. 12. The energy diagram of the kinetic paths for the uncatalyzed and catalyzed reaction of a chemical reaction is given in Figure 2
. Which of the following statements is true about the forward
reaction of these reactions? (Identify uncatalyzed and catalyzed reaction by your own.) a) The activation energy of the catalyzed reaction is 155 kJ. b) The activation energy of the catalyzed reaction is 140 kJ. c) The activation energy of the uncatalyzed reaction is 95 kJ. d) The activation energy of the uncatalyzed reaction is 130 kJ. e) None of the above The catalyzed reaction is (2) (lower activation energy: 95 – (-45) = 140).
13. The rate constant of a reaction is k
1
at 25 o
C and is k
2
at 50 o
C. If the activation energy of this reaction is 35.2 kJ/mol, then the ratio k
2
/k
1
is most close to __________ (with 2 significant figures). (
R
= 8.314 J/K
⋅
mol) a) 85 b) 3.0 c) 0.33 d) 0.012 e) None of the above. Specify your answer: __________ ln (k
2
/k
1
) = -(35200 J/mol /8.314 J/K/mol)(1/298 – 1/323) = 1.0996; k
2
/k
1
= e
1.0996 = 3.00 14. For the reaction below, the disappearance rate of HNO
3
(
aq
) at a given time and concentration is 0.040 M/s. What is the rate of disappearance of H
2
(
g
)? 5 H
2
(
g
) + 2 HNO
3
(
aq
) →
N
2 (
g
) + 6H
2
O (
l
) a) 0.016 M/sec b) 0.10 M/sec c) 0.020 M/sec d) 0.20 M/sec e) None of the above 0.04 mol-HNO
3
/L x (5 mol-H
2
/ 2 mol-HNO
3
) = 0.10 mol-H
2
/L 15. Given that the vapor pressure of CCl
4
at this temperature is 531 mm Hg at 65 °C. Which of the following two solutions does have a HIGHER vapor pressure at 65 °C? Note that I
2
has no vapor pressure. Molar mass of CCl
4
is 153.82 g/mol and molar mass of I
2
is 253.8 g/mol. Solution (1): Pure iodine (105 g) is dissolved in CCl
4 (325 g). (325/153.82)/[(325/153.82)+(105/253.8)]=0.836
Solution (2): Pure iodine (10.5 g) is dissolved in CCl
4.
(45.1 g). (45.1/153.82)/[(45.1/153.82)+(10.5/253.8)]=0.876
Solution (2) has a higher solve mole fraction, so the solution has a higher vapor pressure. (Raoult’s law) a) Solution (1) b) Solution (2) c) same vapor pressure for both d) None of the above 16. The osmotic pressure of the following two solutions at 25 o
C is studied. Solution (1): 18.0 g of sugar (molar mass: 180 g/mol) in 1-L aqueous solution. Solution (2): 18.0 g of unknown compound in 1-L aqueous solution. Note that both sugar and the unknown compound are non-electrolytes. If Solution (1) has a HIGHER osmotic pressure than Solution (2), then the molar mass of the unknown compound is ____ the molar mass of sugar. a) greater than b) equal to c) less than d) None of the above. Solution (1): Solute concentration: (18.0/180) mol /1 L = 0.100 M; Solution (2) has a concentration lower than 0.100 M (lower osmotic pressure) and the unknown substance should have a molar mass greater than 180 g/mol that produces fewer solute particles.
17. Liquid A has normal boiling point of 83 o
C and Liquid B has normal point of 67 o
C. Which of the following statements is TRUE about these two liquids? a) Liquid A has a higher vapor pressure than Liquid B has at 25 o
C. b) Liquid A has a higher evaporation rate than Liquid B has at 25 o
C. c) Liquid A has a higher standard enthalpy of vaporization than Liquid B has. (higher boiling point, greater IMF)
d) None of the above.
Section 2.
Multiple part quetions
:18 points; 6 points each for Q18, Q19 and Q20; no need to show your work. 18. The phase diagram of a pure substance is given in Figure 3
.
(Identify solid, liquid and gas state by your own.) Answer the following questions: Part 1 and Part 2.
Part 1
. Which of the following statements is TRUE about this substance? a) At 350 K, the gas of the substance can compressed into liquid. (false, above critical temperature) b) At 1.25 atm, the substance can sublime from solid to gas. (false, above triple point)
c) Its normal melting point is about 250 K.
(false, it is normal boiling point)
d) At 1 atm and 400 K, the substance is in the gas state. e) None of the above Part 2
. At around 1 atm, which of the following physical states has the GREATEST density? a) gas b) liquid c) solid d) same density for all solid (lower density) can be compressed into liquid (higher density) by crossing liquid/solid boundary 19.
The cubic unit cell of a compound in Figure 4
consists of A- and B- atoms. Answer the questions below: Part 1 and Part 2. Part 1
. Which of the following statements is true? a) A-atoms form BCC structure. b) B-atoms form BCC structure. (8 at corners; 1 at center)
c) A-atoms form FCC structure. d) B-atoms form FCC structure. e) None of the above Part 2
. The empirical formula of this compound is _________. a) A
4
B
3 b) A
2
B
3
c) A
3
B d) A
3
B
2
e) A
3
B
4
f) None of the above. Specify your answer: ___________ A-atoms: 12 x ¼ = 3; B-atoms: 8 x 1/8 +1 = 2 20. A proposed mechanism for the reaction of O
3
and chlorine atom (Cl) in atmosphere is given as follows. Answer the following questions. Step 1: O
3
(g) + Cl (g) OCl (g)
+ O
2
(g)
(slow)
Step 2: OCl
(g)
+ O (g) Cl (g) + O
2
(g) (fast) Part 1
. In the above mechanism, the intermediate is _________. a) O b)
OCl c) Cl d) O
2
e) O
3
f) no intermediate in this reaction OCl is produced and then is consumed right away. Part 2
. In the above mechanism, the catalyst is _________. a) O b)
OCl c) Cl d) O
2
e) O
3
f) no catalyst in this reaction Cl enters the reaction a reactant in Step 1 and comes out as a product in Step 2. Part 3
. The overall reaction order for the rate law predicted by the mechanism is _________ order. a) 3
rd
b) 2
nd
c) 1
st
d) 0
th
e) none of the above Step 2 is the rate determined step; its rate law is Rate = k
2
[OCl[O] Step 1 is the rate determining step; its rate law experssion is Rate = k
2
[O
3
l[Cl]. It is a bimolecular elementary step so the reaction order is 2
nd
order Figure 4
. The unit cell consists of A atoms (smaller atoms) atoms and B atoms (larger atoms).
occupy all edges
occupy all corners;
one sits at center
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Section 3. Short response:
12 points;
3 points for each; show your work to receive partial credits for Part 2, 3 and 4.
1. Consider the following reaction: 4 SO
3
(
g
) 4 SO
2
(
g
) + 2 O
2 (
g
) (Rate = k [SO
3
]
x
. The chemical kinetics data based on the time-dependent concentration of SO
3
at 400 K are given below: Time (sec) [SO
3
] M (= mol/L) 0.0000 1.6000 1.0000 1.3454 2.0000 1.1314 3.0000 0.9514 4.0000 0.8000 5.0000 0.6727 Part 1. From the plots in Figure 5
, determine the reaction order x of SO
3
? a) 0
th
b) 1
st
c) 2
nd
d) 3
rd
e) 4
th
f) none of the above ln [SO
3
] vs. time shows a straight line. Part 2.
Determine the value of the corresponding rate constant k, including the unit, of this reaction. From the slope of the plot of ln [SO
3
] vs. time. (3 sec, -0.05); (5 sec, -0.4)
Slope = (-0.4 – (-0.05))/(5-3) = -0.173 sec
-1
; k = 0.173 sec
-1
. Use any 2 points from the above data: ln (0.6727/1.600) = - k (5.000); k = 0.1733 sec
-1
. Part 3.
Determine the half-life of this reaction in sec. Use the formula (if you remember): t
½
= ln(2)/k
ln(2)/0.1733 = 4.000 sec, or From the above data, It takes 4.0000 sec for the reactant concentration drops from 1.6000 M to 0.8000 M (50% remained). Thus, the half-life is 4.0000 sec. Part 4
. Determine the concentration of SO
3
remained at 8.00 sec. 8.00 sec is 2 half-lives from t = 0, so, the concentration remained is 1.6000 x ½ x ½ = 0.4000 M, or Use the integrative rate equation: ln ([SO
3
]/1.6000) = - 0.1733 x 8; [SO
3
] = 0.400 M -1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0
1
2
3
4
5
6
7
8
ln [SO
3
]
time (sec)
0.5
0.7
0.9
1.1
1.3
1.5
1.7
1.9
2.1
2.3
2.5
0
1
2
3
4
5
6
7
8
1
[SO
3
]
time (sec)
Figure 5
Section 4. Open ended questions: 23 points. Show your detailed calculation and/or explanation to receive partial credits for each question.
1. A pure metal XYZ (atomic mass 203 g/mol) crystallizes in a face-centered cubic lattice. The length of each side is 417 pm. Part 1. (4pt) Determine the atomic radius of the metal in pm. Part 2. (4pt) Determine the density of metal XYZ in g/ cm
3
. 4r =
√
2
𝑎𝑎
→ 𝑟𝑟
=
√2
4
× 417 = 147 𝑝𝑝𝑝𝑝
1 pm = 1 pm x (10
-12
m/ 1 pm) x (100 cm/ 1 m) = 10
-10
cm Density = 4
×(
203
𝑔𝑔
6
.
022
×
10
23
)
(
417
× 10
−10
𝑐𝑐𝑐𝑐
)
3
= 18.6 𝑔𝑔
/
𝑐𝑐𝑝𝑝
3
2. The rate law expression is Rate = k [ClO
2
]
x
[OH
-
]
y
for the reaction between ClO
2
(aq) and OH
-
(aq): 2 ClO
2
(aq) + 2 OH
-
(aq) →
ClO
3
-
(aq) + ClO
2
-
(aq) + H
2
O (liq). At a certain temperature, the following initial rates were obtained for the concentrations listed: Experiment Initial Rate (mol·L
–1
·sec
–1
) [ClO
2
]
0
(mol·L
–1
) [OH
-
]
0
(mol·L
–1
) 1 1.512 0.69 0.17 2 0.504 0.23 0.17 3 4.032 0.23 0.34 Part 1. (4pt) Determine the reaction order x in the rate law: Rate = k [ClO
2
]
x
[OH
-
]
y
. Part 2. (4pt) Determine the reaction order y in the rate law: Rate = k [ClO
2
]
x
[OH
-
]
y
. Part 3. (4pt) Determine the rate constant k and its units in the rate law: Rate = k[ClO
2
]
x
[OH
-
]
y
. Part 4. (3pt) which of the above 3 experiments has the greatest activation energy? a) Experiment 1 b) Experiment 2 c) Experiment 3 d) same for all the experiments From Experiment 1 and 2 1.512
0.504
=
𝑘𝑘
(0.69)
𝑥𝑥
(0.17)
𝑦𝑦
𝑘𝑘
(0.23)
𝑥𝑥
(0.17)
𝑦𝑦
→
3.000 =
�
0.69
0.23
�
𝑥𝑥
= (3.00)
𝑥𝑥
→ 𝑥𝑥
= 1
From Experiment 2 and 3 4.032
0.504
=
𝑘𝑘
(0.23)
𝑥𝑥
(0.34)
𝑦𝑦
𝑘𝑘
(0.23)
𝑥𝑥
(0.17)
𝑦𝑦
→
8.000 =
�
0.34
0.17
�
𝑦𝑦
= (2.00)
𝑦𝑦
→ 𝑦𝑦
= 3
From any experiment, such as Experiment 1, 1.512 =
𝑘𝑘
(0.69)(0.17)
3
→ 𝑘𝑘
=
1.512
(0.69)(0.17)
3
= 446 𝑀𝑀
−3
𝑠𝑠𝑒𝑒𝑐𝑐
−1
The activation energy cannot be changed by concentration and temperature. Note that catalysts provide a way to change the activation energy.
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A session.masteringchemistry.com/myct/itemView?assignmentProblemlD=D182936227
Homework #3
roblem 3.49
Correct
Part C
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Express your answer as a positive value using three significant figures and include the appropriate ur
HA
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Units
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Previous Answers
v Correct
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Heat of Neutralization
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elemsboomfchg lom
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2. Temp. of warm water
°C
22.0
39,0
30.3
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°C
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°C
50.0 g x 4.184 J/K-g) =
02),
5. Heat gained by cooler water (temp. increase x
50.0 g x 4.184 J/K-g) =
30,3 22.0)x
13626J
s0.0gmpi
S0.0gy
6. Heat gained by the calorimeter [(4) – (5)] =
7. Heat capacity of calorimeter:
heat gained by the calorimeter
temperature increase
J/K
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22.2
22.2.
°C
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Temp. of HCI
AT determined from your curve after adding HC1
°C
to the NaOH
Heat gained by solution (temperature increase x ON
100 g x 4.184 J/K-g) =
9977.8J
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1.057 qt
1 ml.
First, the water needs to be heated from 25 °C to 100 °C. The heat associated with this process is calculated using the specific heat of water from the appendix.
= mcAT
= 1.43 x 107 g H,O x
9
4.184 J
g.°C
x (100°C – 25°C) x I kJ
* 1000 J
%3D
= 4.481 x 106 k.J
Once the water reaches 100 °C, it must then be vaporized. The heat associated with…
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Ancient Romans built often out of bricks and mortar. A key ingredient in their mortar was quicklime (calcium oxide), which they
produced by roasting limestone (calcium carbonate).
1. Write a balanced chemical equation, including physical state symbols, for the
decomposition of solid calcium carbonate (CaCO3) into solid calcium oxide and
gaseous carbon dioxide.
x10
2.
ose 20.0 L of carbon dioxide gas
produced by this reaction, at a
temperature of 370.0 °C and pressure of exactly 1 atm. Calculate the mass of
calcium carbonate that must have reacted. Round your answer to 3 significant digits.
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Br
1. PPh3
2. nBuLi
NHỊNH,
ܠ
K0H
Heat
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How much heat is absorbed when 34.5 g of H2O (MM = 18.02 g/mol) at −7.00°C is heated to 16.0°C?
(picture attached)
Answer Options:
A. 2824 kJ
B. 2.81 kJ
C. 14.3 kJ
D. 62.2 kJ
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