CHEMISTRY 202
Hour Exam I
September 24, 2015
Dr. D. DeCoste
Name ______________________________
Signature ___________________________
T.A. _______________________________
This exam contains 33 questions on 11 numbered pages. Check now to make sure you have a
complete exam. You have two hours to complete the exam. Determine the best answer to the first
30 questions and enter these on the special answer sheet. Also, circle your responses in this exam
booklet.
Show all of your work and provide complete answers to questions 31, 32 and 33.
1-30
(60 pts.)
_____________
31
(20 pts.)
_____________
32
(20 pts)
_____________
33
(20 pts.)
_____________
(120 pts)
_____________
Total
Useful Information:
Always assume ideal behavior for gases (unless explicitly told otherwise).
PV = nRT
R = 0.08206 Latm/molK = 8.3145 J/Kmol
STP = standard temperature and pressure = 0°C and 1.00 atm
K = °C + 273
υ rms =
ZA = A
3RT
M
N
V
RT
2πM
N A = 6.022 x 1023
λ=
1
2 ( N / V )(πd 2 )
Z=4
N 2 πRT
d
M
V
Solubility Rules:
1. Most nitrate salts are soluble.
2. Most salts of sodium, potassium, and ammonium cations are soluble.
3. Most chloride salts are soluble. Exceptions: silver, lead(II), and mercury(I) chloride.
4. Most sulfate salts are soluble. Exceptions: calcium, barium, and lead (II) sulfate.
5. Most hydroxide salts can be considered insoluble. Soluble ones: sodium, potassium, and calcium
hydroxide.
6. Consider sulfide, carbonate, and phosphate salts to be insoluble.
CHEMISTRY 202
Hour Exam I
Fall 2015
Page No. 1
1. What is the molar mass of ammonium sulfate?
a)
b)
c)
d)
e)
2.
113.164 g/mol
114.112 g/mol
116.274 g/mol
130.258 g/mol
132.154 g/mol
A binary compound has the formula Z 4 O 10 and is found to be 43.64% Z by mass. What is the
identity of Z?
a) P
3.
c) B
d) N
e) S
A metal ion has 23 electrons in an ionic compound that is 30.06% oxygen by mass. Which metal
is in the compound?
a) Vanadium
4.
b) Si
b) Chromium
c) Manganese
d) Iron
e) Cobalt
A Group(I) or Group(II) metal reacts with water to form the hydroxide of the metal ion along with
hydrogen gas. If 4.61 g of water is required to react with exactly 10.0g of a metal, what is the
identity of the metal?
a) Li
b) Na
c) K
d) Ca
e) Ba
--------------------------------------------------------------------------------------------------------------------------5, 6. Ethane (C 2 H 6 ) reacts with oxygen gas to produce carbon dioxide gas and water. Suppose you
react 100.0 g of ethane with 100.0g of oxygen gas. Answer the following questions:
5.
Determine the maximum mass of carbon dioxide that could be produced.
a) 19.65 g
6.
b) 39.29 g
c) 78.59 g
d) 137.5 g
e) 157.2 g
Determine the mass of left-over reactant.
a) 6.044 g
b) 71.43 g
c) 73.15 g
d) 86.58 g
e) No limiting reactant
--------------------------------------------------------------------------------------------------------------------------7.
A 6 mole sample of copper metal is reacted with excess sulfur to produce a 5 mole mixture of
copper(I) sulfide and copper(II) sulfide. What is the mole ratio of copper(I) sulfide to copper(II)
sulfide in the product mixture?
a) 1:4
8.
b) 4:1
c) 2:1
d) 1:2
e) 1:1
A 1.00-g sample of a chloride salt of a Group 2 metal (alkaline earth) is treated with excess silver
nitrate. All of the chloride is recovered as 3.01 g of silver chloride. What is the identity of the
metal?
a) Ba
b) Sr
c) Ca
d) Be
e) Mg
CHEMISTRY 202
Hour Exam I
9.
A gaseous compound known to contain only carbon, hydrogen, and nitrogen is mixed with exactly the
volume of oxygen gas required for complete combustion to CO 2 , H 2 O, and N 2 . It is found that reacting
9 total volumes of reactant mixture (compound plus O 2 ) produces 4 volumes of CO 2 , 6 volumes of H 2 O,
and 2 volumes of N 2 , all at the same temperature and pressure. What is the molecular formula of the
compound?
a)
b)
c)
d)
e)
10.
C2H6N2
C 4 H6 N2
CH 3 N
C 4 H 12 N 4
Only an empirical formula can be determined with the information given.
Determine the volume of water required to add to 15.0 mL of 3.25 M NaOH to make a solution
with a concentration of 0.125 M NaOH.
a) 26.0 mL
11.
Fall 2015
Page No. 2
b) 187 mL
c) 254 mL
d) 375 mL
e) 390. mL
You dissolve 100.0-g samples of different ionic compounds in separate beakers to make five
solutions with equal volumes. Which of the following has the highest concentration in terms of
molarity?
a) lithium chloride
b) potassium sulfate
c) sodium nitrate
d) magnesium nitrate
e) sodium carbonate
----------------------------------------------------------------------------------------------------------------------12, 13.
Consider mixing 150.0 mL of 0.125M solution of calcium nitrate with 150.0 mL of a 0.125M
solution of sodium phosphate to produce a solid. Answer the following questions.
12.
Determine the concentration of the phosphate ion after the precipitation reaction is complete.
a) 0.0104M
13.
b) 0.0208M
c) 0.0417M
d) 0.0625M
e) 0 M
Determine the mass of solid formed.
a) 0.844 g
b) 1.94 g
c) 2.53 g
d) 3.88 g
e) 5.82 g
----------------------------------------------------------------------------------------------------------------------14. A 5.00-g sample of silver nitrate is heated in an evacuated container with a volume of 1.75 L. The
sample decomposed to Ag 2 O(s), nitrogen dioxide gas, and oxygen gas. Assuming complete
decomposition, determine the pressure in the container after cooling to room temperature (25°C).
Assume the Ag 2 O(s) takes up negligible volume.
a) 0.514 atm
b) 0.720 atm
c) 0.822 atm
d) 1.03 atm
e) 2.06 atm
CHEMISTRY 202
Hour Exam I
Fall 2015
Page No. 3
--------------------------------------------------------------------------------------------------------------------15, 16. Consider two 1.00 mol samples of He(g) each in identical 1.00 L containers. Sample 1 has a
temperature of 136.5°C and Sample 2 has a temperature of –136.5°C.
Determine the following ratios in the two containers.
15.
Z 2 /Z 1
a) 0.333
16.
b) 0.577
c) 1.00
d) 1.73
e) 3.00
(force per impact with the walls) 2 /(force per impact with the walls) 1
a) 0.333
b) 0.577
c) 1.00
d) 1.73
e) 3.00
-------------------------------------------------------------------------------------------------------------------17. A gas at 323°C has the same rate of effusion as H 2 (g) at 25°C. Which of the following could be
that gas?
a) SF 6 (g)
18.
c) SO 2 (g)
d) O 3 (g)
e) He(g)
Consider a gas mixture consisting of xenon and neon at 3.14 atm and 25.0°C. The density of this
mixture is 11.15 g/L. Determine the partial pressure of the xenon.
a)
b)
c)
d)
e)
19.
b) C 2 H 2 (g)
1.14 atm
1.26 atm
1.57 atm
1.88 atm
More information is needed to answer this question.
You have a sample of gas in a closed container of fixed volume. Which of the following changes would
double the root mean square velocity of the gas?
I.
II.
Heating the gas such that the temperature is doubled.
Pumping out 75% of the gas at constant temperature such that the pressure is 25% of its original
value.
III. Heating the gas such that the pressure is quadrupled.
IV. Pumping in more gas at constant temperature such that the pressure is quadrupled.
a) I
20.
b) III, IV
c) III
d) I, II
e) None
Baking soda (sodium bicarbonate) reacts with acetic acid (HC 2 H 3 O 2 ) in vinegar to produce
sodium acetate, carbon dioxide gas, and water. In lecture we reacted 10.5 g of baking soda with
and excess of acetic acid and collected the carbon dioxide in a balloon. Disregarding water vapor
pressure, determine the size of the balloon at room conditions (1.00 atm and 25°C).
a) 0.765 L
b) 1.53 L
c) 3.06 L
d) 6.12 L
e) 12.2 L
CHEMISTRY 202
Hour Exam I
Fall 2015
Page No. 4
--------------------------------------------------------------------------------------------------------------------21-24.
Indicate which of the graphs below best represents each plot described in questions 21-24.
Note: the graphs may be used once, more than once, or not at all.
a)
b)
d)
e)
c)
21.
Collision frequency (Z A ) (y) vs. T (K) (x) for 1 mole of an ideal gas at constant P.
e
22.
Mean free path (λ) (y) vs. T (K) (x) for a 1.0 mole of an ideal gas at constant V.
b
23.
Average kinetic energy (y) vs. T (°C) (x) for a an ideal gas
d
24. Pressure vs. molar mass for 100.0-g samples of different ideal gases at constant V, T.
e
--------------------------------------------------------------------------------------------------------------------25, 26. Consider an equimolar (equal number of moles) mixture of N 2 and O 2 gases in a rigid steel
container. At equilibrium 25.0% of the nitrogen gas has reacted to produce NO(g) as follows:
N 2 (g) + O 2 (g)
25.
Assuming the molecular diameters of N 2 , O 2 , and NO are essentially the same, determine the
intermolecular collision frequency ratio initially and at equilibrium (Z initial : Z equilibrium )
a) 1:4
26.
2NO(g)
b) 4:1
c) 1:2
d) 2:1
e) 1:1
Determine the value of the equilibrium constant, K, at this temperature.
a) 0.111
b) 0.250
c) 0.333
d) 0.444
e) 0.667
-----------------------------------------------------------------------------------------------------------------------27. Consider a sample of pure UF 6 (g) initially at a pressure of 5.00 atm at 357°C. The UF 6 (g) decomposes as
shown:
UF 6 (g)
UF 2 (g) + 2F 2 (g)
After the system has reached equilibrium the pressure is 10.00 atm at 357°C. Determine the value
of K p for this decomposition at 357°C.
a) 3.60
b) 5.00
c) 13.3
d) 25.0
e) 88.7
CHEMISTRY 202
Hour Exam I
Fall 2015
Page No. 5
--------------------------------------------------------------------------------------------------------------------28, 29. Ammonium carbamate decomposes as follows:
NH 4 OCONH 2 (s)
2NH 3 (g) + CO 2 (g)
In an experiment carried out in a 10.0-L container at 25°C a certain amount of ammonium
carbamate is placed in an evacuated container and allowed to come to equilibrium. When the
system has reached equilibrium the total pressure in the flask was observed to be 0.114 atm.
28.
Determine the value of K p for this reaction at 25°C.
a)
b)
c)
d)
e)
29.
1.85 x 10-4
2.19 x 10-4
1.48 x 10-3
2.89 x 10-3
3.25 x 10-3
Determine the minimum mass of ammonium carbamate required for the system to reach equilibrium.
a) 1.21 g
b) 1.82 g
c) 2.97 g
d) 3.63 g
e) This cannot be determined with the information given.
--------------------------------------------------------------------------------------------------------------------30. Consider the “pop” bottles at the beginning of lecture in which we react hydrogen gas and oxygen gas to
produce water. Which of the following correctly describes what would happen if we carried out the
reaction at a higher temperature?
a)
b)
c)
d)
e)
Equilibrium would be shifted to the right and the value of K would increase.
Equilibrium would be shifted to the right and the value of K would decrease.
Equilibrium would be shifted to the left and the value of K would decrease.
Equilibrium would be shifted to the left and the value of K would increase.
Equilibrium would be shifted but the value of K would remain constant.
CHEMISTRY 202
Hour Exam I
31.
Fall 2015
Page No. 6
Ammonia gas [NH 3 (g)] reacts with oxygen gas to produce H 2 O(l) and either NO(g) or NO 2 (g),
depending on the supply of oxygen. If oxygen gas is limiting, NO(g) is one of the products. If
oxygen gas is in excess, NO 2 (g) is a product.
Suppose you react ammonia and oxygen gases in a container fitted with a piston in such a ratio that
the reactant mixture at 25.0°C and 1.00 atm has the same density as air at these conditions
(consider air to be 21.0% oxygen gas and 79.0% nitrogen gas by volume).
a.
Determine the balanced equation for the reaction that occurs and defend your answer with
numbers.
The two possible reactions are:
4NH 3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(l)
4NH 3 (g) + 7O 2 (g) → 4NO 2 (g) + 6H 2 O(l)
If we have the same T, P, and density, a given volume of the mixture must have the same mass as
that same volume of air (and these will have the same number of moles of gas). Assuming we
have 1 mole of air we have 0.210 mole O 2 and 0.790 mole N 2 , which is 28.85 g. So we have:
28.85 g = 17.034x + 32.00(1–x), where x = mol NH 3 . Solving we get x = 0.2128 mol
So, the {mol NH 3 /mol O 2 } ratio = 0.2128/0.7872, which equals 4:14.44. We obviously have an
excess of O gas, so the second reaction is the reaction that occurs.
We can see how much is in excess with (useful for part b):
Initial
Change
End
b.
4NH 3 + 7O 2 → 4NO 2 + 6H 2 O
0.213 0.787 0
0
-0.213 -0.373 +0.213 +0.320
0
0.414 0.213 0.320
Determine the ratio of the volume of the container before and after the reaction. Assume the
liquid water that is produced takes up negligible volume. Show all work.
In this case, the pressure and temperature are constant so the volume will change in the same ratio
as the total moles of gas.
1
volume before
moles gas before 0.213 + 0.787
=
=
= 1.59 =
0.414 + 0.213
0.627
volume after
moles gas after
CHEMISTRY 202
Hour Exam I
31.
Fall 2015
Page No. 7
(con’t)
c.
Specify the temperature range so that the gaseous mixture after the reaction is complete
has the same density as the initial reactant gaseous mixture at 25.0°C and 1.00 atm. Show all
work.
From our work previous, we assumed 1 mole total of reactant gas for a mass of 28.85 g. We can
use PV=nRT to solve for V (getting 24.45 L) to get a density of 1.18 g/L
After the reaction we have 0.414 mol O 2 and 0.213 mol NO 2 for a total of 23.05 g. The volume is
1/1.59 of the original volume, or 15.38 L. Thus the density is 1.50 g/L.
To lower the density we need to increase the volume and we can do this by increasing the
temperature. Thus, the density and temperature are inversely related.
We want to decrease the density by a factor of 1.50/1.18 = 1.27, so we raise the temperature by this
factor. Thus, T = (1.27)(298) = 378 K = 105°C.
Thus, the temperature needs to be 105°C.
Below this, the density is too great, below is, the density is too small.
CHEMISTRY 202
Hour Exam I
32.
Fall 2015
Page No. 8
In this problem you will differentiate between a law and a theory, and apply your results.
a.
Starting with the ideal gas law equation, derive an equation (law) relating the moles of gas
to the Kelvin temperature at a given pressure and volume. Show all steps.
PV = nRT
PV
= constant = nT
RT
n 1 T 1 = constant = n 2 T 2
n1 T1 = n2 T2
b.
Sketch a plot of n (y) vs. T (K) (x) at constant P and V. Explain how your graph supports
the law you derived in part a.
This graph shows that as the temperature of a gas increases
at constant pressure and volume, the number of moles of
gas decreases. That is, there is an inverse relationship
c.
Use the tenets of the kinetic molecular theory to explain the law you derived in part a. Use
complete sentences.
As the temperature of a sample increases, the average kinetic energy of the sample increases.
As the average kinetic energy of the sample increases, the particles of a gas move more
quickly.
Particles with a greater kinetic energy could give rise to a greater pressure at constant volume
(due to increased number and force of collisions) or a greater volume at constant pressure
(while the force per collision is greater, the number of collisions will decrease keeping the
pressure constant)
But since we are told that the pressure and volume of the samples are equal, there must be a
fewer number of moles in the sample of gas at the higher temperature.
CHEMISTRY 202
Hour Exam I
32.
Fall 2015
Page No. 9
(con’t)
d.
A hot-air balloon is open at the bottom and top. Even after a hot-air balloon reaches
maximum volume it will ascend when the air inside is heated. Use the law (part a) and
theory (part c) to explain why this is true. Use complete sentences.
In order for the balloon to ascend, the density of the (hot) air inside the balloon must be less dense
than the surrounding air. In other words, the mass of the hot air in the balloon must have less mass
that the same volume of air at the cooler temperature.
Since the hot-air balloon is open, it is a constant pressure system and we are told that the balloon
has reached maximum volume so the volume is not changing either. The only way, then to change
the density is to change the mass (in this case to decrease both). And the only way to decrease the
mass of air in the balloon is for some of the air to escape. As discussed in part a and c, for a given
P and V, a gas sample at a higher temperature must have a fewer number of moles of gas, thus the
mass is less, thus the density is less.
e.
You heat the air in a hot-air balloon to 65°C. Determine the volume (in liters) required for
this balloon to be able to lift 25.0 kg if the surrounding conditions are 25°C and 1.00 atm.
Assume air is 21.0% oxygen gas and 79.0% nitrogen gas by volume. Show all work.
To lift 25.0 kg (25,000 g) we must displace 25,000 g of air. A 1.0-L sample of air at 25°C and
1.00 atm contains 0.0409 mol (21.0% O 2 and 79.0% N 2 ) with a mass of 1.18 g. A 1.0-L sample of
air at 65°C and 1.00 atm contains 0.0361 mol (21.0% O 2 and 79.0% N 2 ) with a mass of 1.04 g.
Thus a 1.0-L balloon has a lift of 1.18g–1.04g = 0.14g
We need a balloon with a volume of 25,000/0.14 = 179,000 L
CHEMISTRY 202
Hour Exam I
33.
Fall 2015
Page No. 10
Consider a gas-phase reaction (that is, all species are gases) that has reached equilibrium.
To this system we add an inert gas (such as helium) at constant temperature.
a.
Which of the following describes what happens? Fill in the circles next to any
statement that applies.
o There is no shift because adding an inert gas never shifts the equilibrium
position.
o If the system is at constant volume and K = K p , then the equilibrium position
will shift.
o If the system is at constant volume and K ≠ K p , then the equilibrium position
will shift.
o If the system is at constant pressure and K = K p , then the equilibrium position
will shift.
o If the system is at constant pressure and K ≠ K p , then the equilibrium
position will shift.
b. Explain your answer to part a. As part of your explanation provide an example of a chemical
equation in which K = K p , and an example of a chemical equation in which K ≠ K p and justify
your choices. Full credit is reserved for explanations that address all of the conditions given
in part a, that use the actual (real) chemical equations, and that correctly incorporate partial
pressures (and/or partial concentrations) of reactants and products. You may continue your
answer on the next page if needed.
K = K p for CO(g) + H 2 O(g)
K=
[CO 2 ][H 2 ]
[CO][H 2 O]
Kp =
K ≠ K p for N 2 (g) + 3H 2 (g)
K=
[NH 3 ] 2
[N 2 ][H 2 ]3
Kp =
CO 2 (g) + H 2 (g) [Equal number of moles of gas on each side]
P(CO 2 ) • P(H 2 )
n
; since P =
RT, the RT factors cancel
P(CO) • P(H 2 O)
V
2NH 3 (g) [Different numbers of moles of gas on each side]
P(NH 3 ) 2
; the RT factors do not cancel
P(N 2 ) • P(H 2 ) 3
In order to shift equilibrium we must have a Q value that is different from the K value.
The first statement is not true because adding an inert gas can potentially shift the equilibrium
position (as we shall see).
The second and third statements are not true because volume is constant. Adding an inert gas will
not affect the moles of any reactants or products, nor (because volume is constant) will it affect the
volume of the container. Thus, the concentrations of the gases (therefore partial pressures) will not
be affected. Because of this, the value of Q will not have been changed.
CHEMISTRY 202
Hour Exam I
33.
Fall 2015
Page No. 11
(con’t). Continue your answer on this page if needed.
For the system to be at constant pressure, the volume will have to change as the number of moles
of gas change. So, if we add an inert gas, the volume of the container will increase.
If we change the volume, there will not be a shift in equilibrium in the first reaction on the
previous page (K = K p ), but there will be one for the second (K ≠ K p ).
As stated, adding an inert gas will cause the volume to increase. This will lower the partial
pressures (or concentrations) of each species. In the first case this change cancels in the
equilibrium expression, and in the second it does not.
For example, suppose at equilibrium, [CO] = a, [H 2 O] = b, [CO 2 ] = c, [H 2 ] = d. K = cd/ab. Now
suppose we increase the volume by a factor of 10. In this case, [CO] = a/10, [H 2 O] = b/10, [CO 2 ]
= c/10, [H 2 ] = d/10. Q = [(c/10)(d/10)]/[(a/10)(b/10)] = cd/ab = K. So the equilibrium is not
shifted.
In the second reaction, suppose at equilibrium [NH 3 ] = x, [N 2 ] = y. [H 2 ] = z. K = x2/yz3. If we
increase the volume by a factor of 10, [NH 3 ] = x/10, [N 2 ] = y/10. [H 2 ] = z/10. The new conditions
give us Q = [(x/10)2]/[(y/10)(z/10)3] = [x2/yz3][100] ≠ K.