CHEMISTRY 202
Hour Exam II
October 27, 2015
Dr. D. DeCoste
Name ______________________________
Signature ___________________________
T.A. _______________________________
This exam contains 32 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/or provide complete answers to questions 31 and 32.
1-30
(60 pts.)
_____________
31
(20 pts.)
_____________
32
(40 pts.)
_____________
(120 pts.)
_____________
Total
Useful Information:
• Always assume ideal behavior for gases (unless explicitly told otherwise).
• STP = standard temperature and pressure = 0°C and 1.00 atm
• 760 torr = 1.00 atm
• R = 0.08206 Latm/molK = 8.314 J/Kmol
• K = °C + 273
• N A = 6.022 x 1023
∆E = q + w
∆S = q rev /T
H = E + PV
G = H − TS
Here are some of the formulas we used/derived in studying thermodynamics. An individual formula
may or may not apply to a specific problem. This is for you to decide!
∆S = nRln(V 2 /V 1 )
∆S = ∆H/T
C v = (3/2)R
∆S = nCln(T 2 /T 1 )
∆G = ∆G° + RTln(Q)
∆S surr = −q/T
w = −P∆V
q rev = nRTln(V 2 /V 1 )
q = nC∆T
ln(K) = −
∆H   1 
∆S 
  +
R T 
R
K
ln  2
 K1

∆H 
 = −
R

1 1
 − 
 T2 T1 
C p = (5/2)R
CHEMISTRY 202
Hour Exam II
1.
Fall 2015
Page No. 1
Given that ΔH° fusion for Fe(s) = 13.6 kJ/mol and the following data:
Fe 2 O 3 (s)
–826.0
ΔH (kJ/mol)
0
f
Al 2 O 3 (s)
–1676.0
Determine ΔH° rxn for the thermite reaction (at constant temperature) we saw in lecture as
represented by the equation: Fe 2 O 3 (s) + 2Al(s) → Al 2 O 3 (s) + 2Fe(l)
a) –822.8 kJ
2.
b) –836.4 kJ
c) –850.0 kJ
d) –863.6 kJ
e) –877.2 kJ
You pour 500.0 mL of water at 90.°C into a one liter Thermos initially containing air at 1 atm
25°C and seal the Thermos. You are to estimate to the nearest degree the final temperature of
the water in the Thermos. Make the following assumptions:
I.
II.
III.
The heat capacity of water does not change with temperature and is 4.18 J/g°C.
The density of water is exactly 1g/mL and does not change with temperature.
Air has the following heat capacities (assume they do not change with temperature):
C v = 20.71 J/Kmol, C p = 29.03 J/Kmol.
IV. The Thermos is a perfect insulator.
a) 50.°C
3.
b) 60.°C
c) 70.°C
d) 80.°C
e) 90.°C
At 25°C, the following enthalpies of reaction are known:
2C 2 H 2 (g) + 5O 2 (g) → 4CO 2 (g) + 2H 2 O(l)
C(s) + O 2 (g) → CO 2 (g)
2H 2 (g) + O 2 (g)→ 2H 2 O(l)
∆H = –2600.0 kJ
∆H = –394 kJ
∆H = –572 kJ
Calculate ΔH for the following reaction: 2C(s) + H 2 (g) → C 2 H 2 (g)
a) 452 kJ
4.
b) –452 kJ
c) 1634 kJ
d) 226 kJ
Using the information below, calculate ∆H f ° for 1.00 mole of PbO(s).
PbO(s) + CO(g) → Pb(s) + CO 2 (g)
∆H f ° for CO 2 (g) = –393.5 kJ/mol
∆H f ° for CO(g) = –110.5 kJ/mol
a) 283.0 kJ
5.
e) –226 kJ
b) –283.0 kJ
c) –151.6 kJ
∆H° = –131.4 kJ
d) –372.6 kJ
e) 252.1 kJ
An ideal gas is compressed isothermally against a constant pressure until internal and
external pressures reach equilibrium. Which of the following statements (a-d) is true?
a)
b)
c)
d)
e)
ΔG = 0 since the system reaches equilibrium.
ΔG = 0 since the process is isothermal.
ΔG < 0 since the process is spontaneous.
ΔG > 0 since a gas will not spontaneously be compressed.
None of the statements (a-d) is true.
CHEMISTRY 202
Hour Exam II
Fall 2015
Page No. 2
-------------------------------------------------------------------------------------------------------------------6, 7. For substance X ΔH vaporization = 42.8 kJ/mol at its normal boiling point (375°C and 1 atm).
For the process X(l) → X(g) at 1 atm and 375°C, calculate the value of:
6.
ΔS surr
a) 0
7.
b) 114 J/Kmol
c) –114 J/Kmol
d) –66 J/Kmol
e) 66 J/Kmol
ΔS univ
a) 0
b) 114 J/Kmol
c) –114 J/Kmol
d) –66 J/Kmol
e) 66 J/Kmol
--------------------------------------------------------------------------------------------------------------------8.
For the isothermal expansion of an ideal monatomic gas against a constant pressure of 2 atm,
which of the following statements (a-d) is true?
a) Because the expansion is isothermal, ΔH = 0, and because ΔS = ΔH/T, ΔS = 0.
b) ΔS = nC p ln(T 2 /T 1 ) and because T 2 = T 1 , ΔS = 0.
c) ΔS = nC p ln(V 2 /V 1 ) and because P 1 V 1 = P 2 V 2 , ΔS = nC p ln(P 1 /P 2 ). Since pressure is
constant, P 2 = P 1 , so ΔS = 0.
d) ΔS = q rev /T and because the expansion is isothermal and ΔS is a state function, q rev = 0,
so ΔS = 0.
e) None of the above statements (a-d) is true.
9.
Consider the following system at equilibrium at 25°C: PCl 3 (g) + Cl 2 (g)
PCl 5 (g).
For the reaction as written, ΔG° = –92.50 kJ. When the temperature is raised the ratio of
the partial pressure of PCl 3 (g) to the partial pressure of PCl 5 (g) will
a)
b)
c)
d)
10.
Which of the following statements is true for the vaporization of a liquid at a given pressure?
a)
b)
c)
d)
e)
11.
decrease.
increase.
not change.
This is impossible to answer without additional information.
ΔG is positive at all temperatures.
ΔG is negative at all temperatures.
ΔG is negative at high temperatures and positive at low temperatures.
ΔG is positive at high temperatures and negative at low temperatures.
The answer depends on the nature of the liquid.
Determine the signs of ΔH, ΔS and ΔG for the freezing of liquid water at 2°C and 1 atm.
a)
b)
c)
d)
e)
ΔH
–
–
+
+
–
ΔS
–
–
+
–
+
ΔG
–
+
+
0
0
CHEMISTRY 202
Hour Exam II
12.
For how many of the following processes is the entropy change of the system equal to q/T?
I.
II.
III.
IV.
An ideal gas expands into a vacuum.
Hydrogen gas reacts with oxygen gas to produce water.
Water evaporates isothermally at room temperature and pressure.
Water freezes at 273K and 1 atm.
a) 0
13.
Fall 2015
Page No. 3
b) 1
c) 2
a)
b)
c)
d)
e)
IV. G gas decreases
V. S gas decreases
VI. H gas decreases
IV, V
II, IV
I, V
II, III, IV
IV, V, VI
SO 2 (g) and O 2 (g) react to form SO 3 (g). The partial pressures at equilibrium are as follows:
P(SO 2 ) = 0.559 atm, P(O 2 ) = 0.101 atm, and P(SO 3 ) = 0.331 atm. Calculate ΔG° at 1000.0
K for the reaction 2SO 2 (g) + O 2 (g)
2SO 3 (g).
a) 0
15.
e) 4
Which of the following statements are correct if the volume of 1.0 mole of an ideal gas is
decreased at constant temperature?
I. G gas increases
II. S gas increases
III. H gas increases
14.
d) 3
b) 14.7 kJ
c) –14.7 kJ
d) –10.3 kJ
e) 10.3 kJ
Calculate the magnitude of the maximum work produced when 1.00 mol of an ideal
monatomic gas expands isothermally from 2.00 L to 6.50 L at a final pressure of 1.50 atm.
a) 473 J
b) 684 J
c) 915 J
d) 1.16 kJ
e) 2.22 kJ
-----------------------------------------------------------------------------------------------------------------16, 17. Answer the questions based on the following data:
∆H°f (kJ/mol)
S° (J/mol K)
16.
H 2 O(g)
–241.82
188.84
Determine ΔG° for the vaporization of 1.00 mole of H 2 O(l) at 25°C
a) 0
17.
H 2 O(l)
–285.83
69.90
b) –35.4 kJ
c) 35.4 kJ
d) 8.57 kJ
e) –8.57 kJ
As we discussed when considering equilibrium, when water is placed in a sealed container
it will evaporate until the vapor reaches the equilibrium vapor pressure. Determine the
equilibrium vapor pressure of H 2 O(l) at 25°C in torr (1 atm = 760 torr).
a) 3.14 torr
b) 14.3 torr
c) 23.9 torr
d) 31.8 torr
e) 128 torr
---------------------------------------------------------------------------------------------------------------------
CHEMISTRY 202
Hour Exam II
18.
Sodium bromide dissolves in water according to the equation: NaBr(s)
Na+(aq) + Br–(aq).
Determine the value of the equilibrium constant for this process at 25.0°C given the following
data.
ΔG 0f (kJ/mol)
Na+(aq)
–262
Br–(aq)
–121
NaBr(s)
–360
a) 9.35 x 10-5
19.
Fall 2015
Page No. 4
b) 1.01
c) 2.56
d) 80.4
e) 1.08 x 104
We all love the “pop” bottles that begin most of the lectures. A mixture of hydrogen gas
and oxygen gas remains unreacted until it is exposed to burning methane from a Bunsen
burner. Then the reaction occurs very rapidly (at constant pressure).
Given the data provided, how many of the following statements help to explain this
behavior?
2H 2 (g) + O 2 (g) → 2H 2 O(l)
∆G° = –474 kJ
∆H° = –572 kJ
∆S° = –327 kJ
I.
The mixture remains unreacted because the reactants are thermodynamically more
stable than the products.
II. The mixture remains unreacted because the reaction has a small equilibrium constant.
III. The mixture remains unreacted because the negative value for ∆S° slows down the
reaction.
IV. The mixture remains unreacted until the heat from the Bunsen burner raises the
temperature of the system and this makes the reaction more thermodynamically
favorable.
a) 0
20.
b) 1
c) 2
d) 3
e) 4
Each of the following chemical reactions represented below is spontaneous at some
temperature:
I.
II.
III.
IV.
2KClO 3 (s) → 2KCl(s) + 3O 2 (g)
2H 2 O 2 (l) → 2H 2 O(l) + O 2 (g)
Pb(NO 3 ) 2 (aq) + 2NaCl(aq) → PbCl 2 (s) + 2NaNO 3 (aq)
2SO 3 (g) → 2SO 2 (g) + O 2 (g)
From just this fact (that each is spontaneous) and the balanced chemical equations, how
many of the reactions can you conclude must be exothermic? Note: do not use any
background knowledge about the reactions (that is, the question is not asking which are
actually exothermic but which must be exothermic given the information provided).
a) 0
b) 1
c) 2
d) 3
e) 4
CHEMISTRY 202
Hour Exam II
Fall 2015
Page No. 5
------------------------------------------------------------------------------------------------------------------21-25. Two samples of 1.00 mole of helium gas (considered to be an ideal, monatomic gas) are
in separate containers at the same conditions of pressure, volume, and temperature (V i =
2.00 L and P i = 9.00 atm). Both samples undergo changes in conditions and finish with
V f = 3.00 L and P f = 6.00 atm. They do so in different pathways as described below:
Pathway I
Step 1a: Pressure remains constant, final conditions are 9.00 atm, 3.00 L.
Step 1b: Volume remains constant; final conditions are 6.00 atm, 3.00 L.
Pathway II
Step 2a: Volume remains constant; final conditions are 6.00 atm, 2.00 L.
Step 2b: Pressure remains constant, final conditions are 6.00 atm, 3.00 L.
21.
Which of the following is true and best explains the value of ΔH for Pathway I overall?
a) ΔH = 0 for Pathway I because ΔH is a state function.
b) ΔH = 0 for Pathway I because the process is isothermal.
c) ΔH < 0 because heat is released into the surroundings.
d) ΔH > 0 because heat is taken in from the surrounding.
e) ΔH cannot be determined because both pressure and volume change.
22.
For which step is the magnitude of work (that is, |w|) the greatest?
a) Pathway I step 1a
b) Pathway I step 1b
c) Pathway II step 2a
d) Pathway II step 2b
e) The magnitude of work is equally large in at least two of the steps.
23.
True or false? In step 1a of Pathway I, w = 0.
a) True. This is because the process is isothermal.
b) True. This is because it is working against a constant pressure so ΔP = 0.
c) True. This is because w = –PΔV and ΔV=0.
d) False. In step 1a of Pathway I w > 0.
e) False. In step 1a of Pathway I w < 0.
24.
For which method is the magnitude of heat (that is, |q|) the greatest?
a) Pathway I
b) Pathway II
c) The magnitude of q is equal in each pathway (though opposite in sign) because q is a
state function.
d) The magnitude of q is equal in each pathway (though opposite in sign) and q ≠ 0. This
is true even though q is not a state function, because the process is isothermal.
e) For both pathways q = 0, because the process is isothermal.
25.
Determine the value of ΔE for Pathway I step 1b.
a) 1.37 kJ
b) –1.37 kJ
c) 2.29 kJ
d) –2.29 kJ
e) 0
---------------------------------------------------------------------------------------------------------------------
CHEMISTRY 202
Hour Exam II
Fall 2015
Page No. 6
--------------------------------------------------------------------------------------------------------------26-30. Choose the most appropriate plot for each of the following. A plot may be used once,
more than once, or not at all.
a)
d)
b)
c)
e)
26.
|ΔG| for a reaction vs. number of moles of product formed
a
27.
|w| vs. the number of steps in the isothermal compression of an ideal gas
d
28.
|w| vs. |q| for the isothermal expansion of an ideal gas
a
29.
H vs. V at constant temperature for 1.0 mole of an ideal gas
c
30.
G vs. S for 1.0 mole of an ideal gas at constant T
e
CHEMISTRY 202
Hour Exam II
31.
Fall 2015
Page No. 7
We discussed in lecture how, for small temperature changes, we can assume the values of
ΔH and ΔS are independent of temperature. Also, when determining boiling or melting
points, we can often assume this because the changes in ∆H and ∆S somewhat cancel out.
Let’s test these assumptions.
a.
Given the following data (at standard conditions of 25°C and 1 atm), and assuming ∆H and
∆S are independent of temperature, calculate the normal (at 1 atm) boiling point for
bromine. Show all work and briefly explain what you are doing and why. [6 pts.]
∆H°f (kJ/mol)
S° (J/mol K)
Br 2 (l)
0
152.0
Br 2 (g)
31.0
245.0
At the boiling point the system between liquid and vapor is at equilibrium, so ∆G = 0.
Since ∆G = ∆H –T∆S, ∆H –T∆S = 0, thus T = ∆H/∆S.
Since we are assuming ∆H and ∆S are independent of temperature we can use T = ∆H°/∆S°.
T=
b.
[31,000 J - 0 J]
= 333 K
[245 J/K - 152 J/K]
Now that you have determined the boiling point of bromine, calculate ∆H for the
vaporization of Br 2 (l) at this temperature and 1 atm. The heat capacity for Br 2 (l) =
75.69 J/molK and the heat capacity for Br 2 (g) = 36.02 J/Kmol. Assume these heat
capacities are independent of temperature. Show all work. In addition, briefly
justify the relative values of ΔH vaporization at the different temperatures (that is,
which is larger and why?). [6 pts.]
ΔH° at 298K = 31.0 kJ (from above) which is for Br 2 (l) → Br 2 (g)
ΔH° for Br 2 (l) at 333K to Br 2 (l) at 298K = (1 mole)(75.69 J/molK)(298K–333K) = –2.65 kJ
ΔH° for Br 2 (g) at 298K to Br 2 (l) at 333K = (1 mole)(36.02 J/molK)(333K–298K) = 1.26 kJ
So, ΔH at 333K for Br 2 (l) → Br 2 (g) = 31.0 kJ + (–2.65 kJ) + 1.26 kJ = 29.6 kJ
Thus, ΔH at 333K is less than ΔH at 298K which makes sense because at the higher
temperature, less energy is required to vaporize the liquid since the liquid is already at a higher
temperature.
CHEMISTRY 202
Hour Exam II
31.
Fall 2015
Page No. 8
(con’t)
c.
Now that you have determined the boiling point of bromine, calculate ∆S for the
vaporization of Br 2 (l) at this temperature and 1 atm. The heat capacity for Br 2 (l) =
75.69 J/molK and the heat capacity for Br 2 (g) = 36.02 J/Kmol. Assume these heat
capacities are independent of temperature. Show all work. [4 pts.]
ΔS° at 298K = 93.0 J/K (from part a) which is for Br 2 (l) → Br 2 (g)
ΔS° for Br 2 (l) at 333K to Br 2 (l) at 298K = (1 mole)(75.69 J/molK)(ln{298K/333K}) = –8.4 J/K
ΔS° for Br 2 (g) at 298K to Br 2 (l) at 333K = (1 mole)(36.02 J/molK)(ln{333K/298K}) = 4.0 J/K
So, ΔS at 333K for Br 2 (l) → Br 2 (g) = 93.0 J/K + (–8.4 J/K) + 4.0 J/K = 88.6 J/K
d.
T=
e.
Determine the normal boiling point for bromine from the ∆H and ∆S values
calculated in parts b and c. Show all work. [2 pts.]
[∆H] [29,600 J ]
=
= 334 K
[∆S]
[88.6 J/K]
Briefly comment on our assumptions. Are they reasonable for the vaporization of
bromine? Explain. [2 pts.]
The values for ΔH and ΔS did change slightly but not dramatically. The calculated boiling
temperature was within a rounding error. Yes, the assumptions are reasonable.
[4-5% error for ΔH, ~5% error for ΔS, ~0.3% error for T]
CHEMISTRY 202
Hour Exam II
32.
Fall 2015
Page No. 9
On the second day of lecture (back in August) I mentioned that we will be studying
thermodynamics this semester and that the knowledge and understanding we gain will
provide information on “why” processes happen and will allow us to make predictions
about whether a process will occur. Now is the time for you to show what you know about
this for two scenarios that are described on the following pages.
In the first portion you are to discuss the scenario in words (without equations).
Make sure to address the following:
•
•
•
•
•
Discuss what happens during the process.
o For example, if one of the scenarios was “You add a drop of red food
coloring to water”, you should respond with “The food coloring will spread
in the water until the solution is equally red throughout. As long as the food
coloring and water began at the same temperature, the temperature will not
change.”
List which factor(s) make the process more thermodynamically favorable and
discuss why each particular factor is thermodynamically favorable.
List which factor(s) (if any) make the process less thermodynamically favorable
and discuss why each particular factor is thermodynamically unfavorable.
If there is at least one thermodynamically unfavorable factor, explain how the
process you have described can be spontaneous.
Make sure to use the terms ΔS, ΔS surr , and ΔS univ in your discussion. Clearly
specify the system and the surroundings.
In the second portion you are to use the appropriate data and equations to calculate
ΔS, ΔS surr , ΔS univ and explain how these support the fact that the process you have
described in the first portion is spontaneous.
NOTE: You will be provided with all of the data you need, but you may not need all
of the data you are provided.
LIMIT EACH ANSWER TO THE SPACE PROVIDED.
THINK ABOUT WHAT YOU WILL WRITE BEFORE
YOU START WRITING.
CHEMISTRY 202
Hour Exam II
32.
Fall 2015
Page No. 10
Address both of these scenarios. Please see page 9 for what is required.
a. You take liquid water from your freezer that is set to 0.°C and place the water outside
on a winter day in which the temperature is –15°C.
i. What happens and why? Please address all bullet points listed on page 9. [10 pts.]
The system is the water and the surroundings are the rest of “outside”.
The water will freeze (turn to ice) at 0.°C and the ice will eventually reach a temperature of –15°C.
This is thermodynamically favorable because: 1) the process of freezing is exothermic, which will
increase ΔS surr , and 2) thermal equilibrium increases ΔS univ (heat is released as the temperature of
the ice decreases).
This is thermodynamically unfavorable because 1) when a liquid turns into a solid, ΔS decreases
since a solid is more ordered than a liquid and 2) when the ice cools its entropy (ΔS) decreases as
well.
Overall ΔS univ must be positive for the process to occur so the ΔS surr must be greater in magnitude
overall than that of ΔS. This makes sense because the process is exothermic and at low
temperatures exothermicity is thus a more important driving force (the temperature is below the
freezing point of water).
ii. The magnitude for the standard (25°C and 1 atm) change in enthalpy of fusion
water is 6.03 kJ/mol and the freezing point of water is 0.°C, 1 atm. The heat
capacity of ice is 37.5 J/Kmol, and the heat capacity of liquid water is 75.3 J/Kmol.
Assume ΔH, ΔS, and the heat capacities are independent of temperature.
Determine ΔS, ΔS surr , and ΔS univ for the process that occurs when you place
1.00 mole of 0.°C water outside when the temperature is –15°C and explain
how they support what you explained above. [10 pts.]
Freezing of water at 0.°C:
ΔS = (–6030J/273K) = –22.1 J/K
ΔS surr = –(–6030J/258K) = 23.4 J/K
Cooling of ice:
heat released = (1 mole)(37.5 J/Kmol)(–15K) = –562.5 J = q
ΔS = (1 mole)(37.5 J/Kmol)(ln{258/273}) = –2.12 J/K
ΔS surr = –(–562.5J/258K) = 2.18 J/K
Overall:
ΔS = –24.22 J/K
ΔS surr = 25.58 J/K
ΔS univ = 1.36 J/K; since ΔS univ is greater than zero, the process is spontaneous.
[Note: both processes, freezing and cooling, are spontaneous]
CHEMISTRY 202
Hour Exam II
32.
Fall 2015
Page No. 11
(con’t).
c. Consider the two-bulb system as shown below. You have a sample of N 2 gas at 75.0°C
in the left bulb and you have a sample of He gas at 25.0°C in the right bulb. You open
the stopcock between the two bulbs. Assume this system is perfectly insulated.
i. What happens and why? Please address all bullet points listed on page 9. [10 pts.]
The system is the gases in the two-bulb container. The surroundings are outside of the container.
The gases will mix and spread throughout the container so that they are evenly distributed (in
terms of pressure) and will eventually reach the same temperature.
This is thermodynamically favorable because: 1) the entropy of each gas will increase by taking up
a larger volume (more microstates), 2) the temperature of the helium will increase, thus increasing
its entropy. [could also mention thermal equilibrium]
This is thermodynamically unfavorable because 1) the temperature of the nitrogen will decrease,
thus decreasing its entropy.
Overall ΔS univ must be positive for the process to occur and since ΔS surr = 0 (perfectly insulated)
ΔS must be positive (the mixing and increase in temperature of the helium outweighs the decrease
in temperature of the nitrogen).
ii. In the left bulb (3.00 L) you have 1.00 mole N 2 gas at 75.0°C and in the right bulb
(1.00 L) you have 3.00 moles He gas at 25°C. For N 2 , C v = 20.71 J/Kmol and C p
= 29.03 J/Kmol; for He: C v = 12.47 J/Kmol and C p = 20.80 J/Kmol. Assume the
heat capacities are independent of temperature.
Determine ΔS, ΔS surr , and ΔS univ for the process that occurs when you open the
stopcock and explain how they support what you explained above. [10 pts.]
Solve for T f : (1.00 mol)(20.71 J/Kmol)(75–T f ) = (3.00 mol)(12.47 J/Kmol)(T f –25)
T f = 42.8°C = 315.8K
Nitrogen gas:
Decreasing temperature: ΔS = (1 mole)(20.71 J/Kmol)(ln{315.8K/348K})= –2.01 J/K
Spreading out in the container: ΔS = (1 mole)(8.314 J/Kmol)(ln{4.00L/3.00L})= 2.39 J/K
Helium gas:
Increasing temperature: ΔS = (3 mole)(12.47 J/Kmol)(ln{315.8K/298K})= 2.17 J/K
Spreading out in the container: ΔS = (3 mole)(8.314 J/Kmol)(ln{4.00L/1.00L})= 34.6 J/K
Overall:
ΔS = 37.15 J/K
ΔS surr = 0
ΔS univ = 37.15 J/K ; since ΔS univ is greater than zero, the process is spontaneous.