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Chemistry 33000
Physical Chemistry I: Thermodynamics
Fall, 2022
CUNYFirst Class 23729
WITH ANSWERS
Section L-25-LEC Regular
Exam III: Thermodynmics (T.D.) of Phase Equilibria and Liquid-Liquid Solutions
The attached exam consists of five
questions, most of which include several independent
parts that require verbal explanations and/or numerical calculations. The parts within a problem can
often be answered most readily in the sequence they are asked, but if you miss one section it is
usually possible to complete the next
.
Try to budget your time according to the number of points
allotted to each question
; expect a class average of ~60 and a curve to estimate the corresponding
letter grades.
Please write legibly
ON THE EXAM SHEET
and with sufficient coherence that I may infer
your reasoning in those instances where your numerical result is incorrect. Note any assumptions or
approximations that you make.
If you need to clarify how to interpret a question, ask me!
Before you begin to sweat and scribble, write your name
on the exam booklet.
Hand in the
exam questions with your responses and your formula sheet
when you have completed your work.
Here are some physical constants and relationships that you may find useful.
Good luck!
R = 8.31 J-mol
-1
-K
-1
= 0.082 L-atm-mol
-1
-K
-1
= 0.083 L-bar-mol
-1
-K
-1
= 1.99 cal-mol
-1
-K
-1
J = (Pa - m
3
) / (mol - K)
J = (kg-m
2
)/s
2
1 J = 1 N-m
g = 9.8 m/s
2
Standard Temperature & Pressure (STP) = 273 K, 1 atm for gases
STP (º) = 298 K, 1 bar for chemical formation reactions and phase changes
1 atm = 760 torr = 1 x 10
5
Pa
1 bar
100 cm = 10 dm = 1 m
T (K) = t (
0
C) + 273
1 calorie = 4.18 J
U = q + w
dU = TdS - pdV
dG = -SdT +Vdp
∮
dX = 0 for any state function X
C
p
- C
v
= nR for ideal gases
K
f
for water = 1.86 K-kg-mol
-1
dq
P
= nC
P
dT
dq
V
= n C
V
dT
K
f
for naphthalene = 6.94 K-kg-mol
-1
Atomic masses: H=1; C=12; O=16; Na=23; Cl=35; Ca=40
2022-EIII-ans
11/27/2022
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12/12/2023
2
This page has been intentionally left blank.
Be encouraged to tear off the cover page and refer to it as you work the problems.
Also be encouraged to use this page as scratch paper.
1.
2022-EIII-ans
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3
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Pure Substances:
Phase Diagrams for Compact T.D. Information
(29 points, 4 parts)
Plots of pressure vs. temperature, pressure vs. composition, and temperature vs. composition
provide a compact but information-rich picture of the phases and chemical mixtures that are
present under laboratory conditions of interest. Consider the following data for oxygen (O
2
),
which is familiar to us as a life-supporting gas but can also be present in liquid or solid phases
under appropriate circumstances. Consider the following properties:
a: triple point: 54.3 K, 1.14 torr
b: critical point: 154.6 K and 37,828 torr
c: normal melting point: 54.8 K, 760 torr
d: normal boiling point: 90.3 K, 760 torr
a: triple point: 54.3 K, ln p = ln (1.14 torr) = 0.13 torr
b: critical point: 154.6 K, ln p = ln (37,828 torr) = 10.5 torr
c: normal melting point: 54.8 K, ln p = ln (760) = 6.6 torr
d: normal boiling point: 90.3 K, ln p = ln (760) = 6.6 torr
(or mark directly on the diagram)
(3 pts) 1b.
If ln p = 7.6 torr and T = 60 K, which phase(s) are present
? Explain and mark the
phase diagram.
e: liquid only in this region
(9 pts) 1c.
Consider how pressure depends on temperature
for liquid and gas phases
of O
2
that
are in equilibrium with one another. First, answer this question qualitatively
by quoting a
relationship
that describes this dependence and explaining why it has a positive or negative
value. Then, make a quantitative estimate
by using the given data for points a-d on the O
2
phase
diagram. Do a numerical calculation
and explain briefly
.
Qualitatively
, dp/dT is positive since
∆
S and
∆
V are each positive.
∆
S is positive
since the liquid-gas conversion goes toward a state of greater disorder.
∆
V is positive
since gases spread out to occupy a larger molar volume than liquids (or solids).
Quantitatively
, use the data to calculate dp/dT:
p
(
d
)
−
p
(
a
)
T
(
d
)
−
T
(
a
)
=
(
760
−
1.14
)
(
90.3
−
54.3
)
=
758.86
36
= + 21.08
Alternatively, use the data to calculate
p
(
b
)
−
p
(
d
)
T
(
b
)
−
T
(
d
)
or
p
(
b
)
−
p
(
a
)
T
(
b
)
−
T
(
a
)
, also >
0
2022-EIII-ans
12/12/2023
(12 pts) 1a.
These data are sufficient to construct the
phase diagram shown on the right. (Note:
the y-axis
shows ln p
in order to accommodate the large range of
values that are relevant for the three phases of diatomic
oxygen.) Use the data to put a, b, c, and d labels on the
marked locations of the diagram, including the numerical values that support your choices.
4
(5 pts) 1d.
Which values of pressure will ensure that O
2
(s) can sublime
to O
2
(g)? Explain or
make a sketch to justify your answer.
ln p must be below the triple point value of 1.15 torr, as illustrated by the arrow on the graph.
2022-EIII-ans
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5
2.
Pure Substances: Condensed Phases in Equilibrium with Gases
(16 points, 2 parts)
Prior to the discovery that freon-12 (CF
2
Cl
2
) was harmful to the Earth’s ozone layer, this
compound was frequently used as the dispersing agent in cans of hair spray (as well as in
refrigerators, air conditioners, and fire extinguishers). Its enthalpy of vaporization at its normal
boiling point of 244.0 K is 20.25 kJ mol
-1
.
(4 pts) 2a.
Predict the signs of
∆
vap
H and
∆
vap
S
for freon, rationalizing your choices with a
brief explanation for each quantity you designate.
A transformation from liquid to gas always requires input of heat energy, so
∆
vap
H is
endothermic and the value is > 0.
A transformation from liquid to gas always increases disorder, so
∆
vap
S is > 0.
(12 points) 2b.
Estimate the pressure that a can of hair spray that used freon-12 had to withstand
at 313 K (40
°
C), the estimated temperature of a can that has been standing in the sunlight.
You may assume that
∆
vap
H is constant over the described temperature range and equal to its
standard value of 20.25 kJ - mol
-1
.
[adapted from assigned Problem 4.2b. with apologies for the unavoidable blurriness]
2022-EIII-ans
12/12/2023
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3.
Thermodynamics of Chemical Solutions
(31 points, 4 parts)
(6 pts) 3a.
A liquid mixture of toluene (C
7
H
8
, T) and benzene (C
6
H
6
, B) has a toluene mole fraction
of 0.336. At 30
°
C, P*
(toluene)
= 36.7 torr and P*
(benzene)
= 118.2 torr as shown on the graph below.
Assuming that the mixture behaves ideally, find the total pressure
and the partial vapor pressure
of
each component
above the solution at this temperature. (Make a calculation or use the graph.)
P
B
= X
B
P
B
* = (0.664) (118.2 torr) = 78.5 torr
P
T
=
X
T
P
T
*
= (0.336) (36.7 torr)
=12.3 torr
Thus P
TOT
= P
B +
P
T
=
90.8 torr
OR: add the values of P
B
and P
T
that you can read off the graph for the given mole fraction X
B
=
0.664.
(7 pts) 3b.
A different mixture has a total vapor pressure of 100 torr. What is its solution
composition
?
P
TOT
= 100 torr = X
B
p
B
* + X
t
p
T
*
= X
B
p
B
* + (1- X
B
) p
T
*
= 118.2 X
B
+ 36.7 – 36.7 X
B
63.3
= 81.5 X
B
0.78
=
X
B
and thus X
T
= 1- X
B
= 0.22
2022-EIII-ans
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7
(6 pts) 3c.
A third mixture of toluene and benzene has a vapor that is equimolar
in the two
components. What are the solution composition and total vapor pressure for this mixture
? (The
graph of Problem 3a is replicated for your viewing convenience.)
If the vapor has y
B
= y
T
= 0.5, then p
i
= y
i
p
TOT
has the same value for benzene and toluene. Thus p
B
= p
T
where the lines cross and can be read from the graph as p
i
≈
29 torr (each tick mark on the
y-axis is 40 torr). The corresponding solution composition read from the graph is X
B
≈
0.23
(each tick mark on the x-axis is 0.25). The total vapor pressure is then 2 p
B
or 2 p
T
= 58 torr.
Solution compositions (X
i
) relate to P
i
*, whereas vapor compositions (y
i
) relate to P
TOT
!
(Analytical methods of solution are also possible.)
(12 pts) 3d.
Finally, return to the mixture of benzene and toluene that is described in Problem 3a,
for which the mole fraction of toluene is 0.336. Again assuming ideal behavior, calculate
∆
G,
∆
H, and
∆
S
(for the system), when 1 mole of each liquid is mixed at 30
°
C. Do an
analogous calculation
for benzene and toluene in the gas phase
, noting any similarities and
differences
.
∆
mix
G = nRT ( X
b
ln X
b
+ X
t
ln X
t
) = (2) (8.314 J/mol-K) (303 K) [(.664) ln (.664) + (.336) ( ln
(.336)]
=
5038
[ - 0.272
- 0.366
]
(6 pts)
= -3217 J/mol
∆
mix
S = -
∆
mix
G /T = +10.62 J / mol-K
∆
mix
H =
∆
mix
G + T
∆
mix
S = 0
The calculation would be identical for a mixture of ideal gases with the same mole fractions, except
that in practice, T would be higher unless we chose p to retain T at 303 K.
2022-EIII-ans
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4.
Colligative Properties
(10 points)
The addition of 5.00 g of an unknown solid compound to 250 g of liquid naphthalene
lowered the freezing point of this solvent by 0.780 K. Find the molar mass of the compound
.
The cryoscopic constant for naphthalene is 6.94 K-kg-mol
-1
.
[Adapted from assigned Ex. 5.6b]
∆T
≡
(T
f
* - T
f
) = K
f
m
B
(or K
f
b
B
per your textbook), where K
f
= the cryoscopic
constant for the solvent, m
B
or b
B
= molality
≡
moles of the solute per kg of solvent
m
B
=
∆T
K
(
f
)
=
0.780
K
6.94
K
−
kg
−
mol
−
1
=
0.112
mol B
kgnaphthalenesolvent
Since 5.00 g of the unknown compound is dissolved in 250 g naphthalene, this value
corresponds to 20.00 g per 1.000 kg of that solvent.
Then considering 1 kg naphthalene for a basis in each case,
20.00
g
0.112
mol
= 178 g/mol for
the molecular weight of the unknown compound.
Watch your units, definition of molality (m
B
or b
B
), and distinctions between solute and
solvent!
2022-EIII-ans
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5.
Phase Equilibria: the case of a liquid (
l
) with a gas (g) (
14 points
)
Consider the derivation of the Clapeyron equation that was shown in class and in your text.
Provide a brief verbal statement and/or equation to justify each step
. (The extensive variables
(G, S, V) per mole are related as shown, but the bars above the symbols or ‘m’ subscripts
have been omitted for simplicity.)
dG
g
= - S
g
dT + V
g
dp
combined 1
st
& 2nd Laws of T.D. for gas
;
G = U -TS + pV to transform variables from (S,V) to (T, p)
dG
l
= - S
l
dT
+ V
l
dp
combined First & Second Laws of
T.D. for liquid
; G = U -TS + pV to transform variables from (S,V) to (T, p)
- S
g
dT + V
g
dp
= - S
l
dT
+ V
l
dp
gas and liquid are in equilibrium, so dG
g
= dG
l
or
μ
(
g
)
=
μ
(
l
)
dp (V
g
– V
l
)
= dT (S
g
– S
l
)
separate (sequester) variables (p &
T)
dp
dT
=
[
S
(
g
)
−
S
(
l
)
]
[
V
(
g
)
−
V
(
l
)
]
formulate slope that signifies
P vs. T dependence
=
∆ S
∆V
for
l
→
g
define
∆ X
as X
f
– X
i
for a
T.D. process
e.g.,
∆ S
as S
g
– S
g
for
l
→
g
2022-EIII-ans
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10
=
∆ H
T ∆V
∆ S≡
∫
dqrev
T
=
∫
dH
T
=
∆ H
T
for
a phase transition at constant T
2022-EIII-ans
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-0123456789+
Write an equation for the formation of Cu20(s) from its elements in their standard states.
Express your answer as a chemical equation.
Identify all of the phases (e.g. (s), (I), (g) ) in your answer.
Make sure your answer does not include any spaces!
Enter fractions using the "/" symbol (e.g. 1/2).
Answer:
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7. The solubility, C, of sucrose in water, expressed as grams of sucrose per 100 g of solvent, is
given as below
t/°C
C/g
0
179.2
10
20
203.9
190.5
30
219.5
[ΔΗ
(a) Draw an appropriate graph to determine the molar enthalpy of solution of sucrose in the
temperature range 0 to 40°C.
(b) 50mL of saturated solution of benzoic acid at 25°C requires 12.8 mL of 0.05 M NaOH to
completely neutralize. What is the solubility of benzoic acid in g/L water at 25°C.
so ln,m
40
238.1
= 4.96 kJ/mol, Cbenzoic acid = 1.56g/L]
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AgBr(s) 2 Ag*(aq) + Br (aq)
Ag*(aq) + 2NH3(aq) Ag(NH3)2*(aq) Kf = 1.5 x 107
RS
Ksp = 7.7 x 10-13
When NH3 is added to AgBr(s), a combination of these reactions result:
AgBr(s) + 2NH3(aq) 2 Ag(NH3)2*(aq) + Br (aq)
What mass of AgBr will dissolve in a 5.0 L container of 3.0 M NH3 solution?
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REFER TO IMAGE
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2 When 50 cm' of hydrochloric acid of concentration 2.0 mol dm is added to 50 cm' of
sodium hydroxide solution of concentration 2.0mol dm, the temperature increase is 13.0°C
HCl(aq) + NAOH(aq)
+ NaCl(aq) + H,O(1)
The experiment is repeated using 25 cm of the same hydrochloric acid and 50 cm' of
the same sodium hydroxide solution.
What is the temperature increase?
O A 4.9°C
I B 6.5°C
I C 8.7°C
D 13.0°C
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Calculate the percentage change in Kx for the reaction CH3OH(g) + NOCl(g) ⇋ HCl(g) + CH3NO2(g) when the total pressure is increased from 1.0 bar to 2.0 bar at constant temperature.
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