NAME:
Section Number:
2010
KEY
CHEMISTRY 418, Fall, 2010(10F)
Midterm Examination #2, November 11,
10
Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILY
understand your work; only work on the exam sheet will be considered. Write answers, where appropriate, with reasonable numbers of
significant figures. You may use only the handbook of “Essential Data and Equations for a Course in Physical Chemistry”, a calculator,
and a straight edge.
1. (20 points) Use the van’t Hoff equation for the temperature dependence of the equilibrium constant
to estimate the equilibrium constant for the following reaction at 700 K
I 2 ( gas ) → 2 I ( gas )
Θ
Assume that ∆H rxn for this process is independent of temperature.
DO
NOT
WRITE IN THIS
SPACE
p. 1________/20
p. 2________/20
Answer: The van’t Hoff’ equation is
Θ
d ln K a ∆H rxn
=
dT
RT 2
p. 3________/20
p. 4________/20
Based on the data in Table 5.8,
∆H
Θ
rxn
kJ
kJ
kJ
= 2∆H ( I , gas ) − ∆H ( I 2 , gas ) = 2 × 106.8
− 62.4
= 151.2
mol
mol
mol
Θ
f
Θ
f
Θ
(298.15 K ) = 2∆G Θf ( I , gas ) − ∆G Θf ( I 2 , gas ) = 2 × 70.2
∆G rxn
kJ
kJ
kJ
− 19.3
= 121.1
mol
mol
mol
From van’t Hoff equation:
p. 5________/20
=============
p. 6_______/5
(Extra credit)
=============
TOTAL PTS
Θ
∆H rxn
1

 1
ln K a (T = 700 K ) = ln K a (298.15 K ) −
−
=

R  700 K 298.15 K 
=−
=−
Θ
Θ
∆Grxn
∆H rxn
−
R × 298.15 K
R
1

 1
−
=

700
K
298
.
15
K


121100 J / mol
151200 J / mol  1
1

−
−
 = −13.837

J
J
700 K 298.15 K 

8.3144
× 298.15K 8.3144
mol × K
mol × K
Thus:
K a = exp(−13.837) = 9.787 × 10 −7
As expected, even at elevated temperatures iodine does not easily dissociate into atoms in a gas phase.
/100
NAME:
CHEM 418, Midterm Exam #2, Fall, 2010, page 2
2. (20 points) Fill in the blanks
a)
(4 points). __PHASE______ is a form of matter that is uniform with respect to chemical composition and the state of
aggregation on both microscopic and macroscopic scale.
b)
(4 points). An equilibrium condition where gas, solid, and liquid coexist and all three chemical potentials are equal is
called ____TRIPPLE POINT___________
J
for any liquid is called ___Trouton’s rule___
mol × K
c)
(4 points). A rule of a limited use stating that ∆S mvaporization ≈ 90
d)
(4 points). In living organisms molecules that are composed of a polar segment (hydrophilic part) and a non-polar
segment (lypophilic part) often play major roles in the formation of membranes and surfaces. This class of molecules
is called __amphiphiles___
e)
(4 points). The quantity that was first described by the American chemist Josiah Willard Gibbs that is defined as a
partial molar Gibbs free energy is called __chemical potential___
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NAME:
CHEM 418, Midterm Exam #2, Fall, 2010, page 3
3. (20 points) Prove the following Mawell relationship based on the appropriate Differential Equation of Thermodynamics
(Table 5.3):
 ∂T 
 ∂V 
 =


 ∂P  S  ∂S  P
 ∂H 
 ∂H 
dH = TdS + VdP and dH = 
 dS + 
 dP
 ∂S  P
 ∂P  S
 ∂H 
 ∂H 
Thus T = 

 ; and V = 
 ∂P  S
 ∂S  P
For the left expression: hold entropy constant and differentiate over pressure; for the right expression: hold the pressure
constant and differentiate over entropy:
 ∂  ∂H  
 ∂T 
  ; and
 =  

 ∂P  S  ∂P  ∂S  P  S
 ∂  ∂H  
 ∂V 
 
 =  

 ∂S  P  ∂S  ∂P  S  P
Since enthalpy is a continuous and smooth function of both pressure and entropy, the order of differentiation foes not
affect the second derivative and
 ∂  ∂H  
 ∂  ∂H  
 
 
  =  
 ∂P  ∂S  P  S  ∂S  ∂P  S  P
Thus,
 ∂V 
 ∂T 

 =

 ∂P  S  ∂S  P
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NAME:
CHEM 418, Midterm Exam #2, Fall, 2010, page 4
4. (20 points). Calculate the pressure inside a bubble of CCl4 10 µm in diameter at room temperature.
∆P = 2 ×
2γ
26.43 × 10 −3 N / m
N
=4
= 21144 2 = 21144 Pa
−6
r
5 × 10 m
m
The total pressure inside the bubble is:
P = 1 atm + ∆P = 101325 Pa + 21144 Pa = 122469 Pa
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NAME:
CHEM 418, Midterm Exam #2, Fall, 2010, page 5
5. (20 points). In the absence of moisture, calcium carbonate (araganite), CaCO3, can decompose into CaO and CO2.
CaCO3 ( s, ara ) → CaO ( s ) + CO2 ( gas )
If the reaction was performed by placing 0.1 mole of CaCO3 into a 100 L evacuated cylinder, calculate the mass of
gaseous CO2 present at equilibrium at 400°C. The equilibrium constant for this reaction at 400°C was measured to be
-6
4.369 x 10 .
1)
CaCO3 ( s, cal ) → CaO ( s ) + CO2 ( gas )
Initial moles
Final moles
0.1
0.1-x
Final Pressure 0
0
x
0
x
RT
x
V
0
2) Plug everything in
RT
x
V ×1
Θ
 a(CO2 )a(CaO ) 
 = P
K a = 
=
(
)
a
CaCO
1
3


J
× 673.15 K
mol × K
x mol
3
−3 m
100 L × 10
L
= 4.369 × 10 −6
100000 Pa
8.3144
-6
x = 7.806x10 mol
-4
Molar mass of CO2 is 44.0095 g/mol so the total mass of CO2 present in the cylinder at equilibrium is 3.4355x10 g
-3
For curious minds, the corresponding pressure will be 0.437 Pa or 3.28x10 Torr. This means that the cylinder has to be
evacuated with a very good pump, likely a turbomolecular pump, prior to the experiment.
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NAME:
CHEM 418, Midterm Exam #2, Fall, 2010, page 6
6. (5 points, extra credit). Suppose you have two different crystalline forms A and B of the same compound at equilibrium.
It is known that the density of A is greater than the density of B and that the conversion of A to B is an exothermic process.
If we wish to shift the equilibrium towards crystalline form B, should we use high or low temperature and high or low
pressure? Explain your statements clearly. Statements without explanation will be awarded no points.
Since A is more dense than B, the molar volume of B is greater. Therefore, high pressure will shift equilibrium towards the
more dense form and we will need to decrease the pressure to shift the equilibrium towards B.
Since the conversion of A to B is exothermic (∆H<0), the equilibrium constant will decrease as temperature increases.
Thus, increased temperatures will favor the formation of A and in order to shift the equilibrium towards B, we must
decrease the temperature.
Low temperature and low pressure favor the formation of B.
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