Chemistry 452/456
Enter answers in a Blue Book
July 22, 2011
Midterm Examination Key
Useful Constants:
 1 Newton=1 N= 1 kg m s-2
 1 Joule=1J=1 N m=1 kgm2/s2
 1 Pascal=1Pa=1N m-2
 1atm=101325 Pa
 1 bar=105 Pa
 1L=0.001m3
 Universal Gas Constant R=8.31 J K-1 mol-1=0.0821L atm K-1mol-1
 Avagadro’s Number NA=6.024x1023 mol-1
All answers must be in MKS units (i.e. units of meters, seconds, kilograms, Joules,
Pascals etc.)
Helpful math relationships:
dax n
1)
 anx n 1 ; a is a constant, n is a positive or negative number
dx
x n 1
2)  x n dx 
dx  c where c is a constant and n  1
n 1
dx
3)   ln x  c where c is a constant
x
x
4) ln  xy   ln x  ln y; ln    ln x  ln y; x ln y  ln y x
 y
Part 1 (18 points) Answer THREE out of the following SIX questions. Limit
definitions to less than 200 words. Use equations where helpful or required, but
detailed calculations are not necessary.
Question 1.1. Define Euler’s Criterion for Exactness. Explain how this rule is relevant to
the field of thermodynamics.
A differential, M  x, y  dx  N  x, y  dy is exact if there exist a function Z(x,y) such that
 Z 
 Z 
dZ  
 dy  M  x, y  dx  N  x, y  dy . The criterion for exactness is
 dx  
 x  y
 y  x
M N  2 Z
therefore
. The path integrals of exact differentials are dependent only


y
x xy
on the initial and final states of the path, not on the details of the path itself. Differentials
of state functions are exact so when integrated over specific paths, the state function
change is dependent only on the definitions of the initial and final states. (74 words)
Question 1.2. The entropy has been defined as “time’s arrow”. Explain this definition
using thermodynamic reasoning. Utilize in your argument the statistical interpretation of
entropy.
The equations of mechanics (F=Ma) are time reversible. This means they are as valid
going forward in time as going backwards. In contrast, the entropy of an isolated system
increases from initial to final state, or remains the same. The final state is understood to
be reach later in time, so moving toward the final state is marked by increased entropy.
The reversal of mechanics in an isolated system , which in turn would decrease the
entropy is not impossible, but is nevertheless a highly improbable event. (87 words)
Question 1.3 Explain the Carnot engine’s relevance to the efficiency of heat engines.
Explain also the relationship between entropy and the efficiency of a heat engine. Use a
drawing of the engine if you think it helps to explain its function.
The Carnot engine is an idealized heat engine that operates reversibly. It consists of four
reversible steps: isothermal expansion, followed by an adiabatic expansion, then an
isothermal contraction in volume and back to the initial state via an adiabatic contraction.
The Carnot engine yields the efficiency  of a heat engine operating reversibly:
w
T T
  H L  net , where qH is the heat extracted from the high temperature reservoir
TH
qH
(i.e. TH), TL is the temperature of the low temperature reservoir, and –wnet is the net work
done by the engine. Because the engine is cyclic the net entropy is zero such that:
qH qL
T  TL qH  qL  wnet




 0 so that H
TH
qH
qH
TH TL
Question 1.4 Explain the concept of reversibility as it applies to thermodynamic
pathways.
A system is displaced from its initial state by application of an infinitesimal force. The
system is allowed to re-equilibrate before further application of the force. Thus the
system’s path consists of a sequence of equilibrium states separated by these small
displacements and any step can be reversed by application of an infinitesimal force. The
equation of state is valid along a reversible path. (64 words).
Question 1.5: Assume a protein denatures with a melting temperature of Tm=330K.
dqP
Sketch the differential scanning calorimetry data i.e.
vs. T starting with a
dT
temperature T1<<330K to a temperature T2>>330K. Indicate the regions of the data
where the structured form N is most stable, where the unstructured form D is most stable
and where the D and N forms are in 1:1 equilibrium. Explain the appearance of the data.
Answer:
The heat capacity of D is larger than N because the unstructured form has more degrees
of freedom (motions, hydration sites, etc.) that can absorb heat. At Tm D and N are
equally stable and together with the large number of partially structured forms that exist
at this temperature, have the greatest capacity to absorb heat.
Question 1.6: The Helmholtz (A) energy and Gibbs energy (G) are sometimes called free
energies. In the old literature the Gibbs energy is called the “free enthalpy”. Why are the
Gibbs and Helmholtz energies referred to as “free”? In your explanation, you may want
to recall the definitions of A and G in terms of U, H, and TS. Explain also why the Gibbs
energy G is so often discussed in chemistry and biochemistry texts.
Answer: A=U-TS and G=H-TS. In each case we have an energy U or H (understood to be
heat flow at constant V or P, respectively, diminished by a term TS, temperature times
entropy. In the case of G, it is the amount of the enthalpy change of a system at constant
P and T, that is free to produce useful work. The rest of the enthalpy is lost to entropy,
which is represented in energy units by the TS terms, which accordingly is subtracted
from the enthalpy H. Hence the term “free enthalpy” for the Gibbs energy, refers to the
enthalpy freed for work. The same reasoning applies to A only at constant V and T.
The Gibbs energy is of particular interest in chemistry and biochemistry because it
quantifies reversible non-P-V work, produced at constant P and T, conditions that are
common in chemical and biochemical system. A quantifies reversible work at constant V
and T.
Part 2: (20 points) Answer TWO of the FOUR questions. Answers longer than 200
words are acceptable. Equations may be used to illustrate your points but
numerical calculations are not required. Physical reasoning based on
thermodynamic principles is encouraged.
Problem 2.1 The smelting of iron in a blast furnace occurs when carbon derived from
coal reacts with oxygen to produce carbon monoxide gas CO. At the temperature of the
blast furnace, particles of hematite Fe2O3 are reduced to pure iron in the stack of the
furnace above the hot coal by reacting with the abundant CO gas according to:
Fe2O3  s   3CO  g 
2Fe  s   3CO2  g  .
A chimney located in the top of the blast furnace stack allows hot gases to escape. In
the 19th century, blast furnaces with enormously tall stacks were constructed with the idea
that if hot CO gas could be retained for a longer time in the blast furnace before escaping
through the chimney, more iron would be produced. Which Law of Thermodynamics
makes this a bad idea? Explain.
Answer: Solution: The reaction reaches equilibrium and at the high temperatures
within the blast furnace, the equilibrium likely is far to the right, because the process
gives us lots of iron. That said, allowing more time to have hot gases exposed to the
hematite yields nothing because once the system reaches equilibrium, the second law
says the entropy is maximized and further change is not going to happen. Building
those big blast furnaces turned out to be a waste of time.
Problem 2.2 An ideal monatomic gas with initial temperature T1 expands adiabatically
into a vacuum thereby doubling its volume. Student A predicts that the change in the gas
temperature is T  T1  22 / 3  1 while Student B predicts that T=0. Explain each
student’s reasoning. Which student is correct? Justify your answer.
Answer: The ideal gas expands adiabatically into a vacuum. Adiabatic so q=0. If it
expands into a vacuum no work is done because Pext=0. Therefore,
U  nCV T  q  w  0 . Therefore T=0. Student B is correct. Student A is assumes
incorrectly that the expansion is reversible and adiabatic. Student A uses the
CV
R
T 
V 
1
relationship:  2    1     which is derived in turn from dU  nCV dT  PdV
2
 T1 
 V2 
and the ideal gas law P=RT/V. But the work is zero, so Student A has used the wrong
equation to describe an irreversible expansion accompanied by no work.

an 2 
Problem 2.3. The van der Waals equation of state is  P  2  V  nb   nRT . The
V 

parameter a is due to attractive forces between molecules of the gas. The parameter b is
due to repulsive forces, and represents the amount of the total volume that is occupied by
 U 
 U 
a mole of gas molecules. In the internal energy equation dU  
 dT  
 dV ,
 T V
 V T
an 2
 U 
 U 

where 
for a van der Waals gas. Explain why 

 depends on a, but
2
 V T V
 V T
does not depend on b. Hint: This can be answered with a calculation but physical
reasoning involving the meanings of a and b can also be used.
R
Many students interpreted this question to be a request for a mathematical proof for
an 2
 U 
 P 
 U 
. If this was done by using the relationship 

T 



  P and the
2
 V T V
 V T
 T V
v.d. Waals equation of state, I gave full credit.
But what I had hoped for was some physical reasoning. The parameter b represents the
excluded volume of one mole of gas. Therefore the free volume is V-b. If the free volume
changes we have V=V2-b-(V1-b)=V2-V1. This means that the excluded volume is a
constant and when you calculate the change U with a volume change V the excluded
volume correction cancels. On the other hand a volume change will change internuclear
an 2
 U 
interactions parametrized by a. Therefore 
and there is no dependence on


2
 V T V
b.
Problem 2.4 The Principal of Equipartition states that for every degree of mechanical
freedom possessed by a molecule a factor of R/2 is contributed to the heat capacity CV,
where R is the universal gas constant, i.e. R=8.314 JK-1mol-1. Therefore, a monatomic
gas with three degrees of translational freedom has CV=3R/2, while a diatomic gas with
has seven degrees of freedom and therefore has a heat capacity of CV=7R/2. Consider the
table of heat capacities CP at T=298.15K taken from the back of your text:
Gas
Helium
Nitrogen
Oxygen
Neon
Hydrogen Argon
-1
-1
CP (JK mol
20.79
29.13
29.38
20.79
28.84
20.79
Which of these gases has a heat capacity that is predicted by the Principle of
Equipartition? Explain your reasoning and account for any discrepancies. Assume the
gases behave ideally.
Answer: He, Ne, and Ar are monatomic gases and their heat capacities are all
CV  CP  R  20.79 JK 1mol 1  8.31JK 1mol 1  12.48 JK 1mol 1  3R / 2
The monatomic gases therefore obey the equipartition principle.
Nitrogen, oxygen and hydrogen are diatomic gases and their heat capacities are
CV  CP  R  29.00  0.4 JK 1mol 1  8.31JK 1mol 1  20.69  0.4 JK 1mol 1  5R / 2
The diatomic gases have heat capacities about R less than predicted by the equipartition
principle. This results because at T=298K the two vibrational degrees of freedom do not
absorb heat and so the vibrational heat capacity is zero instead of R.
Part 3 (30 points) Answer TWO of the following FOUR questions.
Problem 3.1 Calculate q, w, U and S if one mole of an ideal monatomic gas initially
at T=400K and P=1 atm expands adiabatically and reversibly until P=0.5 atm
Solution:
 T2 
 
 T1 
CV
R
R / CV
P 
P 
 0.5 
  2   T2  T1  2 
  400 K  
  252 K
P
P
1.0


 1
 1
3
U  w  nCV T  1    8.31JK 1mol 1   252 K  400 K   1845 J
2
qrev  0 S  0
2/3
Problem 3.2) The molar heat capacity at constant pressure (units of JK-1mol-1)for a
T
T2
certain gas is given by CP  22.66  4.381102  1.0835 104 2 If the gas increases
K
K
its temperature from 100K to 500K at constant pressure, calculate the molar enthalpy
change of the gas.
Solution:
T2
H   CP  t  dT 
T1
2

2 T
4 T 
22.66

4.381

10

1.0835

10
dT
100 
K
K2 
500
4.381 102
1.0835  104
2
2
500

100



 5003  1003 
2
3
 9064  5257  4479  9842 J
 22.66  500  100  
Problem 3.3 In this problem, you can chose which two out of three derivatives to
evaluate. For water at T=300K and P=1 atm, The heat capacity CP  75.3JK -1mol-1 , the
coefficient of thermal expansion   3.04  104 K 1 and the isothermal compressibility
  4.46  1010 m 2 N 1 . At T=300K the molar volume of water is
Vm  18.1  106 m3 mol -1 . Calculate two of the following three quantities for water at
T=300K and P=1atm : 1) CV ;
 U 
2) 
 ;
 V T
 H 
3) 
 .
 P T
Solution:
2
CP  CV  TV

 3.04 10 K   74.17 JK 1mol 1
2
 CV  CP  TV
 75.3JK -1mol-1   300K  18.1106 m3mol-1 

4.46 1010 m2 N 1
4
1 2
 U

 V
 H

 P

3.04 104 K 1


T

P

300 K   101325 Pa  2.04 108 Pa  1.01105 Pa  2.04 108 Pa

10
2
1 
4.46 10 m N
V 

6 3
4
1
6 3
  TV   V    300 K  18.110 m  3.04 10 K   18.110 m
V
 18.1106 m3  1  0.09   16.44 10 6 m3
Problem 3.4 Ammonia is a common metabolic by-product. It is also very toxic so
terrestrial animals convert ammonia to urea: NH2CONH2. Consider the synthesis of urea
from its constituent elements:
C  gr   N2  g   12 O2  g   2H 2  g   NH 2CONH 2  s 
The table below shows entropy values at for T=298K and heat capacity values for each
reaction component.
Substance
S 0  JK 1mol 1 
CP0  JK 1mol 1 
C(gr)
N2(g)
O2(g)
H2(g)
Urea(s)
5.69
191.49
205.03
130.59
104.6
8.64
29.12
29.36
28.84
93.13
Calculate the entropy change for this reaction at T=310K. Assume all heat capacities are
constant between 298K and 310K.
T 
S T2   S T1   CP ln  2 
 T1 
 104.6  2  130.59  12  205.03  5.69  191.49  JK 1mol 1
 310  1
1
  93.13  2  28.84  12  29.36  8.64  29.12  ln 
 JK mol
298


1
1
1
1
 456.3JK mol  16.99  0.039 JK mol  457.0 JK 1mol 1
Part 4 (32 points) Perform ONE out of the TWO multi-step calculations
Problem 4.1 The shells of marine organisms contain calcium carbonate CaCO3, largely
in a crystalline form known as calcite. There is a second crystalline form of calcium
carbonate known as aragonite. Physical and thermodynamic properties of calcite and
aragonite are given below.
Properties
(T=298K, P=1atm)
H 0f (kJ/mole)
Calcite
Aragonite
-1206.87
-1207.04
G0f (kJ/mole)
-1128.76
-1127.71
S f0 (J/mole K)
92.88
88.70
CP0 (J/mole K)
Density (gm/mL)
81.88
81.25
2.710
2.930
a) Based on the thermodynamic data given, would you expect an isolated sample of
calcite at T=298K and P=1 atm to convert to aragonite, given sufficient time.
Explain.
Solution:
G  G f  aragonite   G f  calcite    1127.7   1128.8  kJmol 1  1.1kJmol 1
G  0 so the process will not occur.
b) Suppose the pressure applied to an isolated sample of calcite is increased. Can the
pressure be increased to the point that isolated calcite will be converted to
aragonite? Explain.
Solution:
By La Chatelier’s Principle, an increase in pressure will favor the crystalline form
with the higher density, i.e. the smaller molar volume. This would be aragonite.
c) What pressure must be achieved to induce the conversion of calcite to aragonite at
T=298K. Assume both calcite and aragonite are incompressible at T=298K.
Solution:
M
100gmol 1
Vmcalc 

 36.9mL  mol 1  3.69  105 m3mol 1
1
darag 2.710gmL
Vmarag 
M
100gmol 1

 34.1mL  mol 1  3.41  105 m3mol 1
dcalc 2.930gmL1
 Vm  Vmarag  Vmcalc  3.41  105 m3mol 1  3.69  105 m3mol 1  2.8  106 m3mol 1




G P,298.15K  0  G 1atm,298.K  Vm P
P  

G 1atm,298.K
Vm
  1.1  10 Jmol
3
6
2.8  10 m
1
3
 3.93  108 Pa  3930bars
d) Can isolated calcite be converted to aragonite at P=1 atm if the temperature is
increased? Explain..
Solution:
Sm  S f arag  S f calc  88.7JK 1mol 1  92.88JK 1mol 1  4.12JK 1mol 1


 
Assume the enthalpy change and entropy changes are constant with T, then
Gm  H m  T Sm . Because Sm  0 , Gm will become more positive as T increases.
Problem 4.2 18 grams of water ice at 230 K are placed in thermal contact with a large
heating block maintained at a temperature of 350 K. Calculate the entropy change S, the
energy change U, and the enthalpy change H of the water, when 18 grams of water ice
at 230 K are converted to liquid water at 350K. Also calculate the entropy change of the
heating block as a result of this process and the entropy change of the universe as a result
of this process. For ice Cp=37.7 J/K. For liquid water Cp=75 J/K. For water
Hfusion=6.01 kJ/mole. Assume the density of ice is 0.92 gm/mL and approximately
constant between 230K and 273K. The density of liquid water is 1 gm/mL at 273K and
approximately constant up to 350K. Assume only P-V work occurs. The external pressure
is 1 atm.
Solution: Designate a three-step path from water ice at T=230K to liquid water at
T=350K:
ice(230 K )  water (350 K )
B
A
ice(273K )  water (273K )
Step 1: H1  nCPice T  (1mole)(37.7 J / K  mole)(273K  230 K )  1621J
Tfinal
273
S1  nCPice ln
 (1mole)(37.7 J / K  mole) ln
 6.46 J / K
Tinitial
230
U1  qP  Pext V  qP  H1
Step 2: H2  (1mole) H fusion  (1mole)(6010 J / mole)  6010 J
H2 6010 J
S2 

 22.0 J / K
Tfusion 273K
U 2  qP  Pext V  qP  H 2  6010 J ; V  Vliquid  Vice
V  0.018 L  (0.018 L / 0.92)  .0016 L
U 2  6010 J  (.0016L)(1atm)(101J / L  atm)  6010.2 J
Step 3: H3  nCPliquid T  (1mole)(75 J / K  mole)(350 K  273K )  5775 J
Tfinal
350
S3  nCPliquid ln
 (1mole)(75 J / K  mole) ln
 18.63 J / K
Tinitial
273
As before U 3  H 3  5775J
Totals: HTwater  H1  H2  H3  13,406 J
STwater  S1  S2  S3  47.09 J / K
ETwater  HTwater  13,406 J
H sys
13406 J

 38.30 JK 1
S for Surroundings: H surr  
Tsurr
350 K
S for Universe:
Suniverse  Ssurroundings  Swater  38.30 J / K  47.09 J / K  8.79 J / K
F
IJ
G
H K
F
IJ
G
H K
F
IJ
G
HK
F
I
G
HJ
K