1
CH341 Test 2
Physical Chemistry
Fall 2011, Prof. Shattuck
Test 2
Name____________________________
Constants: R = 8.3145 J K-1 mol-1 = 0.083145 L bar K-1 mol-1
R = 0.08206 L atm K-1 mol-1
1 F = 96485 C mol-1
1 atm = 1.01325 bar
1 bar = 1x105 Pa
Part I. Answer four (4) of the following six questions. If you answer more than 4, cross out the
problems you wish not to be graded. 8 points each.
1. Give a statement of the Second Law of Thermodynamics.
Answer: Entropy always increases for a spontaneous process in an isolated system.
2. A solution is prepared by adding 0.100 mol of pure acetic acid to 1.00 L of water. The
solution is neutralized by adding solid NaOH. List the chemical constituents and give the number
of thermodynamic components. (You don’t need to list any reactions or chemical constraints)
Constituents: CH3COOH, CH3COO–, H+, OH–, Na+, H2O
Components: c = 3, CH3COOH, NaOH, H2O
Answer:
3. Of sulfur dioxide and carbon dioxide, which molecule is predicted to have the greater
constant pressure heat capacity? Give estimates for the Cp values for both molecules (neglecting
vibration).
see Chapter 8 Problem 31: Note that CO2 is linear and SO2 is bent. For CO2 with two moments
of inertia, neglecting vibrations, with ½ R for each degree of freedom:
Cv,m(CO2) = 3/2 R +
translation
2
= 5 /2 R
/2 R
rotation
and Cp,m = Cv,m + R = 7/2 R = 29.1 J K-1 mol-1. For SO2 with three rotational moments of inertia:
Cv,m(CO2) = 3/2 R +
translation
3
= 6 /2 R
/2 R
rotation
and Cp,m = Cv,m + R = 8/2 R = 33.3 J K-1 mol-1. Neglecting vibrations, SO2 is predicted to have the
larger constant pressure heat capacity.
4. When p-nitroanisole and pyridine are photolyzed in aqueous solution the reaction is:
NO2
hν
+
N
O
N
+
+
H2 O
CH3
O
CH3
NO2-
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CH341 Test 2
The quantum yield for a solution containing 1.00x10-5M p-nitroanisole and 0.0100 M pyridine
in 1% acetonitrile is 4.65x10-3 at 366 nm. The molar absorption coefficient at 366 nm of pnitroanisole is 1990 M-1 cm-1. Calculate the photochemical rate constant for an optically thin
solution assuming the incident flux is 6.70x10-6 mol L-1 s-1 for a 10.00 cm path length.
see Chapter 5 Problem 5: For an optically thin solution:
jB = 2.303 JoΦB εl
= 2.303 (6.70x10-6 mol L-1 s-1)( 4.65x10-3)( 1990 M-1 cm-1)(10.0 cm)
= 1.43x10-3 s-1
d[B]
The rate law (not asker for in the problem) is then:
= jB [A] = 1.43x10-3 s-1 [PNA]
dt
where [PNA] is the concentration of p-nitroanisole.
5. Calculate the standard internal energy of formation at 298.2 K of liquid methyl acetate,
C3H6O2, from its standard enthalpy of formation, which is -442.0 kJ mol-1 at 298.2 K.
see Chapter 8 Problem 3: The balanced reaction is:
3 C (graph) + 3 H2 (g) + O2 (g) → C3H6O2 (l)
and ∆rng = [Σproducts] – [Σreactants] = [0] – [ 4 mol] = -4 mol
∆rH = ∆rU + ∆rng RT
(1 mol)(-442.0 kJ mol-1) = ∆rU + (-4 mol)(8.314 J K-1 mol-1)(1 kJ/1000 J)(298.2 K)
∆fU° = ∆rU = -442.0 kJ + 9.92 kJ = -432.1 kJ
Since the internal energy of formation is for one mole, the final value may also be written as
∆fU° = -432.1 kJ mol-1. Alternatively, ∆rng is usually taken as unitless.
6. For a closed system and PV work only, describe why the second, lost-work term in the
entropy production always gives the change in entropy production as greater than zero for an
isothermal process.
dStot  1
1  đq P Pext dV
=
−

 + −

dt
T Tsurr dt T T  dt
see Chapter 22 Problem 1: For an isothermal process the temperature gradient is zero and the
first term vanishes. Assume the pressure of the system is greater than the external pressure. The
pressure gradient is then positive, P – Pext > 0. Our experience shows that the system will expand
in a spontaneous process giving dV > 0. The product of (P – Pext)/T and dV is positive. Now,
consider a process with the system pressure less than the external pressure, P < Pext. The system
will contract in a spontaneous process giving dV < 0. However, the pressure gradient is also
negative, P – Pext < 0, so the product of (P – Pext)/T and dV is still positive. In either case, the
entropy increases for these spontaneous isothermal processes, dS > 0. On the other hand, if the
system pressure and the external pressure are equal, the pressure gradient is zero and the entropy
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CH341 Test 2
change is zero, dS = 0, for an isothermal process. With no pressure gradient, the system is at
equilibrium and no process occurs.
Part II. Answer two (2) of the following four questions. If you answer more than 2, cross out the
problems you wish not to be graded. 16 points each.
7. Use a cyclic process in a closed system to argue that the internal energy is conserved.
see text p. 287: We need to show: o∫ dU = 0, so assume the converse and work to an impossibility.
Assume that the internal energy is not a state function, which implies o∫ dU ≠ 0. A cyclic process
for a gas expansion and contraction in a closed system is diagrammed below, assuming U is not a
state function.
U
U1'
U1
(P1,T1)
(P1,T1)
V1
(P2,T2)
V2
V
If the integral for the cyclic process is not zero, the internal energy will be different upon
completing the cycle. This difference in internal energy might then be available to do useful
work. In other words, on taking the system from an initial state at P1, V1, and T1 and returning
the system to that same state, energy might be produced. The net result is the production of
energy with no change in the system or surroundings. Such a process has never been discovered
and would constitute perpetual motion, which we know from experience to be impossible.
Therefore, our original assumption that the internal energy is not a state function must be
incorrect.
8. Consider the transfer of heat between two blocks of metal in an isolated system. Block A is
slightly hotter than block B so the transfer of heat is reversible. Assume that the volume change
of the blocks is negligible so dVA = 0 and dVB = 0, giving:
∂UA
∂UB
dUtot = dUA + dUB =  ∂S  dSA +  ∂S  dSB = 0
 A V
 B V
(isolated, cst. VA&VB)
If the entropy, S, shows the direction for energy dispersal, then the sum of the overall change in
S should be positive for the spontaneous heat transfer:
dSA + dSB > 0
(energy dispersal)
∂UA
∂UB
Prove that  ∂S  >  ∂S  . A summary of the relationships is diagrammed below.
 A V  B V
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CH341 Test 2
hotter
colder
đqA,rev= – đqB,rev
dUA
+
dUB = 0
∂UA
∂UB
= 
 dSA +  ∂S  dSB = 0
∂S
 A V
 B V
= dSA
+
dSB > 0
dUtot =
dUtot
dStot
(isolated)
(isolated, cst. VA&VB)
(isolated, spontaneous)
see Chapter 10 p. 412: The inequality, dSA + dSB > 0, shows an increase in energy dispersal.
Comparing the two changes by subtracting dSA from both sides of this last equation gives:
dSB > –dSA
or equivalently –dSA < dSB
(isolated, spontaneous)
Note since system A is losing reversible heat, dSA < 0, so that –dSA is a positive number. In the
comparison, –dSA < dSB, we are comparing two positive numbers; the magnitude of the change
for A is smaller than the magnitude for the change in B. Summarizing these relationships, if the
overall change in S is to be positive for the dispersal of energy, then the change in the magnitude
of S for the hotter block is less than the change for the colder block. However, if –dSA < dSB, the
sum for the internal energy dUtot still must equal zero showing that:
∂UA
∂UB
 ∂S  >  ∂S 
 A V  B V
(isolated, spontaneous)
In other words, if the magnitude of dSA is smaller, the coefficient in front of dSA must be larger
so that the two terms for A and B cancel. [This last inequality is just what we would expect for
the temperatures of the two bodies.]
9. Consider a spontaneous adiabatic expansion of an ideal gas.
(a). Is the entropy change of the system zero, greater than zero, or less than zero?
(b). Is the entropy change of the surroundings zero, greater than zero, or less than zero?
(c). If the entropy change of this adiabatic process for the system is not zero, why isn’t the
entropy change zero? (Don’t just give an equation or two; explain the reasoning behind your
answers. Specifically address the issue of reversible heat transfer versus the actual heat transfer.)
Answer: (a). ∆S > 0 (the expansion is spontaneous as given in the problem, but ∆Ssurr = 0)
(b). ∆Ssurr = 0 since adiabatic.
(c). The entropy change for the system must be calculated using a reversible process. Since
entropy is a state function, the change in entropy is independent of the path. However, the actual
heat transfer is zero, since the process is adiabatic. An even stronger condition is given by the
Clausius inequality: dS = dqrev/T > dq/T for a spontaneous process.
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CH341 Test 2
10. Calculate q, w, ∆U, ∆H, and ∆S for a reversible adiabatic expansion of 1.00 mol of a
monatomic ideal gas. The gas is initially at 5.00 bar and 298.2 K and expands to a final pressure
of 1.00 bar.
see Chapter 9 Problems 24and 25: For an adiabatic expansion of an ideal gas, q = 0, ∆U = w,
∆U = Cv ∆T, ∆H = Cp ∆T. For a monatomic gas Cv = 3/2 nR and Cp = Cv + nR = 5/2 nR. Using Eq.
9.8.19° to determine the final temperature:
T2
T 
 1
Cp
/nR
P2
= P 
 1
and
T2 = T1 (P2/P1)2/5 = 298.15 K(1.000/5.00)2/5
T2 = 156.62 K
∆U = Cv (T2- T1) = 3/2 (1.000 mol)(8.314 J K-1 mol-1)(156.62 – 298.15 K)
∆U = -1765. J = w
∆H = Cp (T2- T1) = 5/2 (1.000 mol)(8.314 J K-1 mol-1)(156.62 – 298.15 K)
∆H = -2942. J
For a reversible adiabatic process for the system ∆S = qrev/T = 0
Part III. Answer two (2) of the following three questions. If you answer more than 2, cross out
the problem you wish not to be graded. 18 points each.
11. Yeasts convert glucose to ethanol. (a). Calculate the change in enthalpy if one mole of
glucose is converted to ethanol at 298.2 K:
C6H12O6 (s) → 2 CH3CH2OH (l) + 2 CO2 (g)
(b). Thermophilic organisms live at 80.0°C. Calculate the change in enthalpy for the reaction at
80.0°C.
Substance
glucose
ethanol
CO2
∆fH° (kJ mol-1)
-1274.
-277.69
-393.51
see Chapter 8 Problems 7 and 9: (a).
Cp (J K-1mol-1 )
209.
111.5
37.11
∆rH° = [Σproducts] – [Σreactants] = ∑νi ∆fH°
∆rH° = [2(-277.69 kJ mol-1) + 2 (-393.51 kJ mol-1)] – [1 (-1274. kJ mol-1)]
= -68.4 kJ mol-1
(b). Assuming constant heat capacities, ∆rHT2 = ∆rHT1 + ∆rCp ∆T. The difference for the heat
capacities for the reaction is:
∆rCp = [Σproducts] – [Σreactants] = ∑νi Cp,i
= [2 (111.5 J K-1mol-1) + 2 (37.11 J K-1mol-1)] – [1(209. J K-1mol-1)]
= 88.2 J K-1 mol-1
The enthalpy at 80.0°C is then:
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CH341 Test 2
∆rHT2 = -68.4 kJ mol-1 + 88.2 J K-1 mol-1(55.0 K)(1 kJ/1000 J)
= -68.4+ 4.9 kJ mol-1 = -63.5 kJ mol-1 = -64. kJ mol-1
1 ∂H
.
CP  ∂P T
For CO2, µJT = 1.11 K bar-1 and CP for CO2 is 37.11 J K-1 mol-1. Calculate the change in enthalpy
per mole for an isothermal process for a change in pressure of 1 bar. Assume that both µJT and CP
are constant over the pressure range.
12. The Joule-Thompson coefficient is given by µJT = –
see Chapter 9 Problem 14: The process corresponds to the partial derivative (∂H/∂P)T. The
value of this partial derivative is given by the Joule-Thompson coefficient:
∂H
  = – µJT Cp
 ∂P T
Integration assuming a constant µJT and Cp gives: ∆H = – µJT Cp ∆P
Substitution of the values for this problem gives:
∆H = – 1.11 K bar-1(37.11 J K-1mol-1) (1.00 bar) = -41.2 J mol-1
∂H
∂H  α 
13. Show that   = CP +    
 ∂T V
 ∂P TκΤ
[Show all your work with reasons for each step. For example: “since H is a state function,”
“from the definition of heat capacity,” or “dividing by dP gives the desired result” are typical
statements. If you use a partial derivative relationship that has a name, if you choose not to
derive the relationship, just state the corresponding relationship name. (e.g. “from the chain rule”
or “from the Euler chain rule”)]
see Chapter 9 Problem 17: We can get a hint of how to proceed by substituting in the definition
of Cp and also that α/κT = (∂P/∂T)V. Correspondingly we need to show that:
∂H ∂H ∂H  ∂P
  =  +   
 ∂T V  ∂T P  ∂P T∂TV
Since we normally consider H(P,T) and not H(V,T), we can consider the “misplaced” variable
the constant V specification. Notice that given H(P,T) the total differential of H is:
∂H
∂H
dH =   dT +   dP
 ∂T P
 ∂P T
Notice that the partials in this total differential also occur in the relationship we are trying to
derive. Simply “dividing” this last equation by dT and specifying constant V for any new partial
derivative gives:
∂H ∂H ∂P ∂H  ∂P
∂H ∂H  ∂P
  =    +    =  +   
 ∂T V  ∂T P ∂PV  ∂P T∂TV  ∂T P  ∂P T∂TV
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CH341 Test 2
Using the definition of the constant pressure heat capacity, Cp = (∂H/∂T)P, and α/κT = (∂P/∂T)V,
gives:
∂H
∂H  α 
  = CP +    
 ∂T V
 ∂P TκΤ
Notice that it is often helpful to “work backwards” from the original statement of the problem.
Working “backwards” can help give you hints on how to proceed. Just remember to present the
full derivation in the “forward” direction as we did here.
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CH341 Test 2
Extra Credit (4 points)
Name the following elements from the pictograms:
“Silly Con” or Silicon
Copper
Tin
“Knee On” or Neon
Here’s an example:
L
≡ “Nicked L” or nickel