Physical Chemistry I CHEM 4641
Final Exam
13 questions, 30 points
Name: ___________KEY____________
Gas constant: R = 8.314 J·mol-1·K-1 = 0.008314 kJ·mol-1·K-1.
h = 6.626×10-34 J·s
Boltzmann constant k = 1.381×10-23 J/K = 0.6950 cm-1/K
c = 2.998×108 m/s
Units: 1 bar = 105 Pa.
1 GPa = 109Pa
1 Joule = 1 Pa·m3 .
1 L·bar = 100 J. 1L=10-3m3.
Formulae:
A = - kT ln Q
Q = qN or qN/N!
Λ= h / √ 2 π m k B T
B = h2 /(8 π2 μ R2 )
μindist = - kT ln(q/N)
μdist = - kT ln(q)
1
H
2
He
1.01
4.00
3
Li
4
Be
6.94
9.01
11
Na
12
Mg
5
B
20
Ca
7
N
8
O
9
F
10
Ne
10.81 12.01 14.01 16.00 19.00 20.18
13
Al
22.99 24.31
19
K
6
C
14
Si
15
P
16
S
17
Cl
18
Ar
26.98 28.09 30.97 32.06 35.45 39.95
21
Sc
22
Ti
23
V
24
Cr
25
Mn
26
Fe
27
Co
28
Ni
29
Cu
30
Zn
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.70 63.55 65.38 69.72 72.59 74.92 78.96 79.90 83.80
37
Rb
38
Sr
39
Y
40
Zr
41
Nb
42
Mo
85.47 87.62 88.91 91.22 92.91 95.94
55
Cs
56
Ba
57
La
72
Hf
73
Ta
74
W
43
Tc
98
75
Re
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
223
88
Ra
89
Ac
226.0 227.0
58
Ce
54
Xe
76
Os
77
Ir
78
Pt
79
Au
80
Hg
81
Tl
82
Pb
83
Bi
84
Po
85
At
86
Rn
209
210
222
104
Rf
105
Db
106
Sg
107
Bh
108
Hs
109
Mt
110
Ds
111
Rg
112
Cn
113
Nh
114
Fl
115
Mc
116
Lv
117
Ts
118
Og
267
268
269
270
269
278
281
280
285
286
289
289
293
294
294
59
Pr
60
Nd
61
Pm
62
Sm
63
Eu
64
Gd
65
Tb
66
Dy
67
Ho
68
Er
69
Tm
70
Yb
71
Lu
140.1 140.9 144.2
90
Th
53
I
101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
132.9 137.3 138.9 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0
87
Fr
52
Te
91
Pa
92
U
232.0 231.0 238.0
145
150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0
93
Np
94
Pu
95
Am
96
Cm
97
Bk
98
Cf
99
Es
100
Fm
101
Md
102
No
103
Lr
237
244
243
247
247
251
252
257
258
259
262
1. (2 points) The word "reversible" has related but different meanings in thermodynamics and in kinetics.
Explain what "reversible" means in the context of the expansion of a gas, and what "reversible" means in
a reaction mechanism. A few sentences can suffice.
Expansion: An ionfinitessimal change in the driving force (e.g., pressure) changes expansion to
compression.
Mechanism: Reaction goes both ways: reactants to products and products to reactants.
2. (1 point) When graphite begins to melt at 4800K,
does the graphite sink in the liquid or does it float on
top of the liquid? Explain how you know.
Answer: The left-sloping coexistence line, dP/dT < 0,
implies that ΔV<0; the molar volume of graphite is
larger than the molar volume of liquid carbon. That is,
graphite is less dense than liquid carbon. Graphite
floats. Another argument, giving the same conclusion,
is that at a fixed temperature liquid is the high-pressure
phase, so it must be the higher-density phase.
3. (2 points) The vibrational modes of the water molecule have frequencies 1595, 3657 and 3756 cm-1.
Which mode contributes most to the zero-point energy?
Answer: 3756
Which mode contributes most to the heat capacity at 1000 K?
Answer: 1595
4. (3 points) Consider one mole of ideal gas.
CP=38.314 J/(mol K).
A three-step cycle is graphed.
I. Isotherm at 302.0 K from 10.0 L to 20.0 L.
II. Isobar from the end of step I, back to 10.0 L.
III. Isochore from the end of step II to the beginning
of step I, all at 10.0 L.
Calculate ΔU, w and ΔS for step I, the reversible
isotherm.
Answer:
ΔU = 0 for an ideal gas at constant temperature.
w=−∫ P dV = −nRT ∫
ΔS =∫
d q rev
T
OR Δ S =
dV
20
= −RT ln
= −1740 J
V
10
( )
= −∫
dw
n R T /V
dV
20
=∫
dV = n R∫
= R ln
= 5.76 J/K
T
T
V
10
qrev
−w
20
= R ln
= 5.76 J/K
T
10
T
=
( )
( )
5. (4 points) Consider the gas-phase equilibrium
I2 ⇌ 2 I .
Both I2 and I are gases.
Assume ideal-gas behavior.
gas
I
I2
CP at 298 K
ΔfGo298 ΔfHo298
So298
(kJ/mol) (kJ/mol) (J/mol·K) (J/mol·K)
70.2
106.8
180.8
20.8
19.3
62.4
260.7
36.9
a) Calculate the equilibrium constant K at 298K.
b) Given that the pressure of I2 at 298 K is 1.0X10-4 bar, calculate the pressure of I.
c) At 350 K, is K larger or smaller than at 298 K? You need not calculate K at 350 K, but explain how
you know.
d) Explain why the heat capacity of I2 is larger than the heat capacity of I.
Answers:
a) ΔrxnGo298 = 2×70.2 - 19.3 = 121.1 kJ/mol.
Or, ΔrxnGo298 = ΔrxnHo298 - TΔSo = 151.2 - 298×10-3(2×180.8 -260.7) = 151.2 - 0.298 × 100.9
= 151.2 - 30.1 = 121.1 kJ/mol
ln K = -121.1/(0.008314×298) =-48.9 . K = 5.9×10-22.
b) K = (PI/Po)2 / (PI2/Po) so PI= Po [ K × 1.0×10-4]1/2 = 1 bar [ 2.4×10-13] = 2.4×10-13 bar.
c) K is larger at larger T because the reaction is endothermic. A test taker might take that as obvious, this
being a dissociation reaction, or might calculate ΔrxnHo298 = 2×106.8 - 62.4 = 151.2 kJ/mol.
d) The heat capacity of I2 is larger because of rotation (a contribution of R) and of vibration ( about
0.9R).
6. (1 point) Consider the four-state energy level diagram that is sketched at right.
Let T=288K, so kBT=200 cm-1.
Calculate the canonical partition function, q.
4
−E i /(k B T )
Answer: q = ∑ e
= 1 + e−200 /200 + e−400 /200 + e−600 /200
i=0
q = 1+ e−1 +e−2 +e −3 = 1+0.368 +0.135+ 0.050=1.55
7. (2 points) Consider a nitrogen molecule, N2, in a volume of 10-27 m3. Its mass is 4.65×10-26kg.
At what temperature does the translational partition function, qT, equal 1.00×105?
Answer:
V (2 π m k B )3 / 2 T 3/ 2
V
qT = 3 =
Λ
h3
(2 π m k B )3/ 2=(2 π×4.65×10−26 kg×1.381×10−23 J/K )3 / 2 = 8.10×10−72 ( kg⋅m/s)3
(2 π m k B )3/ 2 / h3 = 2.78×10 28 m−3 K −3/ 2
V (2 π m k B )3 / 2 / h3 = 10−27 m 3 ×2.78×10 28 m−3 K −3/ 2 = 27.8 K −3 / 2
q T = 27.86 K −3 / 2 T 3/ 2 so T = (1.00×105/ 27.86 K −3 / 2)2 /3 = (3590)2 / 3 = 234 K
8. Let E be the translational kinetic energy of one gas molecule. The probability distribution is
f (E ) =
2 π E 1/2
(π k B T )3/ 2
e
−E /(k B T )
;
0≤E
a) (2 points) What is the most probable kinetic energy? (Derive a formula.)
Answer:
df
2π
1
E 1/2 −E /(k T )
=
−
e
d E (π k T )3/2 2 E 1/ 2
kT
df
1
= 0 for E = k B T
2
dE
(
)
b) (1 point) For comparison, what is the translational kinetic energy of a molecule that has the average
speed? (Again, the answer is a formula, not a number.)
Answer: average speed =
√
8 kB T
πm
4
so E = 1 m v 2 = π k B T
2
9. (2 points) Rare earth elements produced in a
nuclear reactor may enter liquid sodium coolant.
Simulated diffusion of Ce atoms in liquid Na at
1000 K is graphed at right. (Samin, et al., J. Appl.
Phys., 118, 234902, 2015). Recall that 1 ps = 10-12
s, and 1 Å = 10-10 m. What is the diffusion
coefficient, in m2/s?
Answer:
< r2 > = 6Dt so 6D = slope = 3.34 A2/ps
D = (3.34 /6) (10-10m)2/(10-12 s)
D = 5.57 X 10-9 m2/s
10. (2 points) Marteza Waskasi, et al., (J. Am.
Chem. Soc., 2016, 138, 9251-9257) measured the rate of electron transfer between a porphyrin cation and
a C60 anion. The two were covalently bound. Data are below.
a) At what temperature is the activation energy zero?
b) At that temperature, what is the half-life (t1/2) of P+-C-60 ? Give t1/2 in seconds.
Answers:
a) Ea=0 where d ln(kR)/d(1/T) =0, so at 1000/T=6.1. T = 1000/6.1 = 164 K.
b) ln(kR)=20.65 so kR=9.3×108 s-1. t1/2=ln(2)/kR = 7.5×10-10 s.
11. Consider the second-order reaction A + B → products. The rate law is, rate = k [A] [B].
Initial concentrations are [A]0 = 0.12 M, [B]0 = 0.24 M. The reaction is 50.0% complete in 50.0 s.
[ A ]0 [ B]
= k ([ B ]0− [ A ]0) t .
As you may recall, the integrated rate law is ln
[ A ][ B]0
(
)
a) (2 points) Calculate the rate coefficient, k.
b) (1 point) After how many seconds is the reaction 75.0% complete?
Answers:
[B]0-[A]0 = 0.24 - 0.12 = 0.12 M.
a) At 50% completion, [A]=0.06M and [B] = 0.24 -0.06 = 0.18 M.
ln
k=
(0.12×0.18
0.06×0.24 ) ln (1.5) 0.4055
=
=
= 0.0676 M
0.12 M×50 s
6.0 M s
6M s
−1 −1
s
b) At 75% completion, [A]=0.03 M and [B] = 0.24 - 0.09 = 0.15 M.
ln
t=
(0.12×0.15
0.03×0.24 )
0.0676 M
−1 −1
s
×0.12 M
=
ln (2.5)
0.008112 s
−1
=
0.916
= 113 s
0.008112 s−1
12. (3 points) Consider the net reaction I- + OCl- ⇌ OI- + OCl-, which occurs in water.
Here is a possible mechanism:
k1
OCl- + H2O
HOCl + OHk-1
k2
I- + HOCl
HOI + Clk3
HOI + OHH2O + OIa) Based on the initial-rate data, what is the
reaction order with respect to OCl- ?
b) Write the differential rate law for HOCl:
d [HOCl ]
=⋯
dt
c) What is the steady-state expression for [HOCl]
in terms of OCl-, H2O, OH- and rate coefficients?
[I-]0 (mM) [OCl-]0 (mM) initial rate (mM/s)
2.0
1.5
0.18
4.0
1.5
0.36
4.0
3.0
0.72
Answers:
a) Doubling [OCl-] increases the rate by a factor of two, so the order with respect to OCl- is 1.
d [ HOCl ]
= k 1[ OCl - ]−k −1 [ HOCl ][OH - ]−k 2 [ HOCl ][ I - ]
b)
d t
k 1 [OCl - ]
d [ HOCl ]
= 0 gives [HOCl ]ss =
c) Setting
d t
k−1 [OH - ]+ k 2 [ I - ]
13. (2 points) Guanyi Chen, et al., (Biochemical Engineering Journal, 113, 2016, 86–92) studied
biodiesel production catalyzed by bacteria.
bacteria
triglyceride
biodiesel
The reaction rate is in mg/min: milligrams of biodiesel produced per minute. The concentration of
substrate (triglyceride) is in mg/mL and appears as "Cs" on the graph. Methanol was added as an
inhibitor. The concentration of methanol is given
as percent by volume of the reaction mixture.
methanol
a) In terms of Michaelis-Menten kinetics, what is
the value of Rmax?
b) What sort of inhibitor is methanol:
competitive or non-competitive? How do you
know?
Answers:
a) Rmax = 1/intercept = 1/(0.45 min/mg) = 2.2
mg/min.
b) Inhibition is competitive because Rmax is
unchanged (constant intercept) while KM (slope ×
Rmax) changes. Chen wrote, "higher methanol concentration gives higher slope of lines, but an invariant
rmax , indicating that methanol was a competitive inhibitor of enzyme activity."