METU NORTHERN CYPRUS CAMPUS
CHM 351
1
H
1.0079
3
Li
6,941
11
Na
22.990
19
K
39.098
37
Rb
85.468
55
Cs
132.91
87
Fr
(223)
Physical Chemistry
Final Solutions
15 January 2016
Name, Surname:
ID:
4
Be
9.0122
12
Mg
24.305
20
Ca
40.078
38
Sr
87.62
56
Ba
137.33
88
Ra
(226)
21
Sc
44.956
39
Y
88.906
57
La*
138.91
89
Ac**
(227)
*Lanthanides
** Actinides
22
Ti
47.867
40
Zr
91.224
72
Hf
178.49
104
Rf
(261)
Time Allowed: 180 min
Signature:
6
C
12.011
14
Si
28.086
32
Ge
72.64
50
Sn
118.71
82
Pb
207.2
114
Uuq
(289)
7
N
14.007
15
P
30.974
33
As
74.922
51
Sb
121.76
83
Bi
208.98
115
UUp
(288)
8
O
15.999
16
S
32.065
34
Se
78.96
52
Te
127.60
84
Po
(209)
116
Uuh
(291)
9
F
18.998
17
Cl
35.453
35
Br
79.904
53
I
126.90
85
At
(210)
2
He
4.0026
10
Ne
20.180
18
Ar
39.948
36
Kr
83.798
54
Xe
131.29
86
Rn
(222)
118
Uuo
(294)
23
V
50.942
41
Nb
92.906
73
Ta
180.95
105
Db
(262)
24
Cr
51.996
42
Mo
95.94
74
W
183.84
106
Sg
(266)
25
Mn
54.938
43
Tc
(98)
75
Re
186.21
107
Bh
(264)
26
Fe
55.845
44
Ru
101.07
76
Os
190.23
108
Hs
(277)
27
Co
58.933
45
Rh
102.91
77
Ir
192.22
109
Mt
(268)
28
Ni
58.693
46
Pd
106.42
78
Pt
195.08
110
Ds
(281)
29
Cu
63.546
47
Ag
107.87
79
Au
196.97
111
Rg
(272)
30
Zn
65.41
48
Cd
112.41
80
Hg
200.59
112
Uub
(285)
5
B
10.811
13
Al
26.982
31
Ga
69.723
49
In
114.82
81
Tl
204.38
113
Uut
(284)
58
Ce
140.12
59
Pr
140.91
60
Nd
144.24
61
Pm
(145)
62
Sm
150.36
63
Eu
151.96
64
Gd
157.25
65
Tb
158.93
66
Dy
162.50
67
Ho
164.93
68
Er
167.26
69
Tm
168.93
70
Yb
173.04
71
Lu
174.97
90
Th
232.04
91
Pa
231.04
92
U
238.03
93
Np
(237)
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
Md
(258)
102
No
(259)
103
Lr
(262)
Points will be deducted for incorrect use of SF/DP, missing and/or wrong units.
USE THE ABOVE PERIODIC TABLE FOR ATOMIC/MOLAR MASSES.
Notes:
If you wish to truncate significant figures/decimal points before the last step of a series of
calculations, use at least one more than the required number.
Obey the rules for significant figures when quoting the final answer.
Use units in all steps of a series of calculations. This will help you get the correct units for the
answer.
1. (a) (3 points) Why do molecules in a gas have different speeds at a given temperature?
Molecules are in constant motion. Thus, they collide with one another continually. Translational
kinetic energy is transferred at each collision. Therefore, after a time interval, each molecule will
have a different translational kinetic energy, resulting in a distribution of molecular speeds.
(b) (10 points) Use the Maxwell-Boltzmann equation for distribution of speeds to estimate the
fraction of N2 molecules at 400 K that have speeds between 200 m s 1 and 210 m s 1 .
3/2
 m 
 mv 2 /2 kT 2
The fraction is given by f (v)  4 
v dv . Since the speed range is small,
 e
 2 kT 
we may safely take v  205 m s 1 and dv  v  10 m s 1 . Therefore
m


f (v)  4 

21
1
 2 1.38065 10 J K  298.15 K 
3/2
e  m (205 m s
1 2
) /21.380651021 J K 1 298.15 K
(205 m s 1 ) 2 (10 m s 1 )
m is the mass of one single molecule of nitrogen in kg. Therefore
m
28.014 103 kg mol1
 4.6518 1026 kg molecule1 .
6.02214 1023 molecules mol1
Inserting this value into the equation and evaluating it gives
f (v)  6.87 103
2. (a) (3 points) Why does the ionic strength of a solution increase with increasing charge
numbers of the constituent ions?
As the charge number of an ion increases, interaction with other ions increases geometrically.
Higher interactions will introduce higher deviations from non-ideality. Therefore, the sum of
these interactions will result in a higher ionic strength.
(b) (10 points) The solubility product of silver chloride in water at 25°C is 1.5 1010 . Assuming
that the Debye-Hückel limiting law applies, calculate the mean activity coefficient of the
electrolytes in a solution that is 0.05 M in KNO3 and 0.02 M in KCl.
The mean activity coefficient is given by log     Az | z | I 1/2 . This equation involves ionic
strength, I. So we have to calculate it first. I is given by (with b o  1 mol dm 3 )
I
1
1
zi2 (bi / b o )  (0.05 12 )  (0.05 12 )  (0.02 12 )  (0.02 12 )   0.07

2 i
2
The mean activity coefficient of the electrolytes can now be calculated.
log     A | z z | I 1/2
 [0.5086 |1 (1)  (0.07)1/2 ]  0.13456     0.734
2
3. (5 points) Fill in the blanks (no partial credits)
Diffusion is the migration of matter down a concentration gradient.
Electrical conductivity is migration of electric charge down an electrical potential gradient.
Viscosity is the migration of linear momentum down a velocity gradient.
Thermal conductivity is the migration of energy down a temperature gradient.
4. (15 points) Invertase (an enzyme) catalyses the hydrolysis of sucrose (table sugar). In a series
of experiments, the following data was obtained. Use the data to determine by a graphical method
the Michaelis constant of the reaction. Show your work on your graph paper and insert the result
below. (Hint: Make a suitable table and do a double-reciprocal plot)
[Sucrose]o (M)
0.029
0.059
0.088
0.117
0.175
Rate (M s-1)
0.182
0.266
0.310
0.330
0.362
K M  0.0392 M
5. (a) (2 points) What is effusion? Effusion is the escape of a gas in a container through a small
hole into vacuum.
(b) (3 points) When does an ion reach its drift speed at a given temperature in an electrical field?
When the drag (viscous force due to diffusion) equals the electrical force acting upon the ion.
(c) (2 points) What is chemisorption?
It is the attachment of an adsorbate on a substrate by forming chemical bonds with units on the
surface of the substrate.
(d) (4 points) Consider the equation  j   ideal
 RT ln   , which gives the chemical potential of an
j
ion in a solution as the sum of two terms. The last term accounts for the nonideal behaviour of
ion j. Why do we need to make this correction? Explain in a few sentences.
All molecules in a solution interact. Ions interact appreciable and supersede nearly all other
interactions. These, inevitably, will produce nonidealities, since the interactions will no longer be
directly proportional to the number of ions. Hence a correction is needed to account for the sum
of nonidealities.
3
6. (8 points) Consider the cell Co(s) | Co2 (aq,1.0 m) || Ni 2 (aq, 0.1m) | Ni(s) where the cell
reaction is Co(s)  Ni 2 (aq)  Ni(s)  Co2 (aq) . What is the equilibrium constant of the
reaction at 25°C? Is the reaction spontaneous?
Since the cell reaction is Co(s)  Ni 2 (aq)  Ni(s)  Co2 (aq) , the equilibrium constant is
[Co2 ]eq
RT
o
given by K 
. At equilibrium, Ecell  0 , and therefore Ecell

ln K .
2
F
[Ni ]eq
o
From the stoichiometry of the reaction,   2 . We have to determine Ecell
from tables after
splitting the overall reaction into two half reactions. They are
reduction Ni 2 (aq)  2e  Ni(s) E o  0.257 V
oxidation Co(s)  Co2 (aq)  2e E o  0.280 V
o
 [(0.257 V)  (0.280 V)]  0.023 V , Therefore,
Thus, Ecell
8.3144 J mol1 K 1  298.15 K
0.023 V 
ln K
2  96485 C mol1
Thus, ln K  1.790 . Therefore,
K  5.99
o
 0 V , the reaction is spontaneous.
Since Ecell
7. Decomposition of gaseous N2O5 is a unimolecular first order reaction
N2O5 (g) 
2NO2 (g) + 1/2 O2 (g)
with a rate constant of k  4.8  104 s 1 at 25C.
Because all the components are gases, the reaction kinetics is conveniently investigated by the
measurements of the pressure changes in the reaction chamber. Assuming that the reaction takes
place in a closed chamber (constant volume) at 25°C and the initial pressure of N2O5 is 100 kPa,
(a) (6 points) what percentage of N2O5 decomposes after 60 minutes?
Since the reaction is first order, the relevant integrated rate law is
 [A] 
ln 
  kt
 [A]o 
We can use partial pressures to replace molar concentrations, Thus
ln
p
 kt , where p  partial pressure of N 2O5 at time t.
po
4
Thus, when t  60 min 
p
60 s
 (4.8 104 s 1  3600 s)  1.728 . Therefore,
 3600 s , ln
1 min
po
p
ln
 1.728. Thus, p  17.76 kPa.
100
Let X  percentage of N2O5 that decomposes after 60 minutes. It can be determined from
po  p
(100  17.76) kPa
100 
100  82.24%  82% (2 SF)
po
100 kPa
(b) (4 points) what is the total pressure in the chamber after 60 minutes?
X
pT  pN 2O5  pNO2  pO2
pN2O5  17.76 kPa (above)
pNO2  2  82.24kPa =164.48 kPa
pO2 
1
 82.24 kPa  41.12kPa
2
Therefore, pT  223.36 kPa  223 kPa
8. (10 points) Carbon monoxide is produced during photolysis of acetone when it is irradiated
with radiation at 313 nm with a quantum yield of unity. After 20 min of radiation at this
wavelength, 18.4 cm3 of carbon monoxide is produced at a pressure of 1008 Pa and 22C.
Calculate the number of photons absorbed by acetone.
Since, the primary quantum yield is unity, the number of absorbed photons is equal to the number
of carbon monoxide molecules formed by photolysis.
Let N  number of carbon monoxide molecules. If the gas is assumed to behave ideally under
the given conditions, then
N
PVN A

RT
106 m3
 6.02214  1023 molecules mol1
1 cm3
8.3144 J mol1 K 1  (273.15  22) K
1008 Pa  18.4 cm3 
 4.71 1018 molecules.
Therefore, the number of photons is 4.55  1018 photons, since   1 .
9. (15 points) A certain sample of activated charcoal has a surface area of 1000 m2 g 1 . What
amount of ammonia gas, measured at 25C and 1 bar will be adsorbed at full coverage on the
surface of 45 g of charcoal? Take the diameter of a molecule of ammonia as
3.0 1010 m, and assume that four adjacent ammonia molecules are at the
four corners of a square (as illustrated below).
5
First find the total surface, S, offered by 45 g of charcoal. It is given by
S  45g 1000 m2 g 1  45000 m2 .
It is evident from the diagram that the rectangle contains one ammonia molecule. Therefore, area
covered by one ammonia molecule, A, is given by
A  (3.0 1010 m)2  9.0 1020 m2
Total number of ammonia molecules, N, is given by
S
45000 m 2

 5.00 1023 molecules . So the mass of ammonia, mNH 3 , can now be
A 9.0 1020 m2
calculated from
N
mNH3 
5.00 1023 molecules
17.0306 g mol1  14.1 g
6.022 1023 molecules mol1
6