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METU NORTHERN CYPRUS CAMPUS
CHM 351
MT1I
Solutions
12 December 2015
Name, Surname:
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)
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)
Physical Chemistry
Time Allowed: 150 min
ID:
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. (15 points) The principal components of the atmosphere of the Earth are diatomic gases, such
as O2, N2, which can rotate as well as translate. Given that the total translational kinetic energy
density of the atmosphere is 0.15 J cm3 , what is the total kinetic energy density, including
rotation, at 298 K? The average rotational energy of a linear molecule is kT.
Let N  number density (number of molecules in 1 cm3 of air. Using the gas law,
N 
PN A 1.00 atm  6.02214 1023 molecules mol1

 2.463 1019 molecules cm3
RT
82.05746 mL  atm mol1 K 1  298 K
Let E k , r  total rotational kinetic energy density. Thus,
Ek ,r  2.4631019 molecules 1.38065 1023 J K1  298 K=0.1013 J cm3 . Thus, the total
kinetic energy density, Ek , is
E k  (0.15  0.1013) J cm 3  0.25 J cm 3
2. (12 points) The limiting molar conductivities of NaI, NaNO3 and AgNO3 are 12.69
mS m 2 mol1 , 12.16 mS m 2 mol1 and 13.34 mS m 2 mol1 respectively. What is the molar
limiting conductivity of AgI at the same temperature?
The law of independent migration of ions is to be used to get the molar limiting conductivity of
AgI, which is
om       
Counterions do not change the mobility of the remaining other ion at infinite dilution. So,
eliminating common ions and leaving only the required ions (Ag+ and I-), the final result turns out
to be
 oM (AgI)  13.87 mS m 2 mol1
3. (10 points) Provide molecular interpretations for the dependencies of the diffusion coefficient
and the viscosity coefficient on the temperature, pressure and size of gas molecules.
Diffusion Coefficient:
1. Since  increases as pressure is decreased, D decreases with increasing pressure. This means
that gas molecules diffuse more slowly at higher pressures.
2. The mean speed, vmean , increase with temperature, so D increases with temperature. Thus, in a
given sample molecules will diffuse faster in a hot sample than in a cooler sample, provided the
concentration gradient is the same for both samples.
3. Since  increases with decreasing  , D is greater for small molecules than for large molecules.
2
Viscosity Coefficient:
1.  is independent of pressure, since   1/ p and [A]  p. These will cancel the effect of
pressure out.
2.  increases with temperature since vmean  T 1/2 . Thus viscosity of a hot gas will be higher than
the viscosity of a cool gas.
1/2
1
k  8mT 
3.   vmean  m N can be changed to  

 . Thus, the viscosity coefficient
3
3   k 
decreases with increasing size at constant temperature.
4. (a) (3 points) When is molecularity of a reaction equal to the order of the reaction?
When the reaction is elementary.
(b) (4 points) The kinetic theory assumes that collisions between molecules in a gas are elastic.
What does the term “elastic” describe?
When two molecules collide they may exchange energy. According to the kinetic theory, the total
energy does not change due to this collision. Such a collision is termed as an “elastic collision”.
(c) (5 points) Why isn’t the slowest step in a complex reaction always the rate-determining step?
Usually, the slowest step in a reaction mechanism is the RDS. If there is alternative path to the
products, the slowest step may not be the RDS. Furthermore, concentration of the
reactants/products may have an effect on a normally “faster” step being the RDS.
(d) (5 points) How can a small ion have a large hydrodynamic radius (Stokes radius)?
In solutions, ions may be interacting with other ions or solvent molecules so as to produce an
“effective” radius when moving. Thus a small ion may behave as a large ion when it comes to
motion.
k
 Br2, concentration of Br was followed against time and the
5. For the reaction 2Br 
following table was obtained. Using a graphical method
i)
(15 points) show that the reaction is second order with respect to Br
ii)
(5 points) determine k
5
[Br], 10 M
Time,  s
2.48
1.51
1.04
0.80
0.67
0.56
120
220
320
420
520
620
k
3
i)
ii)
1
vs. t should yield a straight line. You
[Br]
should do this plot and see if you actually get a straight line. In fact, data yields a
1
straight line. First make a table of
vs. t
[Br]
k is the slope of this plot. If you choose your axes wisely and use sufficient precision,
you should get k  2.79 108 dm3 mol 1 s 1
If the reaction is second order, a plot of
6. (14 points) The rate constant for the decomposition of a certain substance is
2.80 103 dm3mol1s 1 at 30C and 1.38 102 dm3 mol1s 1 at 50C. Calculate the Arrhenius
parameters of the substance.
The Arrhenius equation can be used to calculate both parameters, A and Ea.
Let k1  2.80 103 dm3mol1 s 1 and k2  1.38 102 dm3mol 1 s 1 . Using the Arrhenius equation,
ln
 1.38 102 
Ea
k2
1
1

 1
 ln 



K
3 
k1
8.3144 J/mol  K  273.15  30 273.15  50 
 2.80 10 
This equation can be solved for Ea to give Ea  65.0 J mol1
Again using the Arrhenius equation for any set of the data will yield A.
A
2.80 103 dm3 mol1 s 1

e
65.0 kJ/mol1000 J
8.3144 J/molK1 kJ(273.15 30)K
 4.32 108 mol dm 3 s 1
7. (12 points) A solid surface with dimensions of 2.3 mm by 3.0 mm is exposed to argon gas at
90. kPa and 500. K. Determine the number of collisions that argon gas makes with the surface in
15 s.
Let N  number of collisions that the gas makes with the given surface (of surface area A) in a
time interval ( t ) of 15 seconds. It can be calculated from
N  Z w At 
p
 A  t
(2 mkT )1/2
106 m 2
 6.9 106 m 2 t  15 s
1 mm 2
p  90 103 Pa m  39.948 10 3 kg mol 1 / N A  6.6337 10 26 kg molecule1 T  500 K
A  (2.3  3.0)mm 2 
Thus,
N  1.6778 1027 collisions m 2 s 1  6.9 106 m 2 15 s  1.7 1023 collisions
4