# Exams - KFUPM Open Courseware

advertisement ```Written and created by Dr. W. Föِrner, Coordinator
CHEM 311 (071)
First Major Exam
2 HOURS
Wednesday, November 7, 2007
Sec. 1 Dr. Badawi
14 Students
Dr. Föِrner
23 Students
Sec. 2
STUDENT NAME:
STUDENT ID NUMBER:
Question No.
Graded by
Maximum Points
1 (Chap. 7)
Dr. Badawi
17
2 (Chap. 7)
Dr. Badawi
17
3 (Chap. 7)
Dr. Badawi
16
4 (Chap. 8)
Dr. Fِrner
16
5 (Chap. 8)
Dr. Fِrner
17
6 (Chap. 8)
Dr. Fِrner
17
TOTAL
Your Score
100
TOTAL: 37 students: average = 53.4
PHYSICAL CONSTANTS:
R = 8.314 J mol-1 K-1
R = 0.08314 L bar K-1 mol-1
kB = 1.381 x 10-23 J K-1
1 J = 1 kg m2 s-2
1 atm = 1.101325 x 105 Pa
c = 2.998 x 108 m s-1
0K: -273.15oC
(RT)/F = 0.0257 V at 25 oC
R = 0.08206 L atm K-1 mol-1
NA = 6.022 x 1023 mol-1
h = 6.626 x 10-34 J s
F = 96,485 C mol-1
g = 9.81 m s-2
-1/2
3/2
B = 0.51 mol dm (in H2O, 25oC)
ln(x)/log10(x) = 2.303 for all x
1.
The Arrhenius theory of the dissociation of a weak electrolyte into ions offers an
explanation of the decrease in molar conductivity of the solution as the concentration
increases.
(a)
Show that for the equilibrium between a weak electrolyte AB and its ions in
aqueous solution
(

2
)
K = c 0

10
where K is the equilibrium constant, c the initial AB concentration, Λ the molar
conductivity and Λ0 the limiting molar conductivity.
AB _ A+ + Bwhere the equilibrium constant, on the assumption of ideal behavior, is
[ + ][ - ]
K= A B
[AB]
Solution:
concentration of AB before dissociati on : c
_degree of dissociati on :  = /  o
AB _
+
A
+
B
-
initial
c
0
0
change
- c
c
c
equilibriu m c(1 -  )
c
c
2
[ A+ ][ B- ] (c )(c )
( / o )2

K=
=
=c
=c
[AB]
c(1 -  )
1-
1 - ( /  o )
(b)
The equation is one form of Ostwald's dilution law. Show how it can be linearized
(i.e. convert it into a form that will allow experimental values of Λ at various
concentrations to be tested by a straight-line plot). Explain how Λo and K can be
obtained from the plot.
6 points
Solution:
o  
o
c 2
c 2
=
|  o ( o -  )
2 o - 

o ( o -  )
o
o
K=
_ K o2 - K o  = c  2 | 
_
plot
1
K o2 
1 1
c
1 1
1
- =
_ = +
(c )
2
 o K o  o K o2
1
1
vs (c ) _ straight line ; intercept =

o
1
1
(intercept )2
; slope =
=
o =
intercept
K o2
K
(intercept )2
K=
slope
6 points
(c)
The elctrolytic conductivity of a 0.100 M solution of a weak acid (HB) is 1.30 x 10-4
Ω-1 cm-1. If the sum of the limiting ionic conductivities of B- and H+ is 250.0 Ω-1
cm2 mol-1, what is the value of the acid dissociation constant Ka?
Solution:
HB(aq) _ B - (aq) + H + (aq)
Ka=
=

c
=
[ B- ][ H + ]
2
=c
[HB]
1-
1.30  10 -4 S cm-1
= 1.300 S cm2 mol -1
mol
0.100
(10 cm )3
=

o
=
1.300
= 5.200  10 -3
250.0
(5.200  10 -3 )2
= 2.72  10 -6 M
K a = 0.100 M
1 - 5.200  10 -3
-6
K a = 2.72  10 M
5 points
2.
A solution of HBr was electrolyzed in a Hittorf cell. After a current of 4.00 mA had been
passed for 5 h, the mass of HBr in the anode compartment had decreased by 50.0 mg. The
atomic masses of H and Br are 1.008 and 79.904 g/mol, respectively.
(a)
Calculate the transport numbers of H+ and Br- and the amount of electricity (total
charge) carried by each type of ions.
Solution:
t+
t
n
= + = Anode
t - 1 - t+ nCathode
_ t+ =
=
n Anode
n
m /
= Anode = HBr M HBr =
I  t/F
n Anode+ nCathode ntotal
5.00  10- 2 g  96485 As/mol
g
s
(1.008 + 79.904)
 4.00  10-3 A  5 h  3600
mol
h
= 0.82810
t+ = 0.828
t - = 1 - t+ = 1 - 0.82810 = 0.17190 = 0.172
Q = I  t _ Q+ = I +  t = t+  I  t
Q+ = 0.82810  4.00  10-3 A  5 h  3600
s
h
Q+ = 59.62 As
Q- = t -  I  t = 12.38 As
10 points
(b)
If Λo(HBr) is 427.75 Ω-1 cm2 mol-1, what are the molar conductivities and the ionic
mobilities of the H+ and Br- ions?
 +o = t+  o = 0.82810  427.75 S cm2 mol -1
= 354.2 S cm2 mol -1 = 354 S cm2 mol -1
 o- = t -  o = 0.17190  427.75 S cm2 mol -1
= 73.53 S cm2 mol -1 = 73.5 S cm2 mol -1
A cm2

V mol = 3.67  -3 2 -1 -1
u+ = =
10 cm V s
As
F
96485
mol
o
+
u- =
7 points
354.2
 o- = 73.53
F
96485
2
-1 -1
-4
2
-1 - 1
cm V s = 7.62  10 cm V s
3.
(a)
Show that for an electrolyte Na2SO4
3
K sp = 4 s  +_
3
where Ksp is the solubility product, γ± is the mean activity coefficient of the Na+
and SO42- ions in solution and s is the solubility.
Solution:
s = c Na 2 SO4 =
1
c Na+ = c SO422
 +2+_1 =  +2  _
K s = a Na+ a SO42- = c Na+  + c SO42-  2
2
2
= (2s )2 s  +3 _ = 4 s 3  +3 _
8 points
(b)
Assuming that the Debye-Hückel limiting law applies calculate the solubility of
Au2CO3 at 25oC in an aqueous solution which is 0.03 M in Li2SO4 and 0.04 M in
KBr, given that its solubility in pure water at 25oC is 1.11 x 10-4 mol dm-3 (note: in
principle an iteration would be necessary, but you can neglect it).
Solution:
Pure water: I = 0.5 [2 + (2)2] x 1.11 x 10-4 M = 3.33 x 10-4 M
log 10  +_ = - z+ | z - | B I =
= - (1)  | -2 |  0.51 M -1/2  3.33  10-4 M =
= - 0.0186
 +_ = 0.958
3
-4
-12
3
3
K sp = 4  (1.11  10  0.958 ) M = 4.810  10 M
Note: if you neglect Au2CO3 in pure water and use Ksp = 4s3 it is also ok
In solution: I = 0.5{[2 + (2)2] x 0.03 + [1 + 1] x 0.04 }M + 0.5 [ 2 + (2)2] x s
= 0.1300 M+3s
Note: if you neglect s here it is also ok
For the sake of completeness the iteration which can be neglected by students:
Iteration 0 :
I = (0.1300 + 3  1.11  10-4 ) M = 0.13033 M
log 10  +_ = - (1)  | -2 |  0.51 M -1/2 0.13033 M = - 0.3682
 +_ = 0.428
_
1/3
s=
(0.25 K sp )
 +_
(0.25  4.810  10-12 M 3 )1/3
=
0.428
= 2.485  10-4 M
Note: if you neglect iterations 1 and 2 and have only iteration 0 that is also ok
Iteration 1 :
I = (0.1300 + 3  2.485  10-4 ) M = 0.1307 M
log 10  +_ = - (1)  | -2 |  0.51 M -1/2 0.1307 M = - 0.3688
 +_ = 0.428
_
1/3
s=
(0.25 K sp )
 +_
(0.25  4.810  10-12 M 3 )1/3
=
0.428
= 2.485  10-4 M
Iteration 2 :
I = (0.1300 + 3  2.485  10-4 ) M = 0.1307 M
No more change in I, iteration complete.
Result: s = 2.485 x 10-4 M
Since concentrations are given to only 1 significant figure:
s = 2 x 10-4 M
Thus iterations can be neglected.
8 points
4.
Given,
3+
o
Au (aq) + 3 e  Au(s); E = + 1.498 V
+
o
Au (aq) + e  Au(s); E = + 1.692 V
calculate the equilibrium constant at standard conditions (25oC) for the reaction
3+
+
+
H 2 (g) + Au (aq)  2 H (aq) + Au (aq)
Solution:
(1) Au 3+ (aq) + 3 e-  Au(s) ;  G1o = - z1 E 1o F ; z1 = 3
(2) Au+ (aq) + e-  Au(s) ;  G o2 = - z 2 E o2 F; z 2 = 1
(1) - (2) : Au 3+ (aq) + 2 e-  Au+ (aq)
 G o = - zE o F =  G1o -  G o2 ; z = 2
1
_ E o = ( z1 E 1o - z 2 E o2 )
z
=
1
  3  (1.498) - 1  (1.692) V = + 1.401 V
2
for the cell reaction :
o
o
o
E = E reduction - E oxidation= 1.401 V - 0 V = 1.401 V
8 points
o
K = exp [
zF o
E ]
RT
 2  96485 As/mol  1.401 V
 2
= exp 
M
 8.314 J/(mol K)  (273.15 + 25) K 
= exp [109.064 VAs/J] M 2 = 2.322  1047 M 2
With 4 significant figures in the voltages (25 oC seems to have only 2, but has really infinite,
because it is by definition for the standard state), so Ko = 2.322 x 1047 M2
8 points
5.
(a)
Set up a cell for the determination of the solubility product, Ksp, of Hg2Br2.
Solution:
Pt(s) | Br 2 (l) | HBr(aq) | Hg 2 Br 2 (s) | Hg(l)
5 points
(b)
Derive an expression for Ksp in terms of the measured emf and constants.
Solution:
Br (aq) 
1
Br 2 (l) + e
2
1
_ e- + Hg 2 Br 2 (s)  Hg(l) + Br - (aq)
2
1
1
Hg 2 Br 2 (s)  Hg(l) + Br 2 (l)
2
2
equilibriu m with solution :
2+
Hg 2 Br 2 (s) _ Hg 2 (aq) + 2 Br - (aq)
1
1
overall : Hg 22+ (aq) + Br - _ Hg(l) + Br 2 (l) z = 1
2
2
4 points
by defintion Ksp is
K sp = a Hg22+  a Br2
thus :
K sp = a Hg22+  a Br-
Nernst equation :
u
RT 
E = Eo ln
zF 

= Eo +
= Eo +
RT
ln
F
RT
ln
F



a Hg22+  a Br- 
1
a Hg22+  a Br-
o
K sp = E +

u
RT
ln K sp
2F
 2F

_ K sp = exp 
(E - E o )  M 3
 RT

5 points
(c)
At 45.00 oC an emf of -0.980 V is measured for this cell. Given that the standard
electrode potentials (reduction) of the right and left electrodes are 0.850 V and 1.540 V,
respectively, calculate Ksp and give its units.
Solution: Eo = 0.850 V - 1.540 V = -0.690 V
 2  96485 As mol-1  (-0.980 + 0.690) V  3
K sp = exp 
M
-1
-1
 8.314 V A s mol K  (273.15 + 45)K 
= exp  - 21.1566 = 6.48 x 10-10 M 3
Rest of c and d are on next page
2 points
(d)
What can be used as left electrode in the above cell (part a) to determine the pH of an
acid solution?
Solution: Any H+ sensitive electrode, e.g. hydrogen electrode or glas electrode
1 point
6.
Consider the following cell:
Pt(s),H2 (1 bar)|HI(aq, m)|AgI(s)|Ag(s)
(a)
Solution:
Write down equations for the two electrode reactions and the overall cell reaction.
reduction : AgI(s) + e-  Ag(s) + I - (aq)
1
oxidation : H 2 (g)  H + (aq) + e2
1
overall : AgI(s) + H 2 (g)  Ag(s) + H + (aq) + I - (aq)
2
4 points
(b)
Derive the EMF of the cell in terms of known constants, temperature, molality of the
acid solution and the mean activity coefficient.
Solution:
Nernst : E = E o -
RT
ln [ a H + a I - ]
zF
z = 1; ai = mi  i ;  +2 _ =  +  - ; m+ = m- = m HI
E = Eo 5 points
(c)
RT
2RT
2RT
ln [  +2 _ m2HI ] = E o ln m HI ln  +_
F
F
F
Given that at 25oC and for 0.095 m HI the EMF is 0.450 V, calculate the mean activity
coefficient (Eo = 0.2100 V).
Solution:
note :
RT
= 0.0257 V at 25 o C
F
ln  +_ =
=
F
( E o - E) - ln m HI
2RT
(0.2100 - 0.450) V
- ln 0.095
2  0.0257 V
= - 4.6693 + 2.3539 = - 2.3154
_  +_ = e- 2.3154 = 0.0987
4 points
(d) is on the next page
(d)
At 25oC and for an HI solution where the mean activity coefficient can be assumed to
be unity, the EMF is 0.834 V. Calculate the pH value of the acid.
Solution:
If  +_  1 and since :
E = Eo -
ln x
= 2.303 _
log 10 x
2RT
2RT  2.303
ln m HBr = E o log 10 m H +
F
F
= E o + 0.11837 V pH _ pH =
pH =
4 points
E - Eo
0.11837 V
(0.834 - 0.2100) V
0.624 V
=
= 5.27
0.11837 V
0.11837 V
Written and created by Dr. W. Föِrner, coordinator
CHEM 311 (071)
Second Major Exam
5:00 - 6:30 pm
Wednesday, December 12, 2007
Dr. Föِrner
Sec. 1
Dr. Badawi
14 Students
Sec. 2
23 students
STUDENT NAME:
STUDENT ID NUMBER:
Graded by: Questions 1-3 by Dr. Badawi; Questions 4-6 by Dr. Fِrner for both sections
Question No.
Maximum
Points
1: (Chap. 9)
20
2: (Chap. 9)
10
3: (Chap. 9)
15
4: (Chap. 10)
17
5: (Chap. 10)
20
6: (Chap. 10)
18
TOTAL
100
Your Score
TOTAL: students, Average:
PHYSICAL CONSTANTS:
R = 8.314 J mol-1 K-1
R = 0.08314 L bar K-1 mol-1
kB = 1.381 x 10-23 J K-1
1 J = 1 kg m2 s-2
1 atm = 1.101325 x 105 Pa
c = 2.998 x 108 m s-1
0 K = -273.15 oC
(RT)/F = 0.0257 V at 25 oC
R = 0.08206 L atm K-1 mol-1
NA = 6.022 x 1023 mol-1
h = 6.626 x 10-34 J s
F = 96,485 C mol-1
g = 9.81 m s-2
-1/2
3/2
B = 0.51 mol dm (in H2O, 25oC)
ln(x)/log10(x) = 2.303
1.
(a)
If for the opposing reactions
k1
A _Z
k -1
the forward and backward reactions are first order and if the experiment is started with
pure A at a concentration a0, show that
k1 t =
xe
x
ln e
a0 x e - x
where x and xe are the concentrations of Z at time t and at equilibrium, respectively.
Solution
10 points
dx
= k 1 [A] - k -1 [Z] = k 1 ( a0 - x) - k -1 x
dt
= k 1 a0 - ( k 1 + k -1 )x
equilibriu m :
dx
| = 0 = k 1 a 0 - ( k 1 + k -1 ) x e
dt xe
_ ( k 1 + k -1 ) = k 1
_
a0
xe
dx
d( xe - x)
a
a
== k 1 a0 - k 1 0 x = k 1 0 ( xe - x)
dt
dt
xe
xe
( xe - x)
_

xe
d( xe - x)
a
= - k 1 0  d t
x e - x
xe 0
_ ln
k1 t =
t
xe - x
a
= - k1 0 t
xe
xe
xe
x
ln e q.e.d.
a0 x e - x
at the lower limits of the integrals t = 0 and (xe-x)(t=0) = (xe-x(t=0)) = xe-0 = xe
(b)
For the reaction mixture
k1
A _Z
k -1
at 300 K, the concentration of Z at equilibrium is xe = 0.751ao. If the forward and
backward steps with rate constants k1 and k-1, respectively, are first order and if at 400
K k1 is 8.39 x 10-4 s-1:
(i)
(ii)
Find k-1.
Starting with pure A how long would it take to reach a 1:1 mol ratio of the two
molecules?
Solution
(i)
4 points
Kc=
[Z]
0.751 a0
0.751
k
=
=
= 3.016 = 1
[A] (1 - 0.751) a0 0.249
k -1
k -1 =
(ii)
8.39
k1
=
 10-4 s-1 = 2.78  10-4 s-1
K c 3.016
Since xe and k1 are given and a0 cancels out, the correct way to solve (ii) is the
use of a concept similar to that of half life (x = a = a0/2 in a reaction mixture
of a 1:1 molar ratio of reactant and product):
6 points
t 1:1 =
=
xe
xe
ln
k 1 a0 x - a0
e
2
0.751 a0
0.751 a0
ln
- 4 -1
8.39  10 s a0 (0.751 - 0.500) a0
= 981 s = 16.4 m
Sketch (not needed to be done by students)
2.
For the dissociation of methyl iodide
CH3-I(g)  [CH3........I](g)  CH3.(g) + I.(g)
at 500 K the preexponential factor is 2.0 x 1015 s-1 and the activation energy is 168.0 kJ mol-1.
Calculate at 500 K:
(a)
The enthalpy of activation, ΔHo
4 points
Solution: no change in the # of molecules when the activated complex is formed from the
reactant:
Since  nyi = 0 :
i
 H = E a - RT = (168.0 - 8.314  10  500)kJ mol
-3
o
-1
= 163.8 kJ mol-1
(b)
The entropy of activation, ΔSo
Solution:
4 points
2.0  1015 s-1 = e(
k B T  S o /R
)e
h
ln (2.0  1015 ) = ln (e) + ln (

kBT
S
)+ 
h
R
J

 1.381  10- 23  500 K
K
35.23 = 1 + ln 
6.626  10-34 J s





 o
 S
+
 8.314 J K -1 mol-1



35.23 = 1 + 29.97 +
o
 S
8.314 J K -1 mol -1
o
-1
 o
-1
 S = 35.4 J K mol
(c)
The Gibbs energy of activation, ΔGo
Solution:
2 points
o
-3
-1


o
 o
 G =  H - T  S = (163.8 - 500  35.4  10 )kJ mol
= 146.1 kJ mol-1
3.
For an elementary reaction between IO3-, F- and OH- the kinetic data at 25oC are as follows
(a)
[IO3-]
(M)
[F-]
(M)
[OH-]
(M)
rate
M/s
0.250
0.250
0.500
3.50
0.500
0.250
0.500
7.00
0.250
0.500
0.500
7.00
0.250
0.250
1.00
3.50
From the kinetic data, write down the rate equation and calculate the rate constant k
for the reaction.
Solution
8 points
lines 1 and 2 in the table: [IO3-]2 = 2 x [IO3-]1, [F-]2 = [F-]1, [OH-]2 = [OH-]1 then the rate
doubles: v2 = 2 x v1, thus the reaction is 1st order in IO3lines 1 and 3 in the table: [IO3-]3 = [IO3-]1, [F-]3 = 2 x [F-]1, [OH-]3 = [OH-]1 then the rate
doubles again: v3 = 2 x v1, thus the reaction is 1st order in Flines 1 and 4 in the table: [IO3-]4 = [IO3-]1, [F-]4 = [F-]1, [OH-]4 = 2 x [OH-]1 then the rate does
not change: v4 = v1, thus the reaction is 0th order in FThus the rate equation must be: v = k[IO3-]1[F-]1[OH-]0 = k[IO3-][F-]
The rate constant can now be calculated from the data of each one of the lines, for example
line 1:
v1 = k [ IO 3 ] 1 [ F ] 1 thus : k =
v1
[ IO ] 1 [ F - ] 1
3
M
s = 56.0 1
k=
2
Ms
(0.250 ) M 2
3.50
(b)
When two ions, S2- and Ag+, react with each other in a series of solvents with
increasing dielectric constant ε, does the rate of reaction increase or decrease with
increasing ε?
Solution
7 points
ln k = ln k 0 -
z S z Ag

 const = ln k 0 -
(-2)  (+1)

 const
2
= ln k 0 +  const

 increases _
1

decreases _ ln k decreases _ k decreases
4.
Let us denote in a reaction with the overall equation
Co(CN )5 OH 22- + I -  Co(CN )5 I 3- + H 2 O
A = Co(CN)5OH22-, D = I-, E = Co(CN)5I3-, and C = H2O, so that the reaction in short can
be written as A + D  E + C. Assume that the reaction follows the mechanism
A _ B+C
(1)
forward rate constant : k 1
backward rate constant : k -1
B+ D  E
(2)
rate constant : k 2
(a)
Derive the rate equation (rate of formation of E in terms of rate constants and reactant
and product concentrations, not of intermediate concentrations) assuming a steady
state.
Solution
4 points
d[B]
= 0 = k 1 [A] - k -1 [B][C] - k 2 [B][D]
dt
_ [B] =
v=
k 1 [A]
k -1 [C] + k 2 [D]
d[E]
k k [A][D]
= k 2 [B][D] = 1 2
dt
k -1 [C] + k 2 [D]
(b)
If [D] is much larger than [C], determine which one of the reactions will be the ratecontrolling step and what are the orders of the reaction with respect to all reactants and
products and overall in this case? Give the rate equation for this case.
Solution
4 points
k 2 [D] « k -1 [C] _ v =
k 1 k 2 [A][D]
= k 1 [A]
k 2 [D]
_orders : n = n A = 1 ; n D = n E = nC = 0
Since we have only k1 and [A] in the rate equation, equlibrium (1) controls the rate.
(c)
If [C] is much larger than [D], determine which one of the reactions will be the ratecontrolling step and what are the orders of the reaction with respect to all reactants and
products and overall in this case? Give the rate equation for this case.
Solution
4 points
k 2 [D] » k -1 [C] _ v =
k 1 k 2 [A][D] k 1 k 2 [A][D]
=
k -1 [C]
k -1 [C]
= Kk 2 [A][D][C ] -1 ; K =
k1
k -1
orders : n = n A = n D = 1 ; nC = - 1 ; n E = 0
Since we have only k2 and the equlibrium constant in the rate equation, (1) must be
much faster than (2) and thus reaction (2) controls the rate.
(d)
If C (H2O) is the solvent and thus [C] is large and appears experimentally to be a
constant in time, how does that modify the rate equations in (a), (b) and (c) and the
reaction orders in (b) and (c)?
Solution
5 points
k -1 [C] = k -1 ( [C] = const. )
(a)
v=
(b)
_
(c)
k 1 k 2 [A][D]
k -1 + k 2 [D]
v = k 1 [A]
no change in orders
v = K  k 2 [A][D] n A = n D = 1 ; n = 2 ; nC = n E = 0
K=
k1
k1
=
k -1 k -1 [C]
[C] = const.
5.
A likely mechanism for the photolysis of propionic acid is:
C 2 H 5 COOH + hny 
C 2 H 5 + C 2 H 5 COOH
C 2 H 5 + COOH
(i)
 C 2 H 6 + C 2 H 5 COO
(ii) : rate constant k 2
C 2 H 5 COO 
CO2 + C 2 H 5 (iii) : rate constant k 3
C2 H 5 + C2 H 5 
C 4 H 10 (iv) : rate constant k 4
The main products are ethane (C2H6), carbon dioxide (CO2) and butane (C4H10). Any further
reactions of the radical COOH can be negelcted.
(a)
What species is (are) the unstable intermediate(s) in the steady-state treatment?
Solution
2 points
C2H5 and C2H5COO are neither stable products nor reactants, but are formed and consumed
and thus they are the intermediates.
(b)
Write the steady-state equation(s) for the unstable intermediate(s).
Solution
5 points
d[ C 2 H 5 ]
= 0 = I a - k 2 [ C 2 H 5 ][ C 2 H 5 COOH] +
dt
+ k 3 [ C 2 H 5 COO] - 2 k 4 [ C 2 H 5 ] 2 ; (1)
mistake in book : k 4 instead of 2 k 4
(iv) : v =
d[ C 4 H 10 ]
1 d[ C 2 H 5 ]
= k4 [ C2 H 5 ]2 = dt
2
dt
 d[ C 2 H 5 ] 
2
_
 = - 2 k4 [ C2 H 5 ]
dt

(iv)
d[ C 2 H 5 COO]
= 0 = k 2 [ C 2 H 5 ][ C 2 H 5 COOH] - k 3 [ C 2 H 5 COO] ; (2)
dt
(c)
Show, that

[ C 2 H 5 ] =  I a
 2 k4
1/2



where Ia is the rate of light absorption.
Solution
4 points
(1) + (2) : 0 = I a - 2 k 4 [ C 2 H 5 ] 2

_ [ C 2 H 5 ] =  I a
 2 k4
(d)
1/2



q.e.d.
Derive the expression for the rate of formation of CO2:

d[ CO2 ]
= k 2  I a
dt
 2 k4
1/2

 [ C 2 H 5 COOH]

Solution
5 points
Equ.(2) : k 3 [ C 2 H 5 COO] = k 2 [ C 2 H 5 ][ C 2 H 5 COOH] =
1/2

= k 2  I a
 2 k4

 [ C 2 H 5 COOH]

where [ C 2 H 5 ] is the solution of (c)
(iii) :
d[ CO2 ]
= k 3 [ C 2 H 5 COO] =
dt

= k 2  I a
 2 k4
(e)
1/2

 [ C 2 H 5 COOH] q.e.d.

Show that the quantum yield for CO2 is
[
COOH]
 = k 2 C 2 H 51/2 1/2
(2 k 4 ) I a
Solution
4 points
if : bB + other reactants + hny  aA + other products
_=
v
; v=
Ia
d[ CO2 ]
1 
dt
__  =
= k 2  I a
Ia
I a  2 k4
_=
6.
1 d[A]
1 d[B]
=a dt
b dt
1/2

 [ C 2 H 5 COOH] result of (d)

k2
[ C 2 H 5 COOH] q.e.d.
2 k4 I a
Apply the steady-state approximation to the following gas phase reaction mechanism:
k1
HI(g) + HI(g) _ HI * (g) + HI(g)
k -1
k2
*
HI (g)  H(g) + I(g)
(* denotes a molecule in an excited state; do not consider any possible following reactions of
the radicals H and I)
(a)
to show that v, the rate of formation of the final product I, is
2
k k [HI ]
v= 1 2
k -1 [HI] + k 2
Solution
The rate of formation of I (possible following reactions are not considered) is
v=
10 points
d[I]
= k 2 [ HI * ]
dt
The excited state HI* is an intermediate and so its concentration can be assumed to be small
and constant during most of the reaction time (steady-state approximation):
d[ HI * ]
= 0 = k 1 [HI ] 2 - k -1 [ HI * ][HI] - k 2 [ HI * ]
dt
thus ( k -1 [HI] + k 2 )[ HI * ] = k 1 [HI ] 2
2
and [ HI * ] =
k 1 [HI ]
k -1 [HI] + k 2
Now the unknown intermediate concentration is substituted in v:
v=
2
d[I]
k k [HI ]
= k 2 [ HI * ] = 1 2
dt
k -1 [HI] + k 2
(b)
What are the rate and the order of the reaction at very large [HI]?
Solution
4 points
When [HI] is very large, then k2 in the denominator can be neglected against k-1[HI] and thus
the rate becomes
v=
2
2
k 1 k 2 [HI ]
k k [HI ] k 1 k 2
 1 2
=
[HI] = K k 2 [HI]
k -1 [HI] + k 2
k -1 [HI]
k -1
and the reaction is 1st order.
(c)
What are the rate and the order of the reaction at very small [HI]?
Solution
4 points
When [HI] is very small, then k-1[HI] in the denominator can be neglected against k2 and thus
the rate becomes
2
v=
k 1 k 2 [HI ]
k k
 1 2 [HI ] 2 = k 1 [HI ] 2
k -1 [HI] + k 2
k2
and the reaction is 2nd order.
Written and created by Dr. W. Föِrner
CHEM 311 (071)
Final Exam
3 HOURS, 7:30 - 10:30 am
Room 5/201
Monday, January 21, 2008, 7:30 am
Sec. 2
Dr. Föِrner
22 students
Sec. 1 Dr. Badawi
14 students (1 absent)
STUDENT NAME:
STUDENT ID NUMBER:
Part A: 4 Questions (50 Points)
Part B: 20 Multiple Choice Questions (50 points)
MCQ:
Correct Answers
2.5
Points/Answer
Your Score
x 2.5
Total:
Question No.
Graded by
Maximum Points
1: Chap. 8
Dr. Badawi
13
2: Chap. 9
Dr. Badawi
12
3: Chap.18
Dr. Fِrner
13
4: Chap.19
Dr. Fِrner
12
MCQ
50
TOTAL
100
Your Score
PHYSICAL CONSTANTS:
R = 8.314 J mol-1 K-1
R = 0.08314 L bar K-1 mol-1
kB = 1.381 x 10-23 J K-1
1 J = 1 kg m2 s-2
1 atm = 1.101325 x 105 Pa
c = 2.998 x 108 m s-1
0 K = -273.15 oC
RT/F = 0.0257 V at 25 oC
R = 0.08206 L atm K-1 mol-1
NA = 6.022 x 1023 mol-1
h = 6.626 x 10-34 J s
F = 96485 C mol-1
g = 9.81 m s-2
-1/2
3/2
B = 0.51 mol dm (in H2O, 25oC)
ln(x)/log10(x) = 2.303
Part A:
1.
Questions: 4 Questions
Given the cell
Au(s) | AuI(s) | I - (aq) || Au+ (aq) | Au(s)
(a)
Solution:
Write the cell reaction.
2 points
right reduction : Au+ (aq) + e-  Au(s)
_
left oxidation : Au(s) + I - (aq)  AuI(s) + eoverall : Au+ (aq) + I - (aq)  AuI(s) _z = 1
Kc=
(b)
1
1
=
[ Au ][ I ] K sp
+
Given the electrode potentials
Au+(aq) + e-  Au(s);
E0 = 1.6921 V
AuI(s) + e  Au(s) + I (aq); E0 = 0.4251 V
calculate the standard cell potential.
Solution:
2 points
0
0
0
0
0
E = E right - E left = E Au+ /Au - E AuI/Au+I -
= (1.6921 - 0.4251) V = 1.2670 V
(c)
Solution:
What is the equilibrium constant, Kc, for the reaction (25 oC)?
5 points
zFE0
 G0 = - zFE 0 = - RT lnK c _ K c = e RT
 96485 A s mol -1  1.2670 V  - 2
 M
K c = exp 
-1
-1
 8.314 J K mol  298.15 K 
= e49.31642 M - 2 = 2.6173  10 21 M - 2
(d)
Solution:
Assuming γ± = 1, and with the solubility product being Ksp = 1/Kc, what is the
solubility of AuI in water?
4 points
K sp =
1
Kc
= 3.8207  10-22 M 2
2
+
2
K sp = [ Au (aq)][ I (aq)] = [AuI(aq) ] = s
_ s = K sp = 1.9547  10-11 M
2.
(a)
Show, that for a second order reaction
2Cl  Cl 2 rate constant k
the integrated time law is given by
x
a0 ( a0 - x)
Solution:
= kt
where x is the the part of the concentration of Cl which has already reacted (to x/2 Cl2)
at time t, and a0 is the initial concentration of Cl.
3 points
-
d[Cl]
d( a o - x)
= k[Cl ] 2 thus = k( a o - x )2
dt
dt
dx
= k[Cl ] 2 = k( a0 - x )2
dt
x
t
-1 x
dx
0 ( a0 - x )2 = k 0 dt _ - a0 - x |0 = kt
1
1
x
a - ( a0 - x)
- = kt _ 0
= kt _
= kt
a0 - x a0
a0 ( a0 - x)
a0 ( a0 - x)
(b)
Solution:
Derive an expression for the half-life of the reactant Cl for the second order reaction
given in (a). (No credit will be given for the result alone.)
3 points
1
1
a
x( t 1/2 ) = 0 _
= kt1/2 _ t 1/2 = 2 =
a0
a0 k a0 k
2
a0 ( a 0 - )
2
2
a0
2
(c)
Illustrate graphically one of the possible ways of plotting results (the one with a
nonzero intercept) to see if a reaction is second order and, if it is, to determine the rate
constant k.
Solution:
3 points
y-axis: 1. x/[a0(a0-x)] or 2. x/(a0 -x) or 3. 1/(a0-x)
slope:
1. k or 2. a0k or 3. k
intercept:
1. 0 or 2. 0 or 3. 1/a0
For 1.: x/[a0(a0-x)]=kt; For 2.: x/(a0 -x)=a0kt; For 3. 1/(a0-x)=kt+1/a0
Derivation done in (a); asked for is No. 3
(d)
Solution:
For a second order reaction of the type given in (a), the initial concentration of Cl was
1.00 M. After 10 minutes, the concentration was 0.700 M. Calculate the rate constant.
3 points
x: concentration of Cl that has already reacted after time t to yield x/2 Cl2:
[Cl]t = a = ao-x:
a0 = 1.00 M _ a0 - x = 0.700 M
_ x = (1.00 - 0.700)M = 0.300M
k=
x
a0 ( a0 - x)t
=
0.300 M
1.00M  0.700 M  10 min
= 0.0429 M -1 min -1 =
0.0429
M  60 s
= 7.15  10-4 M -1 s-1
3.
An important adsorption isotherm is the BET isotherm:
PP0 = 1 + P
V( P0 - P) V 0 K V 0
where V is the volume of gas adsorbed at pressure P, V0 is the volume of gas that can be
adsorbed in a monolayer, P0 is the saturation vapor pressure of the gas and K is the equilibrium
constant for adsorption
(a)
Show, that for P«P0 the BET equation reduces to the Langmuir isotherm:
=
V
=
V0
KP
1+ KP
Solution:
3 points
1
P
P» P0 _ P0 - P  P0 _ PP0 =
+ |  KV
VP0 V 0 K V 0
_ KP =
V
V0
(b)
Solution:
V
+ KP
_
V0
V
V0
= =
KP
1+ KP
Consider nitrous oxide, NO, at 80.0 K adsorbed on Silica gel:
P/kPa 13.1 31.7
V/cm3 115
228
Use these data and P0=91.5 kPa to calculate K and V0 with the BET isotherm.
3 points
(1)
PP0 = 13.1  91.5 kPa = 0.133 kPa = 1 + 13.1 kPa
3
V( P0 - P) 115 cm3 (91.5 - 13.1)
cm V 0 K
V0
(2)
PP0 = 31.7  91.5 kPa = 0.213 kPa = 1 + 31.7 kPa
3
V( P0 - P) 228 cm3 (91.5 - 31.7)
cm V 0 K
V0
(2) - (1) : 0.080
kPa
cm
3
=
18.6 kPa
V0
_ V 0 = 232.5 cm3 = 233 cm3
_
in (2) :
(c)
1
31.7
= (0.213 ) 232.5 kPa _ K = 0.056 (kPa )-1
K
232.5
In a unimolcular surface reaction NO can decompose into O2 and N2. The rate of
reaction does not depend on the partial pressure of N2, but becomes smaller when the
partial pressure of O2 is increasing. Explain this observation and write down an
expression for the rate based on a Langmuir isotherm.
Solution:
4 points
Besides NO also the product O2 can be adsorbed, but not N2. Thus O2 occupies surface sites for the
adsorption of NO and therefore the rate of reaction becomes reduced when PO2 is increased, because
only adsorbed NO can react.
The isotherm that must be used is the Langmuir isotherm with competition:
 NO =
=
K NO [NO]
1 + K NO [NO] + K O2 [ O 2 ]
K NO P NO
1 + K NO P NO + K O2 PO2
v = k  NO =
k K NO P NO
1 + K NO P NO + K O2 PO2
Thus the rate decreases when PO2 is increased and does not depend on PN2.
(d)
Solution:
The partial pressure of O2 can be controlled by adding O2, the product gas of the NO
decomposition to the mixture. At what conditions for PNO and PO2 does the reaction
become first order in NO?
3 points
When the following condition is fulfilled:
K O2 PO2 « 1 + K NO P NO
then : 1 + K NO P NO + K O2 PO2  K O2 PO2
and : v 
kK NO
P NO
K O2 P O2
then the reaction is first order in NO (and -1 order in O2).
Thus PNO must be small and PO2 must be very large for the reaction to be first order in NO.
4.
The diffusion coefficient for a large sugar molecule in a solvent is 5.31 x 10-10 m2 s-1 at 25oC.
The viscosity of the solvent at 25oC is 7.148 x 10-4 kg m-1 s-1. The density of the diffusing sugar
is 2.13 g cm-3, that of the solvent 1.00 g cm-3. Assume Stokes's law to apply.
(a)
Solution:
Use the law to calculate the radius and the volume of the molecule. Which assumption
on the shape of molecules must be made to make Stokes's law applicable?
6 points
r=
kBT
6  D
J
 298.15 K
K
=
2
kg
6  3.141592654  7.148  10-4
 5.31  10-10 m
ms
s
1.381  10- 23
= 5.755  10-10 m = 0.5755 nm
4
V =  r 3 = 7.9841  10- 28 m3
3
Molecules are assumed to be spherical.
(b)
Solution:
Calculate the mass of a molecule and from that the molar mass of the diffusing
substance.
4 points
m = 2.13
g
cm
3
 7.9841  10- 28 ( 10 2 cm )3 = 1.7006  10- 21 g
M = mN A = 1.7006  10 - 21 g  6.022  10 23 mol -1
= 1024 g mol -1
(c)
Solution:
How far could a molecule diffuse in the solvent at 25oC in 10 hours?
2 points
x = 2Dt = 2  5.31  10-10
2
s
m
 10 h  3600
s
h
= 6.18  10-3 m = 6.18 mm
Part B:
Multiple Choice Questions (50 points):
(20 questions, 2.5 points each)
1.
The Eo = 0.331 V for the cell whose reaction is
H2(g, 1 atm) + 2 AgBr(s)  2 H+ + 2 Br- + 2 Ag(s)
When the emf, E = 0.552 V for this cell and at 25oC, the pH of the solution in the cell is given
by
a.
b.
c.
d.
0
(E - Eo)/log PH2
(0.552 - 0) V / 0.0592 V
(0.552 - 0.331) V / (2x0.0592 V)
e- + AgBr(s)  Ag(s) + BrH2(g)  2H+ + 2e-
Solution: Answer d
H2(g) + 2AgBr(s)  2Ag(s) + 2H+ + 2BrThus: z=2 and aH+=aBr-; Nernst equation:
E = Eo -
RT
RT
ln [ a H + a Br- ] 2 = E o ln a 2H +
2F
F
= Eo -
(E - E o )
F
= - ln a H + = - 2.303  lg a H + = 2.303pH
2RT
2.303
2.
2RT
ln a H +
F
RT
(0.552 - 0.331)
= 0.0592V _ pH =
F
2  0.0592
For the reaction
S2O82- + 2 I-  I2 + 2 SO42a decrease in ionic strength will
a.
Increase the rate of the reaction
b.
Decrease the rate of the reaction
c.
Have no effect on the rate of the reaction
d.
Stop the reaction
Solution: Answer b
lgk = lg k o + 2 Bz A z B I = lg k o + 4B I
Because zA=-2, zB=-1. Thus decrease in I gives decrease in lg(k) and thus in k.
3.
A fundamental postulate of the transition state theory of reaction rates is that
a.
b.
c.
An equilibrium exists between reactants and activated complex
The transition state contains one more degree of freedom than the reactants
The pre-exponential term is independent of temperature
d.
Two molecules must collide
Solution: Answer a
4.
Gaseous, unimolecular decomposition reactions may proceed according to the following
mechanism in which the collision between two normal molecules A produces an activated
molecule A* which, in turn, may be deactivated by collision or decompose into products.
A + A  A* + A
rate constant k1
A + A*  A + A
rate constant k-1
A*  products
rate constant k2
Using the steady state approximation for the quantity d[A*]/dt, the rate law for the reaction A
 products becomes
d[A] k 2 k 1 [A ] 2
=
dt
k -1 [A] + k 2
Under which conditions will the rate of the overall reaction tend to be second order?
a.
b.
c.
d.
Increased surface area in reaction vessel
Addition of inert gas
Low pressure
High pressure
Solution: Answer c because
p A V = n A RT _ p A = [A]RT
1
1 dp A
(RT )2
=
1
RT dt
+ k2
k -1 p A
RT
2
k 2 k1 pA
2
dp A
k1 k 2 pA
k
=
 1 p 2A if k -1 p A » k 2 RT
dt k -1 p A + k 2 RT RT
The condition can be always fulfilled by having a sufficiently small pressure.
5.
Consider the mechanism
k1
A_B
k2
k3
B+AC
If B is a species that is present in only negligibly small amounts at all times, the concentration
of B in terms of the major reagents A and C is
a.
k 1 [A]
b.
k 1 [A]
c.
k 1 [A]
k 2 [A] + k 3
d.
k 1 [A]
k 2 + k 3 [A]
Solution: Answer d
d[B]
k 1 [A]
= 0 = k 1 [A] - k 2 [B] - k 3 [A][B] _ [B] =
dt
k 2 + k 3 [A]
6.
The fraction of the electric current carried by a given ion in a strong electrolyte is called
a.
b.
c.
d.
transference number
ion mobility
ion friction coefficient
limiting equivalence conductance
Solution: Answer a
7.
For the mechanism
CH3COCH3  CH3. + CH3CO.
rate constant k1
CH3CO.  CH3. + CO
rate constant k2
CH3. + CH3COCH3  CH4 + .CH2COCH3 rate constant k3
.CH2COCH3
 CH3. + CH2CO
CH3. + .CH2COCH3  C2H5COCH3
the rate of formation of CH3CO. is given by
a.
k1[CH3COCH3] + k2[CH3CO.]
b.
k1[CH3COCH3] - k2[CH3CO.]
c.
k1[CH3COCH3] - k3[CH3COCH3][CH3.]
rate constant k4
rate constant k5
d.
k1[CH3COCH3] + k3[CH3COCH3][CH3.]
Solution: Answer b
d[ CH 3 CO ]
= k 1 [ CH 3 CO CH 3 ] - k 2 [ CH 3 CO ]
dt
CH3CO. is produced in the first reaction and consumed in the second and appears nowhere else.
8.
In several experiments on the kinetics of the reaction A + B  products, the following observations were made:
(i)
Doubling the initial concentration of A, with B fixed, doubled the rate
(ii)
Doubling the initial concentration of both A and B increased the rate by a factor of 8.
What is the rate law?
a.
Rate = k[A]2[B]
b.
Rate = k[A]2[B]2
c.
Rate = k[A][B]
d.
Rate = k[A][B]2
Solution: Answer d
From (i) : c A = 2 c A _ v = 2v _ v_ c A
From (ii); c A = 2 c A  c B = 2 c B _ v = 8v = 2  22 v
since v_ c A [from(i)] : v _ c 2B _ v = k c A c 2B = k[A][B ] 2
9.
The equilibrium constant, Keq, is 30 for the reaction
HA + B _ HB+ + AThe rate constant k1 for the forward reaction is 107 M-1s-1. The rate constant k2 of the reverse
reaction is
a.
3 x 10-6 M-1s-1
b.
3 M-1s-1
5
-1
-1
c.
3 x 10 M s
d.
3 x 108 M-1s-1
Solution: Answer c
7
K eq =
10.
k1
k
10
5
-1 -1
- 1 -1
_ k2 = 1 =
M s = 3  10 M s
k2
K eq 30
For two reactions, whose entropies of activation are equal, the activation energies are 100,000
and 300,000 J/mol, respectively. At 300 K, the number of molecules activated for the first
reaction as compared to the second will be
a.
c.
Larger
The same
b.
Smaller
d.
Inversely proportional to T
Solution: Answer a
1
2
E a = 100 kJ/mol E a = 300 kJ/mol
Ea
1
N TS _ e RT _ N TS _ e
11.
100 kJ/mol
RT
> e-
300 kJ/mol
RT
2
_ N TS
The BET theory is based on the assumption of
a.
b.
c.
d.
Constant heat of adsorption
The gas is ideal
Multilayer adsorption
No lateral interaction between adsorbed molecules.
Solution: Answer c
12.
Which statement is true for the following cell as it discharges at 25oC
Cr(s) | Cr(aq)3+(m = 0.1m) || Sn(aq)2+(m = 0.01m) | Sn(s)
E
a.
b.
c.
d.
0
Sn2+ /Sn
= - 0.1375 V ; E 0Cr 3+ /Cr = - 0.744 V

The mass of the Sn electrode will decrease
Oxidation occurs at the Sn electrode
Electrons will flow from the Sn electrode to the Cr electrode
The concentration of Cr3+ will increase
Solution: Answer d
2e- + Sn2+(aq)  Sn(s) | x 3
Cr(s)  Cr3+(aq) + 3e- | x 2
2 Cr(s) + 3 Sn2+(aq)  2 Cr3+(aq) + 3 Sn(s); z=6, Nernst equation:
RT [ Cr 3+ (aq) ] 2
E= E ln
zF [ Sn2+ (aq) ] 3
o
=E
o
Sn2+ /Sn(s)
-E
o
Cr 3+ /Cr(s)
RT ( 10 -1 )2
ln
6F ( 10 - 2 )3
= (-0.1375 + 0.744)V -
= + 0.6065 V -
RT
ln ( 10 4 )
6F
1
 0.0257 V  9.2103
6
= + 0.6065 V - 0.03945 V = + 0.5670 V
E>0: Reaction from left to right:
Sn2+ becomes reduced, thus reduction occurs at the Sn electrode: b is wrong
Cr becomes oxidized, thus the Cr3+ concentration increases: d is correct
mass of Sn electrode increases: a is wrong
e- from Cr to Sn: when Cr is oxidized it leaves 3 electrons in the Cr electrode: c is wrong
13.
Which ion has the lowest molar conductivity in water at 25oC?
a.
c.
Li+
Na+
b.
d.
H+
K+
Solution: Answer a: not the small Li+ moves, but [Li(H2O)4]+
14.
The molar conductivity at infinite dilution of a weak acid such as HF
a.
b.
c.
d.
is an undefined quantity.
can best be determined from measurements on dilute solutions of NaF, NaCl, and
HCl.
can be determined by measurements on very dilute HF solutions.
can best be determined by extrapolations of measurements of HCl, HBr, and HI.
Solution: Answer b
HF = HCl + NaF - NaCl
o
15.
o
o
o
The ionic strength of 0.20 M aqueous Al2S3 (assuming that it is completely dissolved) is
a.
c.
15 M
3.0 M
b.
d.
1.5 M
6.0 M
Solution: Answer c
3+
2Al 2 S 3 (s) + n H 2 O  2 Al (aq) + 3 S (aq)
[ Al 3+ ] = 2[ Al 2 S 3 ] 0 ; [ S 2- ] = 3[ Al 2 S 3 ] 0
1
I = ([ Al 3+ ]  32 + [ S 2- ]  (-2 )2 )
2
1
= [ Al 2 S 3 ] 0 (2  9 + 3  4)
2
= 15  [ Al 2 S 3 ] 0 = 3.0 M
16.
Λo for NaCl(aq) at 25oC is 140.8 Ω-1 cm2 mol-1. If the transference number of Na+(aq) is 0.531,
what is the value of λo for Na+(aq)?
a.
b.
c.
d.
61.9 Ω-1 cm2 mol-1
74.8 Ω-1 cm2 mol-1
288 Ω-1 cm2 mol-1
195.3 Ω-1 cm2 mol-1
Solution: Answer b
 oNa = t+  oNaCl = 0.531  140.8 -1 cm2 mol -1
+
= 74.8 -1 cm2 mol -1
17.
The coefficient of viscosity of a liquid can be measured by determining the length of time a
given volume of liquid needs, to drain through the capillary portion of an Ostwald viscometer.
Under identical conditions, the time required for samples of methanol and water to drain were
51.3 and 42.6 s, respectively. Given ρ/(g mL-1) = 0.6934 and 0.9982, respectively. η = 1.002 
10-3 Pa s for H2O, the viscosity of ethanol is
a.
b.
c.
d.
8.382  10-4 Pa s
1.918  10-3 Pa s
5.234  10-4 Pa s
1.200  10-3 Pa s
Solution: Answer a
 H O = A  H O t H O ;  Meth = A  Meth t Meth
2
2
2
 Meth  Meth t Meth
 t
=
;  Meth =  H O Meth Meth
H O  H O tH O
H O tH O
2
2
2
2
 Meth = 1.002  10-3 Pa s 
2
2
0.6934  51.3
0.9982  42.6
= 8.382  10-4 Pa s
18.
The viscosity of carbon dioxide at 25C and 101.325 kPa is 13.8.10-6 kg m-1 s-1.The molecular
diameter of carbon dioxide at 25C is (molar mass of CO2: 44.01 g mol-1)
a.
0.475 nm
b.
1.95 μm
c.
3.85 pm
d.
2.85 mm
Solution: Answer a
1/4
T
(
T)
 = mk B ; d = mk3/4B


 3 d2
m=
M
NA
; kB =
d=
R
NA
; mk B =
RM
2
NA
(MRT/ N 2A )1/4
 3/4 
kg
J

 8.314
 298.15 K
 44.01  10-3
mol
K mol

(6.022  10 23 mol -1 )2


=
kg
 3/4 13.8  10-6
ms
2

kg m2
 3.0083  10-46
2

s

=
1/4






1/4




kg
2.3597  13.8  10-6
ms
4.1647  10
2.3597  3.7148  10 -3
-12
=
kg m
s = 4.75  -10 m
10
kg
ms
= 0.475  10 -9 m = 0.475 nm
19.
The rate law for the reaction A  B is second order in [A]. If the reaction started with 2.0 M
of A, after 20 minutes 0.25 M of A remained. What would be the concentration of A remaining after 40 minutes?
a.
b.
c.
d.
0.13 M
0.25 M
0.33 M
0.0 M
Solution: Answer a
a
a = c A (t) :
t
da
da
1
= - k a 2 ;  2 = - k  dt ; [- ] aa0 = - kt
dt
a
0
a0 a
1
1
- = - kt
a0 a
k=
a=
(2.0 - 0.25)
a0 - a
-1
-1
-1
-1
=
M min = 0.175 M min
20

2.0

0.25
taa0
1
1
a0
+ kt
=
2.0 M
a0
=
= 0.13 M
1 + a0 kt 1 + 2.0  0.175  40
20. The viscosity of a gas:
a.
b.
c.
d.
is inversely proportional to T
is independent of T
is proportional to T1/2
decreases exponentially with increasing temperature
Solution: Answer c
Multiple Choice Key
Question
Answer
1
d
2
b
3
a
4
c
5
d
6
a
7
b
8
d
9
c
10
a
11
c
12
d
13
a
14
b
15
c
16
b
17
a
18
a
19
a
20
c
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