CHEM311, Section 02, Term 071
Quiz No. 1
ID #:
W. Föِrner
___________________________
Name: ___________________________
Always show your way of calculation, not only the result!
1.
Electrolysis of molten KBr generates bromine gas (and potassium metal), which can be used in
industrial bromination processes. How long will it take to convert a 500.0 kg batch of phenol
(C6H5OH) to monobromphenol (hint: in this reaction one Br2 molecule gives one C6H4BrOH
and one HBr molecule) using a current of 20000 A (molar mass of phenol: 94.114 g mol-1; F =
96485 C mol-1)?
(4 points)
2.
The transport numbers for HCl at infinite dilution are estimated to be t+ = 0.821 and t- =
0.179 and the molar conductivity is 426.16 Ω-1 cm2 mol-1. Calculate the mobilities of the
hydrogen and chloride ions.
(3 points)
3.
Calculate the mean activity coefficient γ± for the Ba2+ and SO42- ions in a saturated solution of
BaSO4 (Ksp = 9.2 x 10-11 mol2 dm-6) in 0.2 M K2SO4 (in H2O at 25oC), assuming DHLL to
apply (B is 0.51 M-1/2 for H2O, 25oC).
(3 points)
CHEM311, Section 02, Term 071
Quiz No. 2
ID #:
W. Föِrner
___________________________
Name: ___________________________
Always show your way of calculation, not only the result!
1.
Write the electrode reactions , the overall reaction and the expression for the emf for each of
the following reversible cells.
(a)
(b)
(c)
Pt,H2(1 bar)|HI(aq)|AuI(s)|Au
Ag|AgCl(s)|KCl(aq)|Hg2Cl2(s)|Hg
Ag|AgCl(s)|KCl(aq,c1)||KCl(aq,c2)|AgCl(s)|Ag
(3 points)
2.
Calculate the equilibrium constant at 25oC for the reaction
2Fe3+(aq) + 2I-(aq)  2 Fe2+(aq) + I2(s)
using the standard electrode potentials for Fe3+/Fe2+ (0.771 V) and for I2/I- (0.5355 V). Gas
constant: R = 8.314 J mol-1 K-1, absolute zero temperature: -273oC, Faraday's constant: F =
96485 C mol-1.
(3 points)
3.
The Weston standard cell is (the CdSO4 hydrate solution is saturated)
Cd(Hg)(l) | CdSO4 
(a)
(b)
8
H 2 O(aq, s) | Hg 2 SO4 (s) | Hg(l)
3
Write the cell reaction.
At 25oC the emf is 1.01832 V and the temperature derivative of the emf is
-5.00 x 10-5 V K-1. Calculate ΔGo, ΔHo and ΔSo (F=96500 C mol-1).
(4 points)
CHEM311, Section 02, Term 071
Quiz No. 3
ID #:
W. Föِrner
___________________________
Name: ___________________________
Always show your way of calculation, not only the result!
1.
Some results for the rate of reaction between two substances A and B are shown in the
following table.
[A]/M
[B]/M
rate/M s-1
1.4 x 10-2
2.3 x 10-2
7.40 x 10-9
2.8 x 10-2
4.6 x 10-2
5.92 x 10-8
2.8 x 10-1
4.6 x 10-2
5.92 x 10-6
Deduce the order α with respect to A, the order β with respect to B and the rate constant.(4
points)
2.
The reaction
2NO(g) + Cl2(g)  2NOCl(g)
is second order in NO and first order in Cl2. In a volume of 2 dm3, 5 mol of NO and 2 mol of
Cl2 were brought together, and the initial rate was 2.4 x 10-3 M s-1. What will be the rate when
half of the chlorine has reacted?
(3 points)
3.
The rate of a reaction at 300 K is doubled when the pressure is increased from 1 bar to 2000
bar. Calculate the volume of activation, assuming it to be independent of pressure (1 bar = 105
Pa).
(3 points)
CHEM311, Section 02, Term 071
Quiz No. 4
ID #:
W. Föِrner
___________________________
Name: ___________________________
Always show your way of calculation, not only the result!
1.
Derive the integrated rate equation (in terms of x, being the concentration of product Z,
initially being 0) for an irreversible reaction (no backward reaction) 2A + B  Z, the rate being
proportional to [A]2[B]. The initial concentration of A is 2a0 and that of B is a0. (4 points)
2.
The rate constant for a reaction at 230oC is found to be exactly twice the value at 220oC.
Calculate the activation energy (absolute zero temperature: -273oC ; R = 8.314 J K-1 mol-1).(2
points)
3.
The rate constant for a second order reaction in solution is 3.95 x 10-4 M-1 s-1 at 25oC and the
activation energy is 120.0 kJ mol-1. Calculate, at 25oC, the preexponential factor, A, the
enthalpy of activation, ΔHo, the Gibbs energy of activation, ΔGo, and the entropy of
activation, ΔSo (at 25oC, kBT/h = 6.211 x 1012 s-1; e = 2.718).
(4 points)
CHEM311, Section 02, Term 071
Quiz No. 5
ID #:
W. Föِrner
___________________________
Name: ___________________________
Always show your way of calculation, not only the result!
1.
Nitrogen peroxide reacts with nitric oxide in the gas phase according to the stoichiometric
equation
N2O5 + NO  3NO2
The following mechanism has been proposed.
N2O5  NO2 + NO3
(rate constant k1)
NO2 + NO3  N2O5
(rate constant k-1)
NO + NO3  2NO2
(rate constant k2)
Assume that the steady state treatment can be applied to NO3, and derive an equation for the
rate of consumption of N2O5.
(4 points)
2.
The gase phase reaction
Cl2 + CH4  CH3Cl + HCl
proceeds by a free-radical chain reaction in which the chain propagators are Cl (k2) and CH3
(k3) (but NOT H) and the chain-ending step is 2Cl  Cl2 (k-1), which is the backward reaction
of the initiation (k1). Write the mechanism, identify the initiation reaction and the chainpropagating steps, and obtain an expression for the rate (rate of formation of HCl) of the
overall reaction (using steady state treatment)
use also next page!
(4 points)
3.
A 1000 W mercury vapor flash lamp emits radiation of 253.7 nm wavelength, and the duration
of the flash is 1 μs. Suppose that all of the radiation of a single flash is absorbed by mercury
vapor; how many atoms of excited mercury are formed (speed of light: 2.998 x 108 m s-1,
Planck's constant h: 6.626 x 10-34 J s)?
(2 points)
CHEM311, Section 02, Term 071
W. Föِrner
Quiz No. 6
ID #:
___________________________
Name: ___________________________
Always show your way of calculation, not only the result!
1.
Show that, if V is the volume of gas adsorbed at pressure P, and the Langmuir isotherm is
obeyed, a plot of P/V against P is linear. Explain how, from such a plot, the volume Vo
corresponding to complete coverage and the isotherm constant K can be determined. (3
points)
2.
The following data were obtained for the adsorption of krypton on a 1.21 g sample of a porous
solid (P is the pressure of the gas, V the volume adsorbed, as it would be at STP):
P/Torr
1.11
3.08
V/cm3 (at STP)
1.48
1.88
If the saturation vapor pressure is 19.0 Torr, calculate K and Vo using the BET isotherm. (4
points)
Use next page!!
3.
Suggest explanations for the following observations, in each case write an appropriate rate
equation based on a Langmuir isotherm.
(a)
The decomposition of phosphine (PH3) on tungsten is first order at low pressures and
zero order at high pressures, the activation energy being higher at the higher pressures.
(b)
The rate of the decomposition of ammonia on molybdenum becomes smaller when
the amount of product nitrogen increases.
(3 points)
CHEM311, Section 02, Term 071
Quiz No. 7
ID #:
W. Föِrner
___________________________
Name: ___________________________
Always show your way of calculation, not only the result!
1.
The surface tension of water at 20oC is 7.27 x 10-2 N m-1 and its density is 0.998 g cm-3.
Assuming a contact angle θ of zero, calculate the rise of water at 20oC in a capillary tube of
radius
(a) 1 mm
(b) 10-3 cm.
For the earth's gravitational acceleration, take g = 9.81 m s-2.
2.
(3 points)
The two arms of a U-tube have radii of 0.05 cm and 0.10 cm. A liquid of density 0.80 g cm-3 is
placed in the tube, and the height in the narrower arm is found to be 2.20 cm higher than that
in the wider arm. Calculate the surface tension of the liquid, assuming a contact angle θ = 0.
(3 points)
3.
In a normal adult at rest the average speed of flow of blood through the aorta is 0.33 m s-1. The
radius of the aorta is 9 mm and the viscosity of blood at body temperature, 37oC, is about 4.0 x
10-3 kg m-1 s-1. Calculate, in Torr, the pressure drop along a 0.5 m length of the aorta (π =
3.1416; 1 Torr = 133.322 Pa).
(4 points)
CHEM311, Section 02, Term 071
Quiz No. 8
ID #:
W. Föِrner
___________________________
Name: ___________________________
Always show your way of calculation, not only the result!
1.
For non-associated liquids the fluidity φ (the reciprocal of the viscosity) obeys to a good
approximation an equation of the Arrhenius form :
 = A e- E /RT
a
where A and Ea are constants (gas constant R = 8.314 J K-1 mol-1; absolute temperature zero: 273oC).
(a)
(b)
2.
For the liquid CCl4 the viscosity at 0.0oC is 1.33 x 10-3 kg m-1 s-1 and the activation
energy Ea for fluidity is 10.9 kJ mol-1. Estimate the viscosity at 40.0oC.
The Arrhenius equation does not apply very well to associated liquids like water, but it
can be used over small temperature ranges. At 20.0oC the viscosity of water is 1.002 x
10-3 kg m-1 s-1 and the activation energy Ea for fluidity is 18.0 kJ mol-1. Estimate the
viscosity at 40.0oC.
(4 points)
Calculate the mean square distance traveled by a molecule of H2 at 20oC and 101.325 Pa in 10 s
(D = 1.005 x 10-4 m2 s-1).
(2 points)
3.
The diffusion coefficient for horse hemoglobin in water is 6.3 x 10-11 m2 s-1 at 20oC. The
viscosity of water at 20oC is 1.002 x 10-3 kg m-1 s-1 and the specific volume of the protein is
0.75 cm3 g-1. Assume the hemoglobin molecule to be spherical and to obey Stokes's law, and
estimate its radius and the molecular weight (π = 3.14159265; Boltzmann's constant: 1.381 x
10-23 J K-1; Avogadro's number: 6.022 x 1023 mol-1).
(4 points)
1. 2 mol of electrons are needed for 1 mol Br2, because the oxidation is
2 Br-  Br2 + 2 e-.
1 mol of Br2 reacts with 1 mol of phenol:
C6H5OH + Br2  C6H4BrOH + HBr.
Thus 5.00 x 10-5 g/(94.114 g mol-1) = 5313 mol of phenol react with 5313 mol of Br2.
Each Br2 needs 2 electrons and thus all together n = 10626 mol of electrons are needed. Since
It = nF, we get
2  5313 mol  96485C mol-1 1.00 h
t=

= 14.2 h
-1
20000C s
3600 s
Note, that 1 C = 1 As,
and thus 20000 A = 20000 C s-1.
2.
The ionic molar conductivities are
 + = t+  
= 0.821  426.16 -1 cm2 mol -1 = 349.9 -1 cm2 mol -1
_ - = t-  
= 0.179  426.16 -1 cm2 mol -1 = 76.3 -1 cm2 mol -1
Since t+ = Fu+ and t- = Fu-:
349.9 -1 cm2 mol-1
= 3.63  10-3 cm2 V -1 s-1
u+ =
-1
96485 C mol
76.3 -1 cm2 mol-1
= 7.91  10-4 cm2 V -1 s-1
u- =
-1
96485 C mol
3.
The ionic strength of a solution is given by
1
I =  ci z i2
2 i
where the sum over i runs over all ion types in the solution.
Because the solubility product of barium sulfate is very small, the concentration of Ba2+ can be
neglected,
while for the term for sulfate only the part from potassium sulfate must be considered.
Because K2SO4, c(K+) = 2 c(K2SO4) = 0.4 M,
c(SO42-) = c(K2SO4) = 0.2 M and z for the sulfate ion is -2, that for the potassium ion is +1:
I=


1
0.4  (+1 )2 + 0.2  (-2 )2 M = 0.6 M
2
DHLL : log 10  +_ = - z Ba2+ | z SO42- | B I
= - 2  2  0.51 M -1/2  0.6 M = - 1.58
 +_ = 10-1.58 = 0.026
1.
(a)
left: oxidation: 1/2 H2(g)  H+(aq) + e-
right: reduction: AuI(s) + e-  Au(s) + I-(aq)
overall:
AuI(s) + 1/2 H2(g)  Au(s) + H+(aq) + I-(aq); z = 1
RT
u
o 2RT
E= E ln ( a H + a I - ) = E ln auHI
F
F
o
(b)
left: oxidation: Ag(s) + Cl-(aq)  AgCl(s) + e-
right: reduction:
2e- + Hg2Cl2(s)  2Hg(l) + 2 Cl-(aq)
overall:
2 Ag(s) + Hg2Cl2(s)  2 AgCl(s) + 2 Hg(l); z=2
E = Eo; no concentration dependence, because only solids and liquids.
(c)
left: oxidation: Ag(s) + Cl-(a1)  AgCl(s) + eright: reduction: AgCl(s) + e-  Ag(s) + Cl-(a2)
Note that activities (a=γ-c) must be used instead of concentrations (c)
overall: Cl-(a1)  Cl-(a2); z=1, Eo = 0 (same electrodes)
RT a 2 RT a1
E=ln =
ln
F
a1 F
a2
2.
For Eo: reduction - oxidation:
o
o
o
E = E Fe3+ / Fe2+ - E I 2 / I - = (0.771 - 0.5355) V = 0.2355 V
 G = - RT ln K = - zFE _ ln K = zFE
RT
o
o
c
o
o
o
c
2  96485 C mol-1  0.2355 V
=
= 18.342
-1
-1
8.314 J K mol (25 + 273)K
Note, that z=2, and that the unit of the equilibrium constant must be M-2 and has to be
added:
18.342
7
-2
-2
=
=
9.24

10 M
K e
M
o
c
3.
(a)
left oxidation: Cd(Hg)(l)  Cd2+(aq) + 2e-
right reduction: Hg22+(aq) + 2e-  2 Hg(l)
overall (z=2):
Cd(Hg)(l) + Hg22+(aq)  Cd2+(aq) + 2 Hg(l)
because of the saturated solution, one can also write
8
8
Cd(Hg)(l) + Hg 2 SO4 (s) + H 2 O(l)  CdSO4  H 2 O(s) + 2Hg(l)
3
3
(b) Note, that 1J = 1 VAs and that for the enthalpy, the Gibbs energy and the entropy must
be in the same unit, either both with J or both with kJ
G = - zFE = - 2  96485 C mol-1  1.01832 V = - 196.5 kJ mol-1
S = zF
E
= 2  96485 C mol -1  (-5.00)  10-5 V K -1
T
= - 9.65 J K -1 mol -1
H = G + TS =  - 196.5 - 9.65  298  10-3  kJ mol-1 = - 199.4 kJ mol-1
1. The second and the third line of the table have [B]=const.
Thus the rates in these two lines behave as being
proportional to [A]α
From the second to the third line, the value of [A] becomes divided by 10, that of the rate by
100 = 102.
Thus α = 2.
From the first to the second line [A] becomes doubled and [B] becomes also doubled.
The rate changes by a factor of 59.2/7.4 = 8. This we get from [A]2[B]β only if β=1.
Because then the factor is 22 x 2 = 8.
rate = k[A ] 2 [B] _ k =
rate
[A ] 2 [B]
Thus k can be obtained from every line of the table, e.g. the first one:
7.40  10-9 M s-1
-3
- 2 -1
k=
=
1.6

10
s
M
2
(1.4  10-2 M ) 2.3  10-2 M
2. From the information we know
rate = k[NO]2[Cl2]
In the beginning (2 dm3 = 2 L) the rate is vo. 1 molecule Cl2 reacts with 2 molecules of NO.
Thus, when half of the initial Cl2, 1 mol, has reacted and 1 mol is still left, then 2 mol of NO
have reacted and 3 mol of NO are left.
Thus, writing the initial state (5 mol NO and 2 mol Cl2) state with index 0 and the other one (3
mol NO and 1 mol Cl2) with index 1:
rate0/rate1 = [NO]02/[NO]12 x [Cl2]0/[Cl2]1
rate1 = rate0 (3/5)2 (1/2)
= 2.4 x 10-3 M s-1 (3/5)2 (1/2) = 4.32 x 10-4 M s-1
In the ratios of concentrations, the volumes cancel out and the ratios of the numbers of
moles can be used.
3.
From the pressure dependence we know

V

ln =
P
k 0 RT
k
The pressure change is 1999 bar = 1.999 x 108 Pa. Further

V
ln k 1 - ln k 0 =  P1
RT

V
ln k 2 - ln k 0 =  P 2
RT

k2  V
ln =
( P 2 - P1 )
k 1 RT
V  1.999  10 N m
 V
ln 2 = 
=
8.314 J K -1 mol -1  300 K 1.247  10-5 m3 mol -1

8
-2

-6
3
-1
3
-1

 V = 8.65  10 m mol = 8.65 cm mol
1. On the way to time t, of product Z the concentration x is produced.
For this a concentration 2x of A is used.
Thus at time t, [A]t = 2a0 - 2x. [B]t = a0 - x, because
1 molecule of B to be used. Thus (2A+B  Z):
each molecule of Z formed, requires
Time
[A]
[B]
[Z]
0
2a0
a0
0
T
2a0 - 2x
a0 - x
x
The rate of formation of Z is
d[Z] dx
2
= = k[A ] [B]
dt
dt
Therefore at time t we have:
dx
= k(2 a0 - 2x )2 ( a0 - x)= 4k( a0 - x )3
dt
n+1
dx
x
n
_
= 4kdt ; since  x dx =
+C ,
3
n+1
( a0 - x )
( a0 - x )n+1
we have ( a0 - x ) dx = C,
n+1
n
and thus ( a0 - x )-3 dx =
1
+C
2
2( a0 - x )
The integration constant is fixed by integrating from 0 to t and from 0 to x:
x
t
1
1
dx
=
4k
dt
_
0 ( a0 - x )3 0 2( a0 - x )2 2 a02 = 4kt
2
2
2 a0 x - x 2
a0 - ( a0 - x )
_ 2
= 8kt _ 2
= 8kt
2
2
a0 ( a0 - x )
a0 ( a0 - x )
2. We have T1 = 493 K
and thus 1/T1 = 2.0284 x 10-3 K-1 and
we have T2 = 503 K
and thus 1/T2 = 1.9881 x 10-3 K-1.
The Arrhenius equation is
k = Ae
- Ea
RT
_ ln k = ln A - E a
RT
Thus from a plot of ln k versus 1/T we get
ln k 2 - lnk 1 = ln
1 1
k2
= ln 2 = E a ( - ) ; k 2 = 2 k 1
R T1 T2
k1
ln 2 = E a (2.0284 - 1.9881)  10-3 K
R
8.314 J K -1 mol-1  ln 2
-1
_ Ea =
=
143.0
kJ
mol
0.0403  10-3 K -1
3.
In transition state theory, we have
k=

k B T -  Go
k B T -1 
o

) 
e RT _  G = - RT ln  k  (
h
h


 3.95  10-4 
_  G = - 8.314 J K mol (25 + 273) K  ln 
12 
 6.211  10 
_

o
-1
-1
= 92.4 kJ mol-1
Note that the units in the argument of the ln have to be stript off (this is M-1).
Further, since no gases are involved (in solution), we have

o
 H = E a - RT
= 120.0 kJ mol -1 - 8.314  10 -3 kJ K -1 mol -1  (25 + 273) K
= 117.5 kJ mol -1

o
-   G o 117500 - 92400

H
=
J K -1 mol -1 = 84.2 J K -1 mol -1
 S =
T
298

o
For the preexponential factor A we could use the Arrhenius equation.
However, this assumes that A does not depend on temperature, which in transition state theory
is not the case.
However, usally this incorrect method could also be used, because the difference to the
transition state theory equation result is not very large.
Transition state theory:
84.2
k B T  S o
12
- 1 -1
A= e
e R = 2.718  6.211  10 M s e 8.314
h
= 4.22  1017 M -1 s-1
Note, that because the reaction is second order, in A, M-1 has to be added to the unit (s-1),
just as it had to be removed from arguments of logarithms in other cases.
or Arrhenius equation:
k= Ae
- Ea
RT
; A= k e
A = 3.95  10 M s e
_
-4
-1
-1
Ea
RT
120000 J mol-1
8.314 J K -1 mol-1  298 K
= 4.28  1017 M -1 s -1
The difference in the two methods is only in the third digit and thus both can be used.
1.
The rate of consumption of N2O5 is
-
d[ N 2 O5 ]
= v N 2O5 = k 1 [ N 2 O5 ] - k -1 [ NO2 ][ NO3 ]
dt
Because it is a rate of consumption, the first term, where N2O5 is used, is positive,
and the second one, where N2O5 is formed, is negative.
The overall equation suggests that the rates of consumption of NO and of N2O5 must be
equal, thus instead of the first way one can also use
vN 2O5 = vNO = k 2 [NO][ NO3 ]
In both ways, we need the concentration of the intermediate NO3.
Steady state equation for this intermediate:
d[ NO3 ]
= 0 = k 1 [ N 2 O5 ] - k -1 [ NO2 ][ NO3 ] - k 2 [NO][ NO3 ]
dt
_ [ NO3 ] =
k 1 [ N 2 O5 ]
k -1 [ NO2 ] + k 2 [NO]
This can be now substituted into the first equation for the rate of consumption of N2O5:
v [ N 2 O5 ] = k 1 [ N 2 O 5 ] -
k 1 k -1 [ N 2 O5 ][ NO2 ]
k -1 [ NO2 ] + k 2 [NO]


k -1 [ NO2 ]
= k 1 [ N 2 O5 ]  1 
 k -1 [ NO2 ] + k 2 [NO] 
_
k k [ N 2 O5 ][NO]
= 1 2
k -1 [ NO2 ] + k 2 [NO]
Note, that the same result comes out, if [NO3] is subtituted into the second equation for the
other way of solution.
2.
The mechanism is
Cl2  2Cl
initiation, k1
Cl + CH4  HCl + CH3
chain-propagation, k2
CH3 + Cl2  CH3Cl + Cl
chain-propagation, k3
2Cl  Cl2
termination, k-1
The rate of formation of HCl, which is the rate of reaction, is
v = v HCl = k 2 [Cl][ CH 4 ]
CH4 is a reactant and thus ok in a final rate equation, but Cl is an intermediate.
The steady state equations for Cl and CH3 are
2
k 1 [ Cl 2 ] - k 2 [Cl][ CH 4 ] + k 3 [ Cl 2 ][ CH 3 ] - k -1 [Cl ] = 0
_
k 2 [Cl][ CH 4 ] - k 3 [ CH 3 ][ Cl 2 ] = 0
The two terms in the second equation appear with opposite signs in the first equation.
Thus adding the two equations yields
1/2
 k1 
2
1/2
[
]
[Cl
=
0
_
[Cl]
=

 [ Cl 2 ]
]
k 1 Cl 2 k -1
 k -1 
Therefore, the final rate is
1/2
 k 
v = k 2 [Cl][ CH 4 ] = k 2  1  [ Cl 2 ] 1/2 [ CH 4 ]
 k -1 
3.
The frequency of the radiation is
2.998  108 m/s
15 -1
ny = =
=
1.1817

10
s
 253.7  10-9 m
c
Thus the energy of 1 photon of this radiation is
E = hny = 6.626  10-34 J s  1.1817  1015 s-1 = 7.830  10-19 J
A 1000 W lamp emits 1000 J s-1 (=1000 VA = 1000 VAs/s), thus the number of photons, N,
per second is
1000 J s-1
21 -1

1.277
=
N=
s
10
-19
7.830  10 J
Thus in 10-6s, we get 1.277 x 1015 photons, and
since each photon excites one mercury atom, this is also the number of excited atoms we get in
a microsecond.
1.
The Langmuir isotherm is
KP
=
1 + KP
Where the fraction of occupied surface area (gas volume adsorbed/gas volume that can be
maximally adsorbed) is
  V/ V o
Vo is the gas volume that can be adsorbed, covering completely a mono-layer. Thus
V
KP
=
V o 1 + KP
Inversion of both sides of the equation yields
V o 1 + KP
=
V
KP
When we multiply the equation by P and divide by Vo we get
P 1 + KP
1
P
=
=
+
V
VoK
VoK Vo
Therefore a plot of P/V versus P is linear, if the Langmuir isotherm is obeyed. Then
slope =
1
; intercept =
Vo
_Vo=
2.
1
VoK
1
1
slope
; K=
=
slope
V o  intercept intercept
Insertion of the data into the BET isotherm yields:
BET :
PPo = 1 + P
V( Po - P) V o K V o
where Po is the saturation vapor pressure,
Vo the volume that can be adsorbed in a mono-layer,
K is the adsorption equilibrium constant.
The data inserted into the left hand side and the values of P inserted into the right hand one
yield
1.11  19.0 Torr 2
PP
o
(1)
=
V( Po - P) 1.48 cm3 (19.0 - 1.11) Torr
= 0.7965
Torr
cm
3
=
1
1.11 Torr
+
VoK
Vo
3.08  19.0 Torr 2
PP
o
(2)
=
V( Po - P) 1.88 cm3 (19.0 - 3.08) Torr
= 1.9552
Torr
cm
(2) - (1) : 1.1587
3
Torr
cm
3
=
=
1
3.08 Torr
+
VoK
Vo
1.97 Torr
Vo
_ V o = 1.70 cm3
(1.70 cm3 )-1
1
(2) : K =
= 4.10
3.08 Torr
Torr
(1.9552 )
1.70 cm3
3.
(a) The general rate equation is
v=
kK[A]
1 + K[A]
low pressure : sparsely covered surface : K[A] » 1
__ v = kK[A]
high pressure : fully covered surface : K[A] « 1
_ v= k
In the sparsely covered case the observed rate constant is kK,
thus the observed activation energy is a sum of the real activation energy and the heat of
adsorption, which is negative.
In the fully covered case the observed rate constant is k, thus the observed activation energy is
the real one, and larger than in the latter case, because the negative heat of adsorbtion is not
present.
(b) Competitive adsorbtion, which occurs only on sites, where no N2 is adsorbed.
When the N2 pressure increases, more and more sites become blocked by N2 and cannot be
used by ammonia to be adsorbed and decomposed.
kK NH 3 [ NH 3 ]
v=
1 + K NH 3 [ NH 3 ] + K N 2 [ N 2 ]
1.
The capillary rise is
2
h=
gr
(a) r = 1 mm = 10-3 m:
2  7.27  10-2 N m-1
-2
h=
=
1.49

m = 1.49 cm
10
998 kg m-3  9.81 m s- 2  10-3 m
-3
kg
10 kg
3 kg
note : 0.998 3 = 0.998
=
0.998

=
998
10
3
3
( 10- 2 m )3
cm
m
m
g
(b) r = 10 -3 cm = 10-5 m:
2  7.27  10-2 N m-1
h=
= 1.49 m
-3
-2
-5
998 kg m  9.81 m s  10 m
2.
The heights in the two arms and their difference would be
h1 =
2
2
; h2 =
g r 1
g r 2
h = h1 - h2 =
2  1 1 
 - 
g  r 1 r 2 
2 (2000 - 1000) m-1
-1
0.022 m =
=
0.255
m s2  
kg
-3
-2
800 kg m 9.81 m s
 = 0.086 kg s-2 = 0.086 kg m s- 2 m-1 = 0.086 N m-1
3.
To get the volume rate of flow, we have to multiply the linear speed of flow with the area,
A, of the cross section of the aorta:
A =  r 2 = 3.1416  (9  10-3 )2 m2 = 2.54  10-4 m2
dV
= 2.54  10 -4 m2  0.33 m s -1 = 8.40  10-5 m3 s -1
dt
The pressure drop is then given by the Pouisseuille equation:
8l dV 8  4  10-3 N s m-2  0.5 m  8.40  10-5 m3 s-1
P =
=
4
 R4 dt
3.1416   9  10-3  m4
= 65.2 N m- 2 = 65.2 Pa = 0.49 Torr
1.
(a)
At 0.0oC we have for CCl4
 = Ae
A=
- Ea
RT
Ea
RT
_ A=  e =
1
1.33  10 kg m s
-3
-1
-1
e
1

10900 J mol-1
8.314 J K -1 mol-1  273 K
= 91.6  103 kg-1 m s
Then at 40.0oC:
e
Ea
RT
10900 J mol-1
1 Ea 1 kg m-1 s-1
 = = e RT =
8.314 J K -1 mol-1  (40.0+ 273) K
3 e
 A
91.6  10
_
1
= 7.20  10-4 kg m-1 s-1
(b)
At 20.0oC we have for water
 = Ae
A=
- Ea
RT
Ea
RT
_ A=  e =
1
1.002  10 kg m s
-3
-1
-1
e
1

18000 J mol-1
8.314 J K -1 mol-1  (20+ 273) K
= 16.2  105 kg-1 m s
Then at 40.0oC:
e
Ea
RT
18000 J mol-1
1 Ea 1 kg m-1 s-1
 = = e RT =
8.314 J K -1 mol-1  (40.0+ 273) K
5 e
 A
16.2  10
_
1
= 6.23  10-4 kg m-1 s-1
2.
The random walk equation yields
2
-4
-1
-3
2
x = 2tD = 2  10 s  1.005  10 m s = 2.01  10 m
distance = ( x2 )1/2 = 2.01  10-3 m = 0.045 m = 4.5 cm
3. Stokes's law:
D=
kBT
6r
J
 293 K
kBT
K
r=
=
2
kg
6D
-11 m
6  3.14159  6.3  10
 1.002  10-3
s
ms
1.381  10- 23
= 3.402  10-9 m = 3.4 nm
molecular mass Mmolecular from specific volume Vspecific and molar weight MM:
4 3 4
3
-9
- 25
3
V molecule =  r =  3.14159  (3.402  10 m ) = 1.65  10 m
3
3
-25 3
V molecule 1.65  10 m
-19
=
=
2.20

g
10
M molecule=
-6
3 -1
V specific 0.75  10 m g
MM = M molecule  N A = 2.20  10-19 g  6.022  1023 mol-1 = 132000g mol-1
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