QWAQWA CAMPUS / QWAQWA KAMPUS
CEM 214
UNIVERSITEIT VAN DIE VRYSTAAT
UNIVERSITY OF THE FREE STATE
SEMESTER TEST 2
ASSESSORS:
Sat, 14 April 2012
Ms CE Clark & Prof AS Luyt
TIME : 1 Hour
MARKS: 50
ANSWER ALL QUESTIONS AND SHOW ALL CALCULATIONS WITH AN INK
PEN -- NO PENCILS OR "TIPPEX" ALLOWED / PAPER CONSISTS OF 3 PAGES
AND 7 QUESTIONS. RECORD ALL ANSWERS UP TO TWO (2) SIGNIFICANT
FIGURES.
QUANTUM MECHANICS / 15 marks:
1.
Calculate the energy that will be absorbed when a  electron in CH2=CH-CH=CHCH=CH-CH3 is promoted from the highest occupied level (HOMO) to the lowest
unoccupied level (LUMO). The  electrons in this conjugated chain are delocalized
and, to a rough approximation, they can be considered to be particles in a onedimensional box of length 5.8 Å.
[7]
(h = 6.63 x 10-34J s, me = 9.1 x 10-31 kg)
OR
Consider a particle with quantum number n moving in a one-dimensional box of length
l. Determine the probability of finding the particle in the left quarter of the box if ψn(x)
= A sin (nπx/l).
[7]
1
2.
The energy of a particle in a cubic box with side length aaa is given by:
E nx ,ny
2
2
2
h 2  n x  n y  n z 


8m 
a2

Draw an energy level diagram for nx, ny and nz = 1 or 2. Show the energy values in
h2
terms of
.
8ma 2
[8]
THERMODYNAMICS / 20 MARKS
3. Gastric juice in humans has an acid concentration of about 1.00 x 10-1 M (pH  1) and it
is formed from other body fluids, such as blood, which have an acid concentration of
about 4.00 x 10-8 M (pH  7.4). On the average, about 3.00 dm3 of gastric juice are
produced per day. Calculate the minimum work required to produce this quantity at 37
C, assuming the behaviour to be ideal.
[5]
4. Calculate the entropy change when 1 mol of ice is heated from 250 K to 300 K. Take the
heat capacities (CP,m) of water and ice to be constant at 75.3 and 37.7 J.K-1.mol-1,
respectively, and the latent heat of fusion of ice as 6.02 kJ.mol-1.
[8]
5. Consider the closed system below containing 0.4 mol He at 25 °C. Assume that He acts
like an ideal gas.
a)
Calculate the work done if the plunger/piston moves
reversibly and isothermically from x to y.
[3]
b) Calculate P (in Pa) during the process in (a).
[4]
2
ELECTROLYTIC SOLUTIONS / 15 MARKS
6. The equivalent conductivity of a weak electrolyte was determined experimentally at
different concentrations and the values are given in the table below:
105C / mol dm-3
104 / S m2 equiv-1
18.180
111.111
41.796
77.519
89.853
54.645
From these results, a graph of C vs. 1/ was obtained and the following values were
determined: Slope = 3.1 x 10-8 and y-intercept = -7 x 10-7.
Determine:
a) °
b) Ka
c)  (at 1.818 x 10-4 mol.dm-3)
7.
[8]
Calculate the average ionic activity coefficient, f, of a 0.002 mol dm-3 solution of
Zr2(Cr2O7)5. Note: Assume that zirconium has a valency of +5 in this compound.
[7]
3
MEMORANDUM:
1.
Energy absorbed: CH2=CH-CH=CH-CH=CH-CH3
length of box = 5.8 Å given
6 π-electrons occupy 3 molecular orbitals (MOs)
Thus HOMO is E3 
Thus LUMO is E4 
En 
h 2n 2
8ma2
 h2n2
E  E 4  E3  
 8ma 2


2.

 h2n2
 
2


 n  4  8ma
6.63x10 Js  4  3 
89.1x10 kg 5.8 x10 m 
34
31
2
2
2
10
2



 n 3

 1.25 x10 18 J
The possible energies are E111, E121, E211, E112, E122, E221, E212, E222
E111
h 2  1  1  1  h 2 



  3
8m  a 2   8ma 2 

 h2 

E 211  6
2 
 8ma 
 h2 

E112  6
2 
 8ma 
E121 
 h2 
h 2  1  2 2  1





6
2 

8m  a 2

 8ma 
E122 
 h2 
 h2 
h2  1 2 2  2 2   h2 









9
E

9
E

9
2 
  8ma 2  212
 8ma 2  221
8m 
a2
 



 8ma 
E 222 
 h2 
h2  22  22  22 





12
2 

8m 
a2
8ma



4
18
12h2/8ma2
16
14
Energy levels
12
10
9h2/8ma2
8
6h2/8ma2
6
4
3 h2/8ma2
2
0
3.

C1 = C(blood) = 4.00 x 10-8 M
C2 = C(gastric juice) = 1.00 x 10-1 M
V = 3.00 dm3
T = 37 °C = 310.15 K
n(gastric juice) = C(gastric juice) x V(gastric juice)
= (1.00 x 10-1 mol.dm-3) x ( 3.00 dm3)

= 0.300 mol 
wrev
= nRT ln(C2/C1)
= nRT ln[C(gastric juice)/C(blood)]

 wrev = (0.300 mol)(8.3145 J.K-1.mol-1)(310.15 K) ln[(1.00 x 10-1 mol.dm-3)/( 4.00 x
10-8 mol.dm-3)]

= 11396.86 J = 11.40 kJ

5
4.
The entropy change when 1 mole of ice is heated from 250 K to 273.15 K:
S1 = CP.m ln(273.15 K/250 K) 
= (37.7 J.K-1.mol-1) x (0.0886)
= 3.34 J.K-1.mol-1 
For the melting at 273.15 K
S2 = dq/T = (6020 J.mol-1)/(273.15 K) = 22.04 J.K-1.mol-1

For the heating from 273.15 K to 300 K
S1 = CP.m ln(300 K/273.15 K) 
= (75.3 J.K-1.mol-1) x (0.0938)
= 7.06 J.K-1.mol-1 
The entropy change is
S = (3.34 + 22.04 + 7.06) J.K-1.mol-1 = 32.44 J.K-1.mol-1
5.

n = 0.4 mol
Area (A) = 0.5 m2
x = 60 cm = 0.60 m
y = 20 cm = 0.20 m
V1 = area x distance = 0.5 m2 x 0.6 m = 0.3 m3
V2 = area x distance = 0.5 m2 x 0.2 m = 0.1 m3
T = 298 K
Pext = 1 atm = 101325 Pa
a) w = -nRT ln(V2/V1) 
= -[(0.4 mol)(8.3145 J.K-1.mol-1)(298 K) x ln(0.3/0.1)]
= -1088.821 J


6
b) P = nRT[1/V2 – 1/V1]

= (0.4 mol)(8.3145 J.K-1.mol-1)(298 K)[(1/0.3m3) – (1/0.1m3)]

= -6607.256 Pa
6.
° =
a)
m
3.1x10 8

S m 2 equiv -1 
 c  (7 x10 7 )

= 442.857 x 10-4 S.m2.equiv-1

Ka = slope/(o)2
b)

= [3.1 x 10-8/ (442.857 x 10-4)2] mol.dm-3 
= 1.58  10-5 mol.dm-3
 = ° (at 1.818 x 10-4 mol.dm-3) 
c)
111.111x10 4
=
442.857 x10 4


= 0.251
7.

Zr2(Cr2O7)5  2Zr5+ + 5Cr2O72-




1

Zr 5 z Zr
5
2

1
2
2C  5
2

2



 Cr2 O72 z Cr
O2
 5C  2 
2

2
7

2


= ½ [50C – 20C] = 35C
= 35 x 0.002 M = 0.07 M 
 log 10 f    0.509 z  z  

 0.509 5 x(2) 0.07  1.3467
 f   101.3467  2.222


7