Mid year Exam December, 2007
Part A. Answer all questions (5 marks each).
A1.
State the Pauli exclusion principle.
No two electrons in a single atom may have the same four quantum numbers.
A2.
AUFBAU would predict that the ground state electronic configuration of gold (Au) is
[Xe]6s24f145d9. Why is it actually [Xe]6s14f145d10?
In this configuration, the atom has a half-filled 6s orbital, plus a filled 5d subshell. These are
energetically more stable than a filled 6s orbital and a 9/10ths filled 5d subshell.
A3.
Referring to the attached activity series, in a reaction involving iron (Fe) and copper (Cu),
which will get oxidized in a spontaneous REDOX reaction? How do you know?
The Fe will get oxidized. It is higher up in the activity series and is therefore more easily
oxidized.
A4.
Why do group 7A elements have high electronegativities?
The effective nuclear charge of these atoms are very high because of the large number of
protons are relatively small radius.
A5.
In a triple bond between two carbon atoms, why can there be only one sigma () bond?
This sigma bond is made by the end to end overlap of two sp hybrid orbitals. The othet two
bonds are made by the sideways overlaps of pairs of p-orbitals. Since these p-orbitals are
perpendicular to one another and to the sp hybrids, they can not overlap end to end.
A6.
For the endothermic reaction C(s) + H2O(g) Ý CO(g) + H2(g) initially at equilibrium, indicate how
the equilibrium will shift (left, right or no change) for each of the indicated changes:
Add some C(s)
Add some H2O(g)
Add an inert gas such as helium
Decrease the temperature
Decrease the pressure
A7.
No change
Right
No change
Left
right
The reaction N2(g) + 3 H2(g) Ý 2 NH3(g) has negative values of both So and Ho. Will the
reaction be spontaneous at high or low temperature? How do you know?
The reaction will be spontaneous below some temperature. G=H-TS, and this becomes
more negative (i.e. towards spontaneity) as the temperature is decreased since S<0.
CHEM 1000 A and V
A8.
Midyear Exam
Page 2 of 8
What is the hybridization of the oxygen atom in a water molecule?
sp3. (Remember that the O atom has two lone pairs!)
A9.
Consider two containers of the same volume at the same temperature. One contains one mole
of H2O(g), the other one mole of CO2(g). Which has the higher pressure and why?
CO2(g) has the higher pressure. It is non-polar, therefore does not interact very strongly with
itself.
A10. Nitric oxide (NO(g)) is a catalyst in the production of smog.
a. Show a balanced reaction for the production of NO (g) in a car engine.
N2(g) + 2 O2(g) Ý 2 NO(g)
b. Why are very small concentrations of NO(g) in the urban atmosphere significant?
This gas is a catalyst – very little of it is needed to cause the formation of lots of smog.
A11. In the production if nickel metal, the nickel sulfide ore (NiS(s)) is first roasted in air to produce
nickel oxide (NiO(s)).
a. Why is this done?
This is done to form NiO(s) which can be easily reduced to Ni(s). (NiS(s)) is not easily reduced.)
b. What is the major primary pollutant produced by the roasting step?
SO2(g).
A12. Why are the standard enthalpies of formation of the elements in their standard states equal to
zero?
Enthalpies of formation are the enthalpies for the formation of a compound from the elements,
with everything in their standard states. Since elements could only be formed from the same
elements, the net enthalpy change would be zero in each case.
Part B. Answer question B1 and any five of questions B2-B8. If you answer more than five,
the best five will be used to calculate your total mark (20 marks each)
B1. [10 marks] (a) Liquid oxygen, used in large rockets, is prepared by cooling air to a very low
temperature. How many litres of air must be processed to produce 150 L of liquid oxygen?
Assume the air is initially at 25oC and 750 Torr, and contains 19% oxygen by volume, and that the
liquid oxygen has a density of 1.14 g/mL.
150 L x 1140 g/L = 171000 g liquid oxygen
CHEM 1000 A and V
Midyear Exam
Page 3 of 8
171000 g
 5344 mol O2
32.0 g mol1
Since the air is only 19% oxygen, the number of moles of air that must be processed is
5344 moles O 2
 28125 mol air
0.19 moles O 2 / mole air
V
nRT 28125 mol(0.082L atm K 1mol 1)(25  273)K

 6.96  105 L
P
(750 / 760)atm
(b) [6 marks] Use the van der Waals equation to calculate the pressure exerted by 80 moles of
H2S(g) in a 100 L container at 100oC. For H2S(g), a = 4.48 atm L2 mol-2 and b = 0.0434 L mol-1.
nRT
n
p
 a 
V  nb
V
2
 80 mol 
80 mol(0.082 L atmK 1mol1 )(100  273)K

 4.48 atm L2mol2 

1
100 L  80 mol(0.0434 L mol )
 100 L 
2
 25.3 atm  2.9 atm
 22.4 atm
(c) [4 marks] Under the same conditions, would you expect CO2(g) to exert a higher or lower
pressure than H2S(g)? Why?
Higher, since CO2(g) is linear, thus non-polar.
B2. [20 marks] A solution is made by dissolving 0.100 moles of benzoic acid (C6H5CO2H) in 2 L of
water. The acid reacts with the water according to:
C6H5CO2H(aq) + H2O(l) Ý C6H5CO2‾(aq) + H3O+(aq), for which Kc = 6.4 x 10-5
Calculate the concentrations of all species (except water) at equilibrium (in mol/L).
Initially, [C6H5CO2H ] = 0.100 mol / 2 L = 0.050 M
Initial
Change
Equilibrium
C6H5CO2H(aq)
0.050
-x
0.050-x
C6H5CO2‾(aq)
0
+x
x
H3O+(aq)
0
+x
x
CHEM 1000 A and V
Midyear Exam
Page 4 of 8
Thus, at equilibrium,
[C6H5CO2 (aq) ][H3O (aq) ]
[C6H5CO2H(aq) ]

(x)(x)
 K c  6.4  10 5
0.050  x
x 2  6.4  10 5 (0.050  x)
x 2  6.4  10 5 x  3.2  10 6  0
b  b2  4ac 6.4  10 5  (6.4  10 5 )2  4(1)( 3.2  10 6 )
x

2a
2(1)
6.4  10 5  3.58  10 3
2
 0.00176 or  0.00182

Only the first root gives positive concentrations, thus x = 0.00176
Thus, [C6H5CO2H(aq)] = 0.05 – 0.00172 = 0.0483
[C6H5CO2-(aq)] = x = 0.00172
[H3O+(aq)] = x = 0.00172
Check:
x2
0.001722

 6.2  105
0.050  x 0.050  0.00172
(some round off error)
B3. Consider the reaction and data:
CH4(g) + H2O(g)  CO(g) + 3 H2(g)
Hfo (kJ mol-1)
So (J K-1 mol-1)
CH4(g)
-74.8
186.3
H2O(g)
-241.8
188.8
CO(g)
-110.5
197.7
(a) [3 marks] Calculate Ho for this reaction (kJ mol-1).
Ho = Hfo(CO) + 3Hfo(H2) - Hfo(CH4) - Hfo(H2O)
= -110.5 + 3(0.0) – (-74.8) – (-241.8)
= +206.1 kJ mol-1
(b) [3 marks] Calculate So for this reaction (J K-1 mol-1).
So = So(CO) + 3So(H2) - So(CH4) - So(H2O)
= 197.7 + 3(130.7) – (186.3) – (188.8)
= +214.7 J K-1 mol-1
(c) [3 marks] Calculate Go for this reaction (kJ mol-1) at 25oC
Go=Ho-TSo
= +206100 J mol-1 – (25+273)K (214.7 J K-1 mol-1)
H2(g)
0.0
130.7
CHEM 1000 A and V
Midyear Exam
Page 5 of 8
= +142000 J mol-1
= +142.0 kJ mol-1
(d) [3 marks] Calculate the value of Kp at 25oC.
Kp  e
e
 Go 


 RT 


 (142000 J mol1


1
1
8.314
JK
mol
(25

273)K


 1.28  10 25
(e) [4 marks] If pCH4  pH2O  pH2  1.00 atm , calculate pCO at 25oC
pCOp3H2
pCH4 pH2O
pCO 
 Kp
K ppCH4 pH2O
p3H2

1.28  1025 (1.0)(1.0)
 1.28  10 28 atm
(1.0)3
(f) [4 marks] Calculate the temperature (in oC) at which the reaction becomes spontaneous.
At this temperature,
G  0  Ho  TSo
thus, Ho  TSo
T
Ho 206100 J mol1

 960K  687o C
So 214.7 JK 1mol1
B4. (a) [6 marks] Assign formal charges to the atoms in the two structures below and predict
which is the more likely structure:
H
H
.. ..
..
C C=N
C=C=N
..
..
H
H
Structure A
Structure A
Valence
Electrons Assigned
Formal Charge
Structure B
H
1
1
0
H
1
1
0
C C N
4 4 5
3 5 6
1 -1 -1
H H C C N
Structure B
CHEM 1000 A and V
Valence
1
Electrons Assigned 1
Formal Charge
0
Midyear Exam
1
1
0
4
4
0
Page 6 of 8
4 5
4 6
0 -1
Structure B is more likely since the formal charges are closer to zero.
(b) [8 marks] Use VSEPR theory to predict the shapes of the following two species:
XeOF4
Electrons = 8 + 6 + (4 x 7) = 42. Arranging the terminal atoms around the central atom using
single bonds and completing the octets on the terminal atoms uses 40 electrons. The remaining
pair is a lone pair on the central Xe atom. Thus the configuration is AX5E, which is square
pyramidal. (Note that we should move a lone pair from the O atom to make a double bond with
the Xe, but that this does not change the predicted shape of the molecule)
BrO3‾
Electrons = 7 + (3 x 6) + 1 = 26. Arranging the terminal atoms around the central atom using
single bonds and completing the octets on the terminal atoms uses 24 electrons. The remaining pair
is a lone pair on the central Br atom. Thus the configuration is AX3E, which is trigonal pyramidal.
(c) [6 marks] What type of hybrid orbitals is used by the central atom in each of the following
two species?
NCl3
Electrons = 5 + (3 x 7) = 26. There are therefore three bonding pairs plus one lone pair
on the central atom for a total of four electron groups. Must be sp 3 hybridization.
SF5‾
Electrons = 6 + (5 x 7) + 1 = 42. There are therefore five bonding pairs plus one lone pair on
the central atom for a total of six electron groups. Must be sp 3d2 hybridization.
B5. [20 marks] Calculate the enthalpy change for the reaction:
2 CH4(g)  C2H4(g) + 2 H2(g)
Using the following information:
2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(l)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
C2H4(g) + H2(g)  C2H6(g)
Hfo (H2O(l)) = -285.8 kJ mol-1
Ho = -3119.4 kJ mol-1 (rxn 1)
Ho = -890.3 kJ mol-1 (rxn 2)
Ho = -137.0 kJ mol-1 (rxn 3)
The last piece of data means
H2(g) + ½ O2(g)  H2O(l)
Ho = -285.8 kJ mol-1 (rxn 4)
CHEM 1000 A and V
Midyear Exam
Page 7 of 8
We can add the reactions as follows:
4(reaction 2):
4 CH4(g) + 8 O2(g)  4 CO2(g) + 8 H2O(l)
Ho = 4(-890.3 kJ mol-1) = -3561.2
-2(reaction 3):
2 C2H6(g)  2 C2H4(g) + 2 H2(g)
Ho = -2(-137.0 kJ mol-1) = +274.0
-1(reaction 1):
4 CO2(g) + 6 H2O(l)  2 C2H6(g) + 7 O2(g)
Ho = -(-3119.4 kJ mol-1) = +3119.4
-2(reaction 4):
2 H2O(l)  2 H2(g) + O2(g)
Ho = -2(-285.8 kJ mol-1) = +571.6
Adding the above gives twice the desired reaction:
2 CH4(g)  2 C2H4(g) + 4 H2(g)
Ho = sum of above = 403.8 kJ mol-1
So, for the desired reaction, CH4(g)  C2H4(g) + 2H2(g)
Ho = 403.8 kJ mol-1 / 2 = 201.9 kJ mol-1
B6. [10 marks] (a) Calculate the bond dissociation energy of O2 molecules (in kJ mol-1), given that
the maximum wavelength of photon capable of breaking the bond is 240 nm.
hc 6.63  10 34 Js(3.00  108 m s 1 )

 8.29  10 19 J (per photon)
9

240  10 m
23
1
6.02  10 mol  498,907 J mol1
E
 499 kJ mol1
(b) [10 marks] Calculate the wavelength (in nm) of the lowest energy transition in the Paschen
series of the hydrogen atom. Lines in the Paschen series terminate at m = 3.
1
1
1
 1
1
 R  2  2   0.01097 nm1  2  2   5.33  10 4 nm1

4 
m n 
3
1

 1875 nm
5.33  104 nm1
B7. Balance the following REDOX reaction in a basic solution:
AlH4‾(aq) + H2CO(aq)  Al+3(aq) + CH3OH(aq)
Oxidation: [5 marks]
AlH4‾(aq)  Al+3(aq)
CHEM 1000 A and V
Midyear Exam
Page 8 of 8
AlH4‾(aq)  Al+3(aq) + 4H+(aq) (note that it is the H being oxidized!)
AlH4‾(aq)  Al+3(aq) + 4H+(aq) + 8 eReduction : [5 marks]
H2CO(aq)  + CH3OH(aq)
H2CO(aq) + 2 H+(aq)  + CH3OH(aq)
H2CO(aq) + 2 H+(aq) + 2 e-  CH3OH(aq)
Add: [5 marks]
AlH4‾(aq)  Al+3(aq) + 4H+(aq) + 8 e4 H2CO(aq) + 8 H+(aq) + 8 e-  4 CH3OH(aq)
___________________________________
AlH4‾(aq) + 4 H2CO(aq) + 4 H+(aq)  Al+3(aq) + 4 CH3OH(aq)
Convert to basic solution: [5 marks]
AlH4‾(aq) + 4 H2CO(aq) + 4 H+(aq) + 4 OH-(aq)  Al+3(aq) + 4 CH3OH(aq) + 4 OH-(aq)
AlH4‾(aq) + 4 H2CO(aq) + 4 H2O(l)  Al+3(aq) + 4 CH3OH(aq) + 4 OH-(aq)
B8. (a) [6 marks] Write the four important oxygen-only reactions that take place in the
stratosphere.
O2 + h  O + O
O + O 2  O3
O3 + h  O2 + O
O + O 3  2 O2
(b) [6 marks] Why is there a maximum in the ozone concentration at a particular altitude?
Above this altitude, the O2 concentration is low, so the first reaction in the above scheme is slow.
Below this altitude, the light intensity is low, so again the first reaction is slow. Thus there must be
a local maximum in the O3 concentration.
(c) [6 marks] Beginning with NO(g), show the three reactions that result in the production of
ozone in urban smog.
NO + ½ O2  NO2
NO2 + h  NO + O
O + O 2  O3
(d) [2 marks] Why are single oxygen atoms so reactive?
They are reactive because they have two unpaired electrons (i.e. they are radicals).