Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 1 of 14
Part A. Answer all eight questions (5 marks each).
1. Why does the pH of a buffered solution not change much if small amounts of acid
or base are added to it?
The large quantity of the conjugate base (or acid) re-equilibrates with the weak acid (or base),
restoring the pH to close to its original value.
2. Define the term “rate limiting step.”
The rate limiting step is the slowest step in the mechanism of a multistep reaction.
3. A solution is made by dissolving 2 moles of KCl(s) in 1 mole H2O(l). What is the mole fraction of
K+(aq) in this solution?
2 moles of KCl gives 2 moles of K+ ions and 2 moles of Cl- ions. The mole fraction of K+ is therefore
2/(2+2+1) = 0.40.
4. Chemical reactions may release or absorb energy. Why do we not observe a change in mass,
according to ΔE = Δmc2?
There is a change in mass, but it is too small to be observed directly.
5. Name the four steps in a radical polymerization process.
Generation of a radical initiator
Initiation
Propagation
Termination
6. Name two reasons that the reaction C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) results in an increase of
the entropy of the Universe.
This reaction has a positive ΔS, which tends to increase the entropy of the Universe. It also has a
negative ΔH, which causes the surroundings to heat up, increasing the entropy of the surroundings,
and increasing the entropy of the Universe as well.
7. Name two reasons, related to the lead acid battery, why it is difficult to start your car on a cold day.
According to the Nernst equation, lowering the temperature can lower the battery voltage. Also, the
electrolye becomes very viscous when cold, decreasing the flow of ions through it, decreasing the
current available from the battery.
8. Why is the Henry’s law constant, KH, for SO2 so much larger than that for CO2?
SO2 is bent (draw the Lewis structure to prove this). It is therefore polar. CO2 is linear and therefore
non-polar. Thus, there will be strong forces of attraction between SO 2 and water, hence a high
solubility and a high KH value.
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 2 of 14
Part B. Answer all three questions B1, B2 and B3 (20 marks each).
B1(a) Balance the following nuclear reactions by filling in the blanks:
(i) 2 0 n 
1
(ii)
(iii)
Pu 
239
94
Na 
22
11
22
10
241
95
Ne 
Pa   
226
91
222
89
Am 
0
1
0
1
e  (  )
e  (  )
Ac
(iv) 2 He  2 He  2 He  2 1H
3
(v)
238
92
3
U
4
1
Th    
234
90
B1(b). When an electron and a positron annihilate (destroy) each other, they produce two photons in
the gamma ray part of the spectrum: 01 e  01 e  2 . Assuming both gamma rays have the same
energy as each other, calculate the wavelength of the radiation produced (in nm).
0 
1
e  01 e   2
m  m  (me  me )
 0  2(9.11 10 31 kg)
 1.822  10 30 kg
E  mc 2
 1.822  10 30 kg(3.00  108 m s 1 )2
 1.63  10 13 J (per two photons)
 0.815  10 13 J (per photon)
hc
E

hc 6.63  10 34 Js(3.00  108 m s 1 )
thus,  

 2.42  10 12 m  0.00242 nm
13
E
0.815  10 J
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 3 of 14
B2. (a) Name the following compounds:
H2N
3-isopropylpentane or 2-methyl-3-ethylpentane
NH2
para-diaminobenzene or 1,4-diaminobenzene
Br
O
O
OH
pentanal
3-bromobutanoic acid
OH
2-methylpropan-1-o1 or 2-methyl-1-propanol
B2. (b) Draw the structure or give the correct name for each missing reactant or product in the
following reactions:
+
Pd
H2
or pentane
O
O
Cl
OH
H N
CH3
CH3
+
Cl
HO
+
O
or propyl-4-chlorobutanoate
O
N
CH3
OH
O
or propanoic acid
O
OH
Oxidation
or 2-hexanol or hexan-2-ol
+
CH3
H2O
H2SO4
OH
or 3-hexanol or hexan-3-ol
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 4 of 14
OH
or 2-hexanol or hexan-2-ol
B3. For the reaction 2I−(aq) + S2O82−(aq)  I2(aq) + 2SO42−(aq), the following data were collected:
[I−], mol L-1 [S2O82−], mol L-1
Initial rate, mol L-1 min-1
0.125
0.150
0.0045
0.375
0.150
0.0135
0.125
0.050
0.0015
(a) Determine the rate law for this reaction.
Comparing the first and second experiments, we see that if [I -] is tripled, the rate triples.
Therefore the reaction is first order in [I-].
Comparing the first and third experiments, we see that if [S 2O8-2] is tripled, the rate also triples.
Therefore the reaction is also first order in [I-].
The rate law is therefore rate = k[I-][S2O8-2]
(b) Calculate the rate constant (with the correct units) for this reaction.
Using the first experiment (we could use any of the three), we have:
rate
0.0045 mol L1 min1
k 

 0.24 L mol1min1
2
1
1
[I ][S2O8 ] 0.125 mol L (0.150 mol L )
(c) The decomposition reaction of NOBr is second order in [NOBr], with a rate constant of
25 L mol-1 min-1. If the initial concentration of NOBr is 0.025 mol L-1, find
(i) the time at which the concentration of NOBr will be 0.010 mol L-1
1
1

 kt
[NOBr] [NOBr]0
1
1

[NOBr] [NOBr]0
thus, t 
k
1
1

1
0.010 mol L
0.025 mol L1

 2.4 min
25 L mol1 min1
(ii) the concentration of NOBr after 125 min of reaction.
Final Examination April 2009
CHEM 1000 A and V
R. Burk
1
1

 kt
[NOBr] [NOBr]0

1
 25 L mol min1(125 min)
0.025 mol L1
 3165 L mol1
[NOBr] 
1
 0.000316 mol L1
1
3165 L mol
Page 5 of 14
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 6 of 14
Part C. Answer any four of the seven questions (C1 – C7). If you answer more than four, the
best four will be used to calculate your mark (20 marks each).
C1 (a). A chemical engineer wants to determine if it is possible to make ethanol (C 2H5OH) by
reacting ethylene with water according to C2H4(g) + H2O(l) Ý C2H5OH(l). Use the following data to
determine if this reaction is spontaneous under standard conditions:
C2H4(g)
H2O(l)
C2H5OH(l)
ΔHfo (kJ mol-1)
52.0
-286.0
-278.0
So (J K-1 mol-1)
219.0
70.0
161.0
H0 = Hf0(C2H5OH(l)) – (Hf0(C2H4(g)) + Hf0(H2O(l)))
= – 278.0 – (52.0 + (-286.0)) kJ mol-1
= – 44.0 kJ mol-1
S0 = S0(C2H5OH(l)) – (S0(C2H4(g)) + S0(H2O(l)))
= 161.0 – (219.0 + (70.0)) J K-1 mol-1
= -128 J K-1 mol-1
G0 = H0 - TS0
= – 44,000 J mol-1 – (298 K)(–128 J K-1 mol-1)
= – 5856 J mol-1
Since G0 < 0, this reaction is spontaneous under standard conditions.
C1 (b). Determine the equilibrium constant Kc for the same reaction at 25oC.
 G 0 
K c  exp 

 RT 


5,856 J mol 1
 exp 

1
1
 8.314 J K mol (298 K) 
 exp(2.36)
 10.6
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 7 of 14
C2 (a). Calculate the pH of 0.5 M sodium acetate, CH3COONa. Ka of acetic acid, CH3COOH,
is 1.8 x 10-5.
Sodium acetate will dissociate in water:
CH3COONa(s)  CH3COO-(aq) + Na+(aq)
And the acetate ion will hydrolzye according to:
CH3COO-(aq) + H2O(l)  CH3COOH(aq) + OH-(aq)
Since the acetate ion is the conjugate base of acetic acid, then the equilibrium constant for this
hydrolysis reaction is Kb:
Kb 
Kw
K a(CH3COOH)
1.0 1014

 5.56 1010
5
1.8 10
Thus:
CH3COO-(aq)
0.5
-x
0.5-x
Initial, M
Change, M
Equilibrium, M
CH3COOH(aq)
0
+x
x
OH-(aq)
0
+x
x
At equilibrium:
[CH3COOH (aq) ][OH  (aq) ]
[CH3COO (aq) ]
 Kb
x(x)
 5.56 1010
0.5  x
Here, Kb is very small, so 0.5 – x  0.5.
Thus, x2 = 0.5(5.56 x 10-10)
x2 = 2.78 x 10-10
x = 1.67 x 10-5
pOH = -log10[OH-(aq)] = -log10(x) = -log10(1.67 x 10-5) = 4.78
pH = 14-pOH = 9.22
C2 (b). Calculate the pH of 2.0 x 10-7 M HCl(aq). This acid hydrolyzes fully in water. Kw = 1.0 x 10-14.
The HCl hydrolyzes fully:
HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
Since this is such a low concentration of H3O+(aq) from the HCl, we must take account of the
autohydrolysis of water:
Final Examination April 2009
CHEM 1000 A and V
R. Burk
2 H2O(l) Ý H3O+(aq) + OH-(aq) Kw = 1.0 x 10-14
Thus,


[H3O(aq)
][OH(aq)
]  1.0  10 14
or

[OH(aq)
]
1.0  10 14

[H3O(aq)
]
But the total negative charge must equal the total positive charge:



[H3O(aq)
]  [OH(aq)
]  [Cl(aq)
]

[H3O(aq)
]
1.0  1014
 2.0  10 7

[H3O(aq)
]


[H3O(aq)
]2  2.0  10 7 [H3O(aq)
]  1.0  10 14  0
[H3O

(aq)
2.0  107  (2.0  10 7 )2  4(1)( 1.0  10 14 )
]
2(1)
2.0  107  2.83  107

2
7
 2.41 10 or  4.15  10 8

pH   log10 [H3O(aq)
]   log10 (2.41 10 7 )  6.62
Page 8 of 14
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 9 of 14
C3 (a). What volume of fluorine gas, measured at STP, is produced by electrolysis of molten KF
using a current of 10.0 A for 2.00 hours?
KF(l)  K+ + F2 F-  F2 + 2 eit = nF
n
it 10.0 C s1(2.00  3600)s

F
96,487 C mol1
 0.746 mol e 
0.746 mol e  
1mol F2(g)
2 mol e 
0.373 mol F2(g) 
 0.373 mol F2(g)
22.4 L
 8.36 mol F2(g) at STP
mol
C3 (b). What mass of potassium metal is produced in this same reaction?
K+ + e-  K(s)
0.746 mol e-  0.746 mol K(s)
0.746 mol K(s) x 39.1 g/mol = 29.2 g K(s) produced.
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 10 of 14
C4 (a). The enthalpy of vaporization of mercury, Hg(l) is 59.1 kJ mol-1. Its normal boiling point is
357oC. Calculate the vapor pressure of mercury (in Torr) at 25 oC.
Here, T1 = 357oC = 630 K
p1 = 760 Torr (since this is the boiling point)
T2 = 25oC = 298 K
ln(p2 )  ln(p1 ) 
 ln(760) 
Hvap  1 1 
  
R  T1 T2 
59100 J mol1  1
1 

1
1 
8.314 JK mol  630 K 298 K 
 5.94
p2  e5.94  0.00263 Torr
C4 (b). Water boils at 100oC and has an enthalpy of vaporization of 40.0 kJ mol -1. Calculate the value
of S for the vaporization of water at the boiling point.
Since the system H2O(l) Ý H2O(g) is at equilibrium at the boiling point, we have:
Gvap  Hvap  Tb Svap  0
thus, Hvap  Tb S vap
or, S vap 
Hvap
Tb

40,000 Jmol1
 107.2JK 1mol1
(100  273)K
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 11 of 14
C5 (a). What hybridization is each of the four indicated atoms using in the molecule of Vanillin shown
below?
4
H
O
O
1
OH
2
CH3
3
Atom 1:_sp2______
Atom 2:_sp3______
Atom 3:_sp3______
Atom 4:_sp2______
C5 (b). Use VSEPR theory to predict the shape of the KrF4 molecule.
8 + (7 x 4) = 36 electrons
Making the four bonds and completing the octets on the F atoms uses 32 of these. This leaves 4
electrons, placed as two lone pairs on the Kr atom. It is therefore an AX 4E2 structure, which is square
planar.
C5 (c). Why do atomic radii decrease going from left to right across a period?
The effective nuclear charge increases going left to right
Or
Electrons are being added to the same principal shell, hence have similar energies. However,
protons are also being added, which collectively exert a greater attraction for these electrons, pulling
them inwards, reducing the radius.
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 12 of 14
C6 (a). At 90oC, the equilibrium constant is 0.0068 for the reaction H2(g) + S(s) ∏ H2S(g). If 0.15 mol
H2(g) and 1.0 mol S(s) are heated in a 1.0 L reaction chamber to 90oC, what will be the partial
pressures of H2(g) and H2S(g) at equilibrium?
Note that the sulphur is a solid and is not considered in the equilibrium expression, and that since the
other species are gases, the equilibrium constant is Kp, not Kc. But you are given molar quantities so
you need to convert these to the initial partial pressures:
0.15 mol(0.082L atmK 1mol1 )(90  273)K
pH2 

 4.46 atm
V
1.0 L
nH SRT 0 mol(0.082L atmK 1mol1 )(90  273)K
pH2S  2

 0 atm
V
1.0 L
nH2 RT
Initial, atm
Change, atm
Equilibrium, atm
pH2S( g)
pH2( g)
H2(g)
4.46
-x
4.46 - x
H2S(g)
0
+x
x
 Kp
x
 0.0068
4.46  x
x  0.0068(4.46  x)
x  0.0304  0.0068x
1.0068x  0.0304
x  0.0301
Therefore, pH2 = 4.46 – x = 4.43 atm and pH2S = x = 0.0301 atm
C6 (b). Another 0.15 mol H2(g) are added to the equilibrium mixture from part (a). Calculate the new
equilibrium partial pressures.
Before adding the extra hydrogen,
nH2 
pH2 V
RT

4.43 atm(1.0 L)
 0.149 mol
(0.082LatmK 1mol1 )(90  273)K
Thus, after adding it, we have 0.149 + 0.15 = 0.299 mol H 2, and its new partial pressure is:
pH2 
nH2 RT
V

0.299 mol(0.082LatmK 1mol1 )(90  273)K
 8.90 atm
1.0 L
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 13 of 14
Then the system re-equilibrates:
Initial, atm
Change, atm
Equilibrium, atm
pH2S( g)
pH2( g)
H2(g)
8.90
-x
8.90 - x
H2S(g)
0.0301
+x
0.0301 + x
 Kp
0.0301  x
 0.0068
8.90  x
0.0301  x  0.0068(8.90  x)
x  0.0068x  0.06052  0.0301
1.0068x  0.03012
x  0.0299
Thus, pH2 = 8.90 – x = 8.87 atm and pH2S = 0.0301 + x = 0.0600 atm
C6 (c). More S(s) is added to the equilibrium mixture from part (b). What happens? Why?
Nothing happens. The equilibrium position is not affected by the addition of a solid.
Final Examination April 2009
CHEM 1000 A and V
R. Burk
Page 14 of 14
C7 (a) Lead (Pb) crystallizes in a face centred cubic unit cell with an edge length of 495 pm.
Calculate the radius of a Pb atom (in pm).
If the edge length = l = 495 pm, we can calculate the radius knowing that the diagonal of the face is
l
495 pm
4r. Thus, (4r)2 = l2 + l2. Thus, 16r2 = 2l2, or, r 

 175.0 pm
2 2
2 2
(b) Calculate the density of Pb (in g/cm3).
The mass of the unit cell is that of four atoms:
mass 
4 atoms(207.2 g mol1 )
 1.38  10 21g
6.02  1023 mol1
volume  l3  (495  10 10 cm)3  1.21 10 22 cm3
density 
mass
1.38  10 21g

 11.4g cm3
22
3
volume 1.21 10 cm