CHEM 1000 A and V
Final Exam
April, 2006
Part A. Answer all eight questions (5 marks each).
1. Name five functional groups containing the carbonyl group, C=O.
Aldehyde, ketone, ester, amide and carboxylic acid
2. Which of
168
78
Pt or
201
78
Pt is more likely to decay via   emission and which via  emission? Why?
  emission leads to a decrease in the n/p ratio, so
201
78
Pt will decay via this mode in order for its n/p ratio to
approach the middle of the “island of stability”.  emission leads to an increase in the n/p ratio, so 168
78 Pt will
decay via this mode for the same reason.
3. Reaction A has a rate constant k = 100 s-1. Reaction B has a rate constant k = 200 s-1? What can be said about
the relative values of ∆Go for these two reactions?
Nothing. (The rate of the reaction is not related to its free energy change.)
4. Name a salt that could be added to a solution of the weak acid HOCl(aq) to make an acidic buffer system.
Anything containing the OCl- ion, namely NaOCl, KOCl, etc.
5. Which solution will have the higher boiling point: 2.0 m KCl(aq) or 4.0 m aqueous glucose? Why? Note that
glucose is not ionic and does not dissociate in water.
Although both these solutions have the same molality (4.0 m in dissolved particles), the K+ and Cl- ions will
form ion pairs, effectively reducing the molality and thereby reducing the boiling point elevation. 4.0 m glucose
will therefore have the higher boiling point.
6. Vehicles in the future might use hydrogen to power fuel cells which produce electricity. Name three ways that
hydrogen might be stored in such a vehicle.
1. As a chilled liquid.
2. As a compresed gas
3. Dissolved in a metal as a hydride.
7. Suppose for the reaction SO2(g) + ½ O2(g) ∏ SO3(g), we wish to maximize the equilibrium concentration of
SO3(g).
(a) Should the temperature be high or low?
Low. This is an oxidation which must therefore be exothermic. Lowering the temperature moves the
equilibrium in the exothermic direction, i.e. towards more SO3(g) in this case.
(b) Should the pressure be high or low?
High. In any reaction, raising the pressure will move the reaction towards the side with a lower number
of moles of gas, i.e. towards more SO3(g) in this case.
8. Fill in the electrons on the molecular orbital diagram below for oxygen, O2, and state why the molecule is or
is not paramagnetic.
*2p
2p
2p
2p
*2s
2s
This molecule is paramagnetic because of the two unpaired electrons in the *2p orbitals.
Part B. Answer all four questions (20 marks each).
1. The reaction 2 I-(aq) + S2O8-2(aq) → I2(aq) + 2 SO4-2(aq) was studied at 25oC and the following results were
obtained:
[I-(aq)] (mol L-1)
0.080
0.040
0.080
[S2O8-2(aq)] (mol L-1)
0.040
0.040
0.020
Rate (mol L-1 s-1)
1.25 x 10-5
6.25 x 10-6
6.25 x 10-6
(a) Determine the rate law for this reaction.
Comparing the first and second experiments, the rate is halved when [I-(aq)] is halved. It is therefore first order in
[I-(aq)].
Comparing the first and third experiments, the rate is halved when [S2O8-2] is halved. It is therefore first order in
[S2O8-2].
The rate law is thus rate = k[I-(aq)] [S2O8-2]
(b) Calculate the value and units of the rate constant, k.
rate  k[I  (aq) ][S2O8 2(aq) ]
thus, k 
rate
[I (aq) ][S2O8 2(aq) ]

Using the first experiment (we could use any of the three):
1.25 105 mol L1s 1
k
(0.080 mol L1 )(0.040 mol L1 )
 3.90 103 L mol1s 1
(c) Suppose the rate constant is three times higher at 50oC than at 25oC. Calculate the activation energy of the
reaction (kJ mol-1).
k 25  Ae
k 50  Ae
  Ea 


 R (298K ) 
  Ea 


 R (323K ) 
  Ea 


 R (323K ) 
  Ea 


 R (323K ) 
k 50
Ae
e
 3   E    E 
a
a
k 25




Ae  R (298K )  e  R (298K ) 
 E a 
E a
ln(3) 

R(323K)  R(298K) 


1
1
ln(3)  E a 


1
1
1
1
 8.314JK mol (298K) 8.314JK mol (323K) 
solving, E a  35,167 J mol 1
 35.2 kJ mol1
2 (a) Balance the following nuclear reactions by supplying the missing particle(s).
(i)
73
31
(ii)
(iii)
73
Ga  32
Ge  
205
83
Bi 
212
87
205
82
Pb  
Fr   
208
85
At
141
92
1
(iv) 01 n  235
92 U  56 Ba  36 Kr  3 0 n
(v)
56
26
Fe   
56
26
Fe
(b) Calculate how much energy (in kJ) would be released upon fusion of 1.0 mole of 11 H with 1.0 mole of
H according to the reaction: 11 H + 21 H → 23 He . The masses of the nuclei are: 1H: 1.007825 amu; 2H: 2.0140
amu; 3He: 3.01603 amu.
2
1
m  3.01603  (1.007825  2.0140)  0.005795 amu
kg
kg
1.66 1027
 6.02 1023 mol1  5.79  106
amu
mol
2
E  mc
kg
(3.00 108 m s 1 ) 2
mol
11
 5.2110 J mol1
 5.79 106
 5.21108 kJ mol1
(c) The artist Rembrandt lived from 1642-1672. You would like to have a painting dated to see if it is a genuine
Rembrandt. One gram of living organic material has a 14C activity of 15.3 s-1. One gram of the paint has an
activity of 15.1 s-1. Could the painting be a Rembrandt? The half life of 14C is 5730 y.
 A 
ln 
  kt
A
 0
 A 
ln 

A0 

thus, t 
k
0.693 0.693
here, k 

 1.2110 4 y 1
t1/ 2
5730 y
 15.3 
ln 

15.1 

thus, t 
 109 y
1.2110 4 y 1
The painting cannot not therefore be a genuine Rembrandt.
3. (a) Name the following compounds:
Br
NH2
O
2-hexanone or
methylbutylketone
2-amino-4-bromoheptane
CH3
O
N-propylethanamide
or N-propylacetamide
N
H
HC
CH
HC
CH
1-methyl-4-chlorobenzene
or 4-chlorotoluene
or para-chlorotoluene
or parachloromethylbenzene
Cl
4-methylpent-1-yne
or 4-methyl-1-pentyne
(b) Draw the structure or give the correct name for each missing reactant or product in the following reactions:
+
Cl
Cl2
Cl
or 1,2-dichlorobutane
OH
O
oxidation
or 3-hexanone or
ethylpropylketone
O
H
O
OH
OH
+
H
O
or propylmethanoate
H3C
O
OH
+
N
CH3
H3C
H
or dimethyl amine
N
O
CH3
OH
OH
+
Heat, H2SO4
O
or diisopentylether
4 (a) Use the data in the table below to calculate Go (kJ mol-1) for the reaction C(s)diamond  C(s)graphite at 25oC:
Hfo (kJ mol-1)
So (J K-1 mol-1)
C(s)diamond
2.0
2.0
C(s)graphite
0
6.0
Horxn = Hfo(C(s)graphite) - Hfo(C(s)diamond)
=0 – 2
= -2 kJ mol-1
Sorxn = Sfo(C(s)graphite) - Sfo(C(s)diamond)
=6.0 – 2.0
= +4 J K-1 mol-1
Gorxn = Horxn - TSorxn
= -2000 J mol-1 – (25+273)K(+4 J K-1 mol-1)
= -3,192 J mol-1
= -3.19 kJ mol-1
(b) Is this reaction spontaneous? If so, why do diamonds not appear to spontaneously change to graphite?
The reaction is spontaneous, but it is so slow that it can not be observed.
Part C. Answer any three of the six questions. If you answer more than three, the best three will be used
to calculate your mark (20 marks each).
5.
For a Galvanic cell using the half-reactions:
VO2+(aq) + 2 H+(aq) + e  VO+2(aq) + H2O(l)
Zn+2(aq) + 2 e  Zn(s)
Eo = 1.00 V
Eo = 0.76 V
(a) Determine the overall balanced cell reaction.
2 VO2+(aq) + 4 H+(aq) + 2 e  2 VO+2(aq) + 2 H2O(l)
Zn(s)  Zn+2(aq) + 2 e
________________________________________________
2 VO2+(aq) + 4 H+(aq) + Zn(s)  2 VO+2(aq) + 2 H2O(l) + Zn+2(aq)
(We know the reaction is in this direction because it must be a Galvanic cell, i.e. it must have Eocell>0)
(b) Calculate the standard cell potential (V).
Eocell = 1.00 + 0.76 = 1.76 V
(c) At 10oC, suppose [VO2+(aq)] = 2.0 M, [VO+2(aq)] = 0.010 M, [H+(aq)] = 0.50 M and [Zn+2] = 0.10 M.
Calculate the cell potential under these conditions (V).
Q
[VO 2 (aq) ]2 [Zn 2 (aq) ]
[VO 2  (aq) ]2 [H  (aq) ]4

(0.010) 2 (0.10)
 4.0 105
2
4
(2.0) (0.50)
RT
ln(Q)
nF
J 8.314J K 1mol 1 (273  10)K
 1.76 
ln(4.0 10 5 )
1
C
2(96487 C mol )
J
J
 1.76  (0.12 )
C
C
J
 1.88
C
 1.88 V
E  E0 
6. Calculate the pH and the concentrations (mol L-1) of all species present (except water) in a 1.0 M solution of
carbonic acid, H2CO3(aq). For this acid, Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11.
H2CO3(aq) + H2O(l) ∏ HCO3-(aq) + H3O+(aq)
K a1 
[HCO3 (aq) ][H3O (aq) ]
[H 2CO3(aq) ]
Initial
Change
Equilibrium
H2CO3(aq)
1.0
-x
1.0-x
HCO3-(aq)
0
+x
x
Thus at equilibrium,
x(x)
 K a1  4.3  107
1.0  x
x 2  4.3 107 (1.0  x)
x 2  4.3 107 x  4.3 107  0
a  1; b  4.3  107 ; c  4.3 107
x
b  b 2  4ac
2a
4.3 107  1.85 1013  4(1)(4.3  107 )

2(1)
4.3 107  1.31103

2
4
 6.55 10 or 6.55 104
Thus, x = [HCO3-(aq)] = [H3O+(aq)] = 6.55 x 10-4 mol L-1
pH = -log10[H3O+(aq)] = -log10(6.55 x 10-4) = 3.18
HCO3-(aq) + H2O(l) ∏ CO3-2(aq) + H3O+(aq)
Ka2 
[CO32 (aq) ][H3O  (aq) ]
or, [CO3
[HCO3 (aq) ]
2
(aq)
[OH  (aq) ] 
]
K a 2 [HCO3 (aq) ]
[H 3O

(aq)
]
 K a 2  5.6 1011 mol L1
Kw
1.0 1014

 1.53 1011 mol L1

4
[H 3O (aq) ] 6.55 10
H3O+(aq)
0
+x
x
7. For the reaction 2 NO(g) ∏ N2(g) + O2(g), the value of Kp at 1000oC is 1.78 x 107. If 1.00 atm of NO(g) is placed
in a flask at 1000oC, calculate the equilibrium partial pressures (atm) of all three gases.
Initial
Change
Equilibrium
p(NO(g)), atm
1.00
-2x
1.00-2x
p(N2(g)), atm
0
+x
x
p(O2(g)), atm
0
+x
x
Thus at equilibrium,
p N 2 p O2
p
2
NO
 Kp 
x(x)
(1.00  2x) 2
x  K p (1.00  2x) 2
2
x 2  K p (1.00  4x  4x 2 )
x 2  4K p x 2  4K p x  K p  0
(1  4K p )x 2  4K p x  K p  0
a  1  4K p  1  4(1.78 107 )  7.12 107
b  4K p  7.12 107
c  K p  1.78 107
b  b 2  4ac 7.12 107  (7.12 107 ) 2  4(7.12  107 )(1.78 107 )
x

2a
2(7.12 107 )
7.12 107  0

14.24 107
 0.5
Thus, pNO = 1.00 – 2x = 0
pO2 = pN2 = x = 0.5 atm
(This indicates essentially complete dissociation of the NO to O2 and N2, which could be predicted from the
very large value of Kp for this reaction.)
8(a) Use the ideal gas law to calculate the pressure (atm) exerted by 1.00 mol of N2(g) in a 1.00 L container at
25oC.
nRT 1.00 mol(0.082L atm K 1mol 1 )(25  273)K

V
1.00 L
 24.4 atm
p
(b) Repeat the calculation using the van der Waals equation. For N2(g), a = 1.39 atm L2 mol-2 and b = 0.0391 L
mol-1
RT
n
p
a 
V  nb
V
2
 1.00 mol 
0.082L atm K 1mol1 (25  273)K

 1.39 atm L2 mol2 

1
1.00 L  1.00 mol(0.0391 L mol )
 1.00 L 
2
 25.43 atm  1.39 atm
 24.04 atm
(c) Why are the values calculated in (a) and (b) so different?
The pressure calculated using the van der Waals equation is lower because of the attractive forces
between the nitrogen molecules. (Note that the fact that the nitrogen molecules take up space would tend
to make the pressure higher, not lower than that calculated by the ideal gas law.)
(d) Repeat the calculation for 1.00 mol of Xe(g) in a 1.00 L container at 25oC using the van der Waals equation.
For Xe(g), a = 4.19 atm L2 mol-2 and b = 0.0511 L mol-1
p
RT
n
a 
V  nb
V
2
 1.00 mol 
0.082L atm K 1mol1 (25  273)K

 4.19 atm L2 mol2 

1
1.00 L  1.00 mol(0.0511 L mol )
 1.00 L 
2
 25.75 atm  4.19 atm
 21.56 atm
(e) Why is the calculated pressure of Xe(g) different from that of N2(g) calculated using the van der Waals
equation?
The pressure of Xe is lower because the intermolecular interactions among the Xe atoms are stronger than
those among N2 molecules. (Note that here again, the value of ‘b’, the size of the molecules, is not a factor
since a larger ’b’ value would increase the pressure.)
9. Iridium (Ir) has a face centred cubic unit cell with an edge length of 383.3 pm. Calculate the density of solid
Iridium (g/cm3).
The unit cell contains the equivalent of 4 atoms:
mass unit cell  4 atoms 
192.2 g mol1
 1.28 1021 g
6.02 1023 mol1
A cubic unit cell has a volume that is the cube of the edge length:
Vunit cell  (383.3 1010 cm)3  5.63 1023 cm3
The density is therefore:

munit cell
Vunit cell
1.28 1021 g

 22.7 g cm3
23
3
5.63 10 cm
10(a) The normal boiling point of isooctane (a component of gasoline) is 99.2oC and its enthalpy of vaporization
is 35.76 kJ/mol. Calculate the vapor pressure of isooctane at 25oC.
We set p1 = 1.0 atm, since the normal boiling point is defined as the temperature (99.2oC in this case) at
which the vapour pressure is 1.0 atm. Thus,
H vap  1 1 
  
R  T1 T2 

35760 J mol 1 
1
1
 ln(1 atm) 

1
1 
8.314 J K mol  (273  99.2)K (273  25)K 
ln(p 2 )  ln(p1 ) 
0  2.88
 2.88
p 2  exp(2.88)  0.056 atm
(or 42.8 Torr)
(b) A solution is made by dissolving 3.00 mole of CaCl2(s) in enough water to make 2.00 L of solution. The
density of the solution is 1.25 g/mL.
(i) Calculate the molality of the solution (not the molarity!)
1 L of the solution has a mass of 1,000 mL x 1.25 g/mL = 1,250 g.
The solution is 1.5 M CaCl2, i.e. 1.5 mol CaCl2 per L of solution.
The 1 L of solution therefore contains 1.5 mol x 111.1 g/mol = 166.7 g CaCl2.
The mass of water is therefore 1,250 g – 166.7 g = 1083.3 g water.
Molality is mol/kg water
Thus,
molality 
1.5 mol
 1.38 m
1.0833 kg
(ii) Calculate the freezing point (oC) of the solution. (For water, Kf = 1.86oC kg mol-1)
Molality of ions = 3 x 1.38 = 4.14 mol kg-1
∆Tf = Kf m = 1.86 oC kg mol-1 x 4.14 mol kg-1 = 7.70oC
Thus, the freezing point is 0.0 – 7.70 = -7.70oC
(iii) Calculate the vapor pressure (mm Hg) of the solution at 25oC. (The vapor pressure of pure water at 25oC is
23.8 mm Hg.)
In 1 L of the solution, we have 1.5 moles CaCl2 (= 4.5 moles ions) and 1083.3 g H2O = 1083.3/18.0 = 60.18 mol
H2O
X H2O 
n H2O
n H2O  n solute

60.18
 0.930
60.18  4.5
pso ln  p 0solv X solv  23.8 mm Hg  0.930  22.1 mm Hg