Final Examination April 2010
CHEM 1000 A and V
R. Burk
Page 1 of 12
Part A. Answer all eight questions (5 marks each).
1.
2.
The valence electrons in a Dubnium (Db) atom are 7s26d3.
What are the four quantum numbers for each of these five electrons?
One of the 7s electrons:
n = __7__ l = __0__ ml = ___0_ ms = __+ ½
The other 7s electron:
n = __7__ l = __0__ ml = ___0_ ms = __- ½
One of the 6d electrons:
n = __6__ l = __2__ ml = __ -1_ ms = __+ ½
The second 6d electron:
n = __6__ l = __2__ ml = ___0_ ms = __+ ½
The third 6d electron:
n = __6__ l = __2__ ml = ___1_ ms = __+ ½
A carbon–carbon triple bond is stronger than a carbon–carbon single bond. Why then is
acetylene (H–C≡C–H) more reactive towards oxygen than ethane (H3C–CH3)?
The four electrons in the pi–bonds in acetylene are delocalized, meaning that they are further
away from the carbon nuclei, and thus are easier to remove. An oxidant such as oxygen will
easily pull these electrons away from acetylene.
3.
Why do isotopes along the lower edge of the “island of stability” tend to decay by alpha (),
beta plus (+) or electron capture (EC), but not by beta minus (–)?
Along the lower edge of the island of stability, the n/p ratio is low. Alpha, beta plus (+) or
electron capture (EC) decays result in an increase of this ratio, which places the product
isotope further into this island of stability. More stability = more energy released, so the
products are at a lower energy, which is favourable. The opposite is true for beta minus (–)
decay.
4.
Why can sodium metal not be produced by electrolyzing a solution of NaCl(aq)?
The water in this solution is preferentially reduced because its reduction potential is higher
than that of the sodium ions.
5.
Would the freezing point of 1.0 m CaCl2(aq) be greater or less than that of 3 m aqueous
glucose? Why?
Both solution have a total molality of 3 m (since the CaCl2 dissociates in water), but ion pairing
in the CaCl2 solution reduces this molaltiy slightly, reducing the freezing point depression, and
so the CaCl2 solution will have a higher freezing point than the glucose solution.
6.
Why is sulfuric acid (H2SO4(aq)) a stronger acid than sulfurous acid (H2SO3(aq))?
Final Examination April 2010
CHEM 1000 A and V
R. Burk
Page 2 of 12
This is because the extra oxygen in H2SO4 draws electrons away from the O–H bond,
weakening this bond and causing the acid to be stronger.
7.
In a reaction mechanism, what is the difference between an intermediate and a catalyst?
An intermediate is produced in an early step and consumed in a later step. A catalyst is
consumed in an early step and reproduced in a later step.
8.
Name five functional groups containing the carbonyl (C=O) group.
Aldehyde, ketone, ester, amide, carboxylic acid
Part B. Answer all four questions B1, B2, B3 and B4 (20 marks each).
B1(a) [8] Balance the following nuclear reactions by filling in the blanks:
(i) 0 n 
1
(ii)
235
92
Th 
234
90
(iii) 0 n 
1
(iv)
U
63
28
103
42
Mo  2 01n  ___ 152
50 Sn _________
Pa  __– (or
234
91
0
1
e  _)_________
Ni  11p  _ 63
27 Co ___________
Tc    ___ 99
43Tc _________
99
43
B1(b) 30P, which has a half–life of 150 seconds, decays by positron emission. Given a sample
containing 0.100 g 30P,
(i) [6] Calculate the decay rate (s–1).
rate = kN
=
0.693
N
t1/2
=
0.693  0.100 g 
23
-1

 × 6.02 ×10 mol
150 s  30 g / mol 
= 9.3 ×1018 s-1
Final Examination April 2010
CHEM 1000 A and V
R. Burk
(ii) [6] Calculate the time required for 90% of the
N
ln 
 N0
30P
Page 3 of 12
to decay (s).

 = -kt

  N0
 ln 
N
t= 
 k


  1 

 ln 



  = t =   0.1   = 498 s
  0.693  

 

 

  150 s  
B2. (a) [10] Name the following compounds:
Cl
Cl
Br
2–bromo–2–methylbutane
Br
4–bromo–1,2–dichlorobenzene
O
N
H
CH3
N–methylpentanamide
O
t–butylethylether
NO2
O
4–nitro–2–pentanone
Final Examination April 2010
CHEM 1000 A and V
R. Burk
Page 4 of 12
B2. (b) [10] Draw the structure or give the correct name for each product in the following reactions:
Cl
+
Cl2
Cl
or 2,3–dichloropentane
O
O
Br
OH
+
O
HO
Br
or 2–bromopropylpropanoate
Cr2O3
or 2–pentene
500oC
OH
Air, 300 oC
O
or 2–hexanone
+
H2O
H2SO4
OH
or 3–hexanol
Final Examination April 2010
B3.
CHEM 1000 A and V
R. Burk
Page 5 of 12
(a) [12] For the reaction 2 NO2 → N2O4, the rate constant is 0.027 M–1s–1 at 227oC and is
0.24 M–1s–1 at 277oC. Calculate the activation energy of this reaction (kJ mol–1).
k1 = Ae
 -Ea 
 RT 
 1
k 2 = Ae
  -Ea 
k1  e R(500 K) 
=
k 2   -Ea 
 R(550 K) 
e
 -Ea 
 RT 
 2





k 
-Ea
Ea
ln  1  =
+
 k 2  R(500 K) R(550 K)
1 
 0.027 
 1
ln 
 = -Ea 

 0.24 
 500R 550R 
-2.184
Ea =
1 
 1


 500R 550R 
= 99,868 J mol-1
= 100 kJ
(b) [8] Calculate the rate constant for the same reaction at 325oC.
k
A=
e
=
 -Ea 


 RT 
0.027 M-1s-1


-100000 J mol-1


-1
-1
 8.314 JK mol (500 K) 
(Could have used the data at 550 K instead)
e
= 7.33 ×108 M-1s-1
k 3250 C = Ae
 -Ea 


 RT 
8
-1 -1
= (7.33 ×10 M s )e
= 1.38 M-1s-1


-100000 J mol-1
 8.314 JK -1mol-1 (325+273) K) 


Final Examination April 2010
B4.
CHEM 1000 A and V
R. Burk
Page 6 of 12
(a) [10] Palladium (Pd) metal crystallizes in a face–centred cubic unit cell. The radius of a Pd
atom is 135 pm. Calculate the edge length of the unit cell of palladium.
d
l
l
In a face–centred arrangement of atoms, d = 4r. Thus,
l2 + l2 = d2
2l2 = (4r)2
2l2 = 16r2
l2 = 8r2
l = 81/2r
= 81/2(135 pm)
= 382 pm
(b) [10] Calculate the density of palladium metal.
Mass of the unit cell = mass of 4 atoms = 4(106.4 g mol–1)/6.02 x 1023 mol–1) = 7.07 x 10–22 g
Volume of the unit cell = l3 = (382 x 10–12 m)3 = (382 x 10–10 cm)3 = 5.57 x 10–23 cm3
Density = mass/volume = 7.07 x 10–22 g / 5.57 x 10–23 cm3 = 12.7 g/cm3
Final Examination April 2010
CHEM 1000 A and V
R. Burk
Page 7 of 12
Part C. Answer any three of the six questions C1 – C6. If you answer more than three, the best
three will be used to calculate your mark (20 marks each).
C1. (a) [10] Balance the following redox reaction in acidic solution:
H5IO6(aq) + Cr(s) → IO3(aq) + Cr3+(aq)
Oxidation:
Cr(s) → Cr3+(aq)
Cr(s) → Cr3+(aq) + 3 e–
Reduction:
H5IO6(aq) → IO3(aq)
H5IO6(aq) → IO3(aq) + 3 H2O(l)
H5IO6(aq) + H+(aq) → IO3(aq) + 3 H2O(l)
H5IO6(aq) + H+(aq) + 2 e– → IO3(aq) + 3 H2O(l)
2 Cr(s) → 2 Cr3+(aq) + 6 e–
3 H5IO6(aq) + 3 H+(aq) + 6 e– → 3 IO3(aq) + 9 H2O(l)
__________________________________________________
3 H5IO6(aq) + 2 Cr(s) + 3 H+(aq) → 3 IO3(aq) + 2 Cr3+(aq) + 9 H2O(l)
Check:
Left
Right
H
18
18
I
3
3
O
18
18
Cr
2
2
Charge
+3
+3
Final Examination April 2010
(b)
CHEM 1000 A and V
R. Burk
Page 8 of 12
[10] How long would it take (in seconds) to electroplate all the Cu2+ in 0.250 L of 0.245 M
CuSO4(aq) solution with a current of 2.45 A?
0.250 L × 0.245
mol
= 0.06125 mol Cu2+
L
but, Cu2+ + 2 e– → Cu(s)
Therefore, n = 0.06125 x 2 = 0.1225 mol e– required (= n)
it = nF
nF
i
0.1225 mol(96487 C mol-1 )
=
2.45 C s-1
t=
= 4824 s
C2.
[20] Concentrated ammonia (NH3(aq)) is 14.8 M and has a density of 0.898 g/mL. Calculate the
molality and the mole fraction of ammonia in this solution.
 1000 mL   0.898 g 
mass of 1 L of solution = 1L× 
×
 = 898 g
L

  mL 
mass of NH3 in this solution = 14.8 mol×
17 g
= 251.6 g NH3
mol
mass of water in this solution = 898 g – 251.6 g = 646.4 g H2O
molality of NH3 = mNH3 =
nH2O =
14.8 mol
mol
= 22.9
0.6464 kg
kg
646.4 g
= 35.9 mol
18.02 g mol-1
XNH3 =
nNH3
nNH3 +nH2O
=
14.8
= 0.292
14.8 + 35.9
Final Examination April 2010
C3.
CHEM 1000 A and V
R. Burk
Page 9 of 12
[20] Calculate the pH and the concentrations of all species (except water) in a 0.01 M solution
of H2S(aq) . For this acid, Ka1 = 8.9 x 10–8 and Ka2 = 1.0 x 10–19.
H2S(aq) + H2O(l) Ý H3O+(aq) + HS–(aq)
Initial, M
Change, M
Equilibrium, M
H2S(aq)
0.01
–x
0.01–x
[H3O+ ][HS- ]
= K a1
[H2S]
x(x)
= 8.9×10-8
0.01- x
x(x)
 8.9×10-8
0.01
x 2  8.9×10-10
x  2.98×10 -5
[H2S] = 0.01- x = 9.97×10 -3M
[H3O+ ] = x = 2.98×10 -5 M
[HS- ] = x = 2.98×10 -5 M
pH = -log10 [H3O+ ] = -log10 (2.98×10 -5 ) = 4.53
HS–(aq) + H2O(l) Ý H3O+(aq) + S2–(aq)
[H3O+ ][S2- ]
= K a2
[HS- ]
2-
[S ] =
=
K a2 [HS- ]
[H3O+ ]
K a2 x
x
= K a2
= 1.0×10-19 M
H3O+(aq) HS–(aq)
0
0
+x
+x
x
X
Final Examination April 2010
C4.
CHEM 1000 A and V
R. Burk
Page 10 of 12
[20] At 1020°C, Keq = 167.5 for the reaction CO2(g) + C(s) Ý 2 CO(g). A high-pressure chamber
containing excess graphite powder is charged with 53.1 atm of each of CO2 and CO and then
is heated to 1020°C. Calculate the equilibrium partial pressures of both gases (atm).
Initial, M
Change, M
Equilibrium, M
pCO2
pCO
53.1
–x
53.1–x
53.1
+ 2x
53.1+ 2x
2
pCO
= K eq
pCO2
(53.1+ 2x)2
= 167.5
(53.1- x)
(53.1+ 2x)2 = 167.5(53.1- x)
2819.6 + 212.4x + 4x 2 = 8894.2 -167.5x
4x 2 + 379.9x - 6074.6 = 0
x=
-b ± b2 - 4ac
2a
-379.9 ± 379.9 2 - 4(4)(-6074.6)
2(4)
-379.9 ± 491.4
=
8
= 14.0 or -108.9
=
The negative root would give negative partial pressures, thus we choose x = 14.0
At equilibrium;
pCO2 = 53.1 – x = 53.1 – 14.0 = 39.1 bar
pCO = 53.1 + 2x = 53.1 + 2(14.0) = 81.1 bar
Check:
2
pCO
81.12
=
=168  K eq
pCO2 39.1
Final Examination April 2010
C5.
CHEM 1000 A and V
R. Burk
Page 11 of 12
(a) [8] For the melting of ammonia, NH3(s) → NH3(l), H = 5.65 kJ mol–1 and
S = 28.9 J K–1 mol–1. Calculate the melting point of ammonia (oC).
At the melting point, the two phases are in equilibrium so G = 0.
But, G = H – TS, thus H = TS, or
ΔH
T=
ΔS
5,650 J mol-1
=
28.9 J K -1mol-1
=196 K
= -77o C
(b) [12] Benzene (C6H6) has a vapour pressure of 100 mm Hg at 26.1oC and
Hvap = 33.9 kJ mol–1. Calculate the vapour pressure of benzene at 40 oC (mm Hg).
Here, p1 = 100 mm Hg, T1 = 26.2oC = 299.25 K and T2 = 40oC = 313.15 K
ln(p2 ) = ln(p1) +
= ln(100) +
ΔHvap  1 1 
 - 
R  T1 T2 

33,900 J mol-1 
1
1
-1
-1 
8.314JK mol  299.25 K 313.15 K 
= 5.21
p2 = e5.21
= 183 mm Hg
Final Examination April 2010
C6.
CHEM 1000 A and V
R. Burk
Page 12 of 12
(a) [10] What hybridization is each of the five indicated atoms using in the amino acid Tyrosyl
shown below?
4
3
2
O
5
1
HO
Atom 1:___sp3___ Atom 2:___sp2___ Atom 3:___sp3__ Atom 4:__sp2___ Atom 5:___sp2___
(b) [5] Use VSEPR theory to predict the shape of KrF4.
Electrons = 8 + 4(7) = 36
Arranging the four F atoms around the less electronegative Kr atom and making four single
bonds uses 8 electrons. Completing the octets uses another 24. This leaves 4 electrons to be
placed as two lone pairs on the Kr atom. The molecule therefore has the shorthand notation
AX4E2, which is square planar.
(c) [5] The highest energy electron in a molecule of NO is in a *2p molecular orbital. Is the
bond energy of NO+ higher or lower than that of NO? Why?
Making the NO+ ion requires removing the electron from NO that is in the highest energy MO.
Since this is an antibonding MO, removing it will result in an increased bond order. This in turn
will make the bond energy in NO+ higher than that in NO.