CHEM 1000 A and V
Midyear Exam
December 6, 2008
Part A. 60 marks. Answer each question (5 marks each).
1.
2.
Does the position of the equilibrium in the reaction 2 CO(g) + O2(g) ∏ 2 CO2(g) shift left or right when we:
Increase the pressure
Right
Increase the temperature
Left
Add some O2(g)
Right
Remove some CO2(g)
Right
Remove some CO(g)
Left
Explain how the effective nuclear charge changes going DOWN a group in the periodic table, and
explain why.
The effective nuclear charge decreases going down a group because inner electrons shield valence
electrons.
3.
4.
State whether each of the following processes has a positive or negative entropy change:
(a) C8H18(l) + 12.5 O2(g)  8 CO2(g) + 9 H2O(g)
Positive
(b) CO2(s)  CO2(g)
Positive
(c) H2O(g)  H2O(l)
(d) Na+(g) + Br¯(g)  NaBr(s)
Negative
Negative
(e) H2O(l) at 5°C  H2O(l) at 10°C
Positive
A reaction has negative values of both H and S. Will this reaction be spontaneous at high
temperature, low temperature, both or neither? How do you know?
We know that a reaction is spontaneous if ΔG<0, and we know that ΔG = ΔH – TΔS. A negative value
of ΔH tends to make ΔG negative. However, a negative value of ΔS tends to make ΔG positive, unless
the temperature is low. The reaction is therefore spontaneous at a low temperature.
5.
Which of HNO3(aq) or HNO2(aq) is a stronger oxidant? Why?
HNO3(aq) is a stronger oxidant. The N atom in HNO3(aq) has an oxidation state of +5, whereas it is only
+3 in HNO2(aq).
6.
Why is the H-C-H bond angle in CH3Cl greater than the ideal tetrahedral angle of 109.5o?
Chemistry 1000 A and V
Mid-Year Examination, December, 2008
Page 2 of 7
The Cl atom is very electronegative. This pulls electrons towards itself, lowering the density of electrons in the
C-Cl bond. The repulsion between these electrons and those in the C-H bonds is therefore smaller and the H
atoms move apart, increasing the H-C-H bond angle.
7.
In the van der Waals equation, why is the ‘a’ term for H2O larger than that for CO2?
This is due to the fact that water is a polar molecule, causing intermolecular attractions that are stronger than
those in CO2.
8.
Can H+(aq) ions oxidize Cu(s)? Why or why not?
No, because hydrogen is above copper in the activity series.
9.
Why do deep-sea divers breathe a mixture of oxygen and helium?
Nitrogen is quite soluble in blood. As the diver ascends, the nitrogen comes out of the blood as bubbles causing a
very painful and sometimes fatal condition called the bends. Helium is not vey soluble in blood, and so this
problem is avoided.
10.
11.
Consider the spontaneous reaction carried out at atmospheric pressure: C3H6(g) + 4.5 O2(g) ↓ 3 CO2(g) + 3 H2O(g).
State whether the following variables increase or decrease as the reaction progresses from left to right:
Volume of the system
Increase
Temperature of the surroundings
Increase
Entropy of the system
Increase
Entropy of the surroundings
Increase
Entropy of the Universe
Increase
In a C2 molecule, the highest energy electrons are in a 2p bonding molecular orbital. Is the bond length of C2+
greater or less than the bond length of C2? How do you know?
Since the electron must be removed from this bonding molecular orbital, the bond order of C2+ will be less than
that of C2. Thus the bond length of C2+ must be greater than that of C2.
12.
The electronic configuration of the Mo atom is predicted by AUFBAU to be [Kr] 5s24d4. What do you think is
the actual configuration and why?
It is likely to be [Kr] 5s14d5, since this configuration has all unpaired electrons, i.e. half-filled subshells.
Part B. Answer both questions B1 and B2. (20 marks each)
B1. [6 marks] (a) At a particular temperature and pressure, it takes 4.55 minutes for a 1.5 L sample of He(g) to effuse
through a porous membrane. How long will it take for 1.5 L of F2(g) to effuse under the same conditions?
2
Chemistry 1000 A and V
time F2
time He

Mid-Year Examination, December, 2008
Page 3 of 7
MWF2
rate He

rate F2
MWHe
thus, time F2  time He
MWF2
MWHe
 4.55 min
38.0 g mol1
4.00 g mol1
 14.0 min
(b) [6 marks] A sample of Freon-12 (CF2Cl2) occupies 24.5 L at 298 K and 253.3 kPa. Find its volume at STP.
p1V1 p 2 V2

n1T1 n 2 T2
In this example, n1 = n2, thus
p1V1 p 2 V2
p VT

, or V2  1 1 2
T1
T2
T1p 2
Note that STP means 1 atm (101,325 Pa) and 0oC (273 K). Thus,
V2 
253,300 Pa(24.5 L)(273)K
 56.1 L
298K(101,325 Pa)
(c) [8 marks] Chlorine is produced electrochemically from seawater and collected in a sealed container. If a 15.0 L
container holds 0.580 kg of Cl2(g) at 200oC, calculate the pressure (in atm) using the van der Waals equation. For Cl2(g), a
= 6.49 atm L2 mol-2 and b = 0.0562 L mol-1
n
580 g
 8.17 mol
2(35.5 g mol1 )
nRT
n
p
a 
V  nb
V
2
 8.17 mol 
8.17 mol(0.082 L atm K 1mol 1 )(200  273)K

 6.49 atm L2 mol 2 

1
15.0 L  8.17 mol(0.0562 L mol )
 15.0 L 
2
 21.79  1.93 atm
 19.86 atm
B2. (a) [5 marks] Beginning with a sulphide ore such as NiS(s), show the reactions by which acid rain is
produced.
NiS(s) + 1.5 O2(g) → NiO(s) + SO2(g)
SO2(g) + ½ O2(g) → SO3(g)
SO3(g) + H2O(l) → H2SO4(aq)
(b) [5 marks] Beginning with NO(g), show the reactions that produce ozone at ground level, i.e. in smog.
NO(g) + ½ O2(g) → NO2(g)
NO2(g) + h → NO(g) + O(g)
O(g) + O2(g) → O3(g)
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Chemistry 1000 A and V
Mid-Year Examination, December, 2008
Page 4 of 7
(c) [5 marks] Show the reactions in the catalytic cycle by which Cl atoms destroy ozone in the stratosphere.
Cl(g) + O3(g) → ClO(g) + O2(g)
ClO(g) + O(g) → Cl(g) + O2(g)
or
2 Cl(g) + 2 O3(g) → 2 O2(g) + 2 ClO(g)
2 ClO(g) → Cl2O2(g)
Cl2O2(g) + h → 2 Cl(g) + O2(g)
(d) [5 marks] Show the four oxygen-only reactions in the stratosphere collectively known as the Chapman cycle
O2(g) + h  O(g) + O(g)
O(g) + O2(g)  O3(g)
O3(g) + h  O2(g) + O(g)
O(g) + O3(g)  2 O2(g)
Part C. Attempt all five questions C1 to C5. The best four will be used to calculate your mark. (20 marks each)
C1. (a) [10marks] Ozone (O3) protects us from damaging radiation from the Sun because photons having a
wavelength lower than 270 nm are absorbed causing a bond to break in the O3 molecule. Calculate the bond
energy of this bond (kJ mol-1).
hc
3.00 108 m s 1
34
E  h 
 6.63 10 J s
 7.37 1019 J (per photon)
9

270 10 m
23
1
 6.02 10 mol  443, 473 J mol1  443 kJ mol1
(b) [8 marks] Calculate the frequency (s-1) of the photon emitted by a hydrogen atom when an electron falls
from n=7 to m=3.
1
1
 1
R 2  2

m n 
1 1
 0.01097 nm 1  2  2 
3 7 
4
 9.95 10 nm 1

1
 1005 nm
9.95 104 nm 1

c 3.00 108 m s 1

 2.99 1014 s 1
 1005 109 m
(c) [2 marks] Are the photons in part (b) in the UV, visible or IR portion of the electromagnetic spectrum?
1005 nm photons are in the IR portion of the spectrum.
C2. (a) [5 marks] Two structures for Cl2O are possible, Cl-O-Cl and O-Cl-Cl. Draw the Lewis structures for
these and use formal charge to predict which structure is more likely.
4
Chemistry 1000 A and V
..
: ..
Cl
..
: O..
Valence
Electrons assigned
Formal Charge
Page 5 of 7
..
Cl :
..
..
O
..
Cl
7
7
0
Valence
Electrons assigned
Formal Charge
Mid-Year Examination, December, 2008
O
6
6
0
Cl
7
7
0
..
Cl :
..
..
..
Cl
O
6
7
-1
Cl
7
6
1
Cl
6
7
-1
Cl-O-Cl is therefore the more likely structure because the formal charges are all zero.
(b) [2 marks each] Name the type of hybrid orbitals used by the central atom in each of the following three
structures:
AsF5
SiCl4
Br3¯
5+(5x7) = 40 e¯
4+(4x7) = 32 e¯
Thus, AX5 (5 charge clouds on the As) Thus, AX4 (4 charge clouds on the Si)
Thus, sp3d hybrids
Thus, sp3 hybrids
(3x7) + 1 = 22 e¯
Thus, AX2E3 (5 charge clouds
on the central Br)
Thus, sp3d hybrids
(c) [3 marks each] Use VSEPR to predict the shapes of the following three species. Wrong name = zero marks.
CNO(the N atom is in the centre)
AsF4
4+5+6+1 = 16 electrons
C=N=O, with two lone pairs on each
of the C and O atoms uses all 16 electrons
Thus, AX2
Thus, linear
5 + (4 x 7) = 33 eThus, AX4E
Thus, see-saw
XeOF4
8 + 6 + (4 x 7) = 42 eThus, AX5E
Thus, square pyramid
C3. [20 marks] Balance the following REDOX reaction in a basic solution:
CN¯ (aq) + MnO4¯ (aq)  CNO¯ (aq) + MnO2(s)
Oxidation:
CN¯ (aq)  CNO¯ (aq)
CN¯ (aq) + H2O(l)  CNO¯ (aq)
CN¯ (aq) + H2O(l)  CNO¯ (aq) + 2 H+(aq)
CN¯ (aq) + H2O(l)  CNO¯ (aq) + 2 H+(aq) + 2 e¯
Reduction:
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Chemistry 1000 A and V
Mid-Year Examination, December, 2008
Page 6 of 7
MnO4¯ (aq)  MnO2(s)
MnO4¯ (aq)  MnO2(s) + 2 H2O(l)
MnO4¯ (aq) + 4 H+(aq) + 3e¯  MnO2(s) + 2 H2O(l)
Overall:
3(CN¯ (aq) + H2O(l)  CNO¯ (aq) + 2 H+(aq) + 2 e¯)
+ 2(MnO4¯ (aq) + 4 H+(aq) + 3e¯  MnO2(s) + 2 H2O(l))
=
3 CN¯ (aq) + 3 H2O(l)  3 CNO¯ (aq) + 6 H+(aq) + 6 e¯
2 MnO4¯ (aq) + 8 H+(aq) + 6e¯  2 MnO2(s) + 4 H2O(l)
_________________________________________________________
3 CN¯ (aq) + 2 MnO4¯ (aq) + 2 H+(aq)  3 CNO¯ (aq) + 2 MnO2(s) + H2O(l)
Convert to basic solution:
3 CN¯ (aq) + 2 MnO4¯ (aq) + 2 H+(aq) + 2 OH¯ (aq)  3 CNO¯ (aq) + 2 MnO2(s) + H2O(l) + 2 OH¯ (aq)
3 CN¯ (aq) + 2 MnO4¯ (aq) + 2 H2O(l)  3 CNO¯ (aq) + 2 MnO2(s) + H2O(l) + 2 OH¯ (aq)
3 CN¯ (aq) + 2 MnO4¯ (aq) + H2O(l)  3 CNO¯ (aq) + 2 MnO2(s) + 2 OH¯ (aq)
Check:
Left
C
3
N
3
Mn
2
O
9
H
2
Charge -5
Right
3
3
2
9
2
-5
C4. (a) [10 marks] The reaction Ba+2(aq) + SO4-2(aq) → BaSO4(s) is carried out in a calorimeter. When 2.00 mol
Ba+2(aq) are mixed with 2.00 mol SO4-2(aq) in 4.00 L of water at 25oC, the temperature of the solution rises to
28.1oC. Assuming that the calorimeter itself absorbs a negligible amount of heat, that the final solution has a
density of 1.00 g mL-1, and that the specific heat capacity of the solution is 4.18 J oC-1 g-1, calculate the enthalpy
of reaction (H) per mole of BaSO4(s) formed.
4.00 L = 4,000 mL
4,000 mL x 1.00 g mL-1 = 4,000 g water
q = mCpΔT
= 4,000 g (4.18 J oC-1 g-1)(28.1 – 25.0)oC
= 51,832 J
H = - 51,832 J / 2 mol = - 25,916 J/mol = - 25.9 kJ/mol (negative because heat is lost by the system)
(b) [10 marks] Calculate the standard enthalpy of formation of B2H6(g) given the following data:
6
Chemistry 1000 A and V
Mid-Year Examination, December, 2008
2 B(s) + 3/2 O2(g) → B2O3(s)
B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g)
H2(g) + ½ O2(g) → H2O(l)
H2O(l) → H2O(g)
Page 7 of 7
H = 1273 kJ mol-1
H = 2035 kJ mol-1
H = 286 kJ mol-1
H = +44 kJ mol-1
We want the reaction: 2 B(s) + 3 H2(g) → B2H6(g)
B2O3(s) + 3 H2O(g) → B2H6(g) + 3 O2(g)
H = (2035 kJ mol-1) = +2035 kJ mol-1
2 B(s) + 3/2 O2(g) → B2O3(s)
H = 1273 kJ mol-1
3 H2(g) + 3/2 O2(g) → 3 H2O(l)
H = 3(286 kJ mol-1) = 858 kJ mol-1
3 H2O(l) → 3 H2O(g)
H = 3(+44 kJ mol-1) = +132 kJ mol-1
_______________________________________________________________________
2 B(s) + 3 H2(g) → B2H6(g)
H = +36 kJ mol-1
C5. [20 marks] For the reaction Fe+3(aq) + SCN-(aq) ∏ FeSCN+2(aq), the equilibrium constant Kc = 1100 at 25oC.
Calculate the concentrations of Fe+3(aq), SCN-(aq) and FeSCN+2(aq) at equilibrium if 0.0200 mol Fe+3(aq) and
0.1000 mol SCN-(aq) are added to 1.000 L water at 25oC.
Initial, M
Change, M
Equilibrium, M
At equilibrium,
[FeSCN 2(aq) ]
[Fe3(aq) ][SCN(aq) ]
Fe+3(aq)
0.02
-x
0.02 - x
SCN-(aq)
0.10
-x
0.10 - x
FeSCN+2(aq)
0
+x
x
 1100
Thus,
x
 1100
(0.02  x)(0.1  x)
x  1100(0.02  x)(0.1  x)
x  1100(.002  .02x  .1x  x 2 )
x  2.2  22x  110x  1100x 2
1100x 2  133x  2.2  0
b  b 2  4ac 133  1332  4(1100)(2.2) 133  89.5
x


 0.101 or 0.01977
2a
2(1100)
2200
x = 0.101 would give negative concentrations, thus x = 0.01977
Thus, [Fe+3(aq)] = 0.02 – x = 0.000227
[SCN-(aq)] = 0.10 – x = 0.08023
[Fe(SCN)-] = x = 0.01977
7