Christmas Exam 1997
Answers
Part A.
1. The stronger oxidant will be the one in which the reducible atom (N in this case_) has the higher oxidation state. In HNO3, the ON of the N atom is +5, and in HNO2, the ON of the N atom is +3. Thus, we expect HNO3 to be the stronger oxidant.
2. From AUFBAU, we might expect the valence configuration to be 5s2 4d3. However, the half-filled 5s orbital is more stable than the filled one, so an electron "moves" from the 5s to the 4d orbital.
3. Generally, the ionization potential increases moving from left to right accross a period. However, the electronic configuration of Cd is …5s2 4d10, and that of In is …5s2 4d10 4p1. This 4p electron in In is at a higher energy than any of the 4d electrons, and so requires less energy to remove.
4. Square pyramidal molecules have a lone pair at one corner of the octahedron. This lone pair will repel more than any of the five bonding pairs repel one another. Thus, the shape will not be exactly square pyramidal.
5. An endothermic process will proceed spontaneously as long as the change in entropy is sufficiently positive so that DG is negative.
6. Since both DH<0 and DS>0 are favorable, the reaction will be spontaneous at all temperatures.
7. At high pressures, molecules are forced close together. They then feel the attractive forces of other molecules. These forces draw the molecules closer together, reducing the pressure below that predicted by PV = nRT. Also, the molecular volumes become significant when the molecules are close together.
8. Below the altitude where the maximum occurs, there is not enough light intensity to drive the O2 photolysis reaction required to make O atoms. Above this altitude, there is not enough O2 to combine with the O atoms to make O3.
9. The destruction of O3 by Cl radicals is a catalytic process, whereby the Cl atoms are regenerated in a second reaction and go on to destroy more O3 molecules.
10. Effusion refers to the movement of gas molecules from a region of higher pressure to one of lower pressure, generally through a small opening. Diffusion refers to the movement of gas molecules through other gases at the same pressure.
11. If the pressure is lowered, the equilibrium will shift to increase it, i.e. in the direction that increases the number of moles, i.e. to the LEFT.
12. If the temperature is increased, the equilibrium will shift to decrease it, i.e. in the endothermic direction. Note that this is a combustion reaction, and must therefore be exothermic. The equilibrium thus shifts to the LEFT.
Part B.
1. The balanced reaction for combustion of isopropyl alcohol is:
C3H8O(l) + 9/2 O2(g) ¾ 3 CO2(g) + 4 H2O(l)
The DHo for this reaction is:
DHorxn = 3 DHof(CO2(g)) + 4 DHof(H2O(l)) - DHof(C3H8O(l))
Solving for DHof(C3H8O(l)), we have:
DHof(C3H8O(l)) = 3 DHof(CO2(g)) + 4 DHof(H2O(l)) - DHorxn
but DHrxn = -4011 kJ mol-1, thus:
DHof(C3H8O(l)) = 3 (-393.5) + 4 (-285.8) - (-4011) kJ mol-1
= 1687.3 kJ mol-1
2. The initial concentrations are [CO(g)] = [H2O(g)] = 0.25 mol / 10 L = 0.025 M. To solve this, place values of initial, change and final concentrations of each species in a table:
CO(g) |
H2O(g) |
CO2(g) |
H2(g) |
|
| Initial (M) | 0.025 |
0.025 |
0 |
0 |
| Change (M) | -x |
-x |
+x |
+x |
| Final (M) | 0.025-x |
0.025-x |
x |
x |
At equilibrium, we have:
(Note that Kp = Kc, since Dn = 0)
Thus, we have 
Note that the left hand side is a perfect square. Thus, we can
take the square root of both sides: 
Thus, x = Kp1/2(0.025 - x)
x = 0.025 Kp1/2 - Kp1/2 x
x(1+Kp1/2) = 0.025 Kp1/2
= 0.021
Thus, at equilibrium, [CO(g)] = 0.025 - 0.021 = 0.004
[H2O(g)] = 0.025 - 0.021 = 0.004
[CO2(g)] = 0.021
[H2(g)] = 0.021
3. (b) ClO3- number of valence electrons = 7 + (3 x 6) + 1 = 26. The Lewis structure is therefore:
The structure is of the type AX3E, and must be a trigonal pyramid.
(b) ClF3 number of valence electrons = 7 + (3 x 7) = 28. The Lewis structure is therefore:
The structure is of the type AX3E2, and must be t-shaped.
(c) PCl4F number of valence electrons = 5 + (4 x 7) + 7 = 40. The Lewis structure is therefore:
The structure is of the type AX5, and must be trigonal bipyramidal.
(d) BrICl- number of valence electrons = 7 + 7 + 7 + 1 = 22. The Lewis structure is therefore:
The structure is of the type AX2E3, and must be linear.
4a. The oxidation half reaction is Br-(aq) à BrO3-(aq)
To balance the O, add H2O(l):
Br-(aq) + 3 H2O(l) à BrO3-(aq)
To balance the H, add H+(aq):
Br-(aq) + 3 H2O(l) à BrO3-(aq) + 6 H+(aq)
To balance the charge, add electrons:
Br-(aq) + 3 H2O(l) à BrO3-(aq) + 6 H+(aq) + 6 e-
The reduction half reaction is MnO4-(aq) à MnO2(s)
To balance the O, add H2O(l):
MnO4-(aq) à MnO2(s) + 2 H2O(l)
To balance the H, add H+(aq):
MnO4-(aq) + 4 H+(aq) à MnO2(s) + 2 H2O(l)
To balance the charge, add electrons:
MnO4-(aq) + 4 H+(aq) + 3 e- à MnO2(s) + 2 H2O(l)
To cancel the charge, multiply the reduction half reaction by 2 and add the oxidation reaction:
Br-(aq) + 3 H2O(l) à BrO3-(aq) + 6 H+(aq) + 6 e-
2 MnO4-(aq) + 8 H+(aq) + 6 e- à 2 MnO2(s) + 4 H2O(l)
__________________________________________________
Br-(aq) + 2 MnO4-(aq) + 2 H+(aq) à BrO3-(aq) + 2 MnO2(s) + H2O(l)
To balance in basic solution, add OH- to both sides, forming water:
Br-(aq) + 2 MnO4-(aq) + 2 H+(aq) + 2 OH-(aq) à BrO3-(aq) + 2 MnO2(s) + H2O(l) + 2 OH-(aq)
or, Br-(aq) + 2 MnO4-(aq) + H2O(l) à BrO3-(aq) + 2 MnO2(s) + 2 OH-(aq)
4b. 4.00 g Br-(aq) x (1 mol Br- / 79.9 g mol-1) x (2 mol MnO2 / 1 mol Br-)
x (86.9 g mol-1 / 1 mol MnO2) = 8.70 g MnO2(s)
5. The "wet gas" collected consists of a mixture of nitrogen and water vapor. However, both gases expand to fill the same container, so we need only calculate the volume of one of them. This is done by calculating the volume of the nitrogen gas. The partial pressure of the nitrogen gas = pN2 = 696 - 18.7 = 677.3 mm Hg. Converting to atmospheres, pN2 = 677.3 / 760 = 0.891 atm.
Thus, VN2 (= VH2O) = nN2RT/pN2. Here, nN2 = 1.28 g / (28.0 g mol-1) = 0.0457 mol
Thus, VN2 = 0.0457 mol (0.082 L atm K-1 mol-1)(294 K) / 0.891 atm
= 1.24 L
6a. 
For the F2 molecule, Bond Order = 
6b. For a molecule to exist, the bond order must be greater than zero. The F2- molecule will have an extra electron, which is placed in the s*2p molecular orbital. Its bond order is thus (8-7)/2 = 0.5, and F2- should therefore be possible.
For F2+, an electron has been removed from the p*2p molecular orbital. Its bond order is thus (7-6)/2 = 0.5, and so F2+ is also possible.
For F2+2, another electron must be removed from the p*2p molecular orbital. Its bond order is thus (8-6)/2 = 1, and so F2+2 is also possible.
6c. A paramagnetic species is one which has at least one unpaired electron. F2 has no unpaired electrons, and is therefore not paramagnetic. F2- has one in the s*2p molecular orbital, and so is paramagnetic. F2+ has one in the p*2p molecular orbital, and so is also paramagnetic.
6d. F2- has an extra electron in the s*2p molecular orbital relative to F2. Thus, the bond in F2- has more antibonding character than in F2. Thus, F2- will have a greater bond length and lower bond energy than F2.