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CHEM 1000 A and V
Midterm Test #2
November 11, 2005
Part A. Answer each question with a few sentences or equations where necessary. (5 Marks each)
1.
Under certain conditions, a reaction has Q < Keq. Does this reaction move to the left or the right in order
to attain equilibrium?
Since Q<Keq, the reaction must shift to the right (i.e. towards the products) to attain equilibrium.
2.
The bond length in the superoxide ion, O2¯, is 134 pm. Adding another electron creates the peroxide ion,
O22, which has a bond length of 143 pm. What can be said about the molecular orbital that this second
electron was placed in?
Since the bond length increased, the extra electron must have been added to an antibonding molecular
orbital.
3.
Referring to the activity series on page 5 of this test, can Al+3 ions oxidize Hg? (Answer yes or no!)
No! (Al is higher than Hg on the activity series, and so will itself get oxidized by Hg+2)
4.
Reaction A is twice as fast as reaction B. What can be said about the relative values of H0 of these two
reactions? How do you know?
Nothing! H is a state function, and has nothing to do with the rate of the reaction.
5.
Do the following processes have positive or negative values of S0?
CCl4(l) → CCl4(g)
PCl3(g) + Cl2(g) → PCl5(g)
KCl(s) (100oC) → KCl(s) (50oC)
KBr(s) + H2O(l) → KBr(aq)
CS2(l) → CS2(s)
6.
Positive
Negative
Negative
Positive
Negative
What are the oxidation states of each atom
in a molecule of CF2Cl2?
C +4
F -1
Cl -1
What are the oxidation states of each atom in a
MnO4¯ ion?
Mn +7
O -2
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Part B. Answer any two of the three questions (B1, B2 and B3). If you answer all three, the best two
marks will be used to calculate your overall mark. (20 marks each)
B1. Balance the following REDOX reaction in basic solution: MnO4¯ (aq) + C2O42(aq)  Mn+2(aq) + CO2(g)
Oxidation:
C2O42(aq)  CO2(g)
C2O42(aq)  2 CO2(g)
C2O42(aq)  2 CO2(g) + 2 e
Reduction:
MnO4¯ (aq)  Mn+2
MnO4¯ (aq)  Mn+2 + 4 H2O(l)
MnO4¯ (aq) + 8 H+(aq)  Mn+2 + 4 H2O(l)
MnO4¯ (aq) + 8 H+(aq) + 5 e¯  Mn+2 + 4 H2O(l)
Net Reaction:
5 (C2O42(aq)  2 CO2(g) + 2 e¯ )
2 (MnO4¯ (aq) + 8 H+(aq) + 5 e¯  Mn+2 + 4 H2O(l))
Or,
5 C2O42(aq)  10 CO2(g) + 10 e¯
2 MnO4¯ (aq) + 16 H+(aq) + 10 e¯  2 Mn+2 + 8 H2O(l))
____________________________________________
5 C2O42(aq) + 2 MnO4¯ (aq) + 16 H+(aq)  10 CO2(g) + 2 Mn+2 + 8 H2O(l)
Convert to Basic Solution:
5 C2O42(aq) + 2 MnO4¯ (aq) + 16 H+(aq) + 16 OH¯(aq)  10 CO2(g) + 2 Mn+2 + 8 H2O(l) + 16 OH¯(aq)
5 C2O42(aq) + 2 MnO4¯ (aq) + 16 H2O(l)  10 CO2(g) + 2 Mn+2 + 8 H2O(l) + 16 OH¯(aq)
5 C2O42(aq) + 2 MnO4¯ (aq) + 8 H2O(l)  10 CO2(g) + 2 Mn+2 + 16 OH¯(aq)
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B2. Given the reaction PCl3(g) + Cl2(g)  PCl5(g) and the thermodynamic data:
PCl5(g)
PCl3(g)
Cl2(g)
ΔHfo
(kJ/mol)
 374.9
 287.0
0.0
So (J K-1 mol-1)
364.5
311.7
205.0
(a) Calculate the value of ΔHo for the reaction (in kJ mol-1).
ΔHo = ΔHfo(PCl5(g)) – [ΔHfo(PCl3(g)) + ΔHfo(Cl2(g))]
= –374.9 – [–287.0 + 0.0]
= – 87.9 kJ mol-1
(b) Calculate the value of ΔSo for the reaction (in J K-1 mol-1). Comment on why the value is positive or negative.
ΔSo = So(PCl5(g)) – [So(PCl3(g)) + So(Cl2(g))]
= 364.5 – [311.7 + 205.0]
= – 152.2 J K-1 mol-1
ΔSo is negative because the reaction involves reduction in the number of moles of gas.
(c) Kp = 166 for this reaction at 1000 K. 0.5 atm of PCl3(g) and 0.5 atm of Cl2(g) are introduced into a reaction
vessel at 1000 K. When equilibrium is attained, what are the partial pressures of the three gases (in atm)?
Kp 
p PCl5( g )
p Cl2( g ) p PCl3( g )
 166
At equilibrium, pPCl5 = pCl2 = 0.5 – x and pPCl5 = x
x
 166
(0.5  x)(0.5  x)
166(0.25  x  x 2 )  x
41.5  166x  166x 2  x
Thus, 166x 2  167x  41.5  0
(167)  167 2  4(166)(41.5)
x
2(166)
167  18.2

 0.528or 0.448atm
332
Using x = 0.528 results in negative pressures for the reactants. Thus,
pPCl3 = pCl2 = 0.5 – 0.448 = 0.052 atm and pPCl5 = x = 0.448 atm
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B3. Answer the following questions with respect to the molecular orbital diagram shown below for the NO
molecule:
(a) Calculate the bond orders of NO, NO+, NO¯
NO
bond order = (8 – 3) / 2 = 2.5
NO+
bond order = (8 – 2) / 2 = 3
2p
NO¯
bond order = (8 – 4) / 2 = 2
*2p
2p
+
(b) Which of NO, NO and NO¯ are
paramagnetic? (Answer ‘yes’ or ‘no’ for each.)
NO
Yes
NO+
No
NO¯
Yes
(c) Which of NO and NO+ has the greater bond
length? How do you know?
Ionization of NO to NO+ involves removal of the
electron in the *2p orbital. This causes the bond
order to increase, hence the bond length in NO is
greater than that in NO+.
(d) Which of NO and NO¯ has the greater bond
energy? How do you know?
Ionization of NO to NO¯ involves addition of an
electron to the *2p orbital. This causes the bond
order to decrease, hence the bond energy in NO is
greater than that in NO¯.
2p
*2s
2s