Chemistry 65.100 A, V
Midterm Test #2
November 9, 2001

Part A. Answer each question with a few sentences or equations where necessary. (5 Marks each) 

1.                  What is the oxidation number of the Cl atom in KClO3?
The oxygens are -2 each, for a total of -6. The K is +1. The overall charge is 0, thus the Cl atom must be +5

2.                  Why will the reaction Cl2(g) + Br2(l) ? 2 Cl¯(aq) + 2 Br¯ (aq) not proceed?
Both the Cl2 and the Br2 are being reduced in this reaction. Nothing is being oxidized.

3.         Which of K(s) or Cu(s) would you expect to be a stronger reducing agent? Why?
K is further to the left in the periodic table, and so we expect it to be less electronegative, i.e. less likely to attract electrons to itself, i.e. more likely to lose electrons, i.e. a stronger reducing agent.    

4.         Define the term “state function” and give an example of one.
A state function is one whose value does not depend on the rate of reaction or the path taken from reactants to products, but only on the identity of the reactants and products. Examples are enthalpy (H), entropy (S) and free energy (G).

5.           Under what circumstances can an endothermic reaction be spontaneous?
An endothermic reaction is spontaneous only if DS>0 (and then only if TDS > DH).

6.                  Assume the reaction SO2(g) + ½ O2(g) = SO3(g) is initially at equilibrium. If some additional oxygen is added, which direction (left or right) will the equilibrium shift? Why?
The equilibrium will shift to the right to use up some of the added oxygen, according to le Chatelier’s principle.

Part B. Answer each question. The best two marks will be used to calculate your overall mark. (20 marks each)

 1.         Balance the following REDOX reaction in basic solution: MnO4¯(aq) + SO3¯(aq) ® MnO2(s) + SO4-2(aq)

 For the oxidation half-reaction: 
SO3-(aq) ® SO4-2(aq)
SO3-(aq) + H2O(l) ® SO4-2(aq)
SO3-(aq) + H2O(l) ® SO4-2(aq) + 2 H+(aq)
SO3-(aq) + H2O(l) ® SO4-2(aq) + 2 H+(aq) + e-    [5 marks]

For the reduction half reaction: 
MnO4-(aq) ® MnO2(s)
MnO4-(aq) ® MnO2(s) + 2 H2O(l)
MnO4-(aq) + 4 H+(aq) ® MnO2(s) + 2 H2O(l)
MnO4-(aq) + 4 H+(aq) + 3 e- ® MnO2(s) + 2 H2O(l)     [5 marks]

 Multiply the oxidation half-reaction by 3 so each half reaction has 3 electrons and add them together:
 3 SO3-(aq) + 3 H2O(l) ® 3 SO4-2(aq) + 6 H+(aq) + 3 e-
MnO4-(aq) + 4 H+(aq) + 3 e- ® MnO2(s) + 2 H2O(l)
__________________________________________________
3 SO3-(aq) + MnO4-(aq) + H2O(l) ® MnO2(s) + 3 SO4-2(aq) + 2 H+(aq)      [5 marks]

 To balance the reaction in basic solution, add OH-(aq) to both sides to react with the H+ to make water:

3 SO3-(aq) + MnO4-(aq) + H2O(l) + 2 OH-(aq) ® MnO2(s) + 3 SO4-2(aq) + 2 H+(aq) + 2 OH-(aq)
3 SO3-(aq) + MnO4-(aq) + H2O(l) + 2 OH-(aq) ® MnO2(s) + 3 SO4-2(aq) + 2 H2O(l)
3 SO3-(aq) + MnO4-(aq) + 2 OH-(aq) ® MnO2(s) + 3 SO4-2(aq) + H2O(l)     [5 marks]

 2.  Answer the following questions with respect to the molecular orbital diagram for O2 shown below.
(a) Calculate the bond order for the O2 molecule.
(b) Calculate the bond order of the O2¯ ion.
(c) Calculate the bond order of the O2+2 ion.
(d) Which of O2, O2¯ and O2+2 are paramagnetic?
(e) Why are the two electrons in the p*2p orbital unpaired?
(f) Which has the shorter bond length: O2¯ or O2? Why?


(a) The bond order of the O2 molecule is (8-4)/2 = 2   [4 marks]
(b) To make O2-, we must add an electron to the highest energy orbital, i.e. the p*2p. Then the bond order is (8-5)/2 = 1.5    [4 marks]
(c) To make the O2+2 ion, we must remove the two highest energy electrons, i.e. those in the p*2p orbital. The bond order will then be (8-2)/2 = 3    [4 marks]
(d) Only O2 and O2- are paramagnetic.    [4 marks]
(e) Hund’s rule states that electrons in degenerate (same energy) orbitals must having the maximum unpairing possible.    [2 marks]
(f) The bond order of the O2- ion is less than that of the O2 molecule. Thus, the bond length of the O2- ion is greater.    [2 marks]

3.          Given the reaction 4 NH3(g) + 5 O2(g) ® 4 NO(g) + 6 H2O(g) and the thermodynamic data: 

 

?Hfo (kJ/mol)

So (J K-1 mol-1)

NH3(g)

-46.1

192.3

O2(g)

0

205.0

NO(g)

90.2

210.7

H2O(g)

-241.8

188.7

           
(a)    Calculate the value of DHo for the reaction (in kJ/mol).
DHo = [4 DHfo(NO(g)) + 6 DHfo (H2O(g))] – [4 DHfo(NH3(g)) + 5 DHfo (O2(g))]
= [4(90.2) + 6(-241.8)] – [4(-46.1) + 5(0)]
= -905.6 kJ mol-1     [6 marks]

(b)    Calculate the value of DSo for the reaction (in J K-1 mol-1)
DSo = [4 So(NO(g)) + 6 So (H2O(g))] – [4 So(NH3(g)) + 5 So (O2(g))]
= [4(210.7) + 6(188.7)] – [4(192.3) + 5(205.0)]
= +180.8 J K-1 mol-1      [6 marks]

(c)    Calculate the value of DGo for the reaction (in kJ/mol) at 298 K.
DGo = DHo - T DSo
= -905600 J mol-1 – 298 K (180.8 J K-1 mol-1)
= -959478 J mol-1
= -959.5 kJ mol-1     [6 marks]

(d)    Is this a spontaneous reaction at 298 K? Why or why not?

The reaction is spontaneous at 298 K because DGo < 0.     [2 marks]