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CHEM 1000 A, V Midterm Test #2
November 17, 2006
Part A. Answer each question with a few sentences or equations where necessary. (5 Marks each)
1.
When a particular molecule is positively ionized, the bond length is observed to increase. What
can be said about the molecular orbital from which the electron was removed?
The electron must have been removed from a bonding molecular orbital.
2.
ΔGo for reaction A is -50 kJ mol-1. ΔGo for reaction B is -75 kJ mol-1. Which reaction is faster?
How do you know?
Nothing can be said about the rates of these two reactions. G is a state function and does not
depend on the reaction rate.
3.
For the reaction initially at equilibrium: 2 SO2(g) + O2(g) Ý 2 SO3(g), which way will the equilibrium
shift (left or right) if the temperature is lowered? Why?
This is an oxidation reaction, which generally are exothermic. If the temperature is lowered, the
reaction will always shift in the exothermic direction, which is to the RIGHT in this case.
4.
Why are the concentrations of solid materials not included in equilibrium constant expressions?
The concentrations of solids and liquids do not change during a reaction, so their values are left
out of the expression for the equilibrium constant.
Predict whether ΔH is positive or negative and whether ΔS is positive or negative in each case:
ΔH (Check One!) ΔS (Check One!)
Process
Positive Negative Positive Negative


CH4(g) + 3 O2(g)  CO2(g) + 2 H2O(g)


Returning books to the library. (You are part of the system
in this process!)


Dissolving NaOH(s) in water. (The solution is observed to
get very hot.)


H2O(l) (at 100oC)  H2O(g) (at 100oC)
o
o


H2O(l) (at 100 C)  H2O(l) (at 95 C)
5.
6.
Name two advantages of using hydrogen as a fuel in cars.
Combustion of hydrogen has only water as a waste product, and has a relatively high heat of
combustion.
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Part B. Answer two questions. 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: CN-(aq) +MnO4-(aq)  CNO-(aq) + MnO2(s)
Oxidation:
CN-(aq)  CNO-(aq)
CN (aq) + H2O(l)  CNO (aq)
CN- + H O  CNO- + 2 H+
(aq)
2
(l)
(aq)
(aq)
CN-(aq) + H2O(l)  CNO-(aq) + 2 H+(aq) + 2 e-
Reduction :
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)
Overall:
3(CN-(aq) + H2O(l)  CNO-(aq) + 2 H+(aq) + 2 e-)
+ 2(MnO4-(aq) + 4 H+(aq) + 3 e-  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) + 6 e  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:
Add OH-(aq) to both sides:
3 CN- + 2 MnO - + 2 H+
(aq)
4 (aq)
(aq)
+ 2 OH-(aq)  3 CNO-(aq) + 2 MnO2(s) + H2O(l) + 2 OH-(aq)
Make water:
3 CN- + 2 MnO -
+ 2 H2O(l)  3 CNO-(aq) + 2 MnO2(s) + H2O(l) + 2 OH-(aq)
Cancel waters:
3 CN- + 2 MnO -
+ H2O(l)  3 CNO-(aq) + 2 MnO2(s) + 2 OH-(aq)
(aq)
(aq)
4 (aq)
4 (aq)
3
B2. Use the data below to answer the following questions about the reaction:
C2H4(g) + H2O(g) Ý C2H5OH(l)
ΔHof (kJ mol-1)
So (J K-1 mol-1)
C2H4(g)
52.0
219
H2O(g)
-242
189
C2H5OH(l)
-278
161
a. Calculate ΔHo for the reaction (kJ mol-1).
ΔHo = ΔHof(C2H5OH(l)) – (ΔHof(C2H4(g)) + ΔHof(H2O(g)))
= -278 – (52.0 + (-242)) kJ mol-1
= - 88.0 kJ mol-1
b. Calculate ΔSo for the reaction (J K-1 mol-1).
ΔSo = So(C2H5OH(l)) – (So(C2H4(g)) + So(H2O(g)))
= 161 – (219 + 189) kJ mol-1
= - 247 J K-1 mol-1
c. Calculate ΔGo for the reaction (kJ mol-1).
ΔGo specifically means under standard conditions, i.e. 25 oC (298 K). Thus,
ΔGo = ΔHo - TΔSo
= – 88,000 J mol-1 – 298 K (-247 J K-1 mol-1)
= –14,394 J mol-1
= –14.4 kJ mol-1’
d. Is this reaction spontaneous at 25oC? How do you know?
Yes, since ΔG < 0 under these conditions.
e. Calculate the temperature at which ΔG=0 for this reaction (in oC)
Ho 88,000Jmol1
ΔG=0 when T 

 356K  83o C
o
1
1
S
247JK mol
f. Calculate Kp for the reaction at 100oC.
For this reaction, K p 
1
, since the product is a liquid. Thus, since all species in the
pC2H4 pH2O
 Go 
expression are gases, the expression exp 
 will give us Kp directly:
 RT 
 Go  

( 14,394Jmol1 )
Kp  exp 

 
  103.7
1
1
 RT   8.314JK mol (273  100)K 
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B3. For the reaction 2 NO2(g) Ý N2O4(g), Kc = 4.00 at
If 2.0 moles of NO2(g) are placed in a 1.0 L
flask at 25oC, calculate the equilibrium concentrations (mol L-1) of each gas.
25oC.
In this reaction, although all species are gases, the concentrations will be in mol / L, so the
equilibrium expression is:
Kc 
[N2O4(g) ]
[NO2(g) ]2
 4.00
Initially we have 2.0 moles NO2(g) in a 1.0 L vessel, so its concentration is 2.0 mol/L:
Initial
Change
Equilibrium
2 NO2(g) N2O4(g)
2.0
0
-2x
+x
2.0-2x
x
Thus at equilibrium,
Kc 
[N2O4(g) ]
[NO2(g) ]
2

x
 4.00
(2.0  2x)2
x  4(4  2x  2x  4x 2 )
x  16  32x  16x 2
16x 2  33x  16  0
or, 16x 2  33x  16  0
This is of the form ax2 + bx + c = 0, where a = 16, b = -33 and c = 16. We use the quadratic formula
to find the possible values of x:
x
b  b2  4ac
2a
33  ( 33)2  4(16)(16)
2(16)
 0.78 or 1.28

Only x = 0.78 will give positive concentrations for both species in the reaction. Thus,
[NO2(g)] = 2.0 – 2x = 2.0 – 2(0.78) = 0.44 mol/L
[N2O4(g)] = x = 0.78 mol/L
Check:
x
0.78

 4.01  K c
2
(2.0  2x)
(2.0  2(0.78))2