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CHEM 1000 A, V Midterm Test #3, February 9, 2007.
Part A. Answer each question with a few sentences or equations where necessary. (5
Marks each)
1.
Name the strongest intermolecular force in each of the following pairs of species:
Species
Strongest Intermolecular Force
Two Xe atoms
Dispersion forces
H2O and NH3
Hydrogen-bonding
Cu+2 and CS2
Ion-Induced dipole forces
Two CHF3 molecules
Dipole-dipole forces
Two molecules of
O
Dipole-dipole forces (not H-bonds)
C
H3 C
2.
CH3
Why does H2O(l) have a higher surface tension than CO2(l)?
Since water is s polar molecule, the intermolecular potentials among water molecules
are greater than among the (linear) non-polar CO2 molecules. This stronger force in
water results in a large difference in energy between molecules at the surface and
those in the interior of the liquid.
3.
Why does a 1.00 m solution of NaCl(aq) have a lower boiling point than a 2.0 m
solution of a non-ionic solute?
Although the number of particles in these two solution are the same, the Na+(aq) and
Cl-(aq) ions form ion-pairs, effectively reducing the number of ions in solution, hence
reducing the boiling point elevation.
4.
Why is ice less dense than liquid water?
The water molecules in ice are arranged in an open hexagonal structure, held in place
by H-bonding. This open structure has a lower density than that of liquid water.
5.
A piece of iron anodized with zinc metal will not corrode even if some of the iron metal
is exposed to air and water. Why not?
The oxidation of zinc is electrochemically preferred. The high energy electrons
produced as a result of the zinc oxidation effectively prevent the oxidation of iron,
which oxidizes at a lower potential than zinc.
6.
When electrolyzing a solution of NaCl(aq), why is Na(s) not produced at the cathode?
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The reduction of water occurs at a higher potential than the reduction of aqueous
sodium ions, and so water reduction is favored.
Part B. Answer any three questions. If you answer all four, the best three will be used
to calculate your mark. (20 marks each)
B1. (a) Write the oxidation, reduction and overall reactions in a lead-acid battery. See the
reduction potential table on the last page of this test.
Oxidation: Pb(s) + HSO4-(aq)  PbSO4(s) + H+(aq) + 2 eReduction: PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2 e-  PbSO4(s) + 2 H2O(l)
Overall: Pb(s) PbO2(s) + 2 H+(aq) + 2 HSO4-(aq)  2 PbSO4(s) + 2 H2O(l)
(b) Calculate the standard cell potential for this battery (in Volts).
Eooxidation = -(-0.30) V = +0.30 V
Eoreduction = +1.63 V
Eocell = 0.30 + 1.63 = 1.93 V
(c) A fully charged battery has [H2SO4(aq)] = 10.0 M. Calculate its cell potential at 25oC.
In water, H2SO4(aq)  H+(aq) + HSO4-(aq).
Thus, in 10.0 M H2SO4(aq), [H+(aq)] = [HSO4-(aq)] = 10.0 M
For this reaction, Q 
1
1

 1.0  104

2
2
2
[H (aq) ] [HSO4 (aq) ]
10.0 10.0

2
Thus,
RT
8.314 JK 1mol1(298K)
ln(Q)  1.93V 
ln(1 104 )

 1
nF
2mole (96500C(mole ) )
 1.93V  0.12V  2.05V
E  Eo 
(d) The battery is discharged at 25oC until the cell potential is 1.70 V. Calculate [H2SO4(aq)].
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RT
ln(Q)
nF
RT
Eo  E 
ln(Q)
nF
nF o
ln(Q) 
(E  E)
RT
2mole  (96500C(mole  )1)

(1.93V  1.70V)
8.314 JK 1mol1(298K)
 17.9
E  Eo 
thus,Q  e17.9  6.04  107
but,Q 
1
[H (aq) ] [HSO4 (aq) ]2

2
thus,[H (aq) ]2 [HSO 4 (aq) ]2 
1
 1.66  10 8
Q
[H (aq ) ][HSO4 (aq) ]  1.66  10 8  1.29  10 4
[H (aq) ]  [HSO4 (aq) ]  1.29  10 4  0.0113 M
B2. An object is to be electroplated with gold from a solution of AuCl3(aq). If the
surface area of the object is 1 m2 and the desired thickness of the gold metal is 10.0
m. What current (in Amperes) must be used over a period of 1 hour? The density of
gold metal is 19.3 g cm-3.
Area of gold to be electroplated = 1m2 = 100 cm x 100 cm = 1 x 104 cm2
Thickness of gold = 10.0 m = 10 x 10-6 m = 10 x 10-4 cm = 1.0 x 10-3 cm
Volume of gold = 1 x 104 cm2 x 1.0 x 10-3 cm = 10 cm3
Mass of gold = 10 cm3 x 19.3 g cm-3 = 193 g
Moles of gold = 193 g / 197.0 g mol-1 = 0.98 moles Au
The reduction reaction is Au+3(aq) + 3 e-  Au(s), so the number of moles of electrons
required, n, is 3 x 0.98 = 2.94 moles e-.
i
nF 2.94 mol e ((96500C(mole )1)

 79Cs1  79 A
1
t
1h  3,600sh
B3. (a) Calculate the freezing AND boiling points (oC) of a solution made by dissolving 2.00
moles of MgCl2(s) in enough water to make 1.00 L of solution. The density of the
solution is 1.12 g/mL. For water, Kf = 1.86oC kg mol-1 and Kb = 0.51oC kg mol-1.
You can not assume that molality = molarity in this problem.
1.00 L of solution has a mass of 1.00 L x 1000 mL L -1 x 1.12 g mL-1 = 1120 g
2.00 moles of MgCl2 weighs 2.00 x (24.3 + 2(35.5)) = 190.6 g
Thus, the mass of water is 1120 – 190.6 = 929.4 g.
The molality of the solution is therefore 2.00 mol / 0.9294 kg = 2.15 m MgCl2.
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However, each mole of MgCl2 dissolves to give three ions. Thus, the total
concentration of ions in this solution is 3 x 2.15 = 6.45 m
Tc = 1.86 oC kg mol-1 x 6.45 mol = 12.0oC.
The freezing point of the solution is thus 0.0 – 12.0 = -12.0oC
Tb = 0.51 oC kg mol-1 x 6.45 mol = 3.3oC.
The freezing point of the solution is thus 100.0 + 3.3 = 103.3oC
(b) Calculate the vapor pressure above the solution at 25 oC. The vapor pressure of
pure water at 25oC is 23.8 mm Hg.
In 1.00 L of this solution, we have 929.4 g of water and 2.00 moles of MgCl2
929.4 g / 18.0 g mol-1 = 51.6 moles water
2.00 moles of MgCl2 gives 6.00 moles of ions
X water 
51.6
 0.896
51.6  6.00
Psoln = PosolventXsolvent = 23.8 mm Hg x 0.896 = 21.3 mm Hg
B4. (a) The vapor pressure of CS2(l) is 100 mm Hg at -5.1oC. Its normal boiling
temperature is 46.5oC. Calculate the enthalpy of vaporization of CS2(l) (in kJ mol-1).
ln(P2 )  ln(P1 ) 
Hvap  1 1 
  
R  T1 T2 
rearranging, Hvap


 ln(P )  ln(P ) 
2
2

R
1
1





T1 T2
Here, P1 = 100 mm Hg
P2 = 760 mm Hg (since the normal boiling point is at 1 atm pressure)
T1 = -5.1oC = 268.0 K
T2 = 46.5oC = 319.6 K
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Hvap


 ln(760)  ln(100) 
 8.314 JK mol 
 27,990 Jmol1  28.0kJmol1

1
1



 268.0 319.6 
1
1
(b) Use the phase diagram of HgI2 shown below to answer the following questions. Note that
 and  are two solid forms of HgI2.

(i) What is the normal melting point of HgI2?
259oC
(ii) What is the normal boiling point of HgI2?
354oC

L
1.0
G
P, atm
(iii) Is it possible for the two solid phases to be in
equilibrium with each other and with the liquid phase at
the same conditions?
No. (They could be in equilibrium with the gas phase, but
not with the liquid.)
157
259 354
T,oC
(iv) Describe what happens step by step if some solid HgI 2 at 25oC is placed in an open
beaker at 1 atm pressure and heated slowly to 400 oC.
Below 157oC, only the  phase will exist.
At 157oC the  phase will be in equilibrium with the -phase.
Between 157oC and 259oC, only the -phase will exist.
At 259oC the  phase will be in equilibrium with the liquid phase.
Between 269oC and 354oC, only the liquid phase will exist.
At 354oC, the liquid phase will be in equilibrium with the gas phase
Above 354oC, only the gas phase will exist.