CHEM 1000 A and V
Mid-Year Examination
December, 2009
Mid Year Exam December 12, 2009
Part A. Answer all questions (5 marks each).
A1. Why is light of only certain wavelengths observed in an atomic spectrum?
The light is a result of transitions of electrons from one quantized energy to another. Since the energy levels are
quantized, so are the differences between them, and so are the photon energies.
A2. Why is a molecule with a triple bond in it likely to be very reactive?
A triple bond consists of one sigma-bond plus two pi-bonds. The pi-electrons are delocalized (far from the
nuclei) and so are not tightly bound. Since they are easily removed, this makes the molecule very reactive.
A3. What type of hybrid orbitals is the As atom using to bond with the Cl atoms in AsCl3?
5 + (3 x 7) = 26 electrons. Thus it is AX3E, i.e. the As atoms has four electron groups around it. Thus, it is using
sp3 hybrid orbitals.
A4. Can Ca(s) be used as a reducing agent for Ni+2(aq)? Why or why not?
Yes. Ca is above Ni in the activity series. Thus, Ca is more easily oxidized that Ni. Thus, Ca gets oxidized and
Ni+2 gets reduced.
A5. Explain why a reaction having ΔH < 0 tends to make the entropy of the universe increase.
The heat from the system moves into the surroundings. This increases the entropy of the surroundings. This in
turn increases the entropy of the Universe, since the surroundings are part of the Universe.
A6. Suppose the reaction Zn(s) + Sn+2(aq) Ý Zn+2(aq) + Sn(s) is at equilibrium. Will diluting the solution with water
have an effect on the position of the equilibrium? Why or why not?
Dilution will have no effect since Q = [Zn+2] / [Sn=2], and dilution will not change the value of Q.
1
CHEM 1000 A and V
Mid-Year Examination
December, 2009
A7. Dissolving CuCl2(s) in H2O(l) to make Cu+2(aq) and Cl(aq) ions has ΔSo < 0. Why?
Although the ions in CuCl2(s) have low entropy, making the aqueous ions reduces the entropy of the associated
water molecules even more.
A8. CuCl2(s) spontaneously dissolves in water even though ΔSo < 0. Why?
Because ΔHo < 0
A9. Why are the ionization potentials of phosphorus (P) and sulphur (S) approximately equal?
P has three unpaired p-electrons. Although S has four, and the fourth should be more easily removed since it is
paired with one other, thus decreasing the ionization potential, the S atom also has one more proton, making the
effective nuclear charge greater, increasing the ionization potential. The net effect is no change.
A10. Give two reasons why the van der Waals equation more accurately predicts gas pressure than the ideal gas
law, especially at high pressures.
The van der Waals equation takes account of molecular volumes and intermolecular attractive forces, neither of
which the ideal gas law does.
A11. Write the four reactions collectively known as the Chapman mechanism for ozone in the stratosphere.
O2 + h → O + O
O + O2 → O3
O3 + h → O2 + O
O + O3 → 2 O2
A12. Give two reasons why 1 kg of CH4(g) causes more warming of the atmosphere than 1 kg of CO2(g).
1. CH4 has a lower molecular weight, so 1 kg of it is more moles than 1 kg of CO2
2. CH4 has more modes of vibration, hence absorbs more wavelengths of IR
2
CHEM 1000 A and V
Mid-Year Examination
December, 2009
Part B. Answer all questions on this page (20 marks total).
B1. Ammonium nitrate can be used as an explosive. It decomposes according to:
NH4NO3(s) → N2(g) + 2 H2O(g) + ½ O2(g).
Assuming the gases behave ideally, calculate the total volume of the products at 300oC and 1 atm when 12.0 g
of ammonium nitrate decomposes.
 12 g 

  0.15 mol NH 4 NO3
 80 g / mol 
x 3.5 mol gas / mol NH4NO3 = 0.525 mol gas
V
nRT 0.525 mol(0.082 L atm mol 1K 1)(300  273)K

 24.7 L
p
1 atm
B2. A sample of an unknown gas diffuses in 11.1 minutes. An equal volume of H2(g) in the same apparatus
under the same conditions effuses in 2.42 minutes. Calculate the molecular weight of the unknown gas.
time?? rate H2
MW??


time H2 rate??
MWH2
MW??  time??

MWH2  time H2



2
 time??
MW??  MWH2 
 time H

2
2
2

1  11.1 
1
  2.02 g mol 
  42.5 g mol
2
.
42



B3. Calculate the pressure exerted by 1.00 mol of CH4(g) in a 500 mL vessel at 25.0oC using:
(a) The ideal gas law
nRT 1.00 mol(0.082 L atm K 1mol 1)(25  273)K

 48.9 atm
V
0.500 L
(b) The van der Waals equation. For CH4(g), a = 2.25 L2 atm mol-2 and b = 0.0428 L mol-1
p
p
nRT
n
a 
V  nb
V
2
1.00 mol(0.082 L atm K 1mol 1)(25  273)K
L2atm  1.00 mol 


2
.
25


0.500 L  1.00 mol(0.0428 L mol1)
mol2  0.500 L 
2
 53.45 atm  9.00 atm
 44.5 atm
3
CHEM 1000 A and V
Mid-Year Examination
December, 2009
Part C. Answer any five of questions C1 – C7 (20 marks each). If you answer more than five, the best five
will be used to calculate your total mark.
C1. Nitrogen dioxide decomposes according to 2 NO2(g) Ý 2 NO(g) + O2(g). At a certain temperature, Kp is
extremely small (4.4810-13). If 0.75 atm of NO2(g) is placed in a vessel at this temperature, find the equilibrium
partial pressures of all three gases.
Initial, atm
Change, atm
Equilibrium, atm
p(NO2), atm
0.75
-2x
0.75 – 2x
p(NO), atm
0
+2x
2x
p(O2), atm
0
+x
x
At equilibrium,
p NO 2 p O2
 Kp
p NO2 2
(2x) 2 x
 4.48  10 13
2
(0.75  x)
Here, Kp is so small, we can make the approximation that x << 0.75, so the expression simplifies to
(2x)2 x
 4.48  10 13
(0.75)2
4 x 3  (0.75)2 (4.48  10 13 )
x 3  (0.75)2 (4.48  10 13 ) / 4
x 3  6.3  10 14
x  3.98  10 5
Thus, pNO2 = 0.75 – 2x = 0.75 atm
pNO = 2x = 7.96 x 10-5 atm
pO2 = x = 3.98 x 10-5 atm
Check: (7.96 x 10-5)2(3.98 x 10-5) / (0.75 – 2(3.98 x 10-6))2 = 4.48 x 10-13 = Kp
4
CHEM 1000 A and V
Mid-Year Examination
December, 2009
C2. Given the following information, calculate the ionization energy of lithium (in kJ mol-1):
Lattice energy of LiF(s)
Electron affinity of fluorine
ΔHfo (LiF(s))
Enthalpy of sublimation of lithium metal
Bond dissociation energy of fluorine
1050 kJ mol-1
 328 kJ mol-1
 617 kJ mol-1
161 kJ mol-1
159 kJ mol-1
Li(s) → Li(g)
+161 kJ mol-1
Li(g) → Li+(g) + e-
I1 (Li)
½ F2(g) → F(g)
½ (+159 kJ mol-1)
F(g) + e- → F-(g)
-328 kJ mol-1
Li+(g) + F-(g) → LiF(s)
-(1050) kJ mol-1
_______________________________________
Li(s) + ½ F2(g) → LiF(s)
-617 kJ mol-1
Thus, the ionization potential of lithium is found by difference:
I1(Li) = -617 – (+161 + ½ (159) – 328 – 1050) = 520.5 kJ mol-1
5
CHEM 1000 A and V
Mid-Year Examination
December, 2009
C3. (a) Two possible structures for nitrosyl chloride are Cl=NO and ClN=O.
(i)
Draw complete Lewis diagrams for each of these structures.
::Cl=N:-O:::
(ii)
:::Cl-N:=O::
Calculate formal charges and decide which structure is more likely:
Cl=NO
Enter the
formal charge
of each atom:
More likely
structure
(check one!)
ClN=O
Cl
N
O
Cl
N
O
+1
0
-1
0
0
0
More likely!
(b) What hybridization is each of the four indicated atoms using in the molecule of Aspirin shown below?
1
H
3
H
C C
O
H C
C C O H
C C
H
H
H
O C C
O H
4
Atom 1:__sp2 ___
Atom 2:__sp3_____
2
Atom 3:___sp2____
Atom 4:___sp3____
(Don’t forget that there are lone pairs on atoms 3 and 4!)
6
CHEM 1000 A and V
Mid-Year Examination
December, 2009
C4. Use the data in the table to answer the following questions about the reaction 2 NO2(g) Ý N2O4(g).
ΔHfo, kJ
o
-1
NO2(g)
33.2
240.0
-1
mol
S , J K mol-1
N2O4(g)
9.16
304.2
(a) Calculate ΔHo (kJ mol-1)
ΔHo = ΔHfo(N2O4(g)) - 2 ΔHfo(NO2(g))
= 9.16 – 2(33.2)
= -57.24 kJ mol-1
(b) Calculate ΔSo (J K-1 mol-1)
ΔHo = So(N2O4(g)) - 2 So(NO2(g))
= 304.2 – 2(240.0)
= -175.8 J K-1 mol-1
(c) Calculate ΔGo (kJ mol-1)
ΔGo = ΔHo – TΔSo
= -57,240 J mol-1 – (25 + 273)K(-175.8 J K-1 mol-1)
= -4851.6 J mol-1
= -4.85 kJ mol-1
(d) Calculate Kp at 50oC
Kp  e
 G o 
 RT 


e


4851 J mol1
 8.314 J K 1mol1 ( 50  273 )K 


 6.09
(e) If the total pressure at 50oC is 1 bar, calculate the partial pressure of each gas.
Since the total pressure is 1 bar, then pNO2 + pN2O4 = 1
But, K p  6.09 
p N 2 O4
p NO2
2
Thus, pN2O4 = 6.09(pNO2)2
2
pNO2 + 6.09(pNO2) = 1
6.09(pNO2)2 + pNO2 – 1 = 0
Solving this quadratic, pNO2 = 0.331 atm
pN2O4 = 1 – pNO2 = 1bar - 0.331 bar = 0.669 bar
Check:
p N 2 O4
p NO2
2

0.669
 6.09  K p
0.3312
7
CHEM 1000 A and V
Mid-Year Examination
December, 2009
UV light
C5. (a) Ozone (O3) is destroyed in the upper atmosphere according to the reaction O3(g) 
O2(g) + O(g),
-1
which requires breaking one of the bonds in the ozone molecule. If the bond energy is 445 kJ mol , what is the
maximum wavelength of photon capable of doing this (in nm)?
445 kJ mol-1 / 6.02 x 1023 mol-1 = 7.39 x 10-22 kJ per bond = 7.39 x 10-19 J per bond
c

hc 6.63  10 34 J s (3.00 108 m s 1)
thus,  

 2.69  10 7 m  269 nm
19
E
7.39  10 J
E  h  h
(b) We know about s, p, d and f-orbitals. Electrons will begin to fill g-orbitals after the f-orbitals are filled. Use
Aufbau and your knowledge of quantum numbers to determine the atomic number of the lightest element (as yet
undiscovered) that will contain an electron in a g-orbital.
Orbitals are filled according to the Aufbau principle. We can determine when the first g-orbital will accept an
electron accordingly. The first g-orbital will fill in the n=5 principle shell, and energy-wise, just after the 8s
orbital fills. Thus, count the electrons up to that point and add one:
1s
2s
3s
4s
5s
6s
7s
8s
2p
3p
4p
5p
6p
7p
8p
3d
4d
5d
6d
7d
8d
4f
5f
6f
7f
8f
5g
6g
7g
8g
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f145d10 6p6 7s2 5f14 6d10 7p6 8s2 5g1
And we see that the 5g electron is number 121, i.e. element 121 will be the lightest element containing this
electron.
8
CHEM 1000 A and V
Mid-Year Examination
December, 2009
C6. (a) A sample of a gas is at 50oC in a 50.0 L container at 1.1 atm. The gas is expanded to 75 L and cooled to
0oC. Calculate the new pressure.
p1V1 p2 V2

T1
T2
Thus, p 2 
p1V1T2 1.1 atm(50 L)(273 K)

 0.62 atm
T1V2
(50  273) K (75 L)
(b) 1.00 mol of He(g) under certain conditions exerts a pressure of 5.00 atm. 1.00 mol of Cl2(g) under the
same conditions exerts a lower pressure. Why?
Chlorine is a larger molecule than the very small He atoms. Larger molecules have larger attractive forces
for one another. Hence the Cl2 molecules are pulled together, reducing the pressure they exert.
(c) Calculate the density of Cl2(g) at 10.0 atm and 100oC (in g/L).

p(MW)
10 atm(0.070 kg mol1)

 2.29  102 kg L1  22.9 g L1
1
1
RT
0.082 L atm K mol (100  273)K
9
CHEM 1000 A and V
Mid-Year Examination
December, 2009
C7. Predict the shapes of the following molecules. Wrong name = zero marks.
(a) AsCl52
Electrons = 5 + (5 x 7) + 2 = 42. Making the five bonds uses 10 of these. Completing the octets on the Cl
atoms uses another 30. This leaves two, which are placed as a lone pair on the As atom. Thus the
shorthand notation is AX5E. This is a square pyramid.
(b) PCl5
Electrons = 5 + (5 x 7) = 40. Making the five bonds uses 10 of these. Completing the octets on the Cl
atoms uses the other 30. Thus the shorthand notation is AX5. This is a trigonal bipyramid.
(c) O3
Electrons = 3 x 6 = 18. Making the two single bonds uses 4 of these. Completing the octets on the
terminal atoms uses another 12, leaving two as a lone pair on the central O atom. Thus the shorthand
notation is AX2E, which is bent. (note that moving lone pairs inwards to make double bonds does not
change the shape).
(d) CF3
Electrons = 4 + (3 x 7) + 1 = 26. Making the three bonds uses 6 of these. Completing the octets on the F
atoms uses another 18. This leaves two, which are placed as a lone pair on the C atom. Thus the
shorthand notation is AX3E. This is a trigonal pyramid.
10