CHEM 1000 Mid-Year Exam
December 2002
Part A. 60 marks. Answer each question (5 marks each).
1.
State the Pauli Exclusion Principle.
No two electrons in a single atom may have the same four quantum numbers.
2.
Which of LiF(s) and LiCl(s) has the larger lattice energy and why? Which of KF(s) and SrF2(s) has the
larger lattice energy and why?
ULiF > ULiCl since F¯ is a smaller ion than Cl¯, resulting in a smaller bond length, hence a larger
electrostatic energy
USrF2 > UKF since Sr+2 is doubly charged, whereas K+ is only singly charged, resulting in a larger
electrostatic energy.
3.
In the reaction NO3¯(aq) + 4 Zn(s) + 7 OH¯(aq) + 6 H2O(l)  4 Zn(OH)4¯(aq) + NH3(aq), what atom is being
oxidized and what atom is being reduced? State the oxidation numbers of each before and after
oxidation or reduction.
The N atom gets reduced from +5 in NO3¯ to -3 in NH3.
The Zn atom gets oxidized from 0 in Zn(s) to +3 in Zn(OH)4¯(aq)
4.
The symbol for standard enthalpy of formation is H°f. Explain why the symbol for standard entropy is
S° and not S°f.
The standard enthalpy of formation is the reaction enthalpy upon formation of the compound from the
elements at standard state, i.e. it is a change in enthalpy. A standard entropy is the entropy of a substance
at standard state, and does not represent a change in entropy.
5.
A reaction has a positive value of H and a negative value of S. Will this reaction be spontaneous at
high temperature, low temperature, both or neither? How do you know?
Since G = H - TS, G will be positive at all values of T, and the reaction will therefore not be
spontaneous at high or low temperature
6.
Write the four oxygen-only reactions in the Chapman cycle.
O2 + hυ  O + O
O + O2  O3
O3 + hυ  O2 + O
O + O3  2 O2
7.
Which is larger: the van der Waals constant ‘a’ for Ne(g) or that of Kr(g)? Explain any differences.
Do the same for the ‘b’ constant for these two gases.
aKr > aNe, since Kr is a larger, more polarizable atom
bKr > bNe, since Kr is a larger atom
8.
To maximize the production of CO(g) using the reaction C(s) + CO2(g) = 2 CO(g), should the reaction be
run at high or low pressure? Explain.
In this reaction, Δn = +1. According to Le Chatelier’s principle, decreasing the pressure will cause the
equilibrium position to move to the right, i.e. the products. Thus, the reaction should be run at lower
pressure to maximize the production of CO(g).
9.
Why does a small amount of a chlorofluorocarbon injected into the atmosphere cause the destruction of
a much larger amount of ozone?
The Cl atoms enter in a cyclic destruction process with ozone, i.e. the Cl atoms are catalytic, and are not
consumed by these reactions. Specifically,
Cl + O3  ClO + O2
ClO + O  Cl + O2
10.
PCl5 has 10 valence electrons around the P atom. How is this atom able to break the “octet rule”?
The P atom uses its d-orbitals as well as its valence p-orbitals to make hybrid orbitals, i.e. it has an
“expanded octet”.
11.
In a molecule of NO, the highest energy electron is alone in a *2p molecular orbital. Is the bond energy
of NO greater or less than that of NO? How do you know?
Since the electron will be removed from this antibonding MO during ionization, the bond energy of NO+
will be greater than that of NO.
12.
State whether each of the following processes is endothermic or exothermic:
(a) C8H18(l) + 12.5 O2(g)  8 CO2(g) + 9 H2O(g)
exothermic
(b) CO2(s)  CO2(g)
endothermic
(c) H2O(g)  H2O(l)
exothermic
+
(d) NaBr(s)  Na (g) + Br¯(g)
endothermic
(e) Hg(l) at 25°C  Hg(l) at 50°C
endothermic
Part B. Answer both questions. (20 marks each, partial marks in [ ])
1. Analysis of a newly discovered gaseous compound containing only Si and F shows that it contains 33.01% Si
by mass. At 27oC, 2.60 g of this compound exerts a pressure of 1.50 atm in a 0.250 L vessel. What is the
molecular formula of the gas? Assume ideal gas behaviour.
n
[8]
PV
RT
150
. atm  0.250 L
0.082 Latm K 1mol 1  (27  273) K
 0.0152 mol

2.60 g
 170.6 g mol 1
0.0152 mol
[2]
Thus, the molecular weight is MW 
[5]
Assume a 100 g sample. It therefore contains 33.01 g Si and (100.00 – 33.01) = 66.99 g F
Moles Si = 33.01g / 28.09 g mol-1 = 1.18 mol Si
Moles F = 66.99 g / 18.99 g mol-1 = 3.53 mol F
[5] The empirical formula is therefore SiF3. This compound would have a molecular weight of 28.09 +
3(18.99) = 85.1 g mol-1. The molecular formula, which must have a molecular weight of 170.6 g mol-1, must
therefore be Si2F6.
2. Chlorine gas, Cl2(g), is produced from seawater via the “chlor-alkali” process. The gas is stored in
containers to prevent unwanted and explosive reactions. If a 15.0 L container holds 0.580 kg Cl2(g) at 200oC:
(a) Calculate the pressure assuming ideal behaviour.
n
[6]
580g
 817
. mol
710
. g mol 1
nRT 817
. mol  0.082 Latm K 1mol 1  (200  273) K

V
15.0 L
 211
. atm
P
(b) Calculate the pressure using the van der Waals equation. For Cl2(g), a = 6.49 atm L2 mol-2 and b =
0.0562 L mol-1.
nRT
 n
P
 a 
V  nb  V 
[10]
2
817
.  0.082  (200  273)
. 
 818

 6.49

 15.0 
15.0  817
.  0.0562
 2179
.  193
. atm
 19.9 atm
2
(c) Why are the two calculated pressures so different?
[4]
The van der Waals equation takes account of the fact that the chlorine molecules have volume and
interact.
Part C. Attempt all five questions. The best four will be used to calculate your mark. (20 marks each)
3. At 298 K, Kp = 0.0900 for the reaction H2O(g) + Cl2O(g)  2 HOCl(g).
[15]
(a) Calculate the equilibrium partial pressure of each gas if initially pH2O = pCl2O = 100 Torr and pHOCl =
0.
Initial, Torr
Change, Torr
Equilibrium, Torr
Thus, at equilibrium
H2O(g)
100
-x
100 - x
Cl2O(g)
100
-x
100 - x
HOCl(g)
0
+2x
2x
p 2 HOCl
( 2 x) 2
 K  0.090 
p H 2O p Cl 2O
(100  x)(100  x)
This happens to be a perfect square, so we can take the square root of both sides:
2x
 0.09  0.3
100  x
2 x  0.3(100  x)
2.3 x  30
x  13.0
Thus, pH2O = 100 – x = 87.0 Torr
pCl2 = 100 – x = 87.0 Torr
pHOCl = 2x = 26.0 Torr
Check:
[5]
(2x) 2
2(13.0) 2

 0.090
(100  x)(100  x) (100  13.0) 2
(b) Calculate the value and units of Kc at 100oC.
Kc = Kp(RT)-Δn
For this reaction, Δn = 0, thus Kc = Kp = 0.090
And since Kp is dimensionless, so is Kc.
4. Liquid water is always in equilibrium with H+(aq) and OH¯(aq) ions according to the reaction
H2O(l)  H+(aq) + OH-(aq). Using the data in the table below,
Hof, kJ mol-1
H2O(l)
H+(aq)
OH-(aq)
-286
0
-230
So, J K-1 mol-1
70
0
-11
[5]
(a) Calculate H° and S° for the reaction.
H° = Hf°(H+(aq)) + Hf°(OH-(aq)) - Hf°(H2O(l))
= 0 + (-230) – (-286)
= +56 kJ mol-1
S° = S°(H+(aq)) + S°(OH-(aq)) - S°(H2O(l))
= 0 + (-11) – (70)
= -81 J K-1 mol-1
[5]
(b) Calculate G° at 25°C for the reaction.
G° = H° - TS°
= 56000 J mol-1 – (25+273)K(-81 J K-1 mol-1)
= +80138 J mol-1
[5]
(c) Calculate the value of Kc at 25°C.
Kc = exp (-G°/RT)
= exp(-80138 J mol-1/(8.314 J K-1 mol-1 x 298 K))
= exp(-32.3)
= 8.97 x 10-15
[5]
(d) Calculate the concentrations of H+(aq) and OH¯(aq) at equilibrium.
Kc = [H+(aq)][OH-(aq)] (since H2O(l) is a liquid, it does not appear in the expression)
thus, [H+(aq)] = [OH-(aq)] = Kc1/2 = (8.97 x 10-15)1/2 = 9.47 x 10-8 mol L-1
5.
(a) Use VSEPR theory to predict the shape of SeO3-2.
[6]
Valence electrons = 6 + (3 x 6) + 2 = 26
Making single bonds to the O atoms from the Se atom and completing the octets around the O atoms uses 24
electrons. The last two are placed as a lone pair on the Se atom, resulting in an AX3E configuration. This is
trigonal pyramidal.
(b) Which of CH2Cl2, CHCl3 and CCl4 are polar?
[6]
If you draw the Lewis structures of all three, it is apparent that the CH2Cl2 is bent, the CHCl3 is a trigonal
pyramid and the CCl4 is tetrahedral. Only the CCl4 is symmetrical and only it will be non-polar. The other two
are polar.
(c) Calculate the formal charge on the B atom in BF3 and use this result to explain why the B atom does
not follow the octet rule.
[8]
6.
BF3 is a trigonal planar molecule having 3 + (3 x 7) = 24 electrons. All 24 will be used by forming single
B-F bonds and completing the octets on the F atoms. The B atom is assigned half of the bonding
electrons (3). B is in group 3 and has 3 valence electrons in the free atom. Its formal charge in BF3 is
thus 3-3 = 0, even though it does not have an octet.
Balance the following reaction in a basic solution: As(s) + ClO3¯(aq)  H3AsO3(aq) + HClO(aq)
[5] Oxidation:
As(s)  H3AsO3 (As oxidation number increases from 0 to +3)
As(s) + 3 H2O  H3AsO3
As(s) + 3 H2O  H3AsO3 + 3 H+
As(s) + 3 H2O  H3AsO3 + 3 H+ + 3 e¯
[5] Reduction:
ClO3¯  HClO (Cl oxidation number decreases from +5 to +1)
ClO3¯  HClO + 2 H2O
ClO3¯ + 5 H+  HClO + 2 H2O
ClO3¯ + 5 H+ + 4 e¯  HClO + 2 H2O
[5] Cross multiplying:
4 As(s) + 12 H2O + 12 e¯  4 H3AsO3 + 12 H+
3 ClO3¯ + 15 H+  3 HClO + 6 H2O + 12 e¯
Adding:
4 As(s) + 6 H2O + 3 ClO3¯ + 3 H+  4 H3AsO3 + 3 HClO
[5]
Convert to basic solution:
4 As(s) + 6 H2O + 3 ClO3¯ + 3 H+ + 3 OH¯  4 H3AsO3 + 3 HClO + 3 OH¯
4 As(s) + 6 H2O + 3 ClO3¯ + 3 H2O  4 H3AsO3 + 3 HClO + 3 OH¯
4 As(s) + 9 H2O + 3 ClO3¯  4 H3AsO3 + 3 HClO + 3 OH¯
Check:
As
H
Cl
O
Charge
Left
4
18
3
18
-3
Right
4
18
3
18
-3
7. Natural gas is a mixture of two components: methane (CH4(g)) and ethane (C2H6(g)). A typical mixture might
have XCH4 = 0.915 and XC2H6 = 0.085. Assume you have a 15.50 g sample of natural gas in a volume of 15.00 L
at 20.0°C.
(a) How many total moles of gas are there in the sample?
[8]
ntotal = nCH4 + nC2H6
and nCH4 / nC2H6 = 0.915 / 0.085 = 10.76
or, nCH4 = 10.76 nC2H6
masstotal = massCH4 + massC2H6
= nCH4 x MWCH4 + nC2H6 x MWC2H6
= (10.76 nC2H6) x MWCH4 + nC2H6 x MWC2H6
= nC2H6 (10.76 MWCH4 + MWC2H6) = 15.50 g
thus, nC2H6 = 15.50 g / (10.76 (16.0 g mol-1) + (30.0 g mol-1)) = 0.0767 mol C2H6
0.0767 mol C2H6 = 0.0767 mol x 30.0 g mol-1 = 2.30 g C2H6
Thus, mass CH4 = 15.50 – 2.30 = 13.20 g CH4
Total moles gas = (2.30 g / 30.0 g mol-1) + (13.2 g / 16.0 g mol-1) = 0.902 mol
(b) What is the pressure of the sample (in atm)?
[2]
P = nRT/V = 0.902 mol (0.082 L atm K-1 mol-1)(293 K) / 15.00 L = 1.44 atm
(c) What is the partial pressure of each component (in atm)?
[5]
PCH4 = PtotXCH4 = 1.44 atm (0.915) = 1.32 atm
PC2H6 = PtotXC2H6 = 1.44 atm (0.085) = 0.12 atm
(d) What are the average speeds of the two components?
v
[5]
3RT
MW
v CH 4 
3(8.314 J K 1mol 1 )(293 K)
 676 ms1
1
0.016 kg mol
v C2 H 6 
3(8.314 J K 1mol 1 )(293 K)
 494 ms1
1
0.030 kg mol