Semester One
Examination Period
ANSWERS
Faculty of Science

EXAM CODES: CHM1011 – Mock Exam # 2

TITLE OF PAPER: CHEMISTRY I

EXAM DURATION: 2 hours writing time

READING TIME: 10 minutes

THIS PAPER IS FOR STUDENTS STUDYING AT: (tick where applicable)

 Berwick X Clayton  Malaysia  Off Campus Learning  Open Learning
 Caulfield  Gippsland  Peninsula X Monash Extension  Sth Africa
 Parkville  Other (specify)

During an exam, you must not have in your possession any item/material that has not been authorised for
your exam. This includes books, notes, paper, electronic device/s, mobile phone, smart watch/device,
calculator, pencil case, or writing on any part of your body. Any authorised items are listed
below. Items/materials on your desk, chair, in your clothing or otherwise on your person will be deemed to
be in your possession.

No examination materials are to be removed from the room. This includes retaining, copying, memorising
or noting down content of exam material for personal use or to share with any other person by any means
following your exam.
Failure to comply with the above instructions, or attempting to cheat or cheating in an exam is a discipline
offence under Part 7 of the Monash University (Council) Regulations.


AUTHORISED MATERIALS

OPEN BOOK  YES X NO

CALCULATORS X YES  NO
* Calculators with School of Chemistry/Faculty of Science authorization label only
SPECIFICALLY PERMITTED ITEMS X YES  NO
if yes, items permitted are:
*Molecular Modelling Kits


Candidates must complete this section if required to write answers within this paper


STUDENT ID: __ __ __ __ __ __ __ __ DESK NUMBER: __ __ __ __ __



Data Page
Useful equations
Wave equation: c = 
Einstein equation: E = h
Rydberg equation:
1

= (
1

2 −
1

2)
Bond order = (# bonding electrons - # anti-bonding electrons)
Gases
Ideal Gas Equation: pV = nRT
Non-Ideal Gas Equation: ( +
2
2
)( − ) = nRT
Total Pressure =  Partial Pressures of Component Gases
Physical constants

c = 2.998 x 108 ms-1

h = 6.626 x 10-34 J.s

R = 1.097 x 107 m-1

NA = 6.022  1023

R = 8.314 J/K/mol
= 0.08206 atm·L/mol·K

1 atm = 1.013 x 105 Pa

1 bar = 1.0 x 105 Pa

Kw at 25
oC = 1.0 x 10-14

0 oC = 273.15 K
Thermodynamics
ΔU = q + w
w = -pV
q = mcT

G = H – TS
Go = -RTlnK
G = G + RT ln Q
Equilibria
Henderson-Hasselbach: pH = pKa + log
E°cell = E°red - E°ox
Kinetics
Zero-order reaction: [A]t = [A]o - kt
First-order reaction: [A]t = [A]oexp(-kt)
Second-order reaction (only one reactant A):
1
[]
−
1
[]0
=
Half-life: t1/2 = 0.693/k
Arrhenius equation: k = Ae-Ea/RT
ln (
2
1
) =
−

(
1
2
−
1
1
)



1
2
]acid[
]base[


Question 1 (5 + 1 + 3 + 5 + 4 + 2 + 3 + 7 + 10 = 40 marks)
Sulfur is an essential element for all organisms, as one of the amino acids, cysteine, contains sulfur
in its molecule. It also found in compounds with noxious smells. The following questions relate to
sulfur and some of its compounds.
a) The following are electronic configurations. Refer to these when answering the questions (i) - (v):
A: 1s2 2s2 2p6 3s2 3p6
B: 1s2 2s2 2p6 3s1 3p5
C: 1s2 2s2 2p6 3s3 3p5
D: 1s22s22p63s23p4
E: 1s22s2 2p6 3s2 3p6 4s1

(i) Which of the electronic configurations above represents sulfur in the ground state?

D 1 mark

(ii) Which of the electronic configurations above disobey the Pauli Exclusion Principle?

C 1 mark

(iii) Which of the electronic configurations above represents sulfur in an excited state?

B 1 mark

(iv) Which of the electronic configurations above represent S2-?

A 1 mark

(v) Which of the electronic configurations disobeys Aufbau Principle?

B 1 mark

b) Rank the species in order from largest to smallest radii: Cl, S, Ba, Co, F

Ba, Co, S, Cl, F 1 mark for correct order

c) Draw the following orbital shapes that sulfur contains using the axes below:
a) 2px b) 3dz2 c) 4s





Question 1 continued
d) The sulfur atom has 3p orbitals.
i) Draw the electron density plot for the 3p orbitals using the axes below.


ii) How many nodes will 3p orbital have? Are they planar, radial or both?
2 nodes 1 mark
1 radial and 1 planar 1 mark

iii) What is the angular momentum quantum number, l for 3p?
1 1 mark


e) Elemental sulfur absorbs energy at 264nm.
i) Calculate the frequency at this wavelength (4 significant figures).
=


=
2.998×108
264×10−9
1 mark correct substitution
λ = 1.136 x 1015 Hz (or s-1) 1 mark


ii) Calculate the energy, in Joules, of this wavelength (4 significant figures).

=
ℎ

=
6.626×10−34×2.998×108
264×10−9
1 mark correct substitution
E = 7.525 x 10-19 J 1 mark

Question 1 continued
f) Name the following sulfur-containing salts:
i) CaSO4,
Calcium sulfate 1 mark

ii)Na2S
Sodium sulfide 1 mark


g) A common sulfur containing compound is sulfuric acid, H2SO4. Calculate the wavelength (in
nanometers) of the transition from the n=1 to n=4 levels in atomic hydrogen in sufuric acid. Express
your answer to four significant figures.


1

= (
1

2 −
1

2) = 1.097 × 10
7 (
1
12
−
1
42
) 1 mark correct substitution

1/λ = 1.028 x 107 1 mark
λ = 9.723 x 10-8 m 1 mark











Question 1 continued
h) A compound containing sulfur is SO32-.
(i) Draw a Lewis structure for SO32-, being sure to show lone pairs and any formal charges.




1 mark for lone pairs of electrons
1 mark for correct bonds



(ii) What are the bond angles of SO32-?

109.5°C 1 mark



iii) In the table below, enter the number of electron domains about the central atom, and use
VESPR theory to predict the parent geometry and the molecular geometry.

Species # of Electron
Domains
Parent
Geometry
Molecular
Geometry
SO32-
4
1 mark




tetrahedral
1 mark
trigonal pyramidal
1 mark


Question 1 continued
i) Thiophene, structure on right, is a common precursor to many
compounds, including pharmaceuticals, plasticizers and agrochemicals.
Sketch the valence bond theory picture of thiophene. Make sure you
include all orbitals and labels.


1 mark for each:
1. S – sp3 label
2. S – correct orbital drawing
3. S - lone pair electrons
4. 4 x C– sp2 label
5. 4 x C – correct orbital drawing
6. σ labels (9 in total)
7. H 1s orbital label
8. H1s correct orbital drawing
9. C-C –π bonds correct
10. C-C – π bonds label










Question 2 (1 + 4 + 2 + 1 + 1 + 1 = 10 marks)
Answer the following questions based on the molecular orbital diagram of a homonuclear molecule
formed from an element from the 2nd row of the periodic table:



i) Assuming the atoms and the molecule are NOT charged (i.e. neutral), which element and
molecule does the diagram represent? Label them on the diagram.
atom: oxygen, molecule: O2 1 mark
ii) Label all the atomic and molecular orbitals on the diagram.
see above. 1 mark for correct atomic orbital labels, 1 mark for correct molecular orbitals for 2s, 1
mark for correct σ orbitals for 2p, 1 mark for correct π orbital for 2p
Question 2 continued
iii) Is the molecule diamagentic or paramagnetic? Explain your answer.

paramagnetic 1 mark
unpaired electrons 1 mark

iv) Write the molecular electronic configuration for the molecule.

σ2s2 σ*2s2 σ2pz2 π2px2 π2py2 π*2px1 π2py1
1 mark for correct electronic configuration. No penalty for including or omitting σ1s2 and
σ*1s2.

v) Calculate the bond order for the molecule.

Bond order = ½(8 – 4) = 2 1 mark



vi) Does the molecule exist? Explain your answer.
Yes, because it has a bond order greater than 0.
Question 3 (6 + 2 = 8 marks)

a) Recently, carbon dioxide levels at the South Pole reached 400 μatm. (The 1958 reading was 315
μatm.) At this level, how many CO2 molecules are there in 1.00 L of dry summer air at –35.0 °C?

P = 400 x 10-6 atm 1 mark
V = 1.00 L
R = 0.08206 atm.L/K.mol
T = -35 + 273 = 238K 1 mark
=


=
400×10−6×1.00
0.08206×238
1 mark for correct substitution

n = 2.048 x 10-5 mol 1 mark
number of molecules = number of moles x Avogadro’s number
number of molecules = 2.048 x 10-5 x 6.022 x 1023 1 mark
number of molecules = 1.23 x 1019 1 mark

b) How does this change in winter when the average temperature drops to –64.0 °C?
Pressure, volume are constant.
n1 . T1 = n2 . T2
2.048 x 10-5 x 238 = n2 x 209
n2 = 2.332 x 10
-5 mol 1 mark

number of molecules = 2.332 x 10-5 x 6.022 x 1023
number of molecules = 1.40 x 1019 1 mark

Question 4 (3 marks)

What types of intermolecular forces do you have for the following molecules/atoms?
i) __dispersion forces______1 mark______


ii) __ dispersion forces, polar bonds, H-bonds__1 mark



iii) Ar ___dispersion forces 1 mark_______
Question 5 (1 + 1 + 1 = 3 marks)
a) When a reaction is at equilibrium, which of the following statements is TRUE?
a) G = Go
b) ln Keq = 0
c) Go = 0
d) Q = 0
e) G = 0
Answer: __E 1 mark____

b) For a particular chemical reaction, the values for H and S are both positive. Which of the
following statements is TRUE about this reaction?
a) The reaction is spontaneous at any temperature.
b) The reaction is not spontaneous at any temperature.
c) The reaction is spontaneous when the temperature is high enough to overcome H.
d) The reaction is spontaneous when the temperature is low enough to overcome H.
e) The reaction is spontaneous when the temperature is low enough to overcome S.
Answer: ___C 1 mark___

c) For which of the following processes does the entropy of the system DECREASE?
a) Melting of ice.
b) Dissolving table salt in water.
c) Decomposition of liquid hydrogen peroxide to produce liquid water and oxygen gas.
d) Precipitation of silver chloride from a solution containing silver ions and chloride ions.
e) The entropy of the system increases in all of the above processes.
Answer: __D 1 mark_

Question 6 (4 + 2 + 3 + 2 = 11 marks)
Consider the following reaction and thermodynamic data,
2H2O2 (aq) → 2H2O (l) + O2 (g)
Substance Hf
o, kJ mol-1 So, J mol-1 K-1
H2O2 (aq) -191.2 143.9
H2O (l) -285.8 69.91
O2 (g) 0.000 205.1

a) Determine the value of Go for this reaction at 25 oC.


ΔHr = (2 x -285.8 + 0) – (2 x -191.2) = -189.2 kJ mol-1 1 mark

ΔSr = (2 x 69.96 + 205.1) – (2 x 143.9) = 57.22 J mol-1 K-1 1 mark

ΔG = ΔH - TΔS

ΔG = -189.2 – ((25+273) x -0.05722) 1 mark

ΔG = -206.2 kJ mol-1 1 mark



b) Is this reaction spontaneous? Justify your answer.

yes 1 mark
ΔG < 0 1 mark


c) Express the Second Law of Thermodynamics in written form.
For a reaction to be spontaneous (1 mark), the total entropy (system and surroundings) (1 mark)
must be positive (1 mark). That is, total entropy of universe greater than 0.




d) For each of the pairs below circle the species which has the highest value for its entropy, So.
a) 1 mol of NO(g) or 1 mol of NO2(g)
b) 1 mol of N2O4(s) or 1 mol of N2O4(g)
c) 2 mol of O2(g) or 2 mol of O3(g)
d) 1 mol of NaCl(s) or 1 mol of NaCl(aq)
0.5 mark for each correct answer

Question 7 (2 + 1 + 2 + 2 + 1 = 8 marks)
A sample of ozone (O3) is contained by a weightless, frictionless piston
at atmospheric pressure and a temperature of 25.0°C, shown in the
diagram to the right. Ozone decomposes as described by the following
equation:

2O3 (g)  3O2 (g)

The standard enthalpy of formation for ozone, ΔHf0, is +142.2 kJ mol-1.

a) Calculate Hr0 for the reaction above:


ΔHr = (0) - (2 x 142.2) 1 mark
ΔHr = -284.4 kJ mol-1 1 mark






b) Is this reaction exothermic or endothermic?

exothermic 1 mark

c) Determine H when 0.0400 moles of ozone completely decomposes?

ΔHr = -284.4/2 x 0.04 1 mark
ΔHr = -5.69 kJ 1 mark


d) If 0.0400 moles of ozone completely decomposes in the piston, calculate:
i) Any change to the volume of gas in the piston:

0.0400 moles of ozone will produce 0.0600 mol oxygen
(3) =


=
0.04 × 8.314 × 298
1.013 × 102kPa
= 0.978
(2) =


=
0.06 ×8.314×298
1.013×102kPa
= 1.47 1 mark for calculations
ΔV = V(O2) – V(O3) = 1.47 – 0.978 = 0.492 L 1 mark for answers



ii) The work done to or by the system:

= −∆ = −(1.013 × 105)Pa × (0.492 × 10−3)m3 0.5 mark
w = -49.8 J 0.5 mark




O3
Question 8 (2 + 1 + 4 = 7 marks)
The following question refers to the equilibrium reaction below:
N2 (g) + 2 O2 (g) ⇌ 2 NO2 (g) (Kc = 8.28 × 10-10 at 25 °C)
If a 500 mL flask is initially filled with 0.0400 mol of N2 and 0.0800 mol O2,
a) Determine the concentration of each reactant when each gas first fills the flask.

C=n/V
[N2] = 0.04/0.5L = 0.08M 1 mark
[O2] = 0.08/0.5 = 0.16M 1 mark



b) Write the equilibrium expression, Kc for this reaction.

=
[2]
2
[2][2}2


1 mark

c) Complete the following table and calculate the concentration of NO2 at equilibrium.

Concentration N2(g) O2(g) NO2(g)
Initially 0.08













0.16

0

Change -x -2x +2x
Equilibrium 0.08-x






0.16-2x

2x

1 mark for table
(Use the space below to demonstrate your calculations.)

=
[2]
2
[2][2}2
=
(2)2
(0.08−)(0.16−2)2
= 8.28 × 10−10 1 mark for substitution

Assume [N2] ≈ 0.08M and [O2] ≈ 0.16M
(2)2
(0.08)(0.16)2
= 8.28 × 10−10
x = 6.51 x 10-7
[NO2] at equilibrium is 2 x 6.51 x 10
-7 1 mark
[NO2] = 1.32 x 10
-6 M 1 mark



Question 9 (1 + 1 + 2 + 1 + 2 + 2 = 9 marks)
Consider the following data for the series of hydrogen halide Bronsted acids,
Acid Ka
HF 7.2 × 10-4
HCl 1.0 × 106
HBr 1.0 × 109
HI 3.0 × 109
a) Which of these Bronsted acids would have the weakest conjugate base?

HI 1 mark

b) For the following reaction of hypochlorous acid (HOCl) with water,
HOCl (aq) + H2O (l) ⇌ H3O+ (aq) + OCl- (aq)
what would be the effect of adding sodium hypochlorite (NaOCl) to the reaction at equilibrium?

Reaction will shift left, increasing reactants 1 mark

c) Benzoic acid, C6H5CO2H, is a weak acid (Ka = 6.31 x 10
-5).
i) Calculate the initial concentration (in M) of benzoic acid that is required to produce an
aqueous solution of benzoic acid that has a pH of 2.54.

pH = 2.54 = -log[H3O
+]
[H3O
+] = 2.88 x 10-3 0.5 mark
[H3O
+] = [benzoate] (1:1 mole ratio)
=
[][3
+]
[ ]
= 6.31 × 10−5 =
(2.88×10−3)(2.88×10−3)
[ ]
0.5 mark
[benzoic acid] = 0.132M at equilibrium 0.5 mark
[initial benzoic acid] required: 0.132 + 2.88 x 10-3 = 0.135M 0.5 mark


ii) Calculate [OH-] for this solution at 25.0 oC.
[OH-][H3O
+] = 1.0 x 10-14
[−] =
1.0×10−14
2.88×10−3
= 3.47 × 10−12 1 mark

d) Calculate the pH of a buffer prepared by mixing 0.100 mol of sodium formate and 0.0500 mol of
formic acid in 1.00 L of solution. [HCO2H: Ka = 1.78 x 10
-4]

pKa = -log Ka = 3.75
[base] = 0.1 mol / 1.0 L = 0.1 M
[acid] = 0.05 mol / 1.0 L = 0.05 M

pH = pKa + log
[base]
[acid]
= 3.75 + log
0.1
0.05
1 mark
pH = 4.05 1 mark

e) Would a buffer containing 0.0500 mol of sodium formate and 0.100 mol of formic acid in 1.00 L
solution, be more or less resistant to a change in pH upon the addition of a small amount of conc.
HCl, compared to the buffer prepared in d) above? Why?

less resistant: the buffer with 0.05M base will have less effect than the buffer with 0.10M base

Question 10 (3 + 1 + 2 = 6 marks)
The method of initial rates was used to follow the rate of the following reaction:
BrO3- (aq) + 5 Br- (aq) + 6 H+ (aq) → 3 Br2 (aq) + 3 H2O (l)
The following data were collected:

Expt [BrO3-] (mol L-1) [Br-](mol L-1) [H+] (mol L-1) Initial Rate of consumption
of BrO3- (mol L-1 s-1)
1 0.10 0.10 0.10 1.2 x 10-3
2 0.20 0.10 0.10 2.4 x 10-3
3 0.10 0.30 0.10 3.6 x 10-3
4 0.20 0.10 0.20 9.6 x 10-3

a) What is the order of the reaction with respect to the each of the reactants?


BrO3
- = when conc doubles, rate doubles. Rate order of 1 1 mark

Br- = when conc triples, rate triples. Rate order of 1 1 mark

H+ = when conc doubles, rate quadruples. Rate order of 2 1 mark




b) Write out the rate law for this reaction.

Rate = k[BrO3-][Br-][H+]2 1 mark

c) Calculate the value of the rate constant, k (remember to include units for k in your answer).


Using the first row of values:
1.2 x 10-3 = k (0.1) (0.1) (0.1)2 1 mark
k = 12 M-3 s-1 1 mark











Question 11 (3 + 4 = 7 marks)
The decomposition of hydrogen iodide:
2HI (g) → H2 (g) + I2 (g)
has rate constants of 9.40 x 10-9 L/mol.s at 200C and 1.10 x 10-5 L/mol.s at 300C.

a) Calculate the activation energy (Ea).

ln (
2
1
) =
−

(
1
2
−
1
1
)
ln (
1.10×10−5
9.40×10−9
) =
−
8.314
(
1
(300+273)
−
1
(200+273)
) 1 mark for correct T1 and T2 values
1 mark for correct substitution
Ea = 1.59 x 10
5 J mol-1 1 mark




b) Draw a sketch of the energy profile for this reaction assuming that it is an exothermic process
and include in your diagram the effect of adding a catalyst.


0.5 marks for each label, plot shape, axes names
Question 12 (1 + 7 = 8 marks)

Answer the following based on the electrochemical series below:

Half Reaction E° (volts)
Au3+ + 3e ⇌ Au (s) 1.42
Ag+ + e ⇌ Ag (s) 0.80
Cu2+ + 2e ⇌ Cu (s) 0.34
Pb2+ + 2e ⇌ Pb (s) -0.13
Cr3+ + 3e ⇌ Cr (s) -0.74
Mn2+ + 2e ⇌ Mn (s) -1.03

a) From the table above, which chemical species will be the strongest reductant?

Mn (s) 1 mark



b) You will create an electrolytic cell with Cu2+/Cu and Pb2+/Pb half cells.
i) Write out the full balanced chemical reaction

Pb2+ + Cu (s) → Pb (s) + Cu2+ 1 mark

ii) Draw the cell, ensuring you label the anode, cathode, salt bridge and charges on the
electrodes.


1 mark for each:
correct metal and charge for cathode
correct metal and charge for anode
drawing of salt bridge
sketch of cell

iii) What will be the E° of this cell?


E°cell = E°red - E°ox = -0.13 – (0.34) = -0.47 V 1 mark

END OF PAPER

欢迎咨询51作业君