2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
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Practice Exam One (SAQs only, with answers)
Due No due date Points 50 Questions 11 Time Limit None
Allowed Attempts Unlimited
Instructions
Attempt History
Attempt Time Score
LATEST Attempt 1 11,360 minutes 0 out of 50 *
Submitted Nov 10 at 18:31
Practice Exam
This practice exam will only contain the SAQs of a previous exam completed in CHEM1xx2.
In CHEM1xx2, this represents 50 of the 70 marks possible.
Only this practice exam will contain answers that can be read after your first attempt.
Note that formatting IS NOT required so, for example, CH4(g) is fine to represent CH or 'Delta'
used in the place of . Note that you do have access to the equation editor, but its use is not required.
Please note that this examination is for guidance only. Any learning outcome raised in the course is
examinable, so be prepared to address all material covered in your lectures and tutorials.
This quiz will be available until 15 minutes before the start of the final exam for this unit.
Best of luck!
The First-year teaching team
4(g)
Take the Quiz Again
Not yet marked / 3.5 ptsQuestion 1Unanswered
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
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Your Answer:
The rate of the reaction for the following reaction:
Triples from 25 ºC to 40 ºC, what is the activation energy (in kJ.mol )
of the reaction (Do not provide decimal points in your answer)?
Show all working in the box provided. Marks will be awarded for
calculations and the final answer.
-1
The equation, should be used.
(0.5 marks)
The reaction rate triples between the two temperatures so =
3. (0.5 marks)
R is the Gas Constant = 8.314 J.mol .K
T1 is the lower temperature converted to Kelvin.
T2 is the higher temperature converted to Kelvin.
Re-arrange the equation above with respect to
E = (1 mark)
Note: this equation will provide the answer in J.mol-1 (see the
units of R) so dividing the final answer by 1000 will convert the
final answer to kJ.mol . (0.5 marks)
Final answer is 57 +/- 1 kJ/mol (1 mark)
-1 -1
a
-1
Not yet marked / 2 ptsQuestion 2Unanswered
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
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Your Answer:
A student was asked to draw the reaction mechanism and name the
product for the substitution reaction between 1-bromopropane and
methoxide. Explain what is wrong with their answer (there may be
more than one error). Give reasons for your answers.
Curly arrows- Must start from the nucleophile/electron
source and end at the electrophile (0.5 mark)
Selected Mechanism – the reagent is a primary organohalide
and so the likely mechanism would be an SN2 mechanism,
not the Sn1 mechanism drawn here
Identifying incorrect mechanism (0.5 mark)
Explains why Sn2 over Sn1 (carbocation instability,
sterics) (0.5 mark)
No stereochemistry – The product formed, in either correct
or incorrect mechanism, would not have a stereogenic
centre and so drawing and/or naming the stereochemistry is
incorrect (0.5 mark).
Not yet marked / 2 ptsQuestion 3Unanswered
Explain the trend in stability of the following series. Explain what
factors are involved and what effect they have.
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
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Your Answer:
Hyperconjugation (0.5 marks) – through space interaction of
hydrogen/carbons orbitals filling the empty p-orbital of the
carbocation which increases stability (0.5 marks).
Inductive effect (0.5 marks) – through bond interaction between
carbon-carbon bonds where adjacent carbons donate electron
density to stabilise the carbocation which increases stability (0.5
marks).
Not yet marked / 4 ptsQuestion 4
Your Answer:
Unanswered
What is the name of this class of reaction shown below? Explain and
describe the reaction mechanism for the following, making sure to
discuss any selection criteria:
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
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This is an addition reaction (1/2 marks).
Nucleophilic attack of alkene double bond (1/2 marks) towards
a proton – the electrophile (1/2 marks).
The addition of the H to the alkene, results in two possible
carbocations (1/2 marks). The tertiary carbocation is the more
stable (as compared to a primary carbocation) (1/2 marks).
A lone pair of an oxygen as part of a neutral water molecule acts
as a nucleophile and forms a bond with the tertiary carbocation
(1/2 marks). Finally, another lone pair of an oxygen as part of a
neutral water molecule (1/2 marks) removes a proton from this
protonated alcohol (which breaks the H-alcohol bond) (1/2
marks).
+
0 / 4.5 ptsQuestion 5
Identify the class of reaction for the following reactions from the
reaction scheme shown above (note: some reactions might have the
same answer).
Reaction a) ou Answered
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
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Elimination
Electrophilic
Addition
Correct Answer
Reaction c) ou Answered
SubstitutionCorrect Answer
Reaction d) ou Answered
Acid/BaseCorrect Answer
Reaction f) ou Answered
Grignard ReactionCorrect Answer
Reaction g) ou Answered
EliminationCorrect Answer
Reaction j) ou Answered
ReductionCorrect Answer
Reaction k) ou Answered
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
https://canvas.sydney.edu.au/courses/43874/quizzes/194062 7/15
Other Incorrect Match Options:
Hydrolysis
OxidationCorrect Answer
Reaction m) ou Answered
Acyl nucleophilic
substitution
Correct Answer
Reaction n) ou Answered
Acyl nucleophilic
substitution
Correct Answer
0 / 5 ptsQuestion 6Unanswered
Select the most appropriate reagent to perform the following reactions
shown in the scheme above:

2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
https://canvas.sydney.edu.au/courses/43874/quizzes/194062 8/15
Reaction b) ou Answered
Potassium tert-
butoxide
Correct Answer
Reaction c) ou Answered
DimethylamineCorrect Answer
Reaction d) ou Answered
HClCorrect Answer
Reaction e) ou Answered
MgCorrect Answer
Reaction h) ou Answered
H₂ with a Pd/C
catalyst
Correct Answer
Reaction i) ou Answered
CO₂ (followed by
H⁺/ H₂O workup)
Correct Answer
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
https://canvas.sydney.edu.au/courses/43874/quizzes/194062 9/15
Other Incorrect Match Options:
⁻OH/H₂O with heating
2-Propanone
Concentrated KOH
Br₂
H⁺/Ethanol with heating
Ethanol
HBr
NaOH
Concentrated H₂SO₄
Reaction j) ou Answered
LiAlH₄ (followed by
H⁺/ H₂O workup)
Correct Answer
Reaction k) ou Answered
Na₂Cr₂O₇ / H⁺Correct Answer
Reaction l) ou Answered
SOCl₂Correct Answer
Reaction n) ou Answered
H⁺/H₂O with
heating
Correct Answer
Not yet marked / 4 ptsQuestion 7Unanswered
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
https://canvas.sydney.edu.au/courses/43874/quizzes/194062 10/15
Your Answer:
Describe a series of reagents and reactions that could convert the
starting material to the product shown below:
Reaction with Mg (0.5 mark) to form a Grignard reagent (0.5
mark) then react with propanal (0.5 mark) and then acid/workup
(0.5 mark) to prepare the secondary alcohol (0.5 mark).
Reaction with dichromate/permanganate (0.5 mark) to oxidise
(0.5 mark) to the ketone (0.5 mark)
Not yet marked / 11 ptsQuestion 8Unanswered
The water splitting reaction forms the basis for a carbon-free renewable
energy cycle. This reaction can be achieved in an electrolytic cell by
combining the following two half-cell reactions:
Oxidation: 2H O → O + 4H + 4e E º = -1.23 V
Reduction: 4H O + 4e → 2H + 4OH E º = -0.83 V

This question has eight parts.

2 2
+ -
2
-
2
-
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
https://canvas.sydney.edu.au/courses/43874/quizzes/194062 11/15
Your Answer:
1) Write the water splitting reaction. (1 mark)
2) Calculate the standard cell potential, Eº, for the water splitting
reaction. (1 mark)
3) Describe how the cell potential, E, for this reaction will change if the
pH is decreased equally at each electrode. Justify your answer with
reference to Le Chatelier’s Principle. (1 mark)
4) Calculate the volume of hydrogen gas given off at room temperature
and pressure by passing 1.00 A of current through this cell for 100.0
seconds. (3 marks)
5) It is found experimentally that a voltage higher than the calculated
cell potential is required to drive this reaction. Briefly explain why this is
the case. (1 mark)
6) One drawback of the electrolytic water splitting reaction is that it
requires fresh water. Describe why seawater cannot be used, detailing
any competing chemical processes that might be expected to occur. (2
marks)
[Data: the concentrations of some of the principal ions in seawater are
[Na ] = 0.5 M, [K ] = 0.01 M, [Mg ] = 0.05 M, [Ca ] = 0.01 M, [Fe ] =
0.00002 M, [Cl ] = 0.5 M, [SO ] = 0.03 M, [CO ] = 0.002 M, [Br ] =
0.001 M]
7)Hydrogen fuel cells can be used to convert chemical energy back to
electrical energy by reversing the water splitting reaction. With
increasing temperature, it is found that the voltage of a hydrogen fuel
cell drops but that higher currents can be delivered. By considering
both the thermodynamics and kinetics of this system, suggest reasons
for each of these two observations (i.e. lower voltage but higher
current). (2 marks)

+ + 2+ 2+ 2+
-
4
2-
3
2- -
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
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(1) 2H2O(l) → 2H2(g) + O2(g)
(2) -2.06 V
(3) No change since overall reaction doesn’t depend on [H+] or
[OH-]; ie. deltaG is same at all pHs.
(4) Moles of electrons = It/F = 100/96485 = 1.04 x 10^-3 mol
Moles of H2 gas = ½ x 1.04 x 10^-3 = 5.2 x 10^-4 mol
Volume of H2 gas = nRT/P = 12.7 x 10^-3 L = 12.7 mL
(5) High overpotential at both electrodes since they both involve
the formation of gases.
(6) Cl- oxidation to Cl2 gas (E0 = -1.36 V) will occur to some
degree due to high overpotential, as might Br- oxidation (-1.09
V) and Fe2+ oxidation (-0.77 V) and reduction (-0.44 V)
(7) Lower voltage: deltaS for this reaction is expected to be
negative (3 gas molecules converted to 2 liq molecules) and so
deltaG will decrease on heating. This lower driving force for the
reaction will lead to a drop in the voltage.
Higher current: Slow kinetics of this system include ion transport
through the electrolyte and potentially also the cathode and
anode reactions. Higher temperatures will increase the rate all
of these processes.
Not yet marked / 2 ptsQuestion 9
Your Answer:
Unanswered
Unusually, the water solubility of CaCl (s) decreases with increasing
temperature. What does this tell us about the entropy change for CaCl
dissolution? By considering the species formed in solution, suggest a
reason for this.
2
2
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
https://canvas.sydney.edu.au/courses/43874/quizzes/194062 13/15
deltaS is negative
Ordering of water molecules around Ca2+ (coordination bonds)
and Cl- (ion-dipole forces) is responsible for lowering the
entropy of the water solvent
Not yet marked / 3 ptsQuestion 10Unanswered
The crystal structure rhenium oxide is shown below. Re atoms are grey,
O atoms are white, and the unit cell is shown by dashed lines. The unit
cell is cubic.

This question has three parts.

1) Based on the contents of the unit cell, what is the stoichiometry of
this rhenium oxide, and what is the oxidation state of Re? (1 mark)
2) How many nearest-neighbour O atoms surround each Re atom (i.e.,
what is the coordination number of Re), and what is the shape of the
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
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Your Answer:
ReO coordination environment? (1 mark)
3 ) Does rhenium oxide have a close-packed crystal structure? Briefly
justify your answer. (1 mark)
x
(1) ReO3, Re(VI)
(2) 6, octahedral
(3) No, it has a primitive cubic structure. (If there was an O atom
in the body centre of the cell then it would have fcc-packed O
atoms.)
Not yet marked / 9 ptsQuestion 11Unanswered
This question has 7 parts. Make sure you answer all 7 and clearly label
each part in the box below.
A number of transition metal complex ions have been presented in the
lecture course. Choose an appropriate, complex ion to answer the
questions below.
1. Write the formula for a solid salt of this complex and name it (2
marks)
2. What is the geometry of your complex ion? (1 mark)
3. Write the expression for the stability constant of your complex ion (1
mark)
4. Give the formula for, and name, an isomer of your chosen complex
ion (1 mark)
5. Write the electron configuration for the transition metal. For
example, in a high-spin octahedral complex, the electron
configuration of iron(II) can be described as
(d ) (d ) (d ) (d ) (d ) (1 mark)
6. Is your chosen complex paramagnetic. Propose a very simple
experiment to demonstrate whether your complex (eg in its salt
from part 1) is paramagnetic. (1 mark)
xy
1
xz
1
yz
1
x2-y2
1
z2
0
2022/11/10 15:45 Practice Exam One (SAQs only, with answers): CHEM1012 Fundamentals of Chemistry 1B
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Your Answer:
7. Would you expect your complex to be coloured? Give reasons for
your answer. (2 marks)
1. Lots of examples, eg [trans-[Co(en) Cl ]Br, (NB it needs to be
appropriately chosen - 0 marks if it doesn't have an isomer);
name trans-dichloridobis(ethylenediamine)cobalt(III) bromide
OR trans-dichloridobis(ethane-1,2-diamine)cobalt(III) bromide
2. Here the complex ion is trans-[Co(en) Cl ] with 6 bonds
about the Co(III) so geometry is octahedral
3. Equilibrium is: [Co(H O) ] (aq) + 2en (aq) + 2Cl (aq) ⇌
[Co(en) Cl ] (aq)
so Kstab = { [Co(en) Cl ] (aq)] } / { [[Co(H O) ] (aq)] [en (aq)]
[Cl (aq)] }
4. Eg a geometric isomer [cis-[Co(en) Cl ] ; cis-
dichloridobis(ethylenediamine)cobalt(III)
5. Co(III) has a d configuration, so (d ) (d ) (d ) (d
) (d )
6. The Co(III) in the complex has unpaired electrons so is
paramagnetic. This could be demonstrated by putting its salt, eg
[trans-[Co(en) Cl ]Br, in a plastic vial, tying it to a retort stand
with string and making it swing by bringing a magnet close to
the vial.
7. Yes, I expect the complex to be coloured. In an octahedral
complex the d electrons orbitals are split into 2 groups with
different energy. The d , d and d orbitals do not point toward
the ligand lone pairs so have lower electron-electron repulsion
than the d and d orbitals that do. The difference in these
energies, Delta, is relatively small and corresponds to visible
wavelengths of light. If the complex absorbs eg green light, it will
look red.
2 2
2 2
+
2 6
3+ -
2 2
+
2 2
+
2 6
3+ 2
- 2
2 2
-
6
xy
2
xz
1
yz
1
x2-
y2
1
z2
1
2 2
xy xz yz
x2-y2 z2