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Chemistry
CHEMISTRY
COURSEWORK AND EXAM PREPARATION
VCE AND CAREERS EXPO 2012
Preparation for VCE Units 3&4
Chemistry
• Venue: Caulfield Racecourse
• Date: Saturday 5 May
• Time: 12:00 pm – 12:45 pm
• Lecturer: Mr Brian Ellett
National Educational Advancement Programs (Neap) Pty Ltd ABN 49 910 906 643
96–106 Pelham Street Carlton Victoria 3053 Telephone (03) 8341 8341 Facsimile (03) 8341 8300
This publication is independently produced for students of VCE. Although material may have been reproduced with the permission of the VCAA, the publication
is in no way connected to or endorsed by the VCAA. The notes, handouts and other documents issued at lectures have been specifically researched and
produced by Neap. Reproduction of the whole or part of any document constitutes an infringement of copyright. None of the material may be used or passed on
to other persons without the prior written consent of Neap.
CHEM_EXPO_12.FM
Copyright © Neap 2012
Contents
Neap programs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Overview of Chemistry Units 3&4 . . . . . . . . . . . . . . . . . . . 3
1.1
1.2
1.3
1.4
1.5
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Assessment and levels of achievement. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Achieving A/A+ grades in VCE Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Data sheets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
5
5
6
7
Coursework and exam preparation . . . . . . . . . . . . . . . . . . 17
2.1
2.2
2.3
2.4
Unit 3 – Area of study 1: Chemical analysis . . . . . . . . . . . . . . . . . . . . . . . . . . .
Unit 3 – Area of study 2: Organic chemical pathways . . . . . . . . . . . . . . . . . . .
Unit 4 – Chemistry at work. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Copyright © Neap
CHEM_EXPO_12.FM
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1
VCE Unit 3 Chemistry: Coursework & Exam Preparation
Neap programs
11-Session Intensive Programs
Semester 1 – March 14 – June 3 2012
11 × 2-hour small group sessions covering Unit 3 topics and preparation for the mid-year exam for Biology,
Chemistry or Physics.
Semester 2 – August 1 – October 10 2012
11 × 2-hour small group sessions covering Unit 4 topics and preparation for the end-of-year exam for
Biology, Chemistry, Physics, Maths Methods (CAS) and Specialist Maths.
6-Session Intensive Programs
Semester 1 – Commencing 18 April 2012
Small group sessions for students who need help in, or want to consolidate, specific topics in Maths Methods
(CAS) and Specialist Maths.
Final Revision for Mid-Year Exams
May 29 and June 5 2012
4 or 5-hour lectures covering Unit 3 topics and final preparation for the mid-year exam for Accounting,
Biology, Chemistry, Physics and Psychology.
Winter School
July 2–6 2012
6 hour lectures giving students a head start on Unit 4 topics and concepts. Subjects include Biology,
Chemistry, English, Maths Methods, Psychology and Physics.
September Holiday Programs for End-of-Year Exams
September 24–30 2012
Lectures covering entire Unit 4 or Units 3&4 courses in preparation for the end of year exams. Subjects
offered include: English, Accounting, Legal Studies, Biology, Chemistry, Physics, Psychology, Maths
Methods (CAS) and Specialist Maths.
Final Revision for End-of-Year Exams
October 13, 14, 20 and 21 2012
A 4-hour lecture covering Unit 4 topics and final preparation for the end-of-year exam for English,
Accounting, Biology, Chemistry, Physics, Maths Methods (CAS).
Summer School
Mid January 2013
Lectures giving students a head start on Year 12 topics and concepts.
The value of Neap programs
Neap lectures provide opportunities for students to revise all examinable work, incorporating exam-style
questions with fully-worked solutions. Logically sequenced, comprehensive lecture/study notes and summaries
are provided to accompany each program or lecture. Exam techniques and hints are included in the notes.
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CHEM_EXPO_12.FM
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Overview of Chemistry Units 3&4
Overview of Chemistry Units 3&4
1.1 Introduction
The key aims of this lecture are to
•
revise and consolidate key concepts from Chemistry Units 1 and 2 so that students have a strong
knowledge base upon which to build for Year 12;
•
focus on key concepts and knowledge required for Year 12 so that students are well prepared for the
challenges that they will meet.
The Year 11 Chemistry course encompasses Units 1 and 2: The Big Ideas of Chemistry and Environmental
Chemistry. Success at Year 12 is based upon a thorough understanding of the Year 11 course material, with
the topics organic chemistry, stoichiometry, acids and bases and redox chemistry being particularly
important. The first part of this program will be devoted to reviewing and revising some of these key concepts
from Units 1 and 2. The major topics that will be investigated are
1.
atomic structure and electron subshell configuration;
2.
the mole concept, including empirical and molecular formulas;
3.
stoichiometry and equation writing;
4.
acids and bases, including pH calculations;
5.
redox reactions.
In broader terms, the ability to accurately write the chemical formulas of compounds and subsequently
balanced chemical equations is also essential. An understanding of bonding theory as it relates to ionic,
metallic and molecular compounds is also necessary.
The following shows how the coursework studied at Year 11 is related to that covered in Year 12.
Year 11 topic
Coverage in Year 12
atomic structure
reviewed
electron configuration
reviewed
mole concept, empirical formulas etc.
reviewed
bonding theory (ionic, covalent, metallic)
expected knowledge
intermolecular bonding theory (H bonds etc.)
expected knowledge
organic chemistry
significantly extended into Biochemistry
(proteins, nucleic acids etc.)
stoichiometry
significant topic, with applications across
numerous areas of the course
acids and bases
significantly extended into volumetric
analysis, pH calculations and equilibrium
considerations of weak acids
the atmosphere
relatively minor topic, although the gas laws
are important
redox chemistry
significant topic, with extensive
consideration of all aspects of redox
reactions, with particular emphasis on
galvanic and electrolytic cells and
industrial processes
Copyright © Neap
CHEM_EXPO_12.FM
3
VCE Unit 3 Chemistry: Coursework & Exam Preparation
The Year 12 Chemistry course upon which you are embarking encompasses Units 3 and 4: Chemical
Pathways and Chemistry at Work. Each unit is divided into two Areas of study: Unit 3 investigates Chemical
Analysis and Organic Chemical Pathways while Unit 4 considers aspects of Industrial Chemistry and
Supplying and Using Energy.
Unit 3 Areas of study
Chemical analysis, which looks at the various analytical techniques available to the chemist. The food we
eat, the chemicals and fertilisers used to grow this food, the fuels we use for transport and energy and the
wide range of medications that we use to prolong and enhance our lives all require thorough chemical
analysis to ensure that they will perform their intended function and will not be harmful to either us or the
environment. The techniques that we will investigate include gravimetric and volumetric analyses which can
be performed in the school laboratory as well as more complex modern instrumental techniques involving
spectroscopy and chromatography.
Organic chemical pathways, where we will investigate synthetic organic chemistry and some important
aspects of biochemistry. Knowledge of the homologous series of the alkanes gained from Unit 1 studies will
enable us to branch out into the different functional groups and their chemical properties. The reaction
pathways by which one organic molecule can be synthesised from another will be considered in detail. The
structure and function of proteins and their use as disease markers is an important aspect of our study of
biochemistry, as is the structure and bonding of DNA and its applications in forensic science.
Unit 4 Areas of study
Industrial chemistry, which focuses on the factors that affect the rate and extent of chemical reactions.
Students study energy profiles and how equilibrium laws are applied to homogeneous equilibria. Experiments
are conducted to investigate the effect of temperature, concentration of reagents, pressure and catalysts on the
position of equilibrium of a reaction, and apply Le Chatelier’s principle to explain the results. The industrial
production of one significant chemical selected from ammonia, ethene, sulfuric acid or nitric acid will be
studied in detail.
Supplying and using energy focuses on our use of different energy resources. The extent of energy
resources and the advantages and disadvantages of their use will be investigated. Students will use
calorimeters to measure the energy of chemical reactions. The electrochemical series will be used extensively
to assist in the prediction of redox reactions in aqueous solutions. Students will construct and operate simple
galvanic and electrolytic cells and use the electrochemical series to predict and explain their results.
Faraday’s laws will be used to solve problems involving quantitative calculations for electrolysis reactions.
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CHEM_EXPO_12.FM
Copyright © Neap
Overview of Chemistry Units 3&4
1.2 Assessment and levels of achievement
There are two outcomes which must be met for Unit 3 and two for Unit 4, with each outcome equally
weighted in terms of assessment to ensure that they have been successfully met. School-assessed coursework
in Unit 3 contributes 17% of the total study score and the mid-year examination in June contributes a
further 33%. This split is repeated in the second half of the year, with a final examination in November which
assesses the coursework taught in Unit 4.
In Unit 3, the outcomes are to “evaluate the suitability of techniques and instruments used in chemical
analysis” and to “identify and explain the role of functional groups in organic reactions and construct reaction
pathways using organic molecules”. School-assessed coursework will consist of three tasks across the unit:
an extended experimental investigation worth 50% of the marks available, a written report on one practical
activity from the other Area of study (25%) and another task such as the analysis of data using structured
questions (25%). The extended experimental investigation will involve
•
between three and five hours of practical work, conducted in pairs or small groups;
•
preparation of a risk assessment and management sheet;
•
each individual presenting his/her own report on the task in a format to be decided by your teacher.
In Unit 4, the outcomes are to “analyse the factors that determine the optimum conditions used in the
industrial production of a selected chemical” and to “analyse chemical and energy transformations occurring
in chemical reactions”. As in Unit 3, school-assessed coursework in Unit 4 contributes 17% of the total study
score and the final examination in November 33%. School-assessed coursework will consist of three tasks
across the unit: a summary report including annotations of three related practical activities worth 50% of the
marks available, a written report on one practical activity from the other Area of Study (25%) and another
task such as the analysis of data using structured questions (25%).
1.3 Achieving A/A+ grades in VCE Chemistry
Chemistry, like Mathematics and many other disciplines, is a subject where knowledge is accumulated and
built upon over a number of years. To achieve at a high standard requires a thorough understanding of the
fundamental chemical principles that underpin the course. It is not sufficient to ‘cram’ before the
examinations or SAC tasks and fill your head with facts – this method will yield a B grade at best. It is the
nature of the subject and how it is examined that exam questions require students to reason – to logically
deduce a rational response to information provided in (sometimes) unfamiliar contexts. Some of the
hallmarks of the successful student are listed below.
•
works consistently through the year, revising systematically and reviewing new concepts as they arise
•
asks for assistance when unsure, rather than simply abandoning a concept and hoping that it will
disappear
•
practises question types consistently and comprehensively. These may be from the textbook, revision
guides, old examinations and so on
•
always endeavours to contextualise knowledge, such that they can see how different ideas ‘fit together’
and make sense
•
perhaps most importantly of all, maintains a sensible life balance: family and friends time, sport,
music, social life, relaxation, spiritual renewal and school work. Some sacrifices will be necessary, but
the complete abandonment of all of the pleasures in life will leave you miserable and unproductive.
Copyright © Neap
CHEM_EXPO_12.FM
5
VCE Unit 3 Chemistry: Coursework & Exam Preparation
1.4 Resources
Each of the major publishers has produced textbooks designed to meet the requirements of the recently
accredited Chemistry course. Your teacher is likely to have booklisted one of these texts for your use.
Hogendoorn B, Commons C, Jarrett S, McKenzie C, Moseley W, Porter M, Williamson M, 2007, Heinemann
Chemistry 2, 4th edn, Heinemann, Melbourne.
Jeffery F, 2007, Chemistry smartstudy revision & exam guide for VCE Unit 3, Neap, Melbourne.
Jeffery F, 2007, Chemistry smartstudy revision & exam guide for VCE Unit 4, Neap, Melbourne
Jeffery F, Ellett B, O’Shea P, and Ellis J, 2006, Chemistry Dimensions 1, Pearson Education Australia,
Melbourne.
Jeffery F, Ellett B and O’Shea P, 2007, Chemistry Dimensions 2, Pearson Education Australia, Melbourne.
Lukins N, Elvins C, Lohmeyer P, Ross B, Sanders R and Wilson G, 2006, Heinemann Chemistry 1,
4th edn, Heinemann, Melbourne.
Sharwood J and Gordon J, 2006, Chemistry VCE Units 1 and 2, Nelson, Melbourne.
Sharwood J and Gordon J, 2007, Chemistry VCE Units 3 and 4, Nelson, Melbourne.
Taylor N, Derbogosian M, Ng W, Stubbs A and Stokes R, 2006, Study on Chemistry 1, Wiley and Sons,
Brisbane.
Taylor N, Derbogosian M, Ng W, Stubbs A and Stokes R, 2007, Study on Chemistry 2, Wiley and Sons,
Brisbane.
On the VCAA website you will find a full list of all VCE subjects on offer, including course information, past
examination papers and examiner reports (including solutions). This is a most useful website for all students
and should be accessed regularly: www.vcaa.vic.edu.au.
The website of the Chemistry Education Association is a valuable resource that can be accessed at
www.cea.asn.au.
6
CHEM_EXPO_12.FM
Copyright © Neap
Copyright © Neap
4
Be
9.0
21
Sc
44.9
22
Ti
47.9
79
Au symbol of element
relative atomic mass 197.0
atomic number
29
Cu
63.5
30
Zn
65.4
5
B
10.8
6
C
12.0
7
N
14.0
8
O
16.0
9
F
19.0
beryllium
12
Mg
24.3
magnesium
20
Ca
40.1
calcium
38
Sr
87.6
strontium
56
Ba
137.3
barium
88
Ra
(226)
radium
lithium
11
Na
23.0
sodium
19
K
39.1
potassium
37
Rb
85.5
rubidium
55
Cs
132.9
casesium
CHEM_EXPO_12.FM
87
Fr
(223)
francium
actinium
89
Ac
(227)
lanthanum
57
La
138.9
yttrium
39
Y
88.9
scandium
60
Nd
144.2
seaborgium
106
Sg
(263)
tungsten
74
W
183.8
molybdenum
42
Mo
95.9
chromium
24
Cr
52.0
91
Pa
231.0
protactinium
90
Th
232.0
thorium
uranium
92
U
238.0
praseodymium neodymium
59
Pr
140.9
58
Ce
140.1
cerium
dubnium
105
Db
(262)
tantalum
73
Ta
180.9
niobium
41
Nb
92.9
vanadium
rutherfordium
104
Rf
(261)
hafnium
72
Hf
178.5
zirconium
40
Zr
91.2
titanium
23
V
50.9
neptunium
93
Np
237.1
promethium
61
Pm
(145)
bohrium
107
Bh
(264)
rhenium
75
Re
186.2
technetium
43
Tc
98.1
manganese
25
Mn
54.9
plutonium
94
Pu
(244)
samarium
62
Sm
150.3
hassium
108
Hs
(265)
osmium
76
Os
190.2
ruthenium
44
Ru
101.1
iron
26
Fe
55.8
gold
americium
95
Am
(243)
europium
63
Eu
152.0
meitnerium
109
Mt
(268)
iridium
77
Ir
192.2
rhodium
45
Rh
102.9
cobalt
27
Co
58.9
111
Rg
(272)
gold
79
Au
197.0
silver
47
Ag
107.9
copper
curium
96
Cm
(251)
gadolinium
64
Gd
157.2
berkelium
97
Bk
(247)
terbium
65
Tb
158.9
darmstadtium roentgenium
110
Ds
(271)
platinum
78
Pt
195.1
palladium
46
Pd
106.4
nickel
28
Ni
58.7
name of element
californium
98
Cf
(251)
dysprosium
66
Dy
162.5
112
Uub
mercury
80
Hg
200.6
cadmium
48
Cd
112.4
zinc
einsteinium
99
Es
(252)
holmium
67
Ho
164.9
thallium
81
Tl
204.4
indium
49
In
114.8
gallium
31
Ga
69.7
aluminium
13
Al
27.0
boron
fermium
100
Fm
(257)
erbium
68
Er
167.3
114
Uuq
lead
82
Pb
207.2
tin
50
Sn
118.7
germanium
32
Ge
72.6
silicon
14
Si
28.1
carbon
mendelevium
101
Md
(258)
thulium
69
Tm
168.9
bismuth
83
Bi
209.0
antimony
51
Sb
121.8
arsenic
33
As
74.9
phosphorus
15
P
31.0
nitrogen
nobelium
102
No
(259)
ytterbium
70
Yb
173.0
116
Uuh
polonium
84
Po
(209)
tellurium
52
Te
127.6
selenium
34
Se
79.0
sulfur
16
S
32.1
oxygen
lawrencium
103
Lr
(260)
lutetium
71
Lu
175.0
astatine
85
At
(210)
iodine
53
I
126.9
bromine
35
Br
79.9
chlorine
17
Cl
35.5
fluorine
118
Uuo
radon
86
Rn
(222)
xenon
54
Xe
131.3
krypton
36
Kr
83.8
argon
18
Ar
39.9
neon
10
Ne
20.2
helium
hydrogen
3
Li
6.9
2
He
4.0
1
H
1.0
Overview of Chemistry Units 3&4
1.5 Data sheets
1. Periodic Table of the elements
7
VCE Unit 3 Chemistry: Coursework & Exam Preparation
2. The electrochemical series
E° in volt
F2(g) + 2e
–
–
+2.87
2F (aq)
+
–
H2O2(aq) + 2H (aq) + 2e
+
–
Au (aq) + e
–
Cl2(g) + 2e
+1.68
–
+1.36
2Cl (aq)
+
–
+
–
3+
–
Ag (aq) + e
2H2O(l)
–
+1.09
Ag(s)
+0.80
+
–
H2O2(aq)
–
–
2+
–
4+
–
Cu (aq) + 2e
–
+
–
2+
–
2+
–
2+
–
2+
–
2+
–
Pb (aq) + 2e
Sn (aq) + 2e
Ni (aq) + 2e
Co (aq) + 2e
Fe (aq) + 2e
2+
+ 2e
–
+0.15
–
+0.14
–
2+
Mn (aq) + 2e
H2(g)
0.00
Pb(s)
–0.13
Sn(s)
–0.14
Ni(s)
–0.23
Co(s)
–0.28
Fe(s)
–0.44
3+
H2(g) + 2OH
–
–
Al (aq) + 3e
2+
Mg (aq) + 2e
+
Na (aq) + e
–
–
2+
–
+
–
+
–
K (aq) + e
Mn(s)
Al(s)
Mg(s)
Na(s)
Ca (aq) + 2e
8
H2S(g)
–0.76
Zn(s)
2H2O(l) + 2e
Li (aq) + e
+0.34
2+
+
+0.40
Sn (aq)
S(s) + 2H (aq) + 2e
2H (aq) + 2e
4OH (aq)
Cu(s)
Sn (aq) + 2e
+0.68
+0.54
2I (aq)
O2(g) + 2H2O(l) + 4e
Zn
+0.77
Fe (aq)
–
+1.23
2Br (aq)
O2(g) + 2H (aq) + 2e
I2(s) + 2e
–
2+
Fe (aq) + e
+1.77
Au(s)
O2(g) + 4H (aq) + 4e
Br2(l) + 2e
2H2O(l)
Ca(s)
–
–0.83
–1.03
–1.67
–2.34
–2.71
–2.87
K(s)
–2.93
Li(s)
–3.02
CHEM_EXPO_12.FM
Copyright © Neap
Overview of Chemistry Units 3&4
3. Physical constants
23
Avogadro’s constant (NA) = 6.02 × 10
Charge on one electron = –1.60 × 10
–19
mol
–1
C
–1
Faraday constant (F) = 96 500 C mol
–1
–1
Gas constant (R) = 8.31 J K mol
Ionic product for water (KW) = 1.00 × 10
–14
2
mol L
–2
at 298 K (self ionisation constant)
Molar volume (Vm) of an ideal gas at 273 K, 101.3 kPa (STP) = 22.4 L mol
–1
–1
Molar volume (Vm) of an ideal gas at 298 K, 101.3 kPa (SLC) = 24.5 L mol
–1
Specific heat capacity (c) of water = 4.18 J g
Density (d) of water at 25°C = 1.00 g mL
K
–1
–1
1 atm = 101.3 kPa = 760 mmHg
0°C = 273 K
4. SI prefixes, their symbols and values
SI prefix
Symbol
Value
giga
G
10
mega
M
10
kilo
k
10
deci
d
10
centi
c
10
milli
m
10
micro
µ
10
nano
n
10
pico
p
10
Copyright © Neap
9
6
3
–1
–2
–3
–6
–9
–12
CHEM_EXPO_12.FM
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VCE Unit 3 Chemistry: Coursework & Exam Preparation
1
5. H NMR data
Typical proton shift values relative to TMS = 0.
These can differ slightly in different solvents. Where more than one proton environment is shown in the
formula, the shift refers to the ones in bold letters.
Type of proton
Chemical shift (ppm)
0.9
1.3
R–CH3
R–CH2–R
RCH
CH
1.7
CH3
R3–CH
2.0
O
CH 3
O
2.0
or CH3 C
C
OR
NHR
CH 3
R
C
2.1
O
R–CH2–X (X = F, Cl. Br or I)
R–CH2–OH
3–4
3.6
O
R
3.2
C
NHCH2R
R–O–CH3
or R–O–CH2R
3.3
O
O
C
CH3
4.1
O
R
4.1
C
OCH2R
1–6 (varies considerably under different conditions)
1–5
R–O–H
R–NH2
RHC
4.6–6.0
CH2
OH
7.0
H
7.3
O
R
8.1
C
NHCH 2R
O
R
9–10
C
H
O
R
11.5
C
O
10
H
CHEM_EXPO_12.FM
Copyright © Neap
Overview of Chemistry Units 3&4
6.
13
C NMR data
Type of carbon
R–CH2–OH
Chemical shift (ppm)
8–25
20–45
40–60
36–45
15–80
35–70
50–90
RC
CR
75–95
RC
CR
110–150
R–CH3
R–CH2–R
R3–CH
R4–C
R–CH2–X
RC–NH2
160–185
RCOOH
7. Infrared absorption data
Characteristic range for infrared absorption.
–1
Bond
Wave number (cm )
700–800
750–1100
1000–1300
C–Cl
C–C
C–O
C
C
1610–1680
C
O
1670–1750
O–H (acids)
2500–3300
2850–3300
3200–3550
3350–3500
C–H
O–H (alcohols)
N–H (primary
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amines)
CHEM_EXPO_12.FM
11
VCE Unit 3 Chemistry: Coursework & Exam Preparation
8. 2-amino acids (α-amino acids)
Name
Symbol
alanine
Ala
Structure
CH3
H2N
CH
COOH
NH
arginine
Arg
H2N
CH2
CH2
CH2
CH
COOH
NH
C
NH2
O
asparagine
Asn
H 2N
aspartic acid
C
NH 2
CH
COOH
CH2
COOH
CH
COOH
CH2
SH
CH
COOH
Asp
H2N
cysteine
CH 2
Cys
H2N
O
glutamine
Gln
CH 2
CH 2
CH
COOH
CH2
CH2
H2N
CH
COOH
H2N
CH2
COOH
H 2N
glutamic acid
glycine
C
NH 2
COOH
Glu
Gly
N
histidine
isoleucine
leucine
His
CH
COOH
CH3
CH
CH2
H2N
CH
COOH
CH3
CH
CH3
Leu
CH2
CH
COOH
CH2
CH2
CH
COOH
CH 2
CH 2
CH
COOH
CH2
CH2
NH2
Lys
S
CH 3
Met
H 2N
12
CH3
Ile
H2N
methionine
H
H 2N
H2N
lysine
N
CH 2
CHEM_EXPO_12.FM
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Overview of Chemistry Units 3&4
Name
Symbol
Structure
CH2
phenylalanine
Phe
H2N
CH
H
proline
Pro
serine
Ser
threonine
COOH
COOH
N
CH2
OH
H2N
CH
COOH
CH3
CH
OH
H2N
CH
COOH
Thr
H
N
tryptophan
Trp
CH2
H2N
CH
COOH
OH
CH2
tyrosine
valine
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Tyr
H2N
CH
COOH
CH3
CH
CH3
H2N
CH
COOH
Val
CHEM_EXPO_12.FM
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VCE Unit 3 Chemistry: Coursework & Exam Preparation
9. Formulas of some fatty acids
Name
lauric
Formula
C11H23COOH
myristic
C13H27COOH
palmitic
C15H31COOH
palmitoleic
C15H29COOH
stearic
C17H35COOH
oleic
C17H33COOH
linoleic
C17H31COOH
linolenic
C17H29COOH
arachidic
C19H39COOH
arachidonic
C19H31COOH
10. Structural formulas of some important biomolecules
CH2OH
H
H
O
H
H
O
HOH2C
OH H
H
HO
OH
CH2OH
O
C
OH
H
C
OH
H
C
OH
H
OH
H
H
H
H
OH
sucrose
glycerol
O
HOH2C
OH
H
H
H
H OH
H
deoxyribose
NH2
O
N
C
N
C
HC
HN
CH
C
N
adenine
14
C
N
H
C
NH2
N
C
O
guanine
N
H
O
C
C
N
CH
C
N
NH2
C
CH
N
H
cytosine
CHEM_EXPO_12.FM
C
HN
CH
O
O
C
CH
N
H
thymine
CH3
O–
P
O–
O–
phosphate
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Overview of Chemistry Units 3&4
11. Acid–base indicators
Name
pH range
Ka
Colour change
Acid
Base
thymol blue
1.2–2.8
red
yellow
2 × 10
methyl orange
3.1–4.4
red
yellow
2 × 10
bromophenol blue
3.0–4.6
yellow
blue
6 × 10
methyl red
4.2–6.3
red
yellow
8 × 10
bromothymol blue
6.0–7.6
yellow
blue
1 × 10
phenol red
6.8–8.4
yellow
red
1 × 10
phenolphthalein
8.30–10.0
colourless
red
5 × 10
–2
–4
–5
–6
–7
–8
–10
12. Acidity constants, Ka, of some weak acids
Ka
Name
Formula
ammonium ion
NH4
5.6 × 10
benzoic
C6H5COOH
6.4 × 10
boric
H3BO3
5.8 × 10
ethanoic
CH3COOH
1.7 × 10
hydrocyanic
HCN
6.3 × 10
hydrofluoric
HF
7.6 × 10
hypobromous
HOBr
2.4 × 10
hypochlorous
HOCl
2.9 × 10
lactic
HC3H5O3
1.4 × 10
methanoic
HCOOH
1.8 × 10
nitrous
HNO2
7.2 × 10
propanoic
C2H5COOH
1.3 × 10
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+
CHEM_EXPO_12.FM
–10
–5
–10
–5
–10
–4
–9
–8
–4
–4
–4
–5
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VCE Unit 3 Chemistry: Coursework & Exam Preparation
13. Values of molar enthalpy of combustions of some common fuels at 298 K and 101.3 kPa
16
–1
∆H c (kJ mol )
Substance
Formula
State
hydrogen
H2
g
–286
carbon (graphite)
s
–394
methane
C
CH4
g
–889
ethane
C2 H6
g
–1557
propane
C3 H8
g
–2217
butane
C4H10
g
–2874
pentane
C5H12
l
–3509
hexane
C6H14
l
–4158
octane
C8H18
l
–5464
ethene
C2 H4
g
–1409
methanol
CH3OH
l
–725
ethanol
C2H5OH
l
–1364
1-propanol
CH3CH2CH2OH
l
–2016
2-propanol
CH3CHOHCH3
l
–2003
glucose
C6H12O6
s
–2816
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Coursework and exam preparation
2.1 Unit 3 – Area of study 1: Chemical analysis
Calculations
EMPIRICAL AND MOLECULAR FORMULAS
The empirical formula of a compound shows the simplest numerical ratio of each type of atom or ion present
in the compound.
The molecular formula of a compound shows the actual numbers of atoms present in a molecule of
the compound.
Question 1
A compound of carbon, hydrogen and oxygen is found to contain 52.2% by mass carbon and 13.0% by
mass hydrogen.
Find the empirical formula of the compound.
C
H
O
mass ratio
----------------------------atomic mass
ratio
simplest ratio
MOLE CONCEPT
The mass of one atom is too small for laboratory work. Instead, chemists deal with the mass of a fixed large
number of atoms of an element. The standard number adopted is the number of atoms in 12 grams of carbon.
Experimentally it has been shown that 12 grams of carbon-12 isotope contains approximately 6 × 10
This is called Avogadro’s number. This value is determined experimentally to be
23
NA = (6.02252 ± 0.00028) × 10 .
23
atoms.
A mole of a substance is the amount of substance containing the same number of molecules, ions, atoms or
12
particles as there are atoms in 12 grams of C. The unit for mole is mol.
23
Because all other atomic masses are found by comparison to carbon-12, it logically follows that if 6 × 10
23
carbon-12 atoms weigh 12 grams, then 6 × 10 atoms of any other element would weigh the same as the
relative atomic mass expressed in grams. The molar mass of any substance is the mass of one mole of the
–1
substance expressed in grams. The symbol is M, the unit is g mol .
KMnO4
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RFM = 158 amu
CHEM_EXPO_12.FM
M = 158 g mol
–1
17
VCE Unit 3 Chemistry: Coursework & Exam Preparation
C6H12O6
RMM = 180 amu
M = 180 g mol
–1
To calculate the number of moles of a substance, we make use of the following formulas.
m
•
n = ----M
–1
where n = number of mole in mol, m = mass in grams and M = molar mass in g mol
•
•
N
n = ------NA
23
where N = number of particles and NA = Avogadro’s constant = 6.02 × 10
n = cV
where c = concentration in mol L
•
•
–1
(molarity) and V = volume in L
V
n = ------VM
where V = volume in L and VM = molar volume in L (VM = 22.4 L at STP or 24.5 L at SLC)
pV
n = ------RT
–1 –1
where p = pressure kPa, V = volume in L, R = gas constant (8.31 J mol K ) and T = temperature in K
Question 2
Which of the following contains the smallest number of atoms?
A.
B.
C.
D.
18
45.0 L of SO2 at STP
24
6.0 × 10 molecules of NO2
1.5 mole of O2
128 g of SO2
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pH CALCULATIONS
The pH scale was devised to allow a quantitative measure of the relative acidity or basicity of particular
solutions. It is important to recognise that pH is a mathematical scale based on logarithms, so that a difference
of one unit on the pH scale correlates with a ten-fold change in relative acidity or basicity.
+
–
It can be experimentally proven that the product of the concentrations of H3O and OH ions in any aqueous
solution remains a constant value at a specified temperature. This product is known as the ionisation
constant of water (KW) where
+
–
–14
KW = [H3O ][OH ] = 10
+
–
–7
In neutral solutions, [H3O ] = [OH ] = 10
at 25°C
M at 25°C.
+
–
+
–7
+
–
+
–7
In acidic solutions [H3O ] > [OH ], i.e. [H3O ] > 10
In basic solutions, [H3O ] < [OH ], i.e. [H3O ] < 10
–
–7
–
–7
M and [OH ] < 10
M and [OH ] > 10
M at 25°C.
M at 25°C.
Note that the ionisation constant expression forms the basis of the pH scale.
+
+
pH = –log10[H ] = –log10[H3O ]
or
+
+
–pH
[H ] = [H3O ] = 10
, i.e. at 25°C pH <7 is acidic, pH = 7 is neutral and pH >7 is basic
Question 3
Calculate the pH of each of the following solutions.
a.
1.0 × 10
–4
M HCl
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
b.
10
–6
M NaOH
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
c.
5.0 × 10
–2
M H2SO4
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
EXAM TIP: A common multiple-choice question involves determining the pH of the resulting solution
from a neutralisation reaction. Be careful to note which reactant is in excess, as this will determine
whether the final solution is acidic or alkaline.
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19
VCE Unit 3 Chemistry: Coursework & Exam Preparation
Flowcharts and diagrams
Now that the year is underway you will have begun to study a number of analytical techniques. To understand
these techniques, make use of diagrams and flowcharts. Try to summarise your practical exercises using these
flowcharts. Consider two important analytical techniques: gravimetric analysis and volumetric analysis.
GRAVIMETRIC ANALYSISGravimetric analysis is the process of determining the composition of a sample
by formation of a precipitate and the measurement of mass.
Weigh a sample of the consumer item.
Mix with water to dissolve the soluble ion.
Filter the mixture.
Add excess reagent to form a precipitate.
glass rod
Filter the precipitate and wash with water.
Buchner
funnel
Dry the precipitate in an oven until
no more water is lost.
(weigh to constant mass)
mixture containing
precipitate
to pump
filter flask
Use mass of precipitate to
perform calculations.
VOLUMETRIC ANALYSIS
Volumetric analysis is the process of determining the concentration of one solution by reacting it with another
solution of known concentration.
A standard solution is a solution whose concentration is accurately known.
1. Place weighed
sample in a
volumetric flask.
20
2. Half fill with water.
Shake to dissolve the
sample.
CHEM_EXPO_12.FM
3. Add water to the
calibration line. Shake
again.
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Coursework and exam preparation
In volumetric analysis, amounts in solution are determined by performing a titration.
1. A burette is filled
with standard solution.
2. A pipette delivers a
known volume (aliquot) of
the unknown solution into
the flask.
3. An indicator is added.
4. When the end point is
reached, record the volume
used (titre).
An indicator is a compound whose colour as an acid is different from its conjugate base. An example of an
indicator is phenolphthalein. An indicator must be added to the reaction mixture to detect the equivalence
point. The equivalence point for a titration is when the solutions have been added in the mole ratio shown by
the reaction equation.
Prepare standard solution (acid or base) in the
burette and record the initial volume.
Transfer an aliquot of the unknown
solution into a conical flask by pipette.
Add several drops of suitable
indicator to the flask.
Titrate the unknown solution with the
standard solution until the end point
is reached.
Record the titre (volume) of standard
solution added.
Repeat the titration until three
concordant titres are obtained.
Perform calculations.
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21
VCE Unit 3 Chemistry: Coursework & Exam Preparation
Sample questions
Question 4
2+
The concentration of lead ions (Pb ) in polluted water is determined by adding sodium sulfate solution
(Na2SO4) to exactly 500.0 mL of the water. The equation for the reaction is
Pb(NO3)2(aq) + Na2SO4(aq)→ 2NaNO3(aq) + PbSO4(s)
a.
The PbSO4 precipitate was collected and dried in an oven at 110°C and weighed periodically. The last
five weighings are shown in the table below.
Weighing number
Mass (g)
1
0.9731
2
0.9653
3
0.9614
4
0.9613
5
0.9616
Why is more than one weighing of the precipitate required?
________________________________________________________________________________
________________________________________________________________________________
b.
Calculate the mass of lead in the precipitate.
________________________________________________________________________________
________________________________________________________________________________
c.
Calculate the molarity of Pb
2+
ions in the polluted water sample.
________________________________________________________________________________
________________________________________________________________________________
d.
If the polluted water contained another ion which forms a precipitate with sulfate ions, how would this
2+
affect the calculated result for the concentration of Pb ions?
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
22
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Question 5
Household bleach contains the strong oxidant sodium hypochlorite (NaOCl) as its active ingredient. The
concentration of NaOCl is generally known as the ‘available chlorine’ and is often expressed as the mass of
active ingredient per unit volume of solution, or w/v.
A student wishes to verify the available chlorine content in a sample of commercially available bleach,
quoted as ‘40 g/L available chlorine’.
To achieve this aim, 25.00 mL of the bleach is pipetted into a 250.0 mL standard flask and made up to the
mark with deionised water. Three 20.00 mL aliquots of this diluted solution are pipetted into separate conical
flasks and about 10 mL of acidified potassium iodide solution is added. The solution in the flask immediately
becomes dark brown, as the available chlorine oxidises the iodide ions to iodine, according to the equation
–
–
+
–
OCl (aq) + 2I (aq) + 2H (aq) → I2(aq) + Cl (aq) + H2O(l)
This solution is titrated with a standard solution of 0.513 M sodium thiosulfate solution for each flask. An
average titre volume of 9.57 mL was recorded. The relevant equation is
2–
2–
–
I2(aq) + 2S2O3 (aq) → S4O6 (aq) + 2I (aq)
a.
Calculate the average number of mol of S2O3
2–
in the flask.
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
b.
Hence calculate the average number of mol of I2 reduced by the S2O3
2–
ions.
_________________________________________________________________________________
_________________________________________________________________________________
–
Deduce the number of mol of OCl in the aliquot of diluted solution.
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
c.
–
Calculate the number of mol of OCl in the original sample of bleach.
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
d.
–
Hence deduce the number of mol of chlorine atoms (as Cl ) in the original sample of bleach.
_________________________________________________________________________________
_________________________________________________________________________________
_________________________________________________________________________________
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CHEM_EXPO_12.FM
23
VCE Unit 3 Chemistry: Coursework & Exam Preparation
e.
Calculate the mass of available chlorine in the original sample, and hence determine the w/v ratio.
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
Spectroscopic techniques
There is a range of spectroscopic techniques used by the modern analytical chemist. These techniques can
provide both qualitative information (which chemical substances are present in a sample) and quantitative
information (how much of the chemical of interest is actually present). Several of the spectroscopic techniques
that we will consider involve analysing the energy released or absorbed by molecules, atoms or electrons
under particular circumstances. Indeed, the term spectroscopy refers to the ‘analysis of the spectrum’.
10–12
10–9
10–6
10–3
1
1000
short wave
infrared
wavelength (m)
NOT TO SCALE
VHF
UHF
shortest
wavelength
(highest
gamma X-rays
energy)
rays
ultraviolet visible
light
medium
wave
microwaves
longest
wavelength
(lowest
energy)
The section of the electromagnetic spectrum with which we are most familiar is that which corresponds to
visible light: the rainbow of red, orange, yellow, green, blue, indigo and violet. Radio waves, infrared and
ultraviolet frequencies are also of great importance to the analytical chemist.
MASS SPECTROSCOPY
Mass spectroscopy differs from the other spectroscopic techniques in that it does not use energy from the
electromagnetic spectrum to interact with particles. In this case, atoms or fragments of larger molecules are
ionised by being stripped of electrons, accelerated by an electric field and then deflected by a magnetic field
acting at right angles to the direction of their motion. In this way, the relative masses of the particles can be
determined and this information used to help to identify particular compounds.
ion accelerating
voltage
+
–
beam of
lighter particles
magnetic
field
gas inlet
1. vapouriser produces
a gasous sample
3. accelerator causes
ions to move
2. ioniser produces
positive ions
4. deflector
separates ions
5. detector 6. recorder
produces a mass spectrum
beam of
heavier particles
While the mass spectrometer was an essential tool in determining the relative atomic masses of the elements
by accurately determining the masses and relative abundances of their isotopes, its main use now is in
identifying chemical compounds on the basis of their characteristic mass spectra.
24
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Coursework and exam preparation
Question 6
The mass spectrum of 2-butenoic acid is shown below.
relative
abundance
20
40
60
80
m/z (mass/charge ratio)
a.
Draw the structural formula of 2-butenoic acid.
b.
Explain why a significant peak occurs at 86.
100
_________________________________________________________________________________
_________________________________________________________________________________
c.
Write structural formulas for the fragments that caused peaks at 41, 45 and 69.
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25
VCE Unit 3 Chemistry: Coursework & Exam Preparation
ATOMIC ABSORPTION SPECTROSCOPY
Developed in the 1950s by the CSIRO scientist Alan Walsh, the atomic absorption spectrometer is perhaps
the most widely used analytical instrument in modern laboratories. It is extremely accurate and highly
versatile – capable of testing for the presence of up to 68 elements. AAS is used extensively to test for the
trace amounts of metals in food, drink, blood samples and urine samples.
The fundamental principles of the atomic absorption spectrometer are similar to those of several other
spectrometers: colorimeters, IR spectrometers and UV–visible spectrometers. Light of a particular frequency
which is characteristic of the element being measured is passed through an atomised sample of the substance
to be analysed. Electrons of this element absorb some of this energy and are excited to higher energy levels.
If, for example, a technician wished to test for the presence of mercury in a sample of fish, they would fit the
instrument with a lamp which emits the characteristic frequencies of mercury. Thus, any mercury atoms in the
sample would absorb light of these frequencies and be excited. The amount of light absorbed as the beam passes
through the solution can be recorded and will be directly proportional to the concentration of mercury present.
The lamp emits light
of the required
wavelength (plus
some unwanted
wavelengths).
atomic vapour
in flame
The detector measures
pulsed light.
The light is
chopped into
pulses.
A solution of
sample is
sprayed into
a flame.
The monochromator
and slit select light of
a particular wavelength.
Before the AAS can be used to quantify the amount of an element present it must first be calibrated. In this
procedure, samples of known concentration of the element are passed through the device and their
absorbances recorded. The absorbance of the sample being tested can then be determined from the
calibration curve.
EXAM TIP: Instrumental analysis forms a major part of the new Chemistry course. It is likely that
students will be required to use information obtained from a range of analytical instruments, and
synthesise the data to (for example) identify an unknown molecule.
26
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Coursework and exam preparation
Sample questions
SECTION A
Question 7
If 150 mL of 0.10 M NaOH is mixed with 50 mL of 0.30 M HCl, then the pH of the resultant solution is
A.
1
B.
2
C.
7
D.
13
Question 8
Which of the following processes is involved in analysis by UV–visible spectroscopy?
A.
the absorption of energy by hydrogen nuclei
B.
the absorption of energy by molecules, which changes their vibrational and rotational movements
C.
the absorption of energy as electrons move between energy levelsthe movement of positively charged
particles by combined magnetic and electric fields
SECTION B
Question 9
Many commercial detergents contain phosphorus in the form of sodium polyphosphate. The amount of
phosphorus in a sample of detergent can be determined colorimetrically. 0.500 g of a solid detergent sample
is dissolved in water and the solution made up to 500 mL. 20.0 mL samples of this solution are mixed with
2.00 mL of a standard molybdate solution. A blue colour develops that gives an absorbance reading of 0.130
in a simple colorimeter.
Five standard solutions of sodium polyphosphate were similarly treated with the standard molybdate
solution, and their absorbances are shown below. The graph shows the measured absorbance as a function of
the mass of phosphorus (P) per litre in the standard polyphosphate solutions.
0.3
absorbance
0.2
0.1
0
0
10
30
20
–1
concentration of P in mg L
a.
–1
What is the concentration (in mg L ) of phosphorus (as P) in the solution?
1 mark
What is the percentage by mass of phosphorus (P) in the detergent?
2 marks
b.
Explain why it is necessary to make up a set of standard solutions of sodium pyrophosphate.
1 mark
Total 4 marks
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27
VCE Unit 3 Chemistry: Coursework & Exam Preparation
2.2 Unit 3 – Area of study 2: Organic chemical pathways
Unit 3 Area of study 2, Organic chemical pathways, investigates the structure, bonding and reactions of
selected organic chemicals. Students study homologous series, functional groups, biomolecules and selected
reaction pathways. The roles of organic molecules in the generation of drugs and biofuels are investigated,
along with applications of organic molecules such as DNA in forensic analysis.
Functional group
Name of
compound
–Cl
chloro
–OH (hydroxyl)
alkanol
–NH2
amine
–COOH (carboxyl)
carboxylic acid
–COO R′
ester
–CONH2
amide
System used to name compounds
prefix
–chloro
suffix
–ol
prefix
hydroxy–
suffix
–ylamine
prefix
suffix
–oic acid
prefix
alkyl group ( R′ )
suffix
–oate
suffix
–amide
Question 10
Name the following compounds.
a.
(CH3)2CHCHCH2
________________________________________________________________________________
b.
CH3(CH2)4COOH
________________________________________________________________________________
c.
CH3COO(CH2)3CH3
________________________________________________________________________________
d.
CH3CH(OH)CH3
_________________________________________________________________________
28
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EXAM TIP: In the biochemistry section of the course, the molecules under consideration are often large
and complex. Practice locating and naming functional groups in these molecules, and recall how they can
bond with one another.
2.3 Unit 4 – Chemistry at work
Enthalpy of chemical reactions
The enthalpy of a chemical reaction is a measure of its heat energy. In the previous lecture we defined the
terms exothermic and endothermic in terms of whether energy was lost to, or gained from the surroundings
as a reaction proceeded.
Heat change diagrams are most useful tools in describing such changes in enthalpy.
EXOTHERMIC
ENDOTHERMIC
When we incorporate the concept of enthalpy with the mole ratio of a chemical reaction we generate a
thermochemical equation. These equations enable us to calculate the actual amount of energy lost or gained
by a given amount of reactant as a result of a particular reaction. For example,
CH4(g) + 2O2(g) → 2CO2(g) + 2H2O(l) ΔH = –890 kJ mol
–1
The equation shows us that one mole of methane undergoes complete combustion in oxygen gas to produce
two moles of carbon dioxide and water, and in the process produces 890 kJ of energy.
RULES
•
If you reverse the direction of a reaction then the sign of the Δ H value also changes:
e.g. 2CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH = +890 kJ mol
•
–1
H values are in stoichiometric ratio to the reactants and products as written in the equation
e.g. 2CH4(g) + 4O2(g) → 4CO2(g) + 4H2O(l) ΔH = –1780 kJ mol
–1
•
If equations are added or multiplied by a factor then so are their ΔH values.
•
Symbols of state must be clearly stated as ΔH values change with change of state.
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29
VCE Unit 3 Chemistry: Coursework & Exam Preparation
Question 11
a.
Calculate ∆H for the reaction 2S(s) + 3O2(g) → 2SO3(g), given that
–1
S(s) + O2(g) → SO2(g) ∆H = –297 kJ mol
–1
2SO2(g) + O2(g) → 2SO3(g) ∆H = –196 kJ mol
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
b.
Calculate the amount of energy released when 10.0 g of carbon burns in excess oxygen according to
the following equation.
C(s) + O2(g) → CO2(g) ∆H = –393.4 kJ mol
–1
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
c.
What mass of propanol must be burnt in excess oxygen to produce 500 kJ of energy, given the
following equation.
2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(l) ∆H = –196 kJ mol
–1
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
30
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Question 12
The chemical equilibrium C2H6(g)
C2H4(g) + H2(g) is endothermic in the forward direction.
In order to increase the fraction of ethane converted to ethene and hydrogen at equilibrium, a chemist should
A.
B.
C.
D.
raise both the temperature and the pressure.
raise the temperature and lower the pressure.
lower the temperature and raise the pressure.
lower both the temperature and the pressure.
Question 13
The chemical reaction
–
+
2–
Br2(l) + SO2(aq) + 2H2O(l) → 2Br (aq) + 4H (aq) + SO4 (aq)
proceeds readily and rapidly in aqueous solution. It is proposed that a galvanic cell be made using this
reaction as a cell reaction in the cell shown below.
V
a.
Label this diagram in order to show a cell based on this cell reaction. Label the positive and negative
electrodes and give the chemical composition of each half-cell.
(You may need to use some of the data from the separate data sheet provided.)
b.
Give the half-cell reactions that occur as the cell discharges.
_________________________________________________________________________________
_________________________________________________________________________________
c.
What would be the necessary chemical properties of the electrode materials used?
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VCE Unit 3 Chemistry: Coursework & Exam Preparation
d.
Given the maximum voltage of single cell is 0.89 V, explain how you would recharge a cell that had
been discharged.
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e.
What reaction would occur in the cell when it was being recharged?
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2.4 Solutions
Question 1
C
H
O
mass ratio
----------------------------atomic mass
52.2
---------12.0
13.0
---------1.0
34.8
---------16.0
ratio
4.35
13.0
2.175
simplest ratio
2
6
1
The empirical formula is C2H6O.
Question 2
C
V
45.0
In A, n(SO2) = ------- = ---------- = 2.0 mol
V m 22.4
N(atoms) = 3 × N(SO2) = 3 × n(SO2) × NA = 3.6 × 10
24
In B, N(atoms) = 3 × N(NO2) = 3 × 6.0 × 10
24
25
= 1.8 × 10
In C, N(atoms) = 2 × N(O2) = 2 × n(O2) × NA = 1.8 × 10
24
m 128
In D, n(SO2) = ----- = --------- = 2.0 mol
M 64
N(atoms) = 3 × N(SO2) = 3 × n(SO2) × NA = 3.6 × 10
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Coursework and exam preparation
Question 3
a.
strong acid, ∴complete ionisation
+
∴[H ] = 1.0 × 10
–4
+
–4
pH = –log[H ] = –log(1.0 × 10 ) = 4
b.
soluble hydroxide, ∴complete dissociation
–
–6
∴[OH ] = 10
M
– 14
– 14
10
10
–8
+
- = ------------ = 10
[H ] = --------------–
–6
[ OH ] 10
+
–8
pH = –log[H ] = –log(10 ) = 8
c.
strong, diprotic acid, ∴complete ionisation
+
∴[H ] = 2.5 × 10
–2
= 10
+
–1
–1
pH = –log[H ] = –log(10 ) = 1
Question 4
a.
To ensure the mass of precipitate was constant, indicating that it was dry.
b.
m 0.9614
n(PbSO4) = ----- = ---------------M 303.3
0.9614
n(Pb) = n(PbSO4), m(Pb) = n × M = ---------------- × 207.2 = 0.6568 g
303.3
c.
d.
0.9614
2+
n(Pb ) in 500.0 mL = ---------------303.3
n 0.9614 / 303.3
–3
2+
- = 6.339 × 10 M
∴c(Pb ) = --- = ---------------------------------–
3
V
500.0 × 10
More precipitate would form, so the calculated result would be larger.
Question 5
2–
–3
–3
a.
n(S2O3 ) = c × V = 0.513 × 9.57 × 10
b.
1
2–
–3
n(I2) = --- × n ( S 2 O 3 ) = 2.45 × 10 mol
2
c.
From the first equation, n(OCl ) = n(I2) = 2.45 × 10
d.
250.0
–
–
n(OCl ) in bleach sample = n(OCl ) in 20.00 mL × ------------- = 0.0307 mol
20.00
e.
n(Cl) = n(OCl ) = 0.0307 mol
f.
m(Cl) = n × M = 0.0307 × 35.5 = 1.089 g
1.089 g in 25.00 mL
x g in 1000 mL
–1
x = 43.6 g L
= 4.91 × 10
–
–3
mol
mol
–
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Question 6
a.
O
H
H
H
C
C
C
C
H
H
OH
b.
The relative atomic mass of butenoic acid is 86 and so this is the parent peak for this spectrum. It is
generated by the loss of one electron from the molecule.
c.
The fragments at 41 , 45 and 69 are generated by the fragments
+
+
+
H
H
H
O
H
O
C
C
C
H
H
,
and
C
OH
H
H
C
C
C
C
H
H
respectively.
Question 7
n(NaOH) = c × V = 0.10 × 0.150 = 0.0150 mol
n(HCl) = c × V = 0.30 × 0.050 = 0.0150 mol
∴n(HCl) = n(NaOH)
∴neutral
∴pH = 7
Question 8
C
The absorption of energy by hydrogen nuclei takes place in NMR spectroscopy, so A is incorrect. Infrared
spectroscopy measures the energy taken up by molecules as they vibrate and rotate, so B is incorrect.
Electrons do not absorb energy in UV–visible spectroscopy, so C is correct. Movement of positively charged
particles and fragments of molecules by electric and magnetic fields is a feature of mass spectrometry, so D
is incorrect.
Question 9
a.
[phosphorus] = 13 mg per litre (from graph)
b.
Note that [phosphorus] is the same in the 20.0 mL sample and the 500 mL flask = 13 mg per litre.
If 13 mg phosphorus in 1000 mL
then x mg in 500 mL
∴mass of phosphorus on the original sample = 6.5 mg
–3
6.5 × 10
100
% mass = ------------------------ × --------- = 1.3% m/m
0.500
1
c.
A set of standard solutions were required to generate a calibration curve so that the absorbance of the
fertiliser solution could be converted to a concentration.
Question 10
a.
3-methylbut-1-ene
b.
hexanoic acid
c.
butyl ethanoate
d.
propan-2-ol
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Coursework and exam preparation
Question 11
a.
To obtain the required equation we must multiply the first equation by 2 and add the result to the
second equation.
–1
2 × –297 + (–196) = –790 kJ mol
b.
c.
m 10.0
n(C) = ----- = ---------- = 0.833 mol
M 12.0
energy released = 0.833 × 393.4 = 328 kJ
From the equation, 2 mol of propanol releases 196 kJ of energy
so x mol of propanol releases 500 kJ of energy.
x 500
--- = --------2 196
2 × 500
x = ------------------ = 5.10 mol
196
Mass of propanol = n × M = 5.10 × 60 = 306 g
Question 12
∆H > 0
We want an increase in the forward reaction.
∴raise T and lower P
Question 13
a.
V
cathode (+)
reduction
2–
SO2(g)/SO4 (aq)
+
and H (aq)
–
Br2(l)/Br (aq)
–
b.
Br2(l) + 2e → 2Br (aq)
2–
+
SO2(g) + 2H2O(l) → SO4 (aq) + 4H (aq) + 2e
c.
The electrode materials would need to be inert (unreactive) and good conductors of electricity.
d.
To recharge the cell, an electrical supply of at least 0.89 V would need to be supplied to the cell.
e.
As the cell is recharged, the reaction proceeds backwards.
–
2–
+
2Br (aq) + SO4 (aq) + 4H (aq) → Br2(l) + SO2(g) + 2H2O(l)
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