23/09/2015
Kwong Lee Dow Young Scholars Program
Chemistry Revision 2015
VCE Chemistry Exam
150 minutes (+15 mins reading time)
Tues 10 November 9.00 - 11.45 am
Presenter: Phil Ponder
(Penleigh and Essendon Grammar)
2015 KLD Chemistry VCE Revision Lecture
Useful Websites
• The VCAA website is
www.vcaa.vic.edu.au
• The Chemistry Education Association website is
www.cea.asn.au
• Spectroscopy revision
www.chemicaldetectives.com
• A lot of bright and conscientious students seem to use
www.atarnotes.com
to exchange ideas and materials
2015 KLD Chemistry VCE Revision Lecture
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Analyzing Practice Exams
USE THE EXAMINERS’ REPORTS
These are available from the VCAA and/or CEA websites for all VCAA exams
from 2000 to 2013 that have been sat and marked.
– take careful note of questions that were done poorly and the examiners’
thoughts on why this was the case
(Poorly done questions are often revisited on subsequent exams)
– look carefully at the mark allocation
– look at ‘popular’ but incorrect multiple-choice answers and try to
determine why they were selected (and why they are wrong!)
– examine your own responses critically in the light of the examiners’
expectations
– add noteworthy remarks to your topic summaries as reminders
2015 KLD Chemistry VCE Revision Lecture
Four ways to save marks (SUBS)
Calculations
•  Significant Figures: All of your numerical answers must have
the correct number of significant figures
• Unit: unless a question specifically asks for a particular unit,
you will be penalized if you have no unit or an incorrect unit
Equations
• Balance:
Double check that any equation you write
is balanced for both elements and charge
• Symbols of State: Include and use the terms
(aq), (s), (l) and (g) properly
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Significant Figures
ADDING / SUBTRACTING
- for addition or subtraction, the result is rounded to the
smallest number of DECIMAL PLACES:
eg
if adding:
NB: the periodic table in the DATA BOOK gives atomic masses
to one decimal place.
- so all molar masses should be calculated to one decimal place
2015 KLD Chemistry VCE Revision Lecture
Significant Figures
MULTIPLYING / DIVIDING
- for multiplication and/or division, the result is rounded to the
smallest number of SIGNIFICANT FIGURES in the data used.
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Significant Figures for pH
Unlikely to be tested on the exam, but best to be thorough!
[H+] = 1.45 x 10–4 M
(3 sig figs)
–4
pH = –log10(1.45 x 10 ) = 3.839
(3 sig figs)
The number before the decimal point in a pH refers to the power of 10,
so the 3 sig figs we require come after the decimal point.
pH = 7.85
The 7 refers to the power of 10, so there are only 2 sig figs
(the digits after the decimal point)
+
–7.85
–8
[H ] = 10
= 1.4 x 10 M
(2 sig figs)
Beware It is not acceptable to leave your answer as [H+] = 10–7.85 M
2015 KLD Chemistry VCE Revision Lecture
Formulae
MOLE / STOICHIOMETRY
n = m / Mm
n = V / Vm
n=c×V
n = No. / NA
pV = nRT
n(e–) = Q / F
OTHER
pH= –log[H+]
[H+]= 10–pH
calibration factor = ∆E /∆T
d (density) = m/V
Q = It
E = VIt (hence E=VQ)
E = 4.18 × m × ∆T
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Universal Gas Equation
The Gas Constant, R = 8.31 J K-1 mol-1
• The unit of R is in SI units
• Since R is defined in SI units, all quantities in the
equation should use SI units.
• p should be in Pa, V in m3, n in mol, and T in K.
• P in kPa and V in L only works because 2 conversion
factors of 1000 cancel out (ie. 2 mistakes cancel !).
Beware: Know which unit combinations will work.
2015 KLD Chemistry VCE Revision Lecture
Concentration Unit Conversions
Convert 13.5 %(V/V) ethanol (CH3CH2OH) into mol/L.
(density of pure ethanol is 0.785 g/mL)
Density: There are only 2 things you can do with a density
- calculate m, given V
- calculate V, given m
Note: the unit of density gives
• the formula for density (d = m/V)
• the units for m and V
2015 KLD Chemistry VCE Revision Lecture
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Volumetric Analysis example
Back titration
(Answer: 5.14%)
A sample of a copper ore is obtained in which copper carbonate,
CuCO3, is known to be the only copper containing substance.
5.00 g of the ore is treated with 25.0 mL of 0.250 M H2SO4 and the
resultant mixture allowed to stand until reaction is complete.
All dissolved CO2 is then driven off by heating.
The resultant solution is then titrated with a 0.250 M NaOH solution,
and 17.60 mL of NaOH is required to neutralize the remaining acid.
Assuming that CuCO3 is the only substance present that reacts with
acid, calculate the percentage by mass of copper in the sample of
copper ore.
CuCO3 (aq) + H2SO4 (aq) → CuSO4 (aq) + CO2 (g) + H2O(l)
0.250 M; 0.0250 L
H2SO4 (aq)
+
(excess from first reaction)
2NaOH(aq)
→
Na2SO4 (aq) + 2H2O(l)
0.250 M; 0.0176 L
2015 KLD Chemistry VCE Revision Lecture
Chromatography – GLC (GC)
VCAA past exam question (June 2002)
The diagram represents GC apparatus.
a. Clearly identify, on the diagram, the mobile and
stationary phases, and explain how gas
chromatography is able to separate the
components of a gaseous mixture.
b. A pure substance was suspected of being the
high molecular weight anabolic steroid Stanozolol.
A sample of this substance was dissolved in a
suitable pure solvent and injected into a gas
chromatograph. How many peaks would you
expect to see in the resulting chromatogram?
Justify your result.
c. If a sample of pure
Stanozolol were available,
how could it be used to
confirm the identity of the
suspect sample?
2015 KLD Chemistry VCE Revision Lecture
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Bonding Question
Why is the sodium salt of aspirin much
more soluble in water than aspirin itself?
Na+
Acetylsalicylic acid
(aspirin)
Sodium salt of
acetylsalicylic acid
2015 KLD Chemistry VCE Revision Lecture
Hydrogen Bonding
Hydrogen Bonding is a significantly stronger than usual form
of dipole-dipole bonding that arises in the following situations:
2015 KLD Chemistry VCE Revision Lecture
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Spectroscopy - principles
Atoms and molecules can absorb energies of some wavelengths
and transmit energies of different wavelengths
(ie. the ones not absorbed)
An absorption spectrum is produced by irradiating a sample with
energy of different wavelengths.
The absorbed radiation may cause:
•  electrons to ‘jump’ to higher energy levels: AAS and UV-Vis
•  a change in the vibrational state of a molecule:
IR
http://en.wikipedia.org/wiki/Infrared_spectroscopy
•  a change in the ‘spin’ of particles in the nucleus:
NMR
2015 KLD Chemistry VCE Revision Lecture
Infrared Spectroscopy
• 
• 
• 
• 
interaction of molecules with IR radiation
wavelength absorbed depends on the type of bond
causes a change in the vibrational state of the molecule
used to identify some functional groups, especially
OH (alcohol), OH (acid) and C=O
Copyright: Pearson-Heinemann, 2008"
2015 KLD Chemistry VCE Revision Lecture
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Infrared (IR) Spectroscopy
Ethanol
CH3CH2OH
OH (alcohol)
Ethanoic acid
CH3COOH
OH (acid) and C=O
Ethyl ethanoate
CH3CH2OOCCH3
C=O
!fingerprint"
2015 KLD Chemistry VCE Revision Lecture
Proton (1H) NMR
•  Area under each peak
(at low resolution) is
proportional to the
number of H atoms in
that chemical
environment
•  At high resolution,
H atoms on adjacent C
atoms (in non-equivalent
chemical environments)
1H
NMR of ethyl ethanoate
3 different H environments
B
Not split
(no H’s on
adjacent C
atom)
C
A
Split into Split into
a quartet a triplet
Ratio of low resolution
peak areas is
2:3:3
cause splitting of the
peaks
(‘n+1’ rule)
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Mystery Molecule
2015 KLD Chemistry VCE Revision Lecture
Organic Reaction Pathways
H2 (g)
(esterification)
Copyright: Pearson-Heinemann, 2008"
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Organic reaction pathways
Devise a pathway to prepare ethyl propanoate
from propane and ethene
2015 KLD Chemistry VCE Revision Lecture
Fats and Oils
• Saturated fatty acids form
straight chains.
• Corresponding triglyceride
molecules can pack closely to
each other.
• Tend to be fats (higher m.pts.)
• A double bond puts a kink in
the chain of mono-unsaturated
and poly-unsaturated fatty acids
• Corresponding triglycerides
cannot pack closely to each other
• Tend to be oils (lower m.pts.)
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Proteins
Name the amino acids that would be produced by
hydrolysis of the following section of a polypeptide.
(CH2)4
2015 KLD Chemistry VCE Revision Lecture
Enzymes
Beware!
Biology students
often refer to an
‘optimum temperature’
for an enzyme.
Rate of most reactions
increases as T increases.
Rate decreases due to
denaturing of the enzyme.
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Reaction Rate and Particle Collision Theory
• the rate of a reaction is determined by the
frequency of successful (‘fruitful’) collisions
increased by:
increased by:
• higher reactant concentrations
• higher collision energies
• larger solid surface area
– higher temperature
• higher temperature
• lower activation energy
• higher gas pressure
– catalyst
• note the ‘double-effect’ of a temperature increase
Beware of using “proportion of successful collisions”
2015 KLD Chemistry VCE Revision Lecture
Le Chatelier’s Principle
Remember: at equilibrium :
• rates of forward and back reactions are equal
• amounts and concentrations are constant (but not usually equal)
• the ‘naughty boy’ principle (LCP) states that
if a change is imposed on a system that is at equilibrium,
the system will try to partially oppose the change.
• LCP (Le Chatelier’s Principle) is merely a useful guide as to
what may happen when a change is imposed on a system that is
at equilibrium.
- it does not cause the response!
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Le Chatelier’s Principle
A system at equilibrium will only respond to a change if the
• ratef ≠ rateb
(Rate of forward reaction no longer equals the rate of the back reaction)
• Cf ≠ Kc
(Concentration Fraction no longer equals the Equilibrium Constant)
2015 KLD Chemistry VCE Revision Lecture
2015 KLD Chemistry VCE Revision Lecture
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Le Chatelier’s Principle
colourless
reddish brown
N2O4 (g) ⇌ 2 NO2 (g) ; ∆H = +ve
(ie forward reaction is endothermic)
Kc = 5.5 x 10-3 M (at 25˚C), and increases as T increases
= Kc if the system is at equilibrium
ie
2015 KLD Chemistry VCE Revision Lecture
2015 KLD Chemistry VCE Revision Lecture
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2015 KLD Chemistry VCE Revision Lecture
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Le Chatelier’s Principle
Beware of situations where there is a pressure change at
constant volume
When the pressure change is not a result of a volume change,
LCP may not correctly predict the effect.
Examples
- adding inert gas at constant volume
- heating a gas mixture at constant volume
2015 KLD Chemistry VCE Revision Lecture
Oxidation Numbers
• assigned to each individual atom in an element or compound
• ‘if’ the atom could be an ion, what would be its charge?
• the most electronegative element has negative ox. no.
• if H is positive, it cannot be more than +1
• sign first, then number (to avoid confusion with actual charge, if any)
• atoms in an uncombined element cannot even ‘pretend’ to
have a charge; ox.no. = 0
• no real meaning, just a tool to use in redox reactions
• max. Ox. No. for a main group element = No. of valence e’s
(eg maximum oxidation state for C is +4)
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Oxidation Numbers
Examples
BF3
F2O
H2O2
Na2O2
CaH2
Oxidation No. of
F=
B=
F=
O=
H=
O=
Na =
O=
Ca =
H=
2015 KLD Chemistry VCE Revision Lecture
Electrochemical Series
Write
‘oxidant’
at the top left of the table (above F2 (g)) and
‘reductant’
at the top right
(above 2F–(aq)).
F2 is the strongest oxidant - a substance that causes the oxidation
of something else, and is itself reduced.
BEWARE: a common mistake is to name the substance being oxidised as
‘the oxidant’, or calling the substance being reduced ‘the reductant’.
The Electrochemical Series can be used to predict :
• relative strengths of oxidants and reductants
• half reactions that occur in galvanic and electrolytic cells
• whether or not a redox reaction is likely to occur between 2 reactants.
Note that many of the reductants on the right hand side of the table are
metal elements.
- if a metal element is reacting, it will be oxidised,
and the substance it reacts with will be reduced.
2015 KLD Chemistry VCE Revision Lecture
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Predicting Redox Reactions
BEWARE:
Know the limitations of Electrochemical Series predictions:
• A predicted reaction may seem not to occur if it is very slow
(eg H2O2 reacting with itself)
The Series cannot predict rate of reaction (which is related to EA)
• An oxide coating may make a reactive metal seem unreactive
(eg Al2O3 on Al)
• Non-standard conditions
(eg Cl–(aq) can be oxidised in preference to H2O if [Cl–] is high)
However, in general, predictions can be relied on,
even if conditions are not standard.
2015 KLD Chemistry VCE Revision Lecture
Predicting Redox Reactions
BEWARE:
Don’t blindly use ‘non-standard conditions’ as a
reason for failure to observe a predicted reaction.
Refer more specifically to
• T < 25˚C
or
• concentration < 1M
or
• pressure < 1 atm.
Any (or all) of these could be predicted to slow the reaction rate.
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Predicting Redox Reactions
BEWARE: When predicting electrolysis half reactions, don’t forget:
• to consider H2O in your list of possible reactants if it is an aqueous electrolyte
• that reactive metals such as Li, Na, K, Mg, Ca, Al cannot be produced by
electrolysing an aqueous solution of their cations.
(ie cations with Eo < – 0.83 v cannot be reduced in aqueous solution,
because H2O will reduce in preference)
Likewise, anions with Eo > 1.23 v (eg F–) cannot (theoretically) be
oxidised in aqueous solution.
(note, however, an important limitation of electrochemical series predictions;
if the [Cl–(aq)] is high enough, Cl2 (g) will be produced at the anode in an
electrolytic cell)
• that the subscript for an ion in a molten electrolyte will not be (aq). Use (l)
eg Na+(aq) + e– → Na(s)
Na+ would not be reduced if it were in aqueous solution!
2015 KLD Chemistry VCE Revision Lecture
Physical Constants
The unit for the Faraday constant (C mol–1) allows you to deduce the
relevant formula:
F = Q (in coulomb) / n(e–)
hence
n(e–) = Q / F
You also need
Q = It
(which looks like advice to smokers from the anti-cancer council!)
n(substance)
⇔
Faraday Calculations
n(e–)
⇔
use half equation
use Q = F ×
n(e–)
Q
⇔
I and/or t
use Q = It
Also, since E = VIt ,
you can derive a useful relationship for the energy produced by a cell:
E = VQ = V × F × n(e–).
2015 KLD Chemistry VCE Revision Lecture
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Faraday Law Question
2015 KLD Chemistry VCE Revision Lecture
Calorimetry
• Any calculation that refers to the heating of water may require use of the
Specific heat capacity (c) of water.
You might prefer to memorise the relationship:
E = 4.18 × m × ΔT
BEWARE:
the m in this formula is the mass of water,
not the mass of fuel being burned.
• It may be preferable to look at the unit (J g–1 K–1) and deduce that
4.18 J of energy will be required to heat 1g of water by 1K (1˚C).
The problem can then be solved using ratios.
• Note (i) that a change in temperature measured in oC will have the
same numerical value as when measured in K
eg ΔT = 298 K – 273 K = 25 K
⇒ 25 ˚C – 0 ˚C = 25 ˚C
(ii) 1 g of water is equivalent to 1 mL, using the density of water
(iii) E is in Joule (not kJ)
BEWARE!
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Calorimetry
• A rough value for the Calibration Factor of a calorimeter can easily be
estimated if it is assumed that the water in the calorimeter absorbs most of
the energy.
eg 100 mL of water in the calorimeter will absorb 418 J/˚C.
• The true value of the Calibration Factor will be higher than this
(since other parts of the calorimeter will also absorb heat)
• The Calibration Factor can be more accurately found
(i) using an input of electrical energy,
E = VIt
or
(ii) by performing a reaction with a known ΔH
Calibration Factor = ∆E /∆T
2015 KLD Chemistry VCE Revision Lecture
Calorimetry Question
A bomb calorimeter may be calibrated using a substance with a
well-known heat of combustion.
A commonly used calibrating agent is benzoic acid (C7H6O2),
which has a heat of combustion of 3227 kJ mol-1
2.50 g of pure solid benzoic acid (Molar mass = 122.0 g mol–1)
is placed in a calorimeter and completely reacted with oxygen.
The temperature rise of the calorimeter is observed to be
8.90˚C.
Calculate the calibration factor of the calorimeter in kJ ˚C–1.
2015 KLD Chemistry VCE Revision Lecture
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Ionic product for water
Kw = [H+] × [OH–] = 1.00 × 10–14 M2 at 25 ˚C
Kw will be required for pH calculations in alkaline (basic) solutions.
pH = – log10 [H+]
[H+] = 10–pH
(Remember that H3O+ is interchangeable with H+)
Q.1 Calculate the pH of 0.0050 M Ca(OH)2 solution at 25°C.
The dissociation of Ca(OH)2 is often overlooked:
Ca(OH)2 (s) → Ca2+(aq) + 2OH–(aq).
This indicates that the [OH–] is 2 × [Ca(OH)2].
• The self-ionisation of water equilibrium law applies to aqueous
solutions as well as pure water.
[OH–] = 2 × 0.0050 = 0.010 = 1.0 × 10–2 M
[H+] = 10–14 / 10–2 = 10–12 M
pH = 12
2015 KLD Chemistry VCE Revision Lecture
Ionic product for water
Q.2
A 10,000 L swimming pool has a pH of 9.0.
What volume of 1.0 M HCl is required to bring the pH
down to 8.0?
You will need to assume that
(i)
The HCl added makes an insignificant contribution
to the total volume (a reasonable assumption)
(ii)
OH– is the only base in the pool (unlikely)
(Unsurprisingly, the answer is not 0.090 mL !)
2015 KLD Chemistry VCE Revision Lecture
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Ionic product for water
BEWARE :
A neutral solution does not usually have a pH of 7.
• Because Kw is an equilibrium constant, it will vary with temperature
• Self ionisation of water is an endothermic reaction:
H2O(l) ⇌ H+(aq) + OH–(aq) ; ΔH = +ve
BEWARE :
Kw will only equal 1.00 × 10–14 at 25 ˚C.
At this temperature, [H+] in a neutral solution = [OH–] = 10–7 M,
and hence pH = 7.
Kw will, however increase at T > 25 ˚C, so [H+] > 10–7 M
and pH of a neutral solution will be less than 7.
Likewise, at T < 25 ˚C,
the pH of a neutral solution will be more than 7.
BEWARE :
Because pH is a measure of acidity, it is easy for your brain
to think that if a solution becomes more acidic (higher [H+]),
the pH must be increasing.
• There is an inverse relationship between [H+] and pH.
2015 KLD Chemistry VCE Revision Lecture
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Acidity Constants
• The higher the Ka, the stronger the acid
• All acids in the Data Book tables are weak acids.
(The Ka’s for strong acids are very high)
• There is a relationship between the Ka for the conjugate acid in an acid-base
indicator and the pH of the endpoint of that indicator.
• Any indicator is a mixture of a weak acid (HIn) and its conjugate base (In–),
where these two substances have different colours (see Data Book).
• We can assume that the ‘official’ endpoint for the indicator (when it changes
colour) is roughly when [HIn] ≈ [In–].
For the hydrolysis equation: HIn(aq) ⇌ H+(aq) + In–(aq)
When [HIn] = [In–] (ie at the endpoint)
• At the endpoint,
,
[H+] = Ka
and
pH = –log10Ka
2015 KLD Chemistry VCE Revision Lecture
Acidity Constants
Q. Bromophenol blue is a weak acid (represented as BH) that acts as
an acid-base indicator.
In solution the following equilibrium is established.
BH(aq) ⇌ B–(aq) + H+(aq);
Ka = 6.3 × 10–5 M
yellow
blue
At low pH bromophenol blue exists mainly as the acid, BH, which is yellow in
colour, while at high pH it exists mainly as its conjugate base, B–, which is blue.
An intermediate colour is observed when the concentration of the acid and the
concentration of the conjugate base are similar.
When [BH] = [B–], the mixture appears green.
(i) Calculate the pH at which [BH] = [B–].
(ii) Calculate the ratio [B–] / [BH] when the pH of a solution of bromophenol blue
is 7.
(iii) What colour will the indicator solution appear at pH 7? Justify your answer.
2015 KLD Chemistry VCE Revision Lecture
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Acidity Constants
** If a calculation involves a weak acid, you probably need to look up its Ka **
Change in pH when a weak acid is diluted
• Equilibrium and Le Chatelier’s Principle will be relevant.
• Compare
dilution by a factor of ten of HCl from 1.0 M to 0.10 M,
with
dilution of CH3COOH from 1.0 M to 0.10 M.
Because HCl can be regarded as essentially fully ionised,
[H+] goes from 100 M to 10–1 M, so the pH changes from 0.0 to 1.0
For the weak ethanoic acid, Ka = 1.7 × 10–5 M (at 25˚C). This applies to the equation:
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO–(aq)
Making the usual 2 assumptions: (i) [CH3COOH]at eq’m ≈ [CH3COOH]initially
(ii) [H+] from self-ionisation of water is negligible
leads to:
For 1.0 M CH3COOH
For 0.10 M CH3COOH
Ka = [H+]2 / 1.0 = 1.7 × 10–5
[H+]2 / 0.10 = 1.7 × 10–5
+
∴ [H ] = 0.0041 M and pH = 2.4 (2.38) ∴ [H+] = 0.0013 M and pH = 2.9 (2.88)
2015 KLD Chemistry VCE Revision Lecture
•
•
•
•
•
•
•
•
Why is the pH change smaller for dilution of the weak acid?
Dilution has increased the % hydrolysis of theAcidity
ethanoic acid
Constants
Addition of one of the reactants in the equilibrium has resulted in a net
forward reaction, leading to an increase in the n(H+) in the solution.
However, the greater volume of solution means that the overall [H+] will
decrease, but not by as much as it did for the strong acid, where there were
effectively no equilibrium considerations.
When a system is at equilibrium,
the rate of the forward reaction equals the rate of the back reaction.
Addition of water to the ethanoic acid equilibrium will hardly affect the
frequency of collisions between the reactant particles in the forward reaction,
but will definitely initially decrease the rate of the back reaction
– lower concentration of products will lead to less frequent collisions
between product particles.
With the two rates no longer equal, the system is not at equilibrium
Forward reaction will occur to a greater extent
(thus slowing down the forward reaction)
[CH3COOH] will decrease and the [H3O+] and [CH3COO–] will increase
Eventually the two rates will be equal2015
again.
KLD Chemistry VCE Revision Lecture
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Acidity Constants
of dilution
of dilution
2015 KLD Chemistry VCE Revision Lecture
Fun with Ethanol?
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Concentration Unit Conversion - answer
2.
13.5 mL of ethanol in 100 mL of solution (booze)
d = m/V ⇒ m = d × V = 0.785 × 13.5 = 10.6 g
⇒ 10.6 g of ethanol in 100 mL
⇒ 106 g in 1000 mL (1 L)
⇒ (106 / 46.0) mole in 1 L = 2.30 mol/L (2.30 M)
or
2015 KLD Chemistry VCE Revision Lecture
Volumetric Analysis example - answer
Back titration – solution
CuCO3(aq) + H2SO4(aq) → CuSO4 (aq) + CO2(g) + H2O(l)
0.250 M 0.0250 L
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
(in excess) 0.250 M 0.0176 L
n(NaOH) in titre = C × V = 0.250 × 0.0176 = 4.40 × 10-3 mol
n(H2SO4) in excess = ½ × n(NaOH) = 2.20 × 10-3 mol
n(H2SO4) initially = C × V = 0.250 × 0.0250 = 6.25 × 10-3 mol
n(H2SO4) reacted with carbonate = 6.25 × 10-3 – 2.20 × 10-3
= 4.05 × 10-3 mol
n(CuCO3) = n(H2SO4) initially reacted = 4.05 × 10-3
= n(Cu)
m(Cu) = n × Mm = 4.05 × 10-3 × 63.5 = 0.257 g
% Cu = (0.257 / 5.00) × 100 = 5.14 %
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Chromatography – GLC (GC) - answers
a.
Mobile phase → N2 (or equivalent indication).
Stationary phase → shaded part of U-tube.
Components of a mixture are adsorbed to differing extents
by the stationary phase, so the rate of movement through
the tube is determined by the extent of adsorption.
eg The more strongly adsorbed components will move more
slowly ⇒ bigger Rt
(Note: solubility in a gas is irrelevant!)
b.
Two peaks; one for Stanozolol and one for the solvent.
c.
Check that the retention time for the suspected Stanozolol
is the same as that for pure Stanozolol.
or Mix suspect sample with pure Stanozolol and then verify that
only one peak is obtained (other than that for the solvent)
2015 KLD Chemistry VCE Revision Lecture
Bonding Question - answer
You are strongly advised to answer in point form
• solubility is generally related to the strength of
solute – solvent bonds; the stronger these bonds,
the more soluble the solute
• the main bonding between water molecules and aspirin
molecules will be weak dispersion forces and a small
amount of hydrogen bonding
• the sodium salt will be able to form ion-dipole bonds to
water molecules
• ion-dipole bonds are stronger than hydrogen bonds
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Mystery Molecule answer
• relative emp. formula mass of C3H6O = 36 + 6 + 16 = 58
• Mass Spec: parent peak at 116 ⇒ rel. molec. mass = 116 ⇒ molec. formula C6H12O2
- 2 O’s suggests either a carboxylic acid or an ester (or possibly a diol?)
• IR spectrum is typical of an ester (no OH (acid) trough ⇒ not a carboxylic acid)
• 1HNMR suggests:
- only 2 different hydrogen environments
- 3:1 area ratio ⇒ 9 H’s in one environment (3 identical CH3’s?)
and 3 H’s in the other (a different CH3?);
the bigger shift suggests this CH3 is closer to the O’s
- no splitting ⇒ no H’s on adjacent C’s
• 13CNMR suggests there are 4 different carbon environments
Trial and error gives the ester formed from methanol and dimethylpropanoic acid
Methyl dimethylpropanoate
CH3OOC C(CH3)3
(Base peak (m/z = 57) in mass spectrum is probably due to (CH3)3C+ )
Methylpropan-2-yl ethanoate, (CH3)3C OOCCH3 , would be a worthy try,
but wouldn’t match the NMR shift data (this should have a small singlet at 2.0 ppm)
2015 KLD Chemistry VCE Revision Lecture
Organic Reaction Pathways answer
You will need to have produced ethanol and propanoic acid
prior to the last step.
•
ethene + H2O(g)
•
propane + Cl2
H3PO4
(catalyst)
1- chloropropane
UV
(use fractional distillation to separate this
from the mixture of all the chloropropanes)
1- chloropropane + OH–
Oxidation of 1- propanol
• ethanol + propanoic acid
ethanol
1- propanol (+ Cl–)
Cr2O7
2–
/
H+
(oxidant)
conc. H2SO4
(catalyst)
propanoic acid
ethyl propanoate (+ H2O)
2015 KLD Chemistry VCE Revision Lecture
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Answers to Proteins
L to R
Valine, lysine, serine, threonine, methionine
2015 KLD Chemistry VCE Revision Lecture
Answer to Calorimetry Question
• n(C7H6O2) = 2.50/122.0 = 0.0205 mol
• E = 0.0205 x 3227 = 66.1 kJ
• calibration factor =
66.1 kJ
= 7.43 kJ °C–1
8.90!°C
(NB: 3 sig figs)
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Answer to swimming pool question
At pH = 9.0, [H+] = 10–9 M so
[OH–] = 10–5 M
–
n(OH ) = cV = 10–5 × 10000 = 0.10 mol
At pH = 8.0, [H+] = 10–8 M so
[OH–] = 10–6 M
–
n(OH ) = cV = 10–6 × 10000 = 0.010 mol
Hence
0.10 – 0.010 = 0.090 mol of OH–
needs to be neutralized
Hence
0.090 mol of H+ will be required (0.090 mol of HCl)
V(HCl) = n/c = 0.090 / 1.0 = 0.090 L = 90 mL
2015 KLD Chemistry VCE Revision Lecture
Answer to Bromophenol question
Answer to Bromophenol question
(i)
(ii)
(iii)
[B! ]
[H + ][B! ]
=1
Ka =
= 6.3 × 10–5 ∴ [H+] = 6.3 × 10–5 M
[BH]
[BH]
∴ pH = – log10(6.3 × 10–5) = 4.2
At this pH the solution will be green.
[B! ]
[K a ]
–5
–7
=
= 6.3 × 10 / 10 = 630
+
[BH] [H ]
[B–] = 630 × [BH]
So [blue] = 630 × [yellow]
Hence solution will be blue.
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