Download AGE article for Sept 2013

(‘The Age’ Online - Sept 2013)
The Year 12 VCE Chemistry exam on Tuesday, November 12 (9.00 – 11.45 am) will be the first time
students have had to sit a two and a half hour exam covering all of Units 3 and 4.
Essential preparation for this will involve revision of all the Key Knowledge and its applications as
detailed in the current Study Design. This can be accessed at:
http://www.vcaa.vic.edu.au/Pages/vce/studies/chemistry/chemindex.aspx
Please bear in mind that your teacher may wish to provide you with VCAA exams to practise under
examination conditions. If this is the case, it is most unwise for you to have seen the questions (or their
answers!) in advance, as this will not allow you to identify your weaknesses.
However, if this is not the case, all VCAA exams from 2002 to 2012 and the relevant reports from the
Chief Assessor can be accessed at:
http://www.vcaa.vic.edu.au/Pages/vce/studies/chemistry/exams.aspx
Note that the exams from 2002 to 2007 covered a different Study Design, and some of the questions will
no longer be relevant. If in doubt, consult your teacher. Also note that a few questions relating to
industrial production of a chemical on the 2008 to 2012 November exams will no longer be relevant, as
this material will have been covered on one of the SACs.
2008 saw the introduction of a far more comprehensive data booklet than had previously been supplied
with examinations. You can expect that at least some questions on the November exam will require the
use of data that is provided. Sometimes data is presented at the start of the question, but more often it will
be in the Data Book that accompanies the exam and with which you should now be very familiar.
1. Periodic Table of the Elements
Its main use will be as a source of relative atomic masses for calculations, but students should never
forget that the Group Numbers give clues to the charges on the cations and anions formed by many
common elements. A frequent mistake on the 2008 November exam was to write the formula of
magnesium chloride as MgCl. A quick check of the Periodic Table should reveal that magnesium, a
Group 2 element, would form Mg2+, and chlorine, in Group 17, would exist as Cl– in an ionic compound,
thus combining to make the formula MgCl2.
When using the Periodic Table to find a Relative Atomic Mass, be careful that you don’t use the Atomic
Number by mistake; this can easily happen under the pressure of exam conditions. Mini periodic tables
that fit inside your calculator often have RAMs to two decimal places; use of these will be regarded as
suspicious.
Determining the relative mass (Mr) of potassium hydrogen carbonate (KHCO3) provides an example of
the use of the rules regarding significant figures in chemical calculations. Addition of the relevant
relative atomic masses (RAMs) from the table gives a relative mass of 100.1 [39.1 + 1.0 + 12.0 + (3 ×
16.0)]. This value is accurate to the first decimal place and so will have 4 significant figures (regardless of
the fact that the RAM of H only has 2). This will be important in the likely event that the value is
subsequently used in a calculation involving multiplication or division, where the number of significant
figures in the answer should be the same as the number of significant figures in the bit of data that has the
least number of significant figures (ie. the least accurate bit of data).
Eg.
mass (KHCO3) = 4.06 g
Mr (KHCO3) = 100.1 (Molar mass = 100.1 g mol-1)
n(KHCO3) = 4.06 / 100.1 = 0.0406 mol
Note that the first two zeros in the answer are not significant, and that the answer is rounded off to 3
significant figures.
Significant figures are likely to be assessed somewhere on the exam, and you will lose a mark if you have
not quoted a particular calculated answer to the correct number of significant figures.
2. The Electrochemical Series
The Electrochemical Series will probably need to be accessed more than once. At the beginning of
writing time, it is a good idea to write ‘oxidant’ at the top left of the table, ‘reductant’ at the top right, an
arrow up the LHS to indicate increasing oxidant strength, and an arrow down the RHS to indicate
increasing reductant strength.
This will remind you, for example, that the fluoride ion, F–, is an incredibly weak reductant, meaning that
it ‘hates’ being oxidized. There is no oxidant in the table that can cause the oxidation of F–, so this
process can only be achieved by electrolysis. On the other hand, its conjugate oxidant, F2, is a very strong
oxidant, and ‘loves’ being reduced. Do not pick a fight with this feisty lady; she’ll ‘rip’ electrons off just
about anything!
BEWARE: naming as ‘the oxidant’ the substance being oxidized is a common mistake.
It is noteworthy that half equations for two important strong oxidants are not provided. You should be
able to use your rules to write half equations for acidified MnO4– ions being reduced to Mn2+, and
acidified Cr2O72– ions being reduced to Cr3+. Either of these may well be relevant when considering the
oxidation of an alkanol to an alkanoic acid.
The half equations for a hydrogen-oxygen fuel cell are readily found in the Electrochemical Series. The
overall equation is the same as for the combustion of hydrogen:
2H2 (g) + O2 (g) → 2H2O(l)
Being an exothermic redox reaction, electrical energy can be obtained directly from chemical energy if
the two half reactions can be separated, with oxidation of hydrogen occurring at an anode and reduction
of oxygen at a cathode.
The fuel cell can operate with either an acid or an alkaline electrolyte. For the acid electrolyte, you want
H+ in the half equations (you should be able to find them at Eo = +1.23 v and 0.00 v), and for the alkaline
electrolyte, you want OH– in the half equations (at Eo = +0.40 v and –0.83 v). Note that in both
electrolytes, the predicted cell voltage under standard conditions is 1.23 v, as you would expect, given
that both cells would have the same overall reaction.
If you know the simple rules that apply, the Electrochemical Series can be used to predict the half
reactions that occur in galvanic and electrolytic cells, and also whether or not a redox reaction is likely to
occur between 2 reactants. It is useful to note that many of the reductants on the right hand side of the
table are metal elements. This should remind you that if a metal element is reacting, it will be oxidized,
and the substance with which it reacts will be reduced.
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 the activation energy, EA)
• an oxide coating may make a reactive metal seem unreactive (eg Al2O3 on Al)
• Non-standard conditions may mean that an Eo value is not applicable.
However, in general, predictions can be relied on, even if conditions are not standard, and it is unwise to
use “non-standard conditions” as a reason for not observing a predicted reaction. Lower temperature
and/or pressure than standard conditions might slow a reaction, but higher temperature or pressure would
usually increase the rate of reaction, and make it more likely to be observed.
BEWARE: When using the Electrochemical Series to predict 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. 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 oxidized in aqueous solution.
Note, however, an industrially significant limitation of electrochemical series predictions; if the
concentration of Cl–(aq) is high enough, Cl2 (g) will be produced at the anode in an electrolytic cell.
• that the symbol of state for an ion in a molten electrolyte will not be (aq). Use (l).
eg Na+(l) + e– → Na(l) (Na+ would not be reduced if it were in aqueous solution!)
Be sure to read questions very carefully (highlight important phrases if necessary) and beware of simply
trotting out pre-prepared answers. For example, a question may ask you to explain why a particular redox
reaction is not predicted to occur. Many students will interpret this as if it were asking why a particular
reaction was not observed to occur. This would miss the point of the question, and no credit would be
received for an answer that simply referred to the rate of the reaction being too slow.
3. Physical Constants
Use of some (or all) of the Physical Constants on page 5 will be required, so it is helpful if you have
considered in advance the type of questions that might involve them.
The unit for the Faraday constant (C mol–1) allows you to deduce the relevant formula
F = Q (in coulomb) / n(e–)
You will also need to remember Q = It (which looks like advice to smokers from the anti-cancer council)
and E = VIt , from which can be derived a useful relationship for the energy produced by a cell,
E = VQ = V × F × n(e–).
Any calculation that refers to the heating of water may require use of its Specific heat capacity (c). You
might prefer to memorize the formula:
E = 4.18 × m × ∆T
but BEWARE: the m in this formula is the mass of water. It is a common mistake for students to use (as
m) the mass of the fuel used to heat the water. In some ways it is 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 (1 mL) of water by 1K (1˚C). The problem
can then be solved using ratios.
Because the Gas Constant (R) is given in SI units, you need to be careful regarding the units used for the
variables in the Universal Gas Equation, pV = nRT. Technically, volume (V) should be in m3 and
pressure (p) in Pa (N m-2). However, because R still has the value 8.31 when its unit is kPa L K-1 mol-1,
the equation will also ‘work’ if volume (V) is in litre (L) and pressure (p) in kilopascal (kPa).
The unit of Density (g mL-1) should remind you that d = m/V.
Many students have an unshakeable conviction that a neutral solution must have a pH of 7. It is therefore
essential to recognize the significance of the Ionic product for water (Kw = [H+] × [OH–]). Because this is
an equilibrium constant, it will vary with temperature, and the self ionization is an endothermic reaction:
H2O(l) → H+(aq) + OH–(aq) ; ∆H > 0 (positive)
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. Not a lot more, mind you, but enough for the examiners to set some interesting traps.
4. SI Prefixes
Make sure you can recognize all the prefix symbols, and do not confuse mega (M) with mol L–1 (M) or
even molar mass (usually given the symbol M, although it probably ought to be Mm to avoid confusion,
and to be consistent with the symbols for molar volume, Vm, and relative mass, Mr)
It is not uncommon for the difference between J and kJ to be overlooked when relating specific heat (in J)
to Heat of Combustion of a fuel (usually in kJ). Less familiar prefixes may also need to be allowed for.
5. & 6. NMR Data
This appears daunting, but in many cases chemical shifts will only be used to help you confirm a structure
that has already been deduced from other information. This material was not included in previous study
designs, so exams that predate 2008 will contain no NMR spectroscopy on which to practise.
The best advice is probably to study as many examples as possible of NMR spectra of simple organic
molecules in order to ‘get a feel’ for their interpretation.
7. Infrared Absorption Data
Wave numbers in the ‘fingerprint’ region (< 1500 cm-1) are unlikely to be helpful. Learn to recognize the
characteristic ‘trough’ caused by the carbonyl group (C=O) and the difference between the trough due to
O-H in carboxylic acids compared to that produced by alcohols. The characteristic trough caused by the
C-H bond (which may be masked by the O-H trough in carboxylic acids) will be present in the IR spectra
of most organic molecules and hence will usually be of little use in classifying the compound.
8. α–amino acids
These are the amino acids that are biologically significant, all having an amino group and a carboxyl
group attached to the same carbon atom. The structures given are essentially semi-structural in nature; it
would be good practice to check that you can turn them into:
• full structural formulae, showing all bonds
• molecular formulae (in particular, the ones containing ring structures).
9. Formulas of some Fatty Acids
Prior to the exam, it would be a good idea to ensure you can classify each of these as one of:
• saturated (fitting the formula CnH2n+1COOH)
• mono-unsaturated (2 H’s less than the corresponding saturated fatty acid i.e. CnH2n-1COOH)
• polyunsaturated (4 or more H’s less than the corresponding saturated fatty acid)
10.
Structural Formulas of some Important Biomolecules
A key skill to practise is to show that you can combine 3 fatty acid molecules with a glycerol molecule in
a condensation reaction that produces a triglyceride + H2O, and also the reverse of this process, taking a
triglyceride and hydrolyzing it to form glycerol and 3 fatty acid molecules.
You could consolidate this theme by showing how the sucrose structure can be hydrolyzed into glucose
(6-sided ring) and fructose (5-sided ring).
An important feature of the biomolecules is their solubility in water, a property you should be able to
explain by referring to their ability to form hydrogen bonds with water molecules.
The DNA structure is incredibly complicated, so it is important that you focus on the key aspects of its
building blocks, the 2 backbone chains of alternating deoxyribose and phosphate units, with the double
helix structure maintained by hydrogen bonding between guanine-cytosine (G-C) and adenine-thymine
(A-T) ‘base’ pairs (‘GC3, AT2’ is a handy mnemonic). However, beware of referring to ‘double’ or
‘triple’ bonds between the nitrogen bases, as this might imply that you think the bonding is covalent.
11. Acid-base Indicators
You should be able to use this table to choose an appropriate indicator for a given acid-base titration. This
may involve interpretation of a pH curve for the titration, or consideration of possible hydrolysis reactions
of the chemical species present at the equivalence point.
The table can also be used to determine the colour of the indicator at the endpoint.
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 (one may be colourless). It may be worth considering
the Ka expression for the weak acid, and how the ratio [In–] / [HIn] will vary as the pH varies.
12. Acidity Constants Ka, of some weak acids at 25˚C
Any question involving a weak acid and a calculation will probably require you to look up the appropriate
Ka value for the acid. Typical calculations might involve finding the pH of a solution of known
concentration and/or the percentage ionization (or percentage hydrolysis).
Questions involving the change in pH when a weak acid is diluted are often poorly handled. Students
seem not to recognize that equilibrium and Le Chatelier’s Principle are relevant. An interesting exercise is
to compare the dilution of HCl from 1.0 M to 0.10 M, with the dilution of CH3COOH from 1.0 M to 0.10
M.
The calculation is simple for the strong acid. Because HCl can be regarded as essentially fully ionized,
[H+] goes from 100 M to 10–1 M, so the pH changes from 0 to 1.
For the weak ethanoic acid, however, you will need to look up the Ka value (1.7 × 10–5 M).
This applies to the equation:
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO–(aq)
Making the usual 2 assumptions ([CH3COOH]at eq’m ≈ [CH3COOH]initially, and that [H+] from selfionization of water is negligible) leads to:
[H+]2 / 1.0 = 1.7 × 10–5
∴ [H+] = 0.0041 M and pH = 2.4 for 1.0 M CH3COOH
[H+]2 / 0.10 = 1.7 × 10–5
∴ [H+] = 0.0013 M and pH = 2.9 for 0.10 M CH3COOH
Note that the pH change is smaller for dilution of the weak acid. Dilution has increased the % hydrolysis
of the ethanoic acid, as predicted by Le Chatelier’s Principle. 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. Of
course, 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 was effectively no equilibrium consideration.
Another way of explaining this is to consider the relationship between reaction rates and equilibrium.
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 probably slightly increase the rate of the forward
reaction (more frequent collisions between the reactant particles, and hence mor frequent ‘successful’
collisions), but will definitely decrease the rate of the back reaction (lower concentrations of the products
will lead to less frequent collisions between product particles). With the two rates no longer equal, the
system is out of equilibrium, and the forward reaction will occur to a greater extent until the [CH3COOH]
has decreased enough and the [H3O+] and [CH3COO–] increased enough for the two rates to be equal
again.
13. Molar enthalpy of combustion of some common fuels (at S.L.C.)
A question involving the combustion of a fuel may require a value for the molar enthalpy of combustion.
BEWARE: Heat of Combustion is a concept that seems to result in confusion for many students. At its
simplest level, it is merely the amount of energy produced when a given amount of a fuel is completely
reacted with oxygen. The trick is to look very carefully at the unit of the quantity involved in measuring
the amount of fuel.
For example, you can deduce from Table 13 in the Data Book that the Heat of Combustion of ethane is
1557 kJ/mol. This tells you that 1557 kJ of energy will be released if 1 mole of ethane is completely
reacted with oxygen. However, the Heat of Combustion of ethane could also be expressed as 51.9 kJ/g or
63.6 kJ/L at SLC (25˚C and 1 atm). Are you able to use data from the Data Book to confirm these values?
Note that they are only given to 3 significant figures; do you know why?
The confusion may arise because of the way the data is presented in the Data Book. Heats of Combustion
are given in the form of ΔHc, with the unit in kJ mol-1. Now a ΔH is normally used as part of a specific
thermochemical equation, and in such cases the ‘per mol’ bit does not refer to any one reactant or product
in particular (it doesn’t play favourites!). It is simply signalling that the quoted amount of energy (in kJ)
will be released (if ΔH is –ve) or absorbed (if ΔH is +ve) when the co-efficients in the specific equation
are read as amounts in mole.
For example, the data book can be used to write the thermochemical equation for the combustion of
ethane as:
C2H6 (g) + 7/2 O2 (g) → 2 CO2 (g) + 3 H2O(l) ; ΔH = – 1557 kJ mol-1
But if you choose to eliminate the fraction from the co-efficients, the ΔH value must change, since the
equation will no longer be referring to 1 mole of ethane reacting with 7/2 mole of oxygen, but 2 mole of
ethane reacting with 7 mole of oxygen (so twice as much energy will be released).
2 C2H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O(l) ; ΔH = – 3114 kJ mol-1
Note that the unit is still kJ mol-1.
However, by definition, a ΔHc does play favourites. It is referring to the fuel as the most important part of
the combustion equation and the unit (kJ mol-1) is telling you to write the equation with 1 as the coefficient of the fuel so that the value is ‘per mole’ of fuel. Note also that because it is a form of ΔH, it has
a sign (–ve), which a normal Heat of Combustion value would not.
BEWARE: the Heat of Combustion of any fuel (including mixtures) can be expressed as kJ/g, but the unit
kJ/mol can only be used for a pure substance with a known molar mass.
Combustion reactions of biofuels are highly relevant to both Unit 3 and Unit 4.
BEWARE: It is easy to overlook the oxygen atom in an alkanol when writing a balanced equation for its
combustion. When considering the symbol of state for the alkanol, keep in mind that an aqueous solution
will not usually burn!
When tackling short answer questions such as those that begin "Describe how …" or that ask you to
"Explain why …", you are strongly advised to check how many marks are allocated to the question and to
give your answer in ‘point form’. A classic application of this approach would be the answer to Q.2c on
the VCAA June 2009 examination, which asked for a description of how the purity of a product could be
quantitatively determined by a particular spectroscopic method. Many students gave long-winded,
tortuous and often irrelevant explanations, but the examiners report makes it clear that to obtain full
marks, a student needed to provide at least 3 points out of a possible 4 that would have been present in a
very good answer.
Keep the acronym SUBS in mind every time you perform a calculation or write a chemical equation. The
SU applies to calculations, and is to remind you to check Significant figures and Units. The ‘BS factor’
refers to equations; always check that they are Balanced and show Symbols of state (‘subscripts’).
And here is one last piece of (possibly unwelcome) advice. During this year you will have encountered
situations where ethanol is used as either a biochemical fuel or as a solvent (eg methylated spirits).
However, its even more common use as a legal, but potentially addictive, mood-altering recreational drug
is seldom discussed. Please be aware that any binge drinking of flavoured ethanol beverages is likely to
have a seriously performance-impairing effect on both your revision and your ability to achieve your best
on the exam.
Phil Ponder teaches Chemistry at Penleigh and Essendon Grammar School