Section II (options)
Question 28 Industrial Chemistry (25 marks)
	(a)	(i)	oleum
Key points are that sulfuric acid is very corrosive and that nearly pure acid has quite different properties from aqueous solutions. Make sure your safety precautions refer specifically to sulfuric acid and to transport. ( p 336 and pp 342-3)		(ii)	Sulfuric acid is very corrosive when it comes into contact with organic material including skin and flesh. Hence it needs to be transported in strong containers made of material it does not attack. Hence it is generally transported in steel containers. The reason it does not attack steel is that 98% sulfuric acid is mainly un-ionised. It does not attack glass either, but glass containers are not as strong as steel ones.      Make sure that water does not come into contact with the acid, first because sulfuric acid reacts violently with water, liberating much heat, and secondly because aqueous sulfuric acid attacks steel quite rapidly and so would destroy the container.     Because it is so corrosive people handling it should wear gloves and face masks.
Simply recall of information you should be familiar with. (Le Chatelier's principle  p 117-8, 201, 315; catalysts and equilibrium position p 202; this point should have been repeated in discussions of ammonia and sulfuric acid syntheses)	(b)	(i)	Addition of a catalyst speeds up both the forward and reverse reactions and so equilibrium is reached more quickly. However the equilibrium position reached is the same with as without a catalyst. Increasing the pressure would increase the yield of N2O4. By Le Chatelier's principle if we increase the pressure the reaction moves in the direction that would decrease the pressure, that is in the direction 2NO2 ® N2O4, because in that direction 2 moles ® 1 mole and so pressure would fall.

A routine calculation, but make sure that you show your working. Always read the question carefully so that you do not overlook any part of it such as the and account for bit at the end of this question. ( p 323-4 especially Example 2 and p 329-30 including Exercise 20)

(ii)

Equilibrium constants change as temperature changes. The fact that K increases (that is the reaction moves to the right) as temperature decreases tells us that the reaction as written is exothermic, because as temperature falls the reaction moves in the direction that releases heat (to counteract the decrease in temperature) by Le Chatelier's principle.

Note the use of two of the more difficult Board of Studies verbs, discuss and analyse, so you need more than just a list of bald facts. Seven marks means at least seven significant bits of information and you should spend about 13 minutes on it, so this requires quite a long answer.      Nevertheless the answer given here is probably too long for this time allowance even though it presents only the required information in quite a succinct manner. The first paragraph is included to convince the examiners you know what process you are talking about. Perhaps the last summarising sentence could be omitted.      It is difficult to see how this question could be answered adequately in the time allowed. If caught in the dilemma of too long a question with too short a time, when time has nearly run out jot down the key points and leave it at that. For example      Mercury cathode forms an amalgam, hence mercury cell – no NaCl contamination, no asbestos problems but mercury released.      Synthetic polymers lead to membrane cell – only sodium ions pass through. Hence membrane cell – now no mercury problem. (p 352-7)	(c)	Industrially sodium hydroxide is produced by the electrolysis of concentrated sodium chloride solution. At the anode:   2Cl–  ® Cl2 + 2e– At the cathode:  2H2O + 2e–  ®  H2 + 2OH– Overall:      2Cl– + 2H2O   ®  Cl2 + H2 + 2OH– so a NaCl solution becomes a NaOH solution.      Two processes were developed more or less simultaneously at the end of the 19th century, the diaphragm cell and the mercury cell. The diaphragm cell had an asbestos diaphragm separating the anode and cathode compartments. Sodium ions could pass through it from the anode side to the cathode side and so balance the hydroxide ions being produced there. (Include a diagram such as the one on p 353). This cell had the disadvantages 1. that some chloride diffused through the diaphragm and so the sodium hydroxide was contaminated with small amounts of sodium chloride and 2. that inevitable losses of asbestos into the air and waterways was an environmental hazard (asbestos is highly carcinogenic).      The mercury cell uses a mercury cathode instead of an iron one. This results in the formation of a Na/Hg amalgam: Na+ + e–  ®  Na(dissolved in Hg). Hence these NaOH plants used two vessels, an electrolysis cell using a mercury cathode in which chlorine gas and a Na/Hg amalgam were produced and a second container in which the amalgam was reacted with water to form an NaOH solution: 2Na(in Hg) + 2H2O  ®  H2(g) + 2NaOH(aq) (You don't really have enough time to draw a diagram of this.) This mercury cell method resulted in virtually no NaCl contamination of the product and avoided the problems of using asbestos. However it had the disadvantage that some mercury was inadvertently released into waterways and mercury is a very poisonous substance. On paper all the mercury should be recycled with none escaping; however it was virtually impossible to prevent small losses.      Advances in chemistry saw the development of many synthetic polymers some of which incorporated anionic side-chains that could act as cation exchangers. These advances were incorporated into new designs for the old diaphragm cell in which the asbestos diaphragm was replaced with a teflon cation-exchanging polymer membrane. This allowed sodium ions to pass through but not chloride or hydroxide ions. This so-called membrane cell allowed the production of NaOH free of NaCl contamination, free of asbestos problems and without any mercury releases to the environment. All recently built NaOH plants have used membrane cells. This example clearly illustrates how advances in chemistry, the development of tailor-made synthetic polymers, lead to the development of new and improved technology.
	(d)	(i)	sodium carbonate, Na2CO3
The equation for the re-formation of ammonia is essential. You could include equations for the decomposition of CaCO3 and for the slaking of CaO but the mark allocation does not warrant it. (p 371-3)		(ii)	CaO from the heating kiln is slaked (reacted with water) to form Ca(OH)2. This solution is mixed with the solution of ammonium chloride and heated to drive off ammonia: Ca(OH)2(aq) + 2NH4Cl(aq)  ®  CaCl2(aq) + 2NH3(g) + 2H2O(l) The ammonia is captured and returned to the carbonating tower (in the diagram). This recovery is desirable because it minimises the need to keep purchasing ammonia which is not cheap and it converts otherwise corrosive CaO or Ca(OH)2 to relatively harmless CaCl2.
There is a lot of material to handle here in nine minutes (5 marks). Mention as many environmental effects as you can, especially the first three as they are specific to the Solvay process. For evaluate methods ... you don't really have time to discuss all of them fully. Treat a couple of them briefly and just mention the others. Remember it is discuss so comment on the effectiveness of the ones you treat. ( p 374-5)		(iii)	The environmental issues that may arise include 1. disposal of CaCl2, the other product of the overall reaction 2. loss of ammonia to the atmosphere 3. excess energy use and disposal of waste heat 4. landscape damage by mining limestone 5. transport of reactants (limestone and sodium chloride solid or solution) to the site.      Methods of dealing with CaCl2 disposal include 1. selling it as solid to roads authorities for de-icing roads, but this is not an option in Australia 2.  locating the plant near the coast so the solution can be discharged into the ocean where it gets rapidly diluted; this is quite effective as oceans already contain Ca2+ and Cl– ions so small increases are of little consequence 3. burying solid calcium chloride in suitable locations where leaching will be slow and not dangerous because the environment can tolerate moderate concentrations of Ca2+ and Cl– ions; this is reasonably effective though more expensive than ocean discharge.      To avoid loss of ammonia the plant needs to be designed with this in mind.      To minimise energy use and thermal pollution the plant needs to be designed to recycle heat where-ever possible, that is by using heat released by exothermic reactions as the energy input for endothermic reactions.
Question 29  Shipwrecks, Corrosion and Conservation (25 marks)
( p 406)	(a)	(i)	passivating metals
The verb is compare so a table is quite adequate. There are only 3 marks so three of the five properties given here should suffice along with a list of a few uses. ( p 401-2)
		(ii)	Mild steel (<0.2% carbon)	Stainless steel (Fe with Cr and Ni)
			greyish silver appearance	can be polished to a shiny mirror-like surface
			malleable but relatively soft	malleable but much harder (more difficult to bend)
			corrodes easily	resists corrosion
			good conductor of heat and electricity	poor conductor of heat and electricity
			magnetic	non-magnetic
			used for car bodies, roofing, pipes, nuts and bolts and shipbuilding	used for food-processing machinery, kitchen appliances, cutlery, surgical and dental equipment
Temperature and oxygen concentration are the simplest two factors. Bacterial corrosion is another factor but it is rather complex to outline in three to four minutes (2 marks) so stick with the simple factors. ( p 425-30)	(b)	(i)	Temperature. In very deep water the temperature is generally close to 4oC, the temperature of maximum water density. Rates of chemical reactions decrease as temperature decreases so we would expect corrosion to be much slower in deep water. OR Concentration of dissolved oxygen. Corrosion of iron is reaction with oxygen. If the concentration of dissolved oxygen in water is low then the rate of corrosion will be low. Oxygen concentrations at great depths are much lower than surface concentrations.
You need to know that Mg is a more reactive metal than Fe so Mg will displace iron from solution (that is Mg will reduce Fe2+). Alternatively you need to be able to deduce this from the table of standard electrode potentials at the back of the exam paper. Because the two metals are not in contact, the magnesium does not protect the iron from rusting, so rust will develop on the nail. ( p 41-2, 418-9, 416-7)		(ii)	The magnesium would slowly dissolve, a blackish grey deposit (of iron) would appear on it, and the pale green colour of the solution (due to Fe2+ ions) would slowly fade, and the iron nail would rust (develop a brownish-red coating of hydrated iron(III) oxide. The first reaction occurring is the displacement of iron from the solution by magnesium: Mg(s)  ®  Mg2+(aq) + 2e– Fe2+(aq) + 2e–  ®  Fe(s) leading to the overall reaction Mg(s) + Fe2+(aq)  ®  Mg2+(aq) + Fe(s) The reactions for the rusting of the iron are: Fe(s)  ®  Fe2+(aq) + 2e– O2(aq) + 2H2O(l) + 4e–  ®  4OH–(aq)  Fe2+(aq) + 2OH–(aq)  ®  Fe(OH)2(s)   ...  ®  Fe2O3.xH2O(s)  (rust)
Note the use of two of the more difficult Board of Studies verbs, discuss and analyse, so you need more than just a list of bald facts. Seven marks means at least seven significant bits of information and you should spend about 13 minutes on it, so this requires quite a long answer.     You need to identify the current methods of corrosion protection – paint, sacrificial electrodes, alternative materials – and show how advances in chemistry have lead to these methods or to improvements in the methods. ( p 420-4)	(c)	Current methods of corrosion protection for hulls of ships include painting and sacrificial anodes (blocks of zinc or magnesium bolted to the hull). In other parts of ships galvanising, solid or surface alloys and alternative materials are used in addition to painting.      Painting was the earliest method of protecting ships against corrosion, but it had the disadvantage that it could easily be scratched and so corrosion could occur. Recent advances in chemistry have seen the development of polymers that could be used in marine paints. In these a polymer film that is very impervious to water and oxygen forms on the metal surface. Other advances have lead to additives for these paints that actually react chemically with the metal surface to produce what are called pyroaurites. These are complex mixed metallic hydroxides that incorporate iron ions from the ship's hull and help bind the paint even more firmly to the hull.     As understanding of electrolysis and of electrochemical reactions advanced, the technology of using sacrificial electrodes as a method of corrosion prevention developed. In this method blocks of zinc or magnesium are attached to the hull of ships. These metals being more reactive than iron (having lower electrode potentials) oxidise preferentially to iron and so they dissolve and in so doing stop the iron from dissolving (corroding) or looked at in a different way the electrons released by oxidation of these more reactive metals would cause any Fe2+ ions that form to be converted back to metallic iron: Zn  ®  Zn2+ + 2e–  and Fe2+ + 2e–  ®  Fe . A further advance in chemistry was the discovery that an inert electrode made positive relative to the hull of a ship (or pipeline) could prevent oxidation of iron in the same way, that is by pushing electrons into the iron.      Galvanising is a variant of the sacrificial electrode method. The whole surface of the steel object is coated with zinc. While not practical for hulls of ships in that it is too expensive and welding steel sheets together reduces its effectiveness, it is nevertheless used to protect many fittings within ships.      Advances in chemistry have produced a wide range of alloys including many highly specialised steels. Many of these alloys including stainless steel and alloys of aluminium have been used in shipbuilding partly to improve corrosion resistance. More recently surface alloys have been developed. In surface alloys only the surface layer of the metal is converted to an alloy with superior corrosion-resistant properties while the bulk of the metal remains unchanged.      Chemistry has also produced numerous synthetic polymers often with highly desirable properties for particular purposes, and often very resistant to damage by marine environments. Such materials have been widely used in shipbuilding (though not of the hull) because they often eliminate corrosion problems.      Advances in chemistry such as increased understanding of electrochemistry and the development of special polymers and alloys as discussed have  had a very marked impact upon the technology of preventing corrosion in ships.
	(d)	(i)	To prevent corrosion of the restored cannon
		(ii)	After prolonged submersion in the ocean there was a lot of chloride in pores and fine cracks in the cannon. Step 1 was performed in order to leach most of this chloride out of the cannon; Dilute sodium hydroxide solution was used because (1) it slows down further corrosion of the cannon and (2) the hydroxide ion can replace chloride ion in hydroxy chlorides and this facilitate chloride removal. Periodically replacing the hydroxide solution  removes the displaced chloride and so keeps the chloride concentration near the cannon low; this speeds up diffusion of the chloride out of the cannon.
Five marks means al least five significant facts and you have about nine minutes for the question, so a fair amount of detail is required. Note that you have to justify the procedure as well as describe it. This means explaining why the procedure was carried out. (p 433-4, 437)		(iii)	As shown in the diagram (draw a diagram similar to Fig 12.8 on p 435) the cannon was submerged in a large bath of dilute sodium hydroxide solution. A stainless steel mesh was used as the anode (+ve terminal of the power supply connected to it) while the cathode was the cannon. As current flowed insoluble iron compounds, particularly Fe(OH)2 and Fe(OH)Cl were reduced at the cathode while oxygen gas formed at the anode: At the cathode:  Fe(OH)Cl(s) + 2e–   ®   Fe(s) + OH–(aq) + Cl–(aq) At the anode:   4OH–(aq)   ®   2H2O(l) + O2(g) + 4e–      The justification for carrying out this electrolysis is that it converts  corroded iron back to pure metallic iron and removes residual chloride (left in the cannon after Step 1). The reason for using sodium hydroxide as the electrolyte is that it provides a good conducting solution for electrolysis and it minimises any further corrosion of iron.
Question 32  Forensic Chemistry (25 marks)
The key points are that emission spectra are quite different from one element to another and that the spectrum of one element is not affected by the presence of other elements in the mixture.      There are only three marks for this question so you do not need too much detail. maybe a diagram showing how emission lines arise could be included but it is not really necessary. (p 522-8)	(a)	(i)	sodium and arsenic
		(ii)	An emission spectrum consists of several (or many) bright lines on a dark background. These lines correspond to emissions of radiation at specific wavelengths. These emissions arise as electrons fall from excited energy levels back to lower energy levels and release the energy as radiation.  Each element has a particular set of emission lines that occur at wavelengths that are generally different from those of all other elements. The emission spectrum of a mixture shows the lines of all the elements present; that is the spectra of the elements present do not interfere with one another. Hence by determining which element (or elements) could have produced each line in the spectrum from the mixture all the elements present can be identified.
( p 497-9)	(b)	(i)	X is a base, Y is phosphate, Z is deoxyribose
Point out that the analysis is based on introns not on actual genetic coding  regions, and explain why the DNA of related people shows similarities. There are only two marks (three to four minutes) for this so you cannot include much detail.  ( p 506)		(ii)	Particularly in the non-coding or intron regions of the molecules, everyone's DNA is different from everyone else's. However people who are related do have similarities in their DNA in these intron regions, because they inherit half of their DNA from each parent. A parent and child will have 50% of their introns in common, siblings will have 50% in common while cousins will have 25% in common. Hence a DNA analysis of the introns can identify which, if any, introns two people have in common, and by determining how many they have in common their relationship, if any, can be determined.
You need to point out that the DNA data banks currently being established do not contain genetically useful information, because that it the key factor in determining the ethics of setting up such banks. Discuss means that you must mention both sides of the ethical debate.  ( p 506-7)		(iii)	The commonest form of DNA data bank is a forensic one that just lists the lengths (number of repeat units) of the specific introns used for forensic analysis. It cannot provide any information about the genetic make-up of the person or identify any genetic abnormalities that the person has. Consequently such DNA data banks invade people's privacy to about the same very limited extent as do data banks of fingerprints.       Most people do not see any ethical problem with this; the benefits in being able to identify perpetrators of crimes by comparing crime scene DNA (or fingerprints) with ones in a data bank outweigh the minimal invasion of privacy that the setting up of such a data bank involves. However there are some civil liberties groups that do see an ethical problem: they see the collection and storage of such information as an invasion of people's privacy and claim that such banks are really requiring that people prove their innocence of a crime (have their DNA profiles eliminate them as suspects) rather than the authorities having to prove someone's guilt. There is particular concern about people (only criminals at the moment) being forced to provide DNA samples which may be used to incriminate them.      Proposals to set up genetic DNA data banks that store information about people's genetic make-up and possible genetic abnormalities (often proposed by insurance companies) do raise much more serious ethical questions, because such data banks would be a real invasion of people's privacy. However no such data banks exist at the moment, at least in Australia.
There are four parts to this question, structure, composition, electrophoresis and assessment of potential. With 7 marks you need to spend between 12 and 13 minutes on the question so quite a lot of information is required. Be careful not to overlook composition as distinct from structure. Do not provide so much information about structure or about electrophoresis that you do not leave enough time for the other aspects of the question. That is, make sure you do some assessment of the usefulness of electrophoresis for analysis.	(c)		Structure Proteins are polymers made up of many amino acids joined together through peptide linkages. The general structure of amino acids is where R is a carbon-containing side group. The peptide or amide linkage is: The general structure of a protein is:

Note that the question is about analysis of proteins, not of amino acids so keep your answer slanted that way. (The CCHSC account deals primarily with amino acids.) ( p 484-8, 492-5)		where R1, R2, R3 R4 and so on can be different or the same side groups. Hundreds and thousands of amino acids are joined in this way to form proteins. This is the primary structure of proteins.      These long chains of amino acids are curled into spirals or folded into pleats and packed together in special ways to make the final structure of the protein. These aspects are called the secondary and tertiary structure of proteins.
		Composition Proteins are composed predominantly of carbon, hydrogen, oxygen and nitrogen with smaller amounts of sulfur. Electrophoresis Electrophoresis is a method of separating charged substances based upon the different signs of their charges and upon their different mobilities. The experimental arrangement is: (include a diagram similar to Fig 14.6 on p 493) Electrodes are attached to the ends of a piece of filter paper or a thin gel that contains an electrolyte solution. The sample to be analysed is placed in the middle of the paper or gel, a voltage is applied and the direction of migration, if any, is noted. The distance each substance migrates in a given time is also recorded.     Proteins can be positively or negatively charged or neutral, depending upon whether the R side groups contain –COOH and/or –NH2 groups These groups can be present in the neutral forms just written or they can be charged as –COO– or NH3+. Depending upon the relative numbers of these charged groups, the protein will be overall positive, neutral or negative and the overall charge can vary as buffers of different pHs are used. Different proteins will migrate in different directions and at different rates in the same buffer and if buffers of different pH are also used these directions and rates can change.      By comparing the direction and distances travelled by an unknown with known proteins the identity of the protein can be determined.      Electrophoresis can also be used to analyse mixtures of amino acids. Proteins can thus also be analysed by first hydrolysing them into amino acids then analysing that mixture. Assessment Electrophoresis is a very effective method of analysing proteins, particularly in identifying an unknown protein by comparison with electrophoresis patterns of known proteins, because it can be quite easily done with equipment that is not particularly expensive. Because comparisons between unknown and known proteins can be made in different buffer solutions (different pHs), the identity of the unknown can be established with great certainly. In addition if the protein is not a known one, considerable information about it can be obtained by hydrolysing it into its constituent amino acids and performing electrophoresis on that mixture.
It is not clear whether the question is looking for disaccharide (as opposed to a mono- or polysaccharide) or non-reducing sugar (as opposed to a reducing one), so it is safest to include both classifications in the answer. ( p 467-9)	(d)	(i)	a non-reducing disaccharide
You can either use the specific test for starch, the iodine test, or test for a reducing sugar (glucose) versus a  non-reducing sugar (starch) using one of the three solutions mentioned. You need to describe the test and give the result you would observe for each compound. ( p 476-7)		(ii)	Add brown iodine solution to aqueous solutions of both sugars. If the solution goes blue then the sugar is starch. If it stays brown the sugar is glucose. OR Add drops of Benedict's solution (copper sulfate in alkaline citrate) or Fehling's solution (copper sulfate in alkaline tartrate) to aqueous solutions of the sugars and gently warm the mixtures. If a reddish brown precipitate (of Cu2O) forms then the sugar is glucose. If there is no reaction then the sugar is starch. OR Add drops of Tollen's reagent (a colourless solution of silver nitrate in aqueous ammonia) to aqueous solutions of the sugars and gently warm the mixtures. If a silver mirror forms on the wall of the test tube or if a black precipitate (of metallic silver) forms, then the sugar is glucose. If there is no reaction, then the sugar is starch.
For 5 marks you should spend 9 minutes on the question and you need at least five key pieces of information. Describe the technique briefly, not forgetting to mention that high pressure is used, and including a simple diagram. Probably there are 3 marks for the description (so not too much detail) and 2 marks for the assessment. Make sure that you assess the two solvents in the experiment shown and do not just give a general assessment. Because the Board of Studies does not have the chromatograms on its website, the following sketch is included: There are some puzzling peaks at about 1 minute in Solvent B. These could be due to (a) pressure fluctuation caused by the injection of the sample, (b) another solvent if the sample had initially been dissolved in a solvent different from B, or (c) impurity in the sample used for that analysis but not in the sample used for Solvent A or (d) to some other cause. The experiment would need to be repeated to see whether this was an anomaly of that particular analysis or a consistent feature of all such analyses in Solvent B.      It is a pity that this artifact should have been left in the diagram in that it just serves to confuse students who should concentrate only on the question asked – about the separation of the six carbohydrates and ignore this feature. ( p 514-6)		(iii)	High performance liquid chromatography is a technique for separating mixtures of compounds  based upon their different solubilities in two liquids, one stationary and one moving. The experimental arrangement is shown schematically in the diagram: (Include a diagram similar to Fig 15.4 on p 515.) The sample to be separated (analysed) is dissolved in a suitable solvent then injected on to the column which contains finely divided solid (generally in the form of small spheres) on which a liquid has been adsorbed (the stationary phase). Solvent at high pressure is pumped through the column and washes the sample down the column. High pressure is used to produce a reasonably fast flow of solvent through the column. Because of the different solubilities of the different compounds in the two solvents (the flowing one and the one adsorbed on the solid particles in the column) the compounds move through the column at different rates. At the bottom of the column they pass through a detector (refractive index monitor or uv/visible light detector) which signals the passage of each compound out of the column and measures the amount of each compound as it passes through. Results are displayed on a computer as a graph called a chromatogram. Assessment Neither solvent is completely effective for the separation of these six compounds, because all the peaks are not fully separated in either solvent. However Solvent A is the better of the two in that all peaks except 1 and 2 are fully separated and even for 1 and 2 the separation is sufficiently good to allow a reasonable separation of the two compounds; we would stop the collection of peak 1 just before the minimum of the graph was reached, and start collecting peak 2 when the graph began to move up that peak. In this way we would obtain pure samples of compounds 1 and 2 but would not get100% recovery of either of them. Solvent 2 is much less effective in that peaks 4 and 6 are not at all well separated even though the separation of peaks 1 and 2 is better than in Solvent 1. Because of the overlap of peaks 4 and 6 in Solvent 2, it would not be possible to obtain pure samples of compounds 4 and 6 from this experiment.