2017 HSC Chemistry Exam Paper Solutions (with Option Topics)

Posted on November 1, 2017 by Michelle Wong

The HSC Chemistry exam was held today (1 November 2017). While the official exam paper has not yet been released by NESA, here are our draft suggested solutions and some explanations in case you wanted to know ASAP!

HSC Chemistry Section I A (Multiple Choice) Solutions

Question Solution Explanation
1 A While a measuring cylinder would be more practical, a burette would be more accurate.
2 D Copper gives a blue-green flame.
3 A 2-chloro-1-fluorobutane: according to the frequently misinterpreted IUPAC organic naming rules as described in the “Blue Book”, the chloro prefix comes before fluoro (alphabetical order).
4 D Condensers should never be stoppered to avoid pressure buildup. Sulfuric acid is necessary for the reaction but does not improve safety per se. The direction of water flow is correct (in the bottom, out the top for both reflux and distillation setups). A heating mantle would avoid the open flame of a Bunsen burner.
5 A HSO4 is amphiprotic as it can donate a proton to form SO42-, and gain a proton to form H2SO4
6 B Nuclear reactors produce neutron-rich radioisotopes. Both Co-60 and Np-239 are neutron rich (compare them to the masses on the periodic table, if the mass if higher than expected it is neutron-rich), but only Np-239 is transuranic.
7 C Water and hydrocarbons are immiscible. Hexane does not decolourise bromine water int he absence of UV light.
8 C Barium does not precipitate with chloride but lead does.
9 D The black deposit is a result of incomplete combustion producing C(s).
10 B The carbon rod of the dry cell and the PbO2 of the lead-acid cell are the cathodes, which have a positive charge.
11 D S is oxidised as it goes from S (oxidation number of 0) to SO2 (oxidation number of +4).
12 D Propene polymerises to give a polymer with no double bonds, and a methyl group on every second carbon.
13 B Citric acid is triprotic, so it will require the most base for neutralisation.
14 A HCl/Cl- cannot act as a buffer as HCl is strong, so the system does not form an equilibrium in solution.
15 B This is tricky! Option B is the only option with an octet for each atom.
16 B Adding more hydroxide would remove carbonic acid and shift the equilibrium to the right, removing CO2 and reducing pressure
17 C 1 mole of O3 weighs 3 x 16 = 48 g. 1 mole of any gas occupies 24.79 L at RTP. Hence the density is 48/24.79 = 1.936 g/L.
18 D Y and Z are on the same side of the equation as their concentrations change in the same direction when the equilibrium shifts. The shift is equimolar for X, Y and Z (0.005 M each), so the mole ratio must be 1:1:1.
19 C 48% of 1.20 g is 0.576 g. Moles of sulfate = moles of barium sulfate, so 0.576 / 96.07 x 233.37 = 1.40 g.
20 C Barium hydroxide is limiting (4 x 10-4 mol barium hydroxide vs 2 x 10-3 mol HCl). Therefore the moles of HCl left over is 2 x 10-3 – 8 x 10-4 = 1.2 x 10-3 mol. The total volume is 70 mL. Therefore H+ concentration is 1.714 x 10-3 M, so pH is 1.8.

HSC Chemistry Section I B (Long Response) Solutions

Question Marks Solution
21(a) 2 Troposphere – causes respiratory irritation

Stratosphere – absorbs UV and turns it into heat

21(b) 3 Boiling point – Ozone has a higher boiling point than oxygen

Density – Ozone has a higher density than oxygen

21(c) 1 Cl(g) + O3(g)   ClO(g) + O2(g)
22(a) 3 AAS graph
22(b) 2 From the graph, 0.58 absorbance means 3.45 ppm (indicate on the graph using dotted lines). Therefore the concentration of zinc is too high to be safe.
23(a) 2 The salt bridge is required to complete the circuit, to avoid a buildup of negative charge in the cathode compartment and hence allowing the cell to continue running.
23(b) 4 Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)

n(Zn) = m / mm = 1 / 65.38 = 0.015295 mol

n(Ag) = 2 x n(Zn)

m(Ag) = n x mm = 3.30 g

Total m(Ag) = 10 + 3.3 = 13.3 g (3 s.f)

24(a) 3 CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq)

n(CH3COOH) = c x V = 0.01255 mol = n(NaOH)

c(NaOH) = n / V = 0.01255 / 0.01930 = 0.6503 M (4 s.f.)

24(b) 2 NaCH3COO is not neutral as the acetate ion is basic and reacts with water to produce OH-:

CH3COO(aq) + H2O(l) ⇌ CH3COOH(aq) + OH(aq)

Hence the pH will be greater than 7.

25 4 Hydrolysis: Water is added to break cellulose up into beta-glucose, using cellulase: H(C6H10O5)nOH(s) + (n – 1)H2O(l) → nC6H12O6(aq)

Fermentation: Glucose is fermented to produce ethanol and carbon dioxide, using yeast: C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)

Distillation: Since fermentation can only produce ethanol concentrations up to 15%, the mixture needs to be distilled to produce higher concentrations before the next step.

Dehydration: Concentrated sulfuric acid catalyses the dehydration of ethanol to produce ethylene: C2H5OH(l) → C2H4(g) + H2O(l)

Addition polymerisation: Ethylene is reacted to produce polyethylene through a reaction such as free radical polymerisation (initiation, propagation, termination)

26(a) 3 Sulfur dioxide causes acid rain (SO2(g) + H2O(l) → H2SO3(aq)) which has many detrimental effects:

Acid rain acidifies waterways, which causes an increase in toxic aluminium, which harms fish populations. Fish eggs will also not hatch.

Acid rain also damages forests, both directly (corrosive damage to leaves) and indirectly (leaches nutrients from soil)

26(b) 4 Coal fired power stations produce SO2 from the combustion of sulfur impurities in coal: S(s) + O2(g) → SO2(g)

Metal smelters smelting sulfide ores also produce SO2 e.g. 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)

Hence the concentration of SO2 is highest near these sources, and lower further away as winds and air currents dissipate the SO2. Most of the SO2 in the atmosphere is produced from industrial sources such as these, so the concentration in other areas are very low (< 0.2 x 1012 molecules per L).

27 5 Acetic acid contains a COOH group in its structure, so it can form many strong hydrogen bonds.

Butan-1-ol contains an OH group in its structure, so it can form strong hydrogen bonds.

Butyl acetate contains an ester functional group, which causes its structure to be polar so it can form weaker dipole-dipole forces.

Hence due to these polar intermolecular forces, they have similarly strong intermolecular forces even though their dispersion force strength is different due to different molar masses (dispersion force strength is proportional to molar mass so acetic acid has the weakest dispersion forces, butan-1-ol is next weakest while butyl acetate has the strongest dispersion forces).

The same intermolecular force strength would require similar heat energy to break, hence the boiling points are similar.

28(a) 7 Advantages:

Since ethanol is a liquid fuel, the same fuel infrastructure (current petrol stations, petrol tankers etc.) can be used to deliver the ethanol to consumers.

Ethanol can be produced from renewable sources like starch-based crops, unlike petrol which is derived from non-renewable petroleum which will likely diminish in the future.

Disadvantages:

Ethanol gives lower mileage, which means consumers will need to refuel more often

Blends containing over 15% ethanol can’t be used in current cars in many areas of the world including Australia, so expensive engine modifications would be required for greater use.

28(b) 3 C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Therefore 1360 kJ per 2 mol CO2.

m(CO2) in 2 mol = n x mm = 2 x 44.01 = 88.02 g = 0.08802 kg

Therefore 1360 kJ / 0.08802 kg = 15450 kJ/kg (4 s.f.)

29 4 Site Y would be preferred.

Sources of contamination for X:

  • Limestone: Calcium ions cause an increase in hardness, which is bad for household use as it leads to scale and blockages in plumbing, and is not treated by the methods listed. Limestone also causes high pH (7.3 from the table) but this will be adjusted during pH control.
  • Farming (cows): Pathogens from manure will likely enter the water. Chlorination will kill some bacteria but it is unlikely to be enough, especially as the sand filters with large pore sizes are used without a flocculant. Farming can also lead to eutrophication with high phosphate (1 ppm from the table), which may cause algal overgrowth and the production of toxins, which are not treated by the listed methods.
  • Houses: May cause contamination with grey water, which introduces pathogens. It is likely that chlorination alone is not enough for sanitised water.

Sources of contamination for Y:

  • Saw mill: Likely adds sawdust to the water. These particles are large, and not very good food for bacteria compared to manure, so it is likely that this can be completely removed by sedimentation and sandbed filters. The low level of bacteria can be controlled by chlorination. The slightly acidic pH would be adjusted during pH control.
30 7 Haber process:

N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH < 0

Temperature:

  • High temperatures push equilibrium in the reverse endothermic direction to remove heat and minimise the disturbance according to Le Chatelier’s principle, hence reducing yield.
  • However, low temperatures can make the rate of reaction unacceptably low.
  • Additionally, high temperatures will damage the catalyst.
  • Therefore temperature must be monitored so it can be kept at 400–500 °C (compromise between rate and yield)

Pressure:

  • High pressures push equilibrium in the forward direction to the side with less gas moles minimise the disturbance according to Le Chatelier’s principle, hence increasing yield.
  • High pressures also concentrate the gaseous reactants, increasing reaction rate
  • However, high pressures are expensive and less safe
  • Therefore pressure is monitored to keep it at a relatively high 200–300 atm, and to ensure any dangerous pressure surges are quickly managed

Catalyst:

  • An iron oxide catalyst is used to increase the rate of reaction
  • The catalyst can be poisoned by impurities in the reactants
  • The catalyst can also be degraded by high temperatures
  • The catalyst is monitored so that it can be quickly replaced if it is degraded or poisoned

Removal of ammonia as it forms via liquefaction:

  • Ammonia has a higher BP than the other two gases, therefore it is selectively removed via liquefaction as it forms
  • The equilibrium shifts to the right in response to minimise the disturbance, therefore it increases yield
  • The liquefaction must be monitored so the temperature stays above the BP of the other two reactants, otherwise they will also condense


HSC Chemistry Section II (Option Topic) Solutions

Question 31: Industrial Chemistry

Question Marks Solution
31(a)(i) 2 2017 HSC Chemistry Exam Paper Solutions
31(a)(ii) 3 1. 150 mL of 10 M sodium hydroxide solution was added to 400 g olive oil.
2. The mixture was mixed with a stick blender for about 3 minutes until the mixture thickens (reaches “trace”).
3. The mixture was carefully poured into a mould and set aside for 48 hours.
4. Individual blocks of soap were cut, and allowed to cure for 3 weeks before use.Testing:Two test tubes were each filled with 2 mL vegetable oil and 15 mL water. A small amount of soap was added to one test tube. Both test tubes were stoppered then shaken.
Result: The tube containing soap produced an emulsion where no visible phase boundary could be seen between the oil and the water, while the other tube contained separate oil and water. This shows that a surfactant was produced.
31(b)(i) 2
31(b)(ii) 4 Q = [SO3]/[SO2][O2]0.5

SO2 O2 SO3
Initial 1 1 0
Change -0.7 -0.35 +0.7
End 0.3 0.65 0.7

Q = 0.7 / (0.3 x 0.650.5) = 2.894

K is 12.1 therefore the system is not at equilibrium.

 31(c)(i) 2 X = sodium hydrogen carbonate

Y = calcium oxide

Note: question says “name”

31(c)(ii) 2 Brine is purified by precipitating out the impurities then filtering e.g. calcium is removed by addition of carbonate: Ca2+(aq) + CO32-(aq) → CaCO3(s)
31(c)(iii) 3 Ammonia is used in the Solvay process as a base to produce hydrogen carbonate:

NaCl(aq) + CO2(g) + NH3(g) + H2O(l)  NaHCO3(s) + NH4Cl(aq)

The ammonia can then be recovered and recycled:

Ca(OH)2(aq) + 2NH4Cl(aq) → CaCl2(aq) + 2NH3(g) + 2H2O(l)

This multi-step process is necessary as the activation energy for the direct conversion of limestone and brine to sodium carbonate is far too high to be industrially viable.

31(d) 7 As time has passed, the production of sodium hydroxide has become more technically and environmentally optimal due to technological advances. Three processes are used for the production of sodium hydroxide: the mercury, diaphragm and membrane processes. All three processes use saturated, purified brine as the starting material and produce chlorine gas, hydrogen gas and aqueous NaOH. Each of the processes produces chlorine gas at an inert anode by the oxidation of chloride ions:

2Cl(aq)  Cl2(g) + 2e

In the mercury process, the cathode is a flowing liquid mercury which reduces sodium ions to sodium metal:

Na+(aq) + e  Na(l)

The sodium dissolves in the mercury to produce an amalgam, which is mixed with water in a separate compartment. The sodium reacts with water to produce NaOH and hydrogen gas, and the mercury is recycled.

2Na(l) + 2H2O(l) → 2NaOH(aq) + H2(g)

Although this process produces high-purity NaOH, it also has high energy requirements and it is a two stage process. Both of these factors are industrially undesirable and require fossil fuel use, which leads to pollution. The most important factor which contributed to the change in production process is the unavoidable release of mercury into the environment. Avoiding mercury pollution was a major factor in changing the production process. Also, mercury is expensive, so replacing the lost mercury incurred a significant cost.

The diaphragm process, which developed at a similar time, has the same anode reaction as the mercury process. However, water is reduced at a steel mesh cathode to produce hydroxide ions and hydrogen gas. The anode and cathode compartments are separated by an asbestos diaphragm which allows Na+ ions to pass through, as they are attracted to the cathode, at which OH ions form. Therefore, a solution of NaOH(aq) is formed.

There are two factors which contributed to the phasing out of this process in many countries. First, even though it has lower energy requirements than the mercury process, it is a single-stage process, so the product was contaminated with about 2% NaCl. This level of salt is satisfactory for most applications. Furthermore, asbestos is carcinogenic, causing asbestosis in exposed workers. To overcome this health risk, the asbestos diaphragm was replaced by a diaphragm composed of a composite of metal oxides and a polymer.

The more modern process is the membrane process. In the membrane process, the anode and cathode reactions are the same as in the diaphragm process, but the anode and cathode compartments are separated by a special, selectively permeable polymer membrane (usually made of polytetrafluoroethylene, PTFE). It allows Na+ ions to pass through, but not anions or water. Therefore, Na+ ions can move from the anode compartment to the cathode compartment, where OH- ions are formed, to produce NaOH of high purity, removing the need for further purification. Hardly any chloride ions diffuse into the cathode electrolyte. Furthermore, because neither asbestos nor mercury are used, the process does not have environmental hazards associated with it. It is also the most energy efficient.

Assessment: Hence technological advances, in particular the development of the PTFE membrane, has overcome the technical and environmental issues associated with the older industrial NaOH production processes.

Question 32: Shipwrecks, Corrosion and Conservation

Question Marks Solution
32(a)(i) 2
  1. 4 identical steel nails were cleaned gently with steel wool and weighed.
  2. Each nail was placed in a test tube, and 20 ml of distilled water was added to test tube A.
  3. 20 mL of 0.01 M HCl was added to test tube B; 20 mL of 0.1 M HCl was added to test tube C; and 20 mL of 1 M HCl was added to test tube D.
  4. All 4 test tubes were left in a cupboard.
  5. After 5 days, each nail was removed from solution, rinsed with distilled water, carefully cleaned of corrosion with steel wool and a toothbrush and dried.
  6. The intact metal of each nail was weighed and the mass lost was calculated.
  7. The experiment was repeated using zinc instead of steel.
32(a)(ii) 3 The expected result is that the rate of corrosion of the metals increases with increasing acid concentration. This will be indicated by increased mass loss of the nails immersed in high acid concentration. The presence of an acid will increase the rate of corrosion as the hydrogen ions can be reduced to hydrogen gas, therefore accelerating the oxidation of the metal as shown below:

Fe(s)    →    Fe2+(aq) +   2e

2H+(aq)   +   2e     H2(g)

A higher acid concentration results in a higher cell potential for the oxidation of the metal, resulting in a faster rate of corrosion. Similarly, a more reactive metal will exhibit a faster rate of corrosion due to a higher oxidation potential.

32(b)(i) 2 Increase the concentration of the electrolyte, increase surface area of electrodes
32(b)(ii)  4 The main reactions that would occur at the anode and cathode are as follows:

cathode:          Cu2+(aq)   +    2e     →     Cu(s)                                                 E° = +0.34 V

anode:             H2O(l)    →     ½ O2(g)    +    2H+(aq)   +    2e                           E° = -1.23 V

Due to the reduction of copper(II) ions to metallic copper atoms at the cathode, the mass of the cathode increases. Additionally, as copper(II) ions give a characteristic blue colour, the loss of copper(II) ions from solution also decreases the intensity of the blue colour in the electrolyte solution. The oxidation of water at the anode results in the formation of oxygen gas, which can be observed as bubbles.

32(c)(i) 2 Scientists predicted that the rate of corrosion would have been slow at this depth as the deep ocean would have a low concentration of oxygen, due to lack of photosynthesis and surface mixing, which should lead to less oxidation of metals. Additionally, the temperatures would also be at a minimum due to lack of sunlight penetration. Lower amounts of heat energy result in less kinetic energy, hence a lower number of successful collisions and hence lower rate.
32(c)(ii) 2 Wooden artefacts from long-submerged wrecks have a weak structure as the cell walls are broken down and their cellular contents are leached out and replaced with sea water (a concentrated aqueous ionic solution). The sea water is what maintains its original form. Immediately drying wooden artefacts will cause the wood to contract and collapse. Furthermore, the evaporation of water will cause salt crystals to form and expand, causing cracks that destroy the structure of the wood.
32(c)(iii) 3 Much of the chloride ions are present as insoluble hydroxyoxychlorides trapped within the metal oxide deposits. Cleaning can be achieved by electrolysis. The artefact is placed in a 0.5 mol L-1 NaOH electrolyte and connected as cathode. A stainless steel anode is used. When a low voltage is applied, the negatively charged chloride ions are repelled from the negatively charged cathode (i.e the artefact) and move toward the anode where they are oxidised.

[with diagram]

32(d) 7 Modern ocean-going vessels are constructed with steel. Steel is an alloy of iron with carbon, with varying amounts of other metals and silicon. Due to the impure nature of steel, along with the constant exposure to high electrolyte concentration, ocean-going vessels have a high tendency to undergo corrosion.

The corrosion occurs in 3 steps.

First, iron is oxidised and oxygen is reduced at 2 separate sites in the iron. The electrons provided from oxidation flow through the iron to an impurity like carbon, and reduce oxygen in the thin film of moisture on the iron surface. Iron acts as the anode and an impurity in the iron will act as the cathode.

Oxidation: Fe(s) → Fe2+(aq) + 2e

Reduction: O2(g) + 2H2O(l) + 4e → 4OH (aq)

Then Fe2+ and OH will combine and form a precipitate: 2Fe2+(aq) + 4OH(aq) → 2Fe(OH)2(s)

Iron(II) is then oxidised to form rust: 4Fe(OH)2(s) + O2(g)→ 2Fe2O3·H2O(s) + 2H2O(l). The overall reaction is 4Fe(s)  +  3O2(g)  + 2H2O(l)  →  2Fe2O3·H2O(s).

It is our understanding of the corrosion process that has led to the development of effective protection methods, which has been aided by the works of Volta and Davy.

Volta demonstrated that metal wires in the solution generated electric current. He constructed the first electric battery by connecting two different metals in an electrolyte solution. He later stacked cells to form what became known as the Voltaic pile. This was the first source of a direct electric current. Davy used the electricity generated by the Voltaic pile to decompose compounds into their constituent elements, and thus developed electrolysis.

These findings provided the basis of our understanding of electron transfer reactions, which contributed to our current understanding of the corrosion process and assisted in the development of a number of protection techniques that reduce the extent of corrosion. These include creating physical barriers and using cathodic protection methods. The use of paints, lacquers and polymer-based coatings prevent the steel from coming into contact with seawater (the electrolyte) and hence prevent corrosion.

Cathodic protection prevents corrosion of steel structures by making the steel structure the cathode in an electrochemical cell, thus ensuring the iron in the steel is protected from oxidation. There are two types of cathodic protection: the sacrificial anode technique which sets up a galvanic cell by attaching a metal that is more active (a stronger reactant) than Fe to the steel hull. This active metal forms the anode. An impressed current system makes the steel hull the cathode (negative terminal) of an electrolytic cell with an inert anode attached to the positive terminal. An external power source applies a continuous low voltage. Thus the corrosion protection methods used today have been largely due to the discoveries of Volta and Davy.

Question 34: Chemistry of Art

Question Marks Solution
34(a)(i) 2 Flammable reagents were cleared from the laboratory before the experiment due to
the need for a naked flame.A platinum wire loop was cleaned thoroughly by dipping into concentrated HCl. The cleanliness of the wire was checked by placing the wire in the flame and checking for a burst of colour.The clean wire loop was then dipped into an aqueous solution of calcium nitrate
and placed in a blue Bunsen flame. The colour of the flame was recorded.
34(a)(ii) 3 Calcium = orange-red (brick red) flame

According to the Bohr model, an atom contains electrons which can only occupy
quantised energy levels or orbits around a central nucleus. Electrons can be excited
to higher orbits by input of energy such as from a flame. The electrons will move to
lower energies and emit photons of energy equal to the energy gap between the
levels. The different allowed transitions result in different, discrete wavelengths of
light. These wavelengths combine to give the orange-red colour observed.

34(b)(i) 2 Pigment: insoluble, provides colour and opacity

Medium: binds pigment to surface

34(b)(ii) 4 Red ochre: iron(III) oxide Fe2O3

The colour of red ochre relates to the transition metal it contains (iron), as well as the rest of the compound.

Transition metal compounds are coloured because they have an incomplete d subshell. In compounds, the energies of the d orbitals split, with an energy gap in the visible light range. Electrons can transition from the lower to higher d orbitals with the absorption of a visible wavelength, and the unabsorbed wavelengths will be observed as colour.

The energy gap depends on a number of factors, including the identity of the metal, the
identity of the other species in the compounds (e.g. ligands), and the oxidation state of the
metal. Red ochre contains iron with an oxidation state of 3, bound to oxygen which changes the energy gap. Other iron compounds have different colours e.g. yellow ochre, Fe2O3.xH2O

34(c)(i) 3 Hund’s rule states that all degenerate orbitals in a subshell must be occupied before the
electrons pair up. In the oxygen atom, this means two of the p orbitals should contain one electron each, so there will be two unpaired electrons in the diagram.
34(c)(ii) 2 As seen from the low 1st, 2nd and 3rd ionisation energies in the graph, the first three electrons in Sc are easily removed. However, the 4th electron requires a very high (>7000 kJ mol-1) ionisation energy, so Sc will predominantly have an oxidation state of 3+..
34(c)(iii) 2 Vanadium(III) compounds are coloured because V3+ has electrons in the 4s and 3d subshells, which are very close in energy, with energy gaps in the visible light range. Electrons can transition from the lower to higher energy orbitals with the absorption of a visible wavelength, and the unabsorbed wavelengths will be observed as colour.

However, scandium(III) compounds are not coloured as Sc3+ does not have any electrons in the 4s or 3d subshells, so transitions will not absorb in the visible range and produce colour.

34(d) 7 In a complex ion, the transition metal ion covalently bonds to ligands. The ligands act as Lewis bases and donate lone pairs to the empty orbitals in the valence shell of the transition metal ion, which acts as a Lewis acid. This forms coordinate covalent bonds.

e.g. pentaamminechlorocobalt(III) ion: CoCl(NH3)5]2+

Co3+ has the electron configuration 1s2 2s2 2p6 3s2 3p6 3d6

[include orbital diagram and Lewis structures]


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