Chemistry Exam Paper Solutions (with Option Topics)

Stratosphere – absorbs UV and turns it into heat

Density – Ozone has a higher density than oxygen

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)

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

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

CH₃COO^–_(aq) + H₂O_(l) ⇌ CH₃COOH_(aq) + OH^–_(aq)

Hence the pH will be greater than 7.

Fermentation: Glucose is fermented to produce ethanol and carbon dioxide, using yeast: C₆H₁₂O_6(aq) → 2C₂H₅OH_(aq) + 2CO_2(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) → C₂H_4(g) + H₂O_(l)

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

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)

Metal smelters smelting sulfide ores also produce SO₂ e.g. 2ZnS_(s) + 3O_2(g) → 2ZnO_(s) + 2SO_2(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).

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.

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.

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.)

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.

N_2(g) + 3H_2(g) ⇌ 2NH_3(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

Q = 0.7 / (0.3 x 0.65^0.5) = 2.894

K is 12.1 therefore the system is not at equilibrium.

Y = calcium oxide

Note: question says “name”

NaCl(aq) + CO2(g) + NH_3(g) + H₂O_(l) ⇌ NaHCO_3(s) + NH₄Cl_(aq)

The ammonia can then be recovered and recycled:

Ca(OH)_2(aq) + 2NH₄Cl_(aq) → CaCl_2(aq) + 2NH_3(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.

2Cl^–_(aq) → Cl_2(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) + H_2(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.

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.

Fe_(s)    →    Fe²⁺_(aq) +   2e^–

2H⁺_(aq)   +   2e^– →     H_2(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.

cathode:          Cu²⁺_(aq)   +    2e^–     →     Cu_(s)                                                 E° = +0.34 V

anode:             H₂O_(l)    →     ½ O_2(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.

[with diagram]

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: O_2(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) + O_2(g)→ 2Fe₂O₃·H₂O_(s) + 2H₂O_(l). The overall reaction is 4Fe_(s)  +  3O_2(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.

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.

Medium: binds pigment to surface

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

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.

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]