2021 HSC Trial Solutions PDF

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HSC Chemistry Trial Exam Marking Guidelines 1. B 2. D 3. A 4. B 5. B 6. A 7. C 8. C 9. D 10. B 11. C 12. A 13. B 14. B 15. D 16. B 17. A 18. A 19. D 20. C

Question 21: (9 marks) (a). Define dynamic equilibrium and give one example of an everyday occurence of it. (2) An everyday occurrence is the dynamic equilibrium in soda cans between gaseous carbon dioxide, CO2(g), and aqueous carbon dioxide, CO2(aq). CO2(g) ⇌ CO2(aq) (b). Ammonia is produced via an exothermic equilibrium reaction as follows: 3H2(g) + N2(g) ⇌ 2NH3(g) Use Collision Theory to explain how an increase in temperature and decrease in volume of the reaction vessel will affect the equilibrium in terms of concentrations, reaction rate and Keq. (4) Collision theory states that: 1. As reactant particles collide with greater total translational energy equal to or above the activation energy (E a), the reaction rate increases. 2. As reactant particles collide more frequently the reaction rate increases. This is because for a given number of collisions, some are ‘successful,’ resulting in bond breaking followed by bond formation. Hence more total collisions = more total successful collisions. 3. Particles must collide with the proper orientation. Atoms that will form bonds are faced towards each other upon the particle collision. The net orientation of the particles in the equilibrium is not affected by a change in temperature or volume. The increase in temperature causes the average kinetic energy of every particle in the equilibrium to increase, and also causes each particle’s average speed to increase. This causes more frequent collisions, and greater translational energy of each particle, increasing the reaction rate (as per statements 1. & 2.) of both the forward and reverse reaction. However, because the forward and reverse reactions are exothermic and endothermic respectively, T causes a shift in equilibrium position and thus a change in concentration of every species, as explained below.

The energy-reaction progress graph is shown with the activation energy of the reverse reaction, Ea(r), greater than the activation energy of the forward reaction, Ea(f). The effect of this is shown below on a Boltzmann distribution.

An increase in temperature causes a shift of the graph to the right and makes the peak more ‘squashed.’ Increased T causes a change in the number of particles that have Ea(r). This change is proportionally greater than the relative change in the number of particles that have Ea(f) as illustrated. Hence, after T increases, the greater proportional increase in the number of particles that have Ea(r) compared to the increase in the no. of particles that have Ea(f) causes a net reverse reaction in the equilibrium system. The equilibrium position shifts towards the left of the above equation as written, increasing the concentration of H2(g) and N2(g). N2(g) is increased by 1/3 the percentage that H2(g) is increased by due to the 3:1 mole ratio between H2(g) and N2(g). NH3(g) decreases in concentration as the equilibrium position is shifted towards the left. Keq = hence Keq decreases when the temperature is increased. The decrease in volume is equivalent to an increase in pressure which causes particles in the equilibrium vessel to collide with each other and the walls of the container at a greater rate. This increases the reaction rate of both the forward and reverse reactions as per statement 2. Because C = n/V the decrease in V causes a net increase in the concentration of every species. However, once V is decreased, the increase in concentration of the particles on the LHS of the equation is greater by a factor of 2 than the increase in concentration of the particles on the RHS of the equation, because LHS:RHS = 4:2 = 2:1 mole ratio when the gaseous species on each side are considered. This causes more frequent collisions between LHS particles – H2(g) and N2(g) – than between RHS NH3(g) particles, so a net forward reaction results, shifting the equilibrium to the

right of the above equation as written. Hence, Keq = increases, and the concentration of each species increases overall. But the shift in equilibrium causes the proportional change in C(ammonia) to be greater than the proportional change in C(H2(g)) and C(N2(g)) after the disturbance is imposed on the system. Solution composed by

Chemical

(c). The Keq for PCl5(g) ⇌ PCl3(g) + Cl2(g) at 25oC is 0.225 In a 3L closed reaction vessel there are 0.09 moles of chlorine gas, 0.6 moles of PCl3(g) and 0.03 moles of PCl5(g). Calculate the equilibrium concentration of PCl5(g) in the system. (3)

Question 22: (4 marks) Use the following information to answer parts (a), (b) and (c) Reaction:

6CO2(g) + 6H2O(g) → C6H12O6(aq) + 6O2(g)

∆H:

2803kJmol-1

∆S:

-212Jmol-1K-1

T:

25oC

∆G:

?

(a). Name the reaction taking place. (1) Photosynthesis (b). Calculate ∆G for the reaction and state whether it is spontaneous or non-spontaneous. (2) ∆G = ∆H - T∆S = 2803-(298.15*-0.212) = +2866kJmol-1 at 25o As ∆G is positive, the reaction is non-spontaneous. (c). Sketch, in general terms, the energy profile diagram for the reaction. (1)

Question 23: (6 marks) (a). Explain the action of soap in cleaning molecules of grease or oil off surfaces. Support your response with appropriate diagrams. (4) 1 Mark: Correct intermolecular forces (dispersion) for adsorbing onto grease. 1 Mark: Correct intermolecular forces (ion-dipole) for hydrophilic heads interacting with water. 1 Mark: Stating that surfactants disrupt hydrogen bonds to allow water to fully wet the grease, acting as a surface tension reducing agent. 1 Mark: Outlining the formation of micelles and how they repel each other (same charged surface) + Use terminology such as emulsion and hydrophilic/hydrophobic. Diagram:

(b). With the use of structural formulae, write a reaction for the saponification of stearin to sodium stearate, which contains 18 carbon atoms per surfactant. (2)

Question 24: (10 marks) (a). Outline the conditions that maximise buffer capacity. (2) 1. There is a high concentration of the weak acid/base and its conjugate base/acid. 2. The concentrations of the acid/base and its conjugate base/acid are equal. (b). By the manipulation of equations, show why it is optimal that pKa = pH for buffers. (3) For example, CH3COOH / CH3COO- buffer. pKa = pH -log[Ka] = -log[H+] log[H+] - log([H+]2/[CH3COOH]) = 0 as Ka = [H+][CH3COO-]/[CH3COOH] and [H+] = [CH3COO-] [1] log([H+]/([H+]2/[CH3COOH])) = 0 100 = [CH3COOH]/[H+] = 1 [CH3COOH] = [H+] therefore [CH3COOH] = [CH3COO-] from [1] This answer references 2. from 24(a). Where buffer capacity is maximised when the concentration of the acid and its conjugate base are equal.

Using Henderson-Hasselbach has max ⅔ for this question. It must be derived (the question essentially asks to derive it).

(c). Since the industrial revolution the pH of the ocean has decreased from 8.2 to 8.1 - raising environmental concerns. The ocean contains a natural buffer system, explain the function of this buffer system in maintaining ocean pH at its desired level and state one environmental concern that arises from increasing ocean acidity (5) A buffer is a solution that resists changes in pH. It is formed between a weak acid or base and its conjugate base or acid, respectively. Increasing Acidity: [a] CO2(g) ⇌ CO2(aq) [1] CO2(aq) + H2O(l) ⇌ H2CO3(aq) [2] H2CO3(aq) + H2O(l) ⇌ HCO3- (aq) + H3O+(aq) [3] Carbon dioxide gas from the atmosphere dissolves in the ocean by [1]. This process is accelerating due to industrial processes which release large volumes of greenhouse gases into the atmosphere, primarily CO2(g). The aqueous carbon dioxide reacts with water to form carbonic acid by [2], which then reacts with water in an acid-base reaction. Where carbonic acid acts as the acid and water the base, to form a bicarbonate ion and hydronium in [3]. Increasing concentrations of H3O+ decreases pH levels due to directly increasing acidity. Buffer Counteracting Increasing Acidity: CaCO3(s) ⇌ Ca2+(aq) + CO32-(aq) [4] CO32-(aq) + H+(aq) ⇌ HCO3-(aq) [5] Shells & limestone in sea = reservoir for CO3 2- anions which remove H+ ions from the ocean. First rxn ([4]) produces CO32-(aq), which shifts [5] to the right, forming HCO3- (aq) which shifts all the equilibriums in [a] to the left, counteracting the increased [H3O+] + acidity/decreased pH. Thus, the pH persists at the same level of ~8.1, acting as a natural buffer system to maintain ocean pH at effective levels. One Environmental Concern: Increasing acidity reduces the amount of carbonate in the ocean, making it harder for corals to build skeletons and for shellfish to build the shells for protection.

Question 25: (7 marks) Solid sodium bicarbonate (NaHCO3) dissolves completely in water. (a). Write an equation for the basic ion reacting with water. (1) HCO3-(aq) + H2O(l) ⇌ H2CO3(aq) + OH-(aq) (b). Given the pKa of carbonic acid at 25oC is 6.35, calculate the pH of a 0.335M solution of sodium bicarbonate. (3) pKa + pKb = 14 pKb of HCO3-= 7.65 Kb(HCO3-) = 2.24*10-8 2.24*10-8 = [OH-]2/(0.335-x) As Kb is to the power of -9, assume 0.335-x = 0.335 Therefore [OH-] = 8.66*10-5 pOH = 4.06 pH = 9.94

(c). Sketch the shape of a weak acid + strong base conductivity curve (values are obsolete). Label axes and the equivalence point. (1)

(d). Explain the shape of the graph you sketched for part (c). (2) e.g. CH3COOH(aq) + NaOH(aq) → CH3COONa(aq)+ H2O(l) - The initial conductivity is low because of the low concentration of H+ ions from the partial dissociation of CH3COOH. - The conductivity increases as Na+ ions are added and CH3COOH molecules react with OH- to form CH3COO- ions. - The soln at equivalence point contains Na+, CH3COO- ions and H2O molecules. - Conductivity increases as more NaOH is added after the equivalence point, because of the presence of excess Na+ and OH- ions.

Question 26: (3 marks) (a). In a calorimetry experiment, butan-1-ol was completely combusted to heat 150.0 mL of water from 20.0oC to 71.0oC. The molar heat of combustion of butan-1-ol is 2670 kJ mol-1 . If only 62.5% of the heat released by the combustion was absorbed by the water, what volume of carbon dioxide gas was produced at 27oC and 110 kPa? (3) Answer: 1.7L

Question 27: (5 marks) A chemist conducted an investigation to determine the concentration of sulfate ions in a carbonated tap water sample. They added 100 mL of 1 M nitric acid and then 100 mL of 1 M calcium nitrate solution to a 500 mL sample of the water. The precipitate formed was weighed, results for four trials are shown: Trial

1

2

3

4

Mass of precipitate (g)

2.03

2.06

2.07

3.21

(a). Justify the chemist’s addition of nitric acid. (1) Nitric acid fully dissociates to produce H+ and NO3 - ions, acidifying the solution. The H+ ions react with the carbonate in the “carbonated tap water” to form carbonic acid, H2CO3, which is a weak acid and dissociates to a very low extent. This ‘locks up’ the carbonates by removing CO 3 2from the solution. Therefore calcium sulfate will be the only precipitate, as the possibility of precipitating CaCO3 is removed due to adding nitric acid. (b). Calculate the concentration of sulfate ions in the water sample. (2) 0.03molL-1 (sig figs do not matter for this Q) (c). Outline how the chemist could maximise the validity and accuracy of the experiment. (2) Filter the solution a few times to improve final accuracy. Wash the precipitate to remove soluble impurities. Extensively dry the precipitate to remove any hydrous mass present which gives an invalid weight of ppte upon weighing.

Question 28: (4 marks) An excess of 0.072M silver nitrate solution was added to 30.0mL of a sodium phosphate solution of unknown concentration. The total volume of silver nitrate added was 60.0mL. The precipitate was then filtered out, rinsed thoroughly and the filtrate collected in a conical flask. The filtrate was titrated with a 0.0531M potassium thiocyanate solution and an appropriate indicator according to the following equation: Ag+ (aq) + SCN(aq) → AgSCN (s) An average titre of 15.03mL potassium thiocyanate was obtained. Calculate the concentration of phosphate ions in the original solution. (4)

Question 29: (7 marks) Mr Mac has the following collection of organic liquids in his laboratory: 1. Butan-1-ol 5. Propanamine 2. Butan-2-ol 6. Ethyl propanoate 3. Methylpropan-2-ol 7. Iodoethane 4. Butanoic acid 8. Bromoethane However, there are no labels on any of the containers of the aforementioned liquids. Write a response, outlining observations for each positive test, to help Mr Mac distinguish between these eight organic liquids. (7) Be careful with the order of identification as it affects the feasibility of tests. Set up a sampling sheet with a few drops of each liquid in each small pocket to begin testing. Butanoic Acid: Butanoic acid can be identified by a test with sodium bicarbonate, NaHCO3. Add a pinch of solid NaHCO3 and effervescence will be observed for a positive test as CO2 is evolved. Now the acid can be excluded from future tests (important for next step to be valid). C4H8O2(l) + NaHCO3(s) → C4H7O-Na+(aq) + CO2(g) + H2O(l) Alcohols: Then test with a pinch of solid sodium, all alcohols (liquids 1,2 and 3) will see bubbles of H2 evolve. Isolate the three positive tests and now the primary, secondary and tertiary alcohols must be distinguished between. First add acidified potassium dichromate, K2Cr2O7, with concentrated H2SO4. No reaction will occur for methylpropan-2-ol, identifying liquid 3. Both butan-1-ol, 1o alcohol, and butan-2-ol, 2o alcohol, will show a positive test through a colour change of orange → green as they are oxidised to butanoic acid and butanone respectively. To distinguish between butan-1-ol and butan-2-ol, add Lucas Reagent, which is ZnCl2 in concentrated HCl(aq), the butan-2-ol sample can be identified by turning cloudy due to the formation of a white precipitate in a somewhat slow (5 minute) reaction. The primary alcohol butan-1-ol will essentially not react, and can take up to 1 hour. Thus effectively distinguishing between all alcohols through the aforementioned tests. First Test (Sodium metal, all isomers positive test): 2C4H9OH(l) + 2Na(s) → 2C4H9O-Na+ + H2(g) Second Test (Acidified dichromate):

Propanamine: Add red litmus to remaining samples, it turns red litmus blue as it is basic. C3H9N(l) + H2O(l) ⇌ C3H10N+(aq) + OH- (aq) Ethyl propanoate: A sweet & fruity odour can positively confirm it is the given ester. Haloalkanes: Finally, warm liquids 7 & 8 with some sodium hydroxide solution in a mixture of ethanol and water. Everything will dissolve in this mixture. The halogen atom is displaced as a

halide ion: R-Hal + OH- → R-OH + Hal-. The mixture is acidified by adding dilute nitric acid. This prevents unreacted hydroxide ions reacting with the silver ions to give a confusing precipitate. Then silver nitrate solution is added to give the rxn: Ag+ (aq) + Hal-(aq) → AgHal(s) Iodoethane will produce a yellow precipitate, AgI. Whereas bromoethane will produce a cream precipitate, AgBr, to positively distinguish between the haloalkanes & determine their presence. Ag+(aq) + I-(aq) → AgI (s) (yellow) Ag+(aq) + Br-(aq) → AgBr (s) (cream)

Question 30: (4 marks) A chemist has taken a sample of water from the Murray Darling Basin and suspects it contains one of the following six anions: Cl-, Br-, I-, OH-, CH3COO- and SO42-. Create a flowchart to distinguish between them, identifying all observations for each positive test so that a viewing chemist can figure out which anion is present. (4)

Solution composed by Liam

Question 31: (7 marks) (a). Calculate the equilibrium concentration of Fe3+ and S2- ions when 135.5mL of 0.235M Fe2(SO4)3 and 87.5mL of 0.0050M Na 2S are mixed together in a beaker at 25oC. (4) The Ksp of iron (III) sulfide at 25oC is 4.75 × 10-88. Fe2(SO4)3(aq) + 3Na2S(aq) → Fe2S3(s) + 3Na2SO4(aq) 2Fe3+ + 3SO4 2- + 6Na+ + 3S2- → Fe2S3(s) + 6Na+ + 3SO42n(Fe2(SO4)3) = cv = 0.235 * 0.1355 = 0.0318425mol n(Na2S) = cv = 0.0050 * 0.0875 = 0.0004375mol 0.0633933 moles of Fe3+ in excess, divide by 0.223L to get concentration. Therefore [Fe3+] = 0.28M and Fe3+ becomes a common ion that affects the solubility of Fe2S3. Ksp of Fe2S3 = [Fe3+]2[S2-]3 4.75 × 10-88 = (0.28)2[S2-]3 Therefore [S2-] = 1.8 × 10-29M Overall: [Fe3+] = 0.28M and [S2-] = 1.8 × 10-29M (b). Evaluate the effectiveness of solely using Ksp to determine the solubility of various salts (3) Solely using Ksp to gauge the solubility of a compound is an ineffective method. This is because a lower Ksp does not always correlate with a compound being less soluble in water at 25oC, due to different dissociations resulting in a different number of aqueous ions in moles. Using the Ksp table on the data sheet, the Ksp for PbI2 is 9.8 × 10-9 and for PbSO4 it is 2.53 × 10-8. From just Ksp, this infers that PbI2 is less soluble than PbSO4 as the former will lie further to the left (solid state). Where ‘x’ is the molar solubility of the salt: PbI2(s) ⇌ Pb2+(aq) + 2I-(aq) therefore Ksp = [x][2x]2 = 4x3 9.8 × 10-9 = 4x3 therefore x = 1.35 × 10-3 molL-1 PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq) therefore Ksp = [x][x] = x2 2.53 × 10-8 = x2 therefore x = 1.59 × 10-4 molL-1 1.35 × 10-3M is greater than 1.59 × 10-4M, thus PbI2 is more soluble in water at 25oC in terms of molar solubility despite having a lower Ksp value. In evaluation, Ksp is minimally effective in determining the solubility of various salts (as very low Ksp’s show a salt is insoluble), when directly comparing solubility Ksp is ineffective and should not be solely used.

Question 32: (6 marks) In the HSC Chemistry course you researched four addition polymers and two condensation polymers. Below are two monomers that form a condensation polymer, kevlar.

(a). A student states that: “Condensation polymers are defined as such because they all produce a small molecule of water” Evaluate the students statement with the use of supporting evidence, you may use the given polymer. (2) For condensation polymerisation to successfully occur, the monomers must have two functional groups, one on each end of the monomer. These functional groups react chemically with the functional groups on neighbouring monomers, creating a different functional group in the process - for example an amide link in nylon-6,6. In this reaction between functional groups, a small molecule is released. In the case of nylon-6,6 and polyethylene terephthalate a water molecule is released. However in the polymer shown, kevlar, HCl will be the small molecule released as the amine (NH2) group and chlorinated carboxylate react to form an amide link, discarding a hydrogen and chlorine atom to form HCl. Therefore the students' statement is false, not all condensation polymers produce a small molecule of water, however they aren’t completely incorrect as many do produce H2O. (b). Choose two addition polymers from your mandatory studies. Sketch each polymer in the left column. In the right column, specify a use for each polymer and explain why this polymer is effective in the chosen circumstance in terms of its chemical properties. (4) Sketch:

Use and Explanation:

Polystyrene

Too many ways to attack this question, just cross check your answer against internet/notes/past answers.

Polytetrafluoroethylene

Too many ways to attack this question, just cross check your answer against internet/notes/past answers.

Question 33: (8 marks) Orica Chemical Plant at Kooragang Island (KI), NSW.

Use the information below to answer question 32. Orica KI specialises in the production of ammonium nitrate (NH4NO3). This compound is used as an explosive, most commonly in the mining sector, and must be stored in optimal conditions as small solid pellets. They are located within close proxi...