HSC Chemistry 2022

[2 marks]. Addition across the double bond gives 1,2-dichloropropane, with a chlorine added to each of the first two carbons:

In condensed form the product is $CH_2Cl\text{-}CHCl\text{-}CH_3$. The carbon-to-carbon double bond opens and one chlorine atom bonds to each of those two carbons, so no double bond remains.

Marker's note. Draw a straight-chain product with every atom shown, including all hydrogens, and keep each carbon to four bonds. This is addition (the double bond opens), not substitution.

[2 marks]. The phosphate ion accepts a proton, so it is a base and its conjugate acid is the hydrogen phosphate ion:

The fluoride and hydrogen fluoride pair also works: $F^-$ is a base with conjugate acid $HF$.

Marker's note. A conjugate pair differs by exactly one proton. Pair the base with the acid it becomes after gaining a proton, for example $PO_4^{3-}$ with $HPO_4^{2-}$ or $F^-$ with $HF$.

(a) [1 mark]

The amount of NO(g) would decrease. The forward reaction is exothermic, so raising the temperature favours the endothermic reverse reaction.

(b) [2 marks]

A catalyst lowers the activation energy for both the forward and the reverse reaction, so it speeds both rates by the same factor. Because the two rates increase equally, the system reaches equilibrium sooner but the equilibrium position is unchanged.

(c) [3 marks]

Removing $H_2O(g)$ lowers the concentration of a product, so there are fewer water molecules available to collide with NO in the reverse reaction. The rate of the reverse reaction falls, while the forward rate is unchanged, so the forward rate is now greater. More NO is produced until the rates become equal again, and the new equilibrium settles at a higher $[NO]$ than just after the removal (although $[NO]$ does not return to its original value, because product was withdrawn).

Marker's note. Describe this with collision theory, not Le Chatelier's Principle. State the effect on the concentration of the named species ($[NO]$), keep the forward reaction rate and reverse reaction rate distinct, and note the catalyst affects both rates equally, including in all-gas systems.

[3 marks]. As molar mass increases along the homologous series, the boiling point increases. A larger molecule has more electrons, which makes its instantaneous and induced dipoles larger, so the dispersion forces between molecules are stronger. Boiling separates molecules from one another, so stronger intermolecular dispersion forces require more energy to overcome, giving a higher boiling point.

Marker's note. Use cause-and-effect language and name the correct intermolecular force (dispersion). Boiling overcomes intermolecular forces between molecules, not the covalent bonds within them, so keep intermolecular and intramolecular terms separate.

[3 marks]. HCN is a weak acid, so it only partially ionises in water:

HCl is a strong acid and ionises completely (a one-way reaction). At the same initial concentration, the partial ionisation of HCN gives a much lower hydronium ion concentration than the complete ionisation of HCl. Since $pH = -\log[H_3O^+]$, a lower $[H_3O^+]$ gives a higher pH, so the HCN solution has the higher pH. No calculation is needed.

Marker's note. Show the HCN equation with an equilibrium arrow (and HCl as one-way). Link the smaller degree of ionisation to a lower hydronium ion concentration and then to a higher pH. Check that the adjective is right - lower hydronium concentration, higher pH.

(a) [2 marks]. Average initial temperature $= (21.0 + 21.2)/2 = 21.1$ degrees C, so $\Delta T = 24.4 - 21.1 = 3.3$ K. The combined mass is $100.7 + 102.0 = 202.7$ g $= 0.2027$ kg, and $c = 4.18 \times 10^3$ J/(kg.K).

(b) [2 marks]. Moles of water formed equals moles of HCl reacted $= 0.1000 \text{ L} \times 0.50 \text{ mol/L} = 0.050$ mol. The released energy is $2.6 \times 10^3$ J $= 2.6$ kJ, and neutralisation is exothermic, so

Marker's note. Combine both masses, average the two initial temperatures, and keep units consistent in $q = mc\Delta T$. The enthalpy of neutralisation is exothermic, so the answer must carry a negative sign.

(a) [2 marks]. The two positional isomers are propan-1-ol and propan-2-ol:

Propan-1-ol is $CH_3CH_2CH_2OH$ (hydroxyl on an end carbon); propan-2-ol is $CH_3CHOHCH_3$ (hydroxyl on the middle carbon).

(b) [2 marks]. Propan-1-ol has three different carbon environments, so its carbon-13 NMR spectrum shows three signals. Propan-2-ol is symmetrical, with the two end carbons equivalent, so it has only two carbon environments and shows two signals. Counting the signals identifies the isomer: three signals means propan-1-ol, two signals means propan-2-ol.

(c) [3 marks]. Both oxidations need an oxidising agent and an acid catalyst. Propan-1-ol is a primary alcohol and oxidises fully to the carboxylic acid propanoic acid; propan-2-ol is a secondary alcohol and oxidises to the ketone propanone:

Marker's note. In (b) use spectroscopy terms (carbon environments, signals). In (c) the primary alcohol oxidises all the way to a carboxylic acid (not stopping at the aldehyde), the secondary alcohol gives a ketone, and the conditions (acidified dichromate or permanganate) should sit above the arrow.

(a) [1 mark]. Iron(III) hydroxide, $Fe(OH)_3$. The acid plus oxygen oxidises the iron to the iron(III) state, and excess hydroxide precipitates the brown $Fe(OH)_3$.

(b) [4 marks]. Molar mass of $Fe_2O_3 = 2(55.85) + 3(16.00) = 159.70$ g/mol.

Each formula unit of $Fe_2O_3$ contains two iron atoms, so multiply by 2:

Marker's note. In (a) show iron is in the +3 state, by name or by a correct formula. In (b) apply the 2 : 1 iron-to-oxide ratio the right way (multiply the moles of $Fe_2O_3$ by 2), show each step clearly, and report to three significant figures.

(a) [3 marks]. Plot the enthalpy of combustion (y-axis, more negative downward) against the number of carbon atoms (x-axis: methanol 1, ethanol 2, propan-1-ol 3, pentan-1-ol 5). Draw a smooth curved line of best fit through the four points, then read off the value at four carbons (butan-1-ol):

The estimate is about $-2100$ kJ/mol.

(b) [2 marks]. In a school laboratory the calorimeter is usually a simple metal can that is poorly insulated, so a large fraction of the heat is lost to the surrounding air and to the apparatus rather than warming the water. The measured energy released is therefore smaller in magnitude (less negative) than the published value, which is obtained with insulated, standard equipment. (Incomplete combustion producing soot would also release less energy than complete combustion and is an acceptable reason.)

Marker's note. In (a) plot the points accurately, draw a genuine curve (not a straight line) and read the value at four carbons with the unit and negative sign. In (b) justify the direction of the difference (heat loss or incomplete combustion makes the school value less negative), not just that there is a difference.

[7 marks]. The compound is 3-methylbutan-2-one, $CH_3\text{-}CO\text{-}CH(CH_3)_2$.

Mass spectrum

The parent molecular ion at $m/z = 86$ matches the molar mass of $C_5H_{10}O$. The base peak at $m/z = 43$ is the acylium fragment $CH_3CO^+$, formed by cleavage next to the carbonyl group.

Infrared spectrum

The strong absorption near 1700 per cm is a carbonyl (C=O) stretch. There is no broad O-H band between about 2500 and 3300 per cm, which rules out a carboxylic acid or an alcohol, so the C=O belongs to a ketone.

Carbon-13 NMR

There are four signals although the molecule has five carbons, so two carbons are equivalent (the two methyls of the isopropyl group). The signal near 220 ppm is a ketone carbonyl carbon, and the signals between about 18 and 40 ppm are CH and CH3 carbons.

Proton NMR

Three hydrogen environments fit the structure. The septet (1H) is the lone CH coupled to six neighbouring hydrogens on two adjacent methyls. The singlet (3H) is the methyl of the $CH_3CO$ group, with no hydrogens on the adjacent carbonyl carbon. The doublet (6H) is the two equivalent methyls of the isopropyl group, each split by the single CH proton. All data point to 3-methylbutan-2-one.

Marker's note. Link every spectrum to a feature of the structure: the parent peak to the molar mass and the base peak to $CH_3CO^+$, the 1700 per cm band to C=O and the missing O-H to ruling out an acid, the carbonyl carbon near 220 ppm, and the proton splitting (septet, singlet, doublet) and integration to the CH, $CH_3CO$ and isopropyl groups. Name and draw the structure.

(a) [3 marks]. The method is unsuitable. Adding iodide precipitates silver iodide, $Ag^+(aq) + I^-(aq) \rightarrow AgI(s)$, which removes free $Ag^+$ from solution. By Le Chatelier's Principle this disturbs the complex equilibrium, so the complex decomposes to replace the lost silver:

The complex keeps breaking down and releasing more silver until effectively all of it has precipitated. The titrant therefore measures the total of the free and complexed silver, not the free silver alone, so it overestimates the free $Ag^+$ concentration. The method does not measure what is intended and is unsuitable.

(b) [4 marks]. Rearrange the equilibrium expression for the ammonia concentration:

The fraction present as free silver is 0.010%, so $\dfrac{[Ag^+]}{[Ag(NH_3)_2^+]} = 1.0 \times 10^{-4}$, which means $\dfrac{[Ag(NH_3)_2^+]}{[Ag^+]} = 1.0 \times 10^4$. Substituting:

Marker's note. In (a) engage with the actual method and judge it: precipitation pulls silver from the complex, so the titration measures total silver, making it unsuitable. In (b) note that the equilibrium concentrations are given, so no ICE table is needed, write a correct $K_{eq}$ expression and use the given percentage as the free-to-complex ratio.

(a) [6 marks]. First standardise the NaOH. The first titre (28.60 mL) is a discordant outlier, so average the three concordant titres: $(27.40 + 27.20 + 27.60)/3 = 27.40$ mL.

A 25.00 mL aliquot has $n(KHP) = 0.24430 \times 0.02500 = 6.1074 \times 10^{-3}$ mol. KHP is monoprotic, so $n(NaOH) = n(KHP)$, giving

Now the drink titration. In 13.10 mL, $n(NaOH) = 0.22290 \times 0.01310 = 2.9200 \times 10^{-3}$ mol. Citric acid is triprotic, so it reacts with NaOH in a 1 : 3 ratio:

Scale up to the 250.0 mL flask (factor of 10), which holds all the acid from the 75.00 mL bottle:

This came from 75.00 mL of drink, so

(b) [2 marks]. Dissolved carbon dioxide forms carbonic acid, which would also react with the NaOH during the titration. A larger volume of NaOH would be needed to reach the endpoint than the citric acid alone requires, so the calculated moles, and hence the calculated concentration of citric acid, would be higher than the true value.

Marker's note. Reject the outlier titre, carry the 25 mL aliquot from the 100 mL standard, apply the 3 : 1 NaOH-to-citric-acid ratio and the 250/75 dilution and aliquot factors, and annotate each step with the quantity calculated. In (b) link the acidic CO2 to extra NaOH used and therefore a higher calculated concentration.

[8 marks]. A suitable school-scale process is the esterification of an alcohol with a carboxylic acid to make an ester.

Selection of reagents. Choose reactants that are cheap, available and relatively low in toxicity. Ethanol and ethanoic acid (acetic acid) meet these criteria. The reaction also needs an acid catalyst that doubles as a dehydrating agent, because water is a product; a small amount of concentrated sulfuric acid is suitable. The reaction produces the ester ethyl ethanoate and water:

Reaction conditions

Heat the alcohol, carboxylic acid and a few drops of concentrated sulfuric acid under reflux. Reflux lets the mixture be heated to speed the reaction while volatile reactants and the product condense and return to the flask rather than escaping, and it avoids the pressure build-up of a sealed vessel.

Hazards and safety precautions

The alcohol and ester are flammable, so heat with a heating mantle or water bath, not an open flame. Concentrated sulfuric acid and the carboxylic acid are corrosive, so use only small quantities, wear safety glasses and gloves, and add the acid carefully. Work in a fume cupboard because the ester and acid vapours are irritating.

Yield and purity

Esterification is an equilibrium reaction, so it does not go to completion. The yield is improved by refluxing for a sufficient time (about 30 to 60 minutes) and by using an excess of one reactant. The crude product is a mixture of ester, unreacted alcohol and acid, sulfuric acid and water, so it must be purified, for example by washing to remove the acids and then distilling to collect the ester at its boiling point.

Marker's note. Use subheadings for each required factor and include a balanced chemical equation written with structural or condensed organic formulae. Tie every safety precaution to a specific chemical or piece of equipment (not just "wear PPE"), and link the yield to manipulating the equilibrium and to a purification step.

[4 marks]. First find the hydroxide concentration from the pH. $pOH = 14 - 7.5 = 6.5$, so

Use the equilibrium expression to find the hypochlorite concentration:

All the chlorine species came from the added NaOCl, so the total chlorine concentration needed is

Use the dilution relationship $c_1V_1 = c_2V_2$ with $c_1 = 2.0$ mol/L, $c_2 = 2.53 \times 10^{-4}$ mol/L and $V_2 = 1.00 \times 10^4$ L:

Marker's note. Convert pH to pOH and then to $[OH^-]$, write a correct $K_{eq}$ expression and solve for $[OCl^-]$, then recognise that both HOCl and $OCl^-$ at equilibrium came from the NaOCl added, so sum them before applying the dilution calculation.

[5 marks]. Find the moles of precipitate and the initial moles of each ion. Total volume $= 0.0800 + 0.0800 = 0.1600$ L.

Each mole of precipitate removes 1 mol $Sr^{2+}$ and 2 mol $OH^-$:

Convert to equilibrium concentrations in 0.1600 L:

Marker's note. Work in moles first, subtract the moles of solid from the initial $Sr^{2+}$ and twice that from $OH^-$, convert to concentrations using the combined volume, and write the $K_{sp}$ expression with the hydroxide term squared.

[4 marks]. Evaporation is endothermic, so $\Delta H$ is positive, and converting liquid to gas increases disorder, so $\Delta S$ is positive. Both samples are well below the boiling point, so the difference between the two containers is whether the system can reach equilibrium, not how fast molecules move.

In the sealed bottle the water vapour cannot escape. As molecules evaporate, the water vapour concentration rises until the rate of condensation equals the rate of evaporation. A dynamic equilibrium is established, $Q = K_{eq}$, and no further net evaporation is observed, so the liquid level barely changes.

In the open beaker the vapour diffuses away into the room, so the water vapour concentration stays low and $Q$ remains below $K_{eq}$. Equilibrium is never reached, so the forward (evaporation) process keeps being favoured and the water slowly evaporates to completion. The enthalpy and entropy changes of vaporisation are the same in both containers; the difference is that only the closed system reaches equilibrium, while the open system loses vapour continuously.

Marker's note. Identify the forward process as endothermic with a positive entropy change, and explain the difference by extent, not speed: the closed system reaches a dynamic equilibrium ($Q = K_{eq}$) while the open system never does because vapour is lost, so it proceeds to completion. The enthalpy of vaporisation is the same for both.