HSC Chemistry 2020

[2 marks]. The alkane is propane, $\text{C}_3\text{H}_8$.

The parent ion (molecular ion) peak is the highest significant m/z, at 44. This equals the relative molecular mass of propane:

No alkane of formula $\text{C}_n\text{H}_{2n+2}$ other than propane has a molecular mass of 44, so the alkane is propane. The tiny peak at 45 is the 13C isotope contribution and is not the parent ion.

Marker's note. Read the spectrum carefully to pick the parent ion at 44 (not the base peak), tie it to the molecular mass of propane, and account for the small 45 peak as the 13C isotope rather than the parent.

[5 marks].

Step 1 - identify the cation by flame test. Dip a clean wire in the solution and hold it in a blue Bunsen flame:

• brick-red flame to calcium
• yellow-green (pale green) flame to barium
• no characteristic colour to magnesium.

Step 2 - test for hydroxide. Measure the pH with universal indicator or a pH meter. A strongly basic reading (pH near 13) indicates hydroxide; a near-neutral reading rules it out. Confirm with a precipitation test by adding a few drops of copper(II) nitrate solution; a pale-blue precipitate of copper(II) hydroxide confirms hydroxide.

Step 3 - distinguish chloride from acetate. If the anion is not hydroxide, add silver nitrate solution. A white precipitate of silver chloride confirms chloride:

No precipitate (and a slightly basic pH reading) indicates acetate.

Matching the flame colour with the anion test confirms the full identity of the salt.

Marker's note. Set the tests out as a logical sequence (a flow chart helps), give the correct flame colours, identify the source ions in any precipitation test, and write a balanced net ionic equation such as the silver chloride one.

[4 marks]

1. Use of catalysts

Both Reactor 1 and Reactor 2 use a catalyst. A catalyst lowers the activation energy, so an acceptable rate is reached at the moderate temperatures shown (200 to 300 degrees C and 50 to 70 degrees C) rather than at higher, more costly temperatures. This cuts energy use and running costs.

2. Recycling of unreacted gases

The flow chart feeds ethene and oxygen into Reactor 1, and Separator 1 sends material on for further processing. Separating and recycling unreacted ethene and oxygen means less fresh reactant must be purchased, improving the economics and reducing waste.

3. Plant location and transport to markets

The chart ends with transport to markets and shows by-products going for sale. Locating the plant near ports, rail or road and near suppliers of ethene and oxygen reduces transport costs for both incoming reagents and the ethane-1,2-diol product, making the process viable.

Marker's note. Analyse the stimulus rather than reciting general knowledge; show clear cause and effect for each factor (for example, catalyst lowers activation energy, therefore lower temperature, therefore lower cost); and treat the water in Reactor 2 as a hydrolysis reagent, not a coolant.

(a) [1 mark]. The ester functional group.

(b) [3 marks]. Complete combustion of each fuel:

Diesel needs more oxygen per molecule for complete combustion (and already contains less oxygen in its structure) than biodiesel, which carries two oxygen atoms of its own. Under the same oxygen supply, diesel is more likely to combust incompletely, depositing carbon (soot):

Because biodiesel reaches complete combustion more readily, it produces less soot, $\text{C}(s)$.

(c) [2 marks]. Mass of biodiesel needed for 2141 MJ at 38 MJ/kg:

Volume at density 0.90 kg/L:

(d) [4 marks].

Advantages:

• Renewable and sustainable. Bioethanol comes from crops that can be regrown, whereas petrol comes from finite crude oil, so bioethanol can be produced indefinitely.
• Lower net carbon dioxide. The carbon dioxide released on combustion was recently absorbed by the crop during photosynthesis, so if green energy is used in production the process approaches carbon neutral, unlike petrol which adds fossil carbon to the atmosphere.

Disadvantages:

• Competition with food. Growing crops for fuel diverts land and crops from food production, which can raise food prices or cause shortages.
• Land and water pressure. Producing enough crops needs large amounts of water and fertiliser; over-fertilising causes runoff and water pollution, and irrigation strains local water supplies.

Marker's note. Draw the correct ester group in (a). In (b) balance every equation, identify soot as C(s), and link incomplete combustion to the oxygen available. In (c) include units and show full working. In (d) explain (give cause and effect), not merely list, two advantages and two disadvantages.

[7 marks]. Plot temperature increase (y-axis, degrees C) against volume of citric acid (x-axis, mL). Draw two lines of best fit: a rising line through the early points (0.0 to about 2.0 mL) and a roughly horizontal line through the plateau points (3.0 to 6.0 mL). Their intersection is the equivalence point.

The lines cross at a citric acid volume of about 2.4 mL. At this point the citric acid exactly neutralises the sodium hydroxide.

From the equation, 1 mol citric acid reacts with 3 mol NaOH:

The NaOH was 8.0 mL:

Marker's note. Label both axes with units and sensible scales, plot the points, and find the equivalence volume from the intersection of two lines of best fit (not from a table value). Show the 3:1 mole ratio explicitly in the calculation.

(a) [1 mark].

(b) [4 marks]. The profile shows the products NO at higher energy than the reactants, so the forward reaction is endothermic and its activation energy is larger than that of the reverse (exothermic) reaction.

Raising the temperature increases the average kinetic energy of the molecules, so a greater fraction of collisions exceeds the activation energy. The rate of both the forward and reverse reactions increases. Because the forward reaction has the larger activation energy, the extra energy raises the forward rate more than the reverse rate, so the forward reaction is favoured and the system shifts right, making more NO.

Since

a shift to the right raises $[\text{NO}]$ relative to the reactants, so $K_{eq}$ increases.

Marker's note. Include all states in (a). In (b) reason through collision theory (more molecules exceeding activation energy, both rates up, forward up more) rather than just quoting Le Chatelier, and conclude that K increases. The Maxwell-Boltzmann distribution may be referenced.

(a) [2 marks]. Propan-2-amine accepts a proton on its nitrogen to form the protonated (ammonium) cation, releasing hydroxide:

(b) [3 marks]. Set up an ICE analysis with $x = [\text{OH}^-]$ formed:

Assuming $x \ll 1.00$:

So $[\text{OH}^-] = 2.09\times10^{-2}$ mol/L (about 0.0207 mol/L).

Marker's note. Draw the amine and its protonated cation clearly (do not let N look like H). Present a clear ICE table, state the simplifying assumption, and include units.

[3 marks]

The method has two flaws.

Inaccuracy from the standard

Solid NaOH is used as though it were a primary standard, but it is not one: it is deliquescent and absorbs water (and carbon dioxide) from the air. The 1.000 g weighed therefore includes water, so the true mass of NaOH is less than recorded. The standard solution is more dilute than calculated, which makes the titres larger than they should be and the calculated acid concentration wrong - an inaccurate result.

Unreliability from the indicator

Bromocresol green changes colour around pH 3.2 to 5.2. The titration of a weak acid with a strong base has its equivalence point in the basic region (well above pH 7), so this indicator changes colour on the steeply rising part well before the equivalence point. Tiny differences in where the colour change is judged give large differences in titre, which is why titres 1 to 4 scatter so widely (16.35 down to 10.10 mL) - an unreliable result.

Marker's note. Explain both faults fully, not just name them: NaOH is not a primary standard because it is deliquescent, and the indicator changes outside the equivalence range so endpoints are inconsistent. Link each fault to the data (low true concentration; scattered titres).

[5 marks]. The starting alcohol is 2-methylpropan-2-ol (a tertiary alcohol), because it gives no reaction with acidified potassium dichromate (tertiary alcohols are not oxidised). The other structures follow from the reactions shown.

• A - 2-methylpropan-2-ol, $(\text{CH}_3)_3\text{COH}$ (the tertiary alcohol, not oxidised by dichromate).
• B - 2-methylpropene, $(\text{CH}_3)_2\text{C}=\text{CH}_2$ (dehydration of A).
• C - 2-chloro-2-methylpropane, $(\text{CH}_3)_3\text{CCl}$ (Markovnikov addition of HCl to the alkene, halogen on the more substituted carbon; also formed from A with ZnCl2/HCl).
• D - 2-methylpropan-2-ol again, $(\text{CH}_3)_3\text{COH}$ (acid-catalysed hydration of the alkene, Markovnikov, OH on the more substituted carbon).
• E - 1-chloro-2-methylpropane, $(\text{CH}_3)_2\text{CHCH}_2\text{Cl}$ (the anti-Markovnikov product, halogen on the less substituted carbon).

Marker's note. Deduce the tertiary alcohol first from the no-reaction with dichromate, then work the sequence. Distinguish Markovnikov from anti-Markovnikov addition, draw complete structures with all hydrogen atoms and correct atoms, and account for all five different structures.

[7 marks]. The compound is propan-2-yl acetate (isopropyl acetate), $\text{CH}_3\text{COOCH(CH}_3)_2$.

Justification:

• IR. The strong absorption near 1750 cm-1 is a C=O (carbonyl) stretch, found in esters and carboxylic acids. The absence of the broad O-H stretch at 2500 to 3300 cm-1 rules out a carboxylic acid, so the compound is an ester. This fits the formula C5H10O2.
• 13C NMR. A peak near 170 ppm is the ester carbonyl carbon; a peak near 70 ppm is a carbon singly bonded to oxygen (the O-CH); a peak near 20 ppm is alkyl carbon. This matches an ester with an O-CH group and methyl carbons.
• 1H NMR. The septet (area 1) at 5.0 ppm is a single proton split by six neighbours, that is a CH between two CH3 groups, the $-\text{OCH(CH}_3)_2$ proton. The doublet (area 6) at 1.2 ppm is the two equivalent CH3 groups, each split by the one CH proton. The singlet (area 3) at 2.0 ppm is an isolated CH3 with no neighbours, the $\text{CH}_3\text{C}=\text{O}$ group of the acetate.

Putting these together gives a $-\text{OCH(CH}_3)_2$ group on the ester oxygen and a $\text{CH}_3\text{CO}-$ group on the carbonyl: propan-2-yl acetate.

Marker's note. Use the data explicitly: assign the carbonyl and rule out the acid from the IR; use the septet-doublet pattern to deduce the $-\text{CH(CH}_3)_2$ group and the singlet for the isolated methyl. Annotate the structure to link each signal to its H and C environment. Methyl 2-methylpropanoate is an accepted near-match for 6 marks.

[4 marks]. This is a back titration. The thiocyanate reacts with the silver left over after the chloride has been precipitated.

Excess silver remaining (1:1 with SCN-):

Total silver added at the start:

Silver that reacted with chloride:

Concentration in the 100.0 mL water sample:

Convert to mass using $M(\text{Cl}) = 35.45$ g/mol:

Marker's note. Follow the back-titration steps: find excess Ag+ from the SCN- titre, subtract from the total Ag+ to get Ag+ reacted with chloride, then convert moles of Cl- to mass. Multiply by the molar mass of Cl (35.45), not of AgCl.

[4 marks]. Boiling point depends on the strength of the intermolecular forces that must be overcome.

Both methanol and propanoic acid have an O-H group, so their molecules form hydrogen bonds, the strongest of these forces. The ester methyl propanoate has no O-H, so it cannot hydrogen bond; its molecules attract by dipole-dipole and dispersion forces only. On hydrogen bonding alone, the ester would boil lowest.

However, dispersion forces increase with the number of electrons (molecular size). Methyl propanoate is a larger molecule than methanol, so its dispersion forces are substantial. The combined dipole-dipole and dispersion forces in the ester are stronger than the (hydrogen bonding plus weaker dispersion) forces in the smaller methanol molecule. So the ester (79.8 degrees C) boils higher than methanol (64.7 degrees C), even though methanol hydrogen bonds.

Propanoic acid is larger again and forms hydrogen-bonded dimers, so it has the highest boiling point (141.2 degrees C). The ester therefore sits between the two, higher than the alcohol but lower than the acid.

Marker's note. Name each force correctly (hydrogen bonding, dipole-dipole, dispersion), keep intermolecular and intramolecular distinct, and link molecular size (dispersion) to the boiling point order, explaining why the ester beats the alcohol despite lacking hydrogen bonding.

(a) [2 marks].

The mole ratio Ca(OH)2 to HCl is 1:2, so

(b) [4 marks]. The calcium ions are in 0.100 L:

The solution is saturated with calcium hydroxide, so use $K_{sp}(\text{Ca(OH)}_2) = 5.02\times10^{-6}$ from the data sheet:

Let $[\text{OH}^-] = x$:

Then:

Marker's note. Write the balanced dissociation and neutralisation equations with states, show the mole ratio in (a), and in (b) recognise this needs the Ksp expression (not an ICE table). Finish with pOH = -log[OH-] then pH = 14 - pOH.

[4 marks]. For both acids, pH falls as concentration rises, because a higher acid concentration gives a higher $[\text{H}^+]$.

Hydrochloric acid (strong). HCl ionises completely at every concentration:

So $[\text{H}^+] = [\text{HCl}]$, and because pH is the log of $[\text{H}^+]$, plotting pH against the log concentration scale gives a straight line. The HCl line also lies lower (more acidic) than acrylic acid at any given concentration.

Acrylic acid (weak). Acrylic acid only partly ionises:

The fraction ionised increases as the acid is diluted (the equilibrium shifts toward ions). So at high concentration only a small fraction ionises, giving a higher pH than the fully ionised HCl, and the curve bends upward and away from the straight HCl line at higher concentration rather than running parallel to it.

Marker's note. Refer explicitly to the trend lines: name which acid each line is, explain the linear strong-acid line from full ionisation and the curved weak-acid line from concentration-dependent ionisation, and include both equations. Be careful naming each acid (avoid ambiguous pronouns).

[4 marks]. First find the equilibrium triiodide concentration from the absorbance:

Use the iodide mass balance to find $[\text{I}_2]_{eq}$:

Then $[\text{I}^-]_{eq} = 2[\text{I}_2]_{eq} = 3.10\times10^{-4}$ mol/L.

So $K \approx 5.6\times10^2$.

Marker's note. Write the equilibrium expression before substituting, distinguish initial from equilibrium concentrations, and show every algebraic step, taking care with the numerator, denominator and indices.

[5 marks]. Two processes release heat: dissolving the NaOH, then neutralising it.

Dissolution of NaOH.

Neutralisation. Amount of HCl:

HCl (0.200 mol) is the limiting reagent against NaOH (0.250 mol), so 0.200 mol reacts:

Total heat released.

Temperature rise. The solution mass is the acid plus the added solid, $103 + 10.0 = 113$ g:

Marker's note. Add the heat from both steps, use HCl as the limiting reagent for neutralisation, use the solution mass (113 g) and the given specific heat capacity (not water's 4.18), rearrange q = mc dT for dT, and keep q and dH distinct.