TheLucidSTEM · Chemistry olympiad feeder round mock · 120 minutes · 60 marks · Calculator allowed
Name: ______________________ Date: ____________ · MARK SCHEME
Data sheet
Relative atomic masses: H 1.008, C 12.01, N 14.01, O 16.00, F 19.00, Na 22.99, Mg 24.31, P 30.97, Cl 35.45, Ca 40.08, Xe 131.29
Constants: gas constant $R = 8.314$ J K-1 mol-1; molar volume of a gas at room temperature and pressure (RTP) $= 24.0$ dm3 mol-1; standard pressure $p^\circ = 1$ bar $= 1.00\times10^{5}$ Pa; $0\ ^\circ\mathrm{C} = 273.15$ K; $1$ dm3 $= 1.00\times10^{-3}$ m3.
Useful relations: $pV = nRT$; $\Delta G^\circ = -RT\ln K$; $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$; $\ln\dfrac{K_2}{K_1} = -\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$.
Structured questions
About 24 minutes per question. Marks for each part are shown in brackets.
1Hydrazine rocket fuel[12 marks]
Many spacecraft thrusters burn liquid hydrazine, N2H4, with liquid dinitrogen tetroxide, N2O4. The two liquids ignite on contact and the only products are nitrogen and water (steam).
Standard enthalpies of formation, $\Delta_f H^\circ$ / kJ mol-1: N2H4(l) $+50.6$; N2O4(l) $-19.5$; H2O(g) $-241.8$.
(a) Write a balanced equation, with state symbols, for the reaction of hydrazine with dinitrogen tetroxide.
Answer: $\mathrm{2N_2H_4(l) + N_2O_4(l) \rightarrow 3N_2(g) + 4H_2O(g)}$
Correct formulae and balancing, with state symbols (l), (l), (g), (g) [1].
Accept any correct multiple. Water as (l) loses the mark: the question says steam.
(b) Give the oxidation state of nitrogen in N2H4 and in N2O4. Use the electrons transferred to show that your equation in (a) has the correct ratio of fuel to oxidant.
Answer: $-2$ in N2H4, $+4$ in N2O4; 4 e- lost per N2H4, 8 e- gained per N2O4, so 2 : 1
N is $-2$ in hydrazine and $+4$ in dinitrogen tetroxide [1].
Each N2H4 loses $2\times2 = 4$ electrons going to N2 (0); each N2O4 gains $2\times4 = 8$ electrons; so two hydrazine molecules are needed for each N2O4 [1].
(c) Calculate the mass of N2O4 needed per gram of N2H4 for a mixture in the exact ratio of your equation.
Answer: 1.44 g of N2O4 per g of N2H4
$M(\mathrm{N_2H_4}) = 32.05$ and $M(\mathrm{N_2O_4}) = 92.02$ g mol-1 [1].
Ratio $= 92.02 / (2\times32.05) = 1.44$ [1].
Using a 1 : 1 mole ratio (answer 2.87) scores 0 for the second mark; allow error carried forward from a wrong equation in (a).
(d) Calculate the standard enthalpy change for your equation, and hence the energy released per kilogram of the propellant mixture (fuel plus oxidant). Give your answer in MJ kg-1.
Answer: $\Delta_r H^\circ = -1049$ kJ mol-1; 6.72 MJ kg-1
$\Delta_r H^\circ = 4(-241.8) - [2(+50.6) + (-19.5)]$ [1]
$= -1048.9 \approx -1049$ kJ mol-1 [1].
Mass of reactants in the equation $= 2(32.05) + 92.02 = 156.1$ g, so energy per kg $= 1048.9 / 0.1561 = 6.72\times10^{3}$ kJ kg-1 $= 6.72$ MJ kg-1 [1].
Common error: dividing by the mass of hydrazine only (16.4 MJ kg-1) loses the last mark. Accept 6.7 MJ kg-1; the sign may be given as released or as $-6.72$.
(e) The hot exhaust leaves the combustion chamber at 3000 K and $1.00\times10^{5}$ Pa. Calculate the volume of gas produced from 1.00 kg of propellant mixture under these conditions, assuming ideal behaviour.
Answer: 11.2 m3
7 mol of gas per 156.1 g, so $n = 7 \times 1000/156.1 = 44.8$ mol per kg [1].
$V = nRT/p = 44.8 \times 8.314 \times 3000 / 1.00\times10^{5} = 11.2$ m3 [1].
Allow error carried forward from (a) and (d). Answers in dm3 (11 200 dm3) are fine; a missing unit conversion (for example 0.0112) loses the mark.
(f) Some thrusters use monomethylhydrazine, CH3NHNH2, instead. With N2O4 it gives carbon dioxide, water and nitrogen. Write a balanced equation for this reaction using the smallest whole numbers.
Answer: $\mathrm{4CH_6N_2 + 5N_2O_4 \rightarrow 4CO_2 + 12H_2O + 9N_2}$
Correct products CO2, H2O and N2 with C and H balanced [1].
Fully balanced, smallest whole numbers 4, 5, 4, 12, 9 [1]. State symbols not needed.
Check: O, $5\times4 = 20 = 4\times2 + 12$; N, $8 + 10 = 18 = 9\times2$. Electron count: each CH6N2 loses 10 e- (C $-2$ to $+4$, two N $-2$ to 0), each N2O4 gains 8, giving 4 : 5.
2Compounds of a noble gas[12 marks]
Xenon reacts with fluorine when the mixture is heated or exposed to ultraviolet light. Depending on the conditions, three different colourless solids A, B and C can be made. Each contains only xenon and fluorine. Their percentages of fluorine by mass are:
| Compound | A | B | C |
|---|---|---|---|
| % F by mass | 22.4 | 36.7 | 46.5 |
(a) Determine the formulae of A, B and C. Show your working for one of them.
Answer: A = XeF2, B = XeF4, C = XeF6
For example A: $77.6/131.29 = 0.591$ mol Xe and $22.4/19.00 = 1.18$ mol F, ratio 1 : 2 [1 for working].
All three formulae correct: A XeF2, B XeF4, C XeF6 [2]; two correct [1].
Check: XeF2 22.4%, XeF4 36.7%, XeF6 46.5% F.
(b) Use electron-pair repulsion theory to predict the shapes of A and B. For each, state the number of bonding pairs and lone pairs around the xenon atom.
Answer: XeF2 linear (2 bp, 3 lp); XeF4 square planar (4 bp, 2 lp)
XeF2: 2 bonding pairs and 3 lone pairs, lone pairs equatorial, so linear [1].
XeF4: 4 bonding pairs and 2 lone pairs, lone pairs opposite each other, so square planar [1].
Shape and electron pairs both needed for each mark. "Tetrahedral" for XeF4 loses the mark.
(c) C reacts completely with water to give HF and an explosive white solid, xenon trioxide. Write the equation. With only a little water, C instead gives a liquid, XeOF4; predict its shape.
Answer: $\mathrm{XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF}$; XeOF4 square pyramidal
$\mathrm{XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF}$ [1].
XeOF4: five bonding positions and one lone pair around Xe (treat Xe=O as one bonding region), so square pyramidal, with O at the apex [1].
(d) State the oxidation state of xenon in XeO3 and predict the shape of the XeO3 molecule.
Answer: $+6$; trigonal pyramidal
Oxidation state $+6$ [1].
Three Xe=O bonding regions plus one lone pair, so trigonal pyramidal (like NH3) [1]. "Trigonal planar" loses the mark.
(e) A dissolves in water and slowly decomposes:
$$\mathrm{2XeF_2 + 2H_2O \rightarrow 2Xe + O_2 + 4HF}$$
Calculate the total volume of gas, measured at RTP, given off when 0.500 g of A decomposes completely. Assume that HF stays in solution.
Answer: 106 cm3
$n(\mathrm{XeF_2}) = 0.500/169.29 = 2.95\times10^{-3}$ mol [1].
Gas $=$ Xe $+$ O2 $= 2.95\times10^{-3} + 1.48\times10^{-3} = 4.43\times10^{-3}$ mol, volume $= 4.43\times10^{-3}\times24.0 = 0.106$ dm3 $= 106$ cm3 [1].
Common error: counting only O2 (35.4 cm3) or only Xe (70.9 cm3) loses the second mark.
(f) Xenon forms several fluorides, but no compound of neon has ever been made. Give the main reason.
Answer: Xe has a much lower first ionisation energy (outer electrons far from the nucleus and well shielded)
Xenon's outer electrons are further from the nucleus and more shielded, so its (first) ionisation energy is much lower than neon's and it can lose electron density to fluorine [1].
"Xenon is bigger" alone is not enough; the idea of ionisation energy or of electrons held less tightly is needed. Also accept: xenon can expand its octet using its larger outer shell.
3Three alkenes and the ozone test[13 marks]
A hydrocarbon P contains 85.7% carbon by mass and has relative molecular mass 84. P decolourises bromine water. With hydrogen and a nickel catalyst it gives Q.
Ozonolysis (ozone, then zinc and water) cuts a C=C bond and turns each alkene carbon into a C=O group. For example, $\mathrm{(CH_3)_2C{=}CH_2}$ gives propanone and methanal.
Ozonolysis of P gives two compounds R and S, each with three carbon atoms. Both give an orange precipitate with 2,4-dinitrophenylhydrazine. R gives a silver mirror with Tollens' reagent; S does not.
Two isomers of P, called T and U, also react with bromine water. Ozonolysis of T gives only S. Ozonolysis of U gives only R.
(a) Show that the molecular formula of P is C6H12.
Answer: Empirical formula CH2; $84/14 = 6$, so C6H12
C $85.7/12.01 = 7.14$, H $14.3/1.008 = 14.2$, ratio 1 : 2, empirical formula CH2 [1].
$M(\mathrm{CH_2}) = 14.03$; $84/14.03 = 6$, so C6H12 [1].
(b) Identify R and S, explaining how the tests tell them apart.
Answer: R propanal, CH3CH2CHO; S propanone, CH3COCH3
Both are carbonyl compounds (2,4-DNPH); R is oxidised by Tollens' so it is an aldehyde: propanal, CH3CH2CHO [1].
S is a ketone: propanone, CH3COCH3 [1].
(c) Give the structural formula and the name of P.
Answer: (CH3)2C=CHCH2CH3, 2-methylpent-2-ene
Join the C=O carbons of propanone and propanal with a C=C: (CH3)2C=CHCH2CH3 [1].
Name 2-methylpent-2-ene [1]. Allow error carried forward from (b).
(d) Give the structures of T and U. One of them exists as two stereoisomers: say which, name the type of isomerism, and draw both forms.
Answer: T (CH3)2C=C(CH3)2, 2,3-dimethylbut-2-ene; U CH3CH2CH=CHCH2CH3, hex-3-ene, shows E/Z isomerism
T: two propanone halves joined, (CH3)2C=C(CH3)2 (2,3-dimethylbut-2-ene) [1].
U: two propanal halves joined, CH3CH2CH=CHCH2CH3 (hex-3-ene) [1].
U shows E/Z (geometric, cis/trans) isomerism; both forms drawn correctly, with the two ethyl groups on the same side (Z) and on opposite sides (E) of the C=C [1].
Saying that P or T shows E/Z loses the last mark: each has one alkene carbon carrying two identical groups.
(e) P reacts with hydrogen bromide. Name the major product and explain, in terms of the intermediate, why it is the major product. Which of the two possible products is chiral?
Answer: Major 2-bromo-2-methylpentane (via the tertiary carbocation); the minor product 3-bromo-2-methylpentane is chiral
Major product 2-bromo-2-methylpentane, because H+ adds to C3 to give the more stable tertiary carbocation at C2 (more alkyl groups pushing electron density onto the positive carbon) [1].
The minor product, 3-bromo-2-methylpentane, is chiral (C3 carries H, Br, an ethyl and an isopropyl group); the major product is not, since C2 carries two methyl groups [1].
(f) How many signals are there in the 13C NMR spectrum of P?
Answer: 6
6 [1]. The two methyl groups on C2 are not equivalent: one is on the same side of the C=C as the hydrogen, the other on the same side as the ethyl group. Answer 5 scores 0.
(g) Q reacts with chlorine in ultraviolet light. How many structural isomers of formula C6H13Cl can be formed from Q by monochlorination?
Answer: 5
Q is 2-methylpentane, CH3CH(CH3)CH2CH2CH3. Different carbons: the two equivalent CH3 on C2, C2, C3, C4 and C5, so 5 structural isomers [1]. Stereoisomers are not counted.
4Phosphorus pentachloride vapour[12 marks]
When phosphorus(V) chloride is heated it vaporises and partly dissociates:
$$\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$$
A student puts 4.170 g of PCl5 into an evacuated rigid flask of volume 1.00 dm3 and heats it to 523 K. All of the solid vaporises. When equilibrium is reached the total pressure in the flask is $1.52\times10^{5}$ Pa. Treat all gases as ideal.
(a) Calculate the amount of PCl5 used and the pressure it would exert at 523 K if it did not dissociate at all.
Answer: 0.0200 mol; $8.71\times10^{4}$ Pa
$M(\mathrm{PCl_5}) = 208.2$ g mol-1, $n = 4.170/208.2 = 0.0200$ mol [1].
$p_0 = nRT/V = 0.0200\times8.314\times523/1.00\times10^{-3} = 8.71\times10^{4}$ Pa [1] (accept $8.70\times10^{4}$ from rounding).
Using $V = 1.00$ m3 (answer 87 Pa) loses the second mark.
(b) Calculate $\alpha$, the fraction of PCl5 that has dissociated at equilibrium.
Answer: $\alpha = 0.745$
Each PCl5 that dissociates makes two molecules, so the total pressure is $p_0(1+\alpha)$ [1].
$\alpha = 1.52\times10^{5}/8.71\times10^{4} - 1 = 0.745$ [1] (accept 0.74 to 0.75).
(c) Write the expression for $K_p$ and calculate its value at 523 K, in bar.
Answer: $K_p = 1.90$ bar
$K_p = \dfrac{p(\mathrm{PCl_3})\,p(\mathrm{Cl_2})}{p(\mathrm{PCl_5})}$ with $p(\mathrm{PCl_3}) = p(\mathrm{Cl_2}) = \alpha p_0$ and $p(\mathrm{PCl_5}) = (1-\alpha)p_0$ [1].
$K_p = \dfrac{\alpha^2 p_0}{1-\alpha} = \dfrac{0.745^2\times0.871}{0.255} = 1.90$ bar [1] (accept 1.85 to 1.95; $1.90\times10^{5}$ Pa also scores). Allow error carried forward from (a) and (b).
(d) At 573 K, $K_p = 11.3$ bar. Calculate the standard enthalpy change of the dissociation, assuming it does not change with temperature. Is the dissociation endothermic or exothermic?
Answer: $\Delta H^\circ \approx +89$ kJ mol-1, endothermic
$\ln\dfrac{11.3}{1.90} = -\dfrac{\Delta H^\circ}{8.314}\left(\dfrac{1}{573}-\dfrac{1}{523}\right)$ [1].
$1.783 = \Delta H^\circ \times 2.007\times10^{-5}$, so $\Delta H^\circ = +8.9\times10^{4}$ J mol-1 $= +89$ kJ mol-1; endothermic, as $K$ rises with temperature [1]. Accept +87 to +91 kJ mol-1.
(e) Calculate $\Delta G^\circ$ and $\Delta S^\circ$ for the dissociation at 523 K.
Answer: $\Delta G^\circ = -2.8$ kJ mol-1; $\Delta S^\circ = +175$ J K-1 mol-1
$\Delta G^\circ = -RT\ln K_p = -8.314\times523\times\ln1.90 = -2.8$ kJ mol-1 [1] ($K_p$ must be in bar here, since $p^\circ = 1$ bar).
$\Delta S^\circ = (\Delta H^\circ - \Delta G^\circ)/T = (88\,800 + 2800)/523 = +175$ J K-1 mol-1 [1]. Accept +170 to +180. A positive sign is expected: one mole of gas becomes two.
(f) A little argon is added to the equilibrium mixture at 523 K. State and explain the effect on $\alpha$ when (i) the volume is kept constant and (ii) the total pressure is kept constant.
Answer: (i) no change; (ii) $\alpha$ increases
(i) No change: the partial pressures of the three reacting gases are unchanged, so the quotient still equals $K_p$ [1].
(ii) $\alpha$ increases: the volume must rise, all three partial pressures fall, and the equilibrium moves to the side with more gas molecules [1].
5How hard is this mineral water?[11 marks]
Hardness in water is caused mainly by dissolved Ca2+ and Mg2+ ions. Both are measured by titration with EDTA, which forms very stable 1 : 1 complexes with metal ions.
- Titration 1. 100.0 cm3 of the water is buffered at pH 10 and titrated with 0.01000 mol dm-3 EDTA using an indicator that responds to both Ca2+ and Mg2+. Mean titre 18.40 cm3.
- Titration 2. A second 100.0 cm3 sample is made strongly alkaline (about pH 12.5) with sodium hydroxide, which precipitates magnesium hydroxide. It is then titrated with the same EDTA using an indicator for Ca2+ only. Mean titre 13.20 cm3.
(a) EDTA4- is a hexadentate ligand. Explain what hexadentate means.
Answer: It forms six coordinate (dative) bonds to one metal ion
One ligand bonds to the central metal ion through six donor atoms (two N and four O), forming six dative bonds [1].
(b) Calculate the amounts of Ca2+ and Mg2+ in 100.0 cm3 of the water.
Answer: Ca2+ $1.320\times10^{-4}$ mol; Mg2+ $5.20\times10^{-5}$ mol
Total metal ions $= 18.40\times10^{-3}\times0.01000 = 1.840\times10^{-4}$ mol; Ca2+ $= 13.20\times10^{-3}\times0.01000 = 1.320\times10^{-4}$ mol [1].
Mg2+ $= 1.840\times10^{-4} - 1.320\times10^{-4} = 5.20\times10^{-5}$ mol [1].
Taking titre 1 as Mg2+ only loses the second mark.
(c) Give the concentrations of calcium and of magnesium in the water in mg dm-3 (the units used on bottle labels).
Answer: Ca 52.9 mg dm-3; Mg 12.6 mg dm-3
Ca: $1.320\times10^{-3}$ mol dm-3 $\times 40.08 = 52.9$ mg dm-3 [1].
Mg: $5.20\times10^{-4}$ mol dm-3 $\times 24.31 = 12.6$ mg dm-3 [1].
Forgetting to scale from 100 cm3 to 1 dm3 (5.29 and 1.26) loses the marks once only.
(d) Total hardness is often quoted as if all the Ca2+ and Mg2+ were present as calcium carbonate. Express the hardness of this water in mg of CaCO3 per dm3.
Answer: 184 mg dm-3
$1.840\times10^{-3}$ mol dm-3 $\times 100.09$ g mol-1 $= 184$ mg dm-3 [1].
(e) $K_{sp}$ of Mg(OH)2 is $5.6\times10^{-12}$ mol3 dm-9 at 298 K, and $K_w = 1.0\times10^{-14}$ mol2 dm-6. Calculate the concentration of Mg2+ left in solution at pH 12.5, and so show that magnesium does not interfere with titration 2.
Answer: $[\mathrm{Mg^{2+}}] = 5.6\times10^{-9}$ mol dm-3, about 0.001% of the original
$[\mathrm{OH^-}] = 10^{-1.5} = 3.16\times10^{-2}$ mol dm-3, so $[\mathrm{Mg^{2+}}] = 5.6\times10^{-12}/(3.16\times10^{-2})^2 = 5.6\times10^{-9}$ mol dm-3 [1].
The original concentration was $5.20\times10^{-4}$ mol dm-3, so only about $1\times10^{-5}$ of it (0.001%) stays dissolved: a negligible effect on the titre [1].
Forgetting to square [OH-] gives $1.8\times10^{-10}$ and loses the first mark.
(f) Part of the hardness can be removed by boiling. Write an equation for what happens to calcium hydrogencarbonate when the water is boiled.
Answer: $\mathrm{Ca(HCO_3)_2(aq) \rightarrow CaCO_3(s) + H_2O(l) + CO_2(g)}$
$\mathrm{Ca(HCO_3)_2 \rightarrow CaCO_3 + H_2O + CO_2}$ [1]. Accept the ionic form $\mathrm{Ca^{2+} + 2HCO_3^- \rightarrow CaCO_3 + H_2O + CO_2}$.
(g) A home water softener swaps every Ca2+ and Mg2+ ion for sodium ions from an ion-exchange resin. Calculate the mass of sodium, in mg, added to each dm3 of this water.
Answer: 84.6 mg
Each 2+ ion is replaced by two Na+ ions (charge balance), so $n(\mathrm{Na^+}) = 2\times1.840\times10^{-3} = 3.680\times10^{-3}$ mol per dm3 [1].
Mass $= 3.680\times10^{-3}\times22.99 = 0.0846$ g $= 84.6$ mg [1].
A 1 : 1 swap (42.3 mg) loses the first mark.
TheLucidSTEM practice paper. Questions are original and written in the style of a first national round; this is not an official olympiad paper.