1

46th

INTERNATIONAL

CHEMISTRY OLYMPIAD

2014

UK Round One

STUDENT QUESTION BOOKLET

* * * * * ■ The time allowed is 2 hours.

■ Attempt all 5 questions.

■ Write your answers in the special answer booklet.

■ In your calculations, please write only the essential steps in the answer booklet.

■ Always give the appropriate units and number of significant figures.

■ You are provided with a copy of the Periodic Table.

■ Do NOT write anything in the right hand margin of the answer booklet.

■ The marks available for each question are shown below; this may be helpful when

dividing your time between questions.

Question 1 2 3 4 5 Total

Marks Available

9 9 13 18 16 65

Some of the questions will contain material you will not be familiar with. However, by logically

applying the skills you have learnt as a chemist, you should be able to work through the

problems. There are different ways to approach the tasks – even if you cannot complete certain

parts of a question, you may still find subsequent parts straightforward.

H 1800.1

eH 2

300.4

iL 3

49.6

eB 4

10.9

lobmys

rebmun ci

motassa

m cimota nae

m

B 518.01

C 610.21

N 710.41

O 800.61

F 900.91

eN

0181.02

aN

1199.22

gM

2113.42

lA 31

89.62

iS 4190.82

P 5179.03

S 6160.23

lC 71

54.53

rA

8159.93

K 91201.93

aC

0280.04

cS 1269.44

iT 22

09.74

V 3249.05

rC

4200.25

nM

5249.45

eF 62

58.55

oC

7239.85

iN 82

17.85

uC

9255.36

nZ

0373.56

aG

1327.96

eG

2395.27

sA 33

29.47

eS 4369.87

rB 53

409.97

rK

6308.38

bR

7374.58

rS 8326.78

Y 9319.88

rZ 04

22.19

bN

1419.29

oM

2449.59

cT 34

uR

4470.101

hR

5419.201

dP

644.601

gA

7478.701

dC

8404.211

nI 9428.411

nS 0596.811

bS 1557.121

eT 25

06.721

I 3509.621

eX

4503.131

sC 55

19.231

aB

6543.731

*aL

7519.831

fH 27

94.871

aT

3759.081

W 4758.381

eR

572.681

sO

672.091

rI 772.291

tP 87

90.591

uA

9779.691

gH

0895.002

lT 18

73.402

bP

282.702

iB 38

89.802

oP 48

tA 58

nR

68

rF 78

aR

88

cA

+ 98

sedinahtnaL*

eC

8521.041

rP 95

19.041

dN

0642.441

mP

16

mS26

4.051

uE

3669.151

dG

4652.751

bT

5639.851

yD

6605.261

oH

7639.461

rE 86

62.761

mT

9639.861

bY

0740.371

uL

1779.471

+sedinitc

Ah

T09

10.232

aP 19

U 2930.832

pN

39

uP

49

mA

59

mC

69

kB

79

fC 89

sE 99

mF

001

dM

101

oN

201

rL

301

1

2 13

14

15

16

17

3 4

5 6

7 8

9 10

11

12

18

NA

= 6

.02

× 1

023

mo

l–1

1

nm

= 1

× 1

0−9

m

0 °C

= 2

73 K

2

3

1. This question is about controlling phosphate levels

Although phosphorus compounds are essential in our diet, too much phosphate ion, PO4

3–, in our blood can be harmful. One treatment for high phosphate levels uses a drug called Fosrenol®, the active ingredient of which is lanthanum carbonate.

Once the chewed tablet is swallowed, the carbonate first reacts with the hydrochloric acid present in the stomach. Then the lanthanum ions react with any phosphate ions present to give a precipitate of lanthanum phosphate. Lanthanum phosphate is essentially completely insoluble in water.

All lanthanum compounds in this question contain lanthanum in the +3 oxidation state.

(a) (i) Give an equation for the reaction between lanthanum carbonate and hydrochloric acid.

(ii) Give an ionic equation for the formation of lanthanum phosphate.

Lanthanum carbonate may be prepared by simply mixing aqueous solutions of lanthanum nitrate and sodium carbonate.

(b) Write a balanced equation for this reaction.

The lanthanum carbonate precipitated by the previous reaction is usually the octahydrate. The first sample of pure anhydrous lanthanum carbonate was prepared by dissolving lanthanum oxide in aqueous trichloroethanoic acid to form lanthanum trichloroethanoate. Heating this solution for six hours gave the desired carbonate, trichloromethane, and carbon dioxide.

(c) (i) Draw the displayed formula for trichloroethanoic acid.

(ii) Give the equation for the reaction between lanthanum oxide and trichloroethanoic acid.

(iii) Give the equation for the formation of lanthanum carbonate.

The lanthanum carbonate contained in the Fosrenol® tablets is prepared by mixing aqueous solutions of lanthanum chloride and ammonium hydrogencarbonate. During this reaction, carbon dioxide gas is given off.

(d) Give an equation for this reaction.

Fosrenol® is sold in different sizes of tablet, “1000 mg”, “750 mg”, and “500 mg”. The number refers to the mass of lanthanum ions contained in the tablet.

(e) (i) Calculate the mass of lathanum carbonate dihydrate contained in the 1000 mg tablet.

(ii) Calculate the mass of phosphate ion removed after taking a single 1000 mg tablet of Fosrenol®.

4

2. This question is about a sodium street lamp

Many of our streets are lit by the extremely energy efficient low-pressure sodium discharge lamps. The light appears mainly yellow, but a spectroscope reveals a number of colours present, each with a particular frequency. Careful analysis of the spectrum enables us to determine the first ionization energy of sodium.

The so-called ‘ground state’ of an atom is where all the electrons are in their lowest possible energy levels.

(a) Give the complete electronic configuration for the ground state of sodium in terms of the relevant s and p sub-shells.

The ‘valence-shell electron’ is the electron easiest to remove from a neutral sodium atom.

(b) What is the valence-shell electron of sodium?

The familiar yellow colour of a sodium flame is actually due to an excited electron in a sodium atom essentially falling from a 3p orbital to the 3s. The wavelength for this transition is 589.8 nm.

The energy, E, (in joules) corresponding to light of wavelength (m) is given by the equation:

where h is Planck’s constant = 6.626 × 10–34 J s

c is the speed of light = 2.998 × 108 m s–1.

(c) Calculate the energy for the 3p to 3s transition in:

(i) J (atom–1)

(ii) kJ mol–1.

5

The atomic emission spectrum of sodium includes a series of lines which all fall on a straight line when (1 / wavelength) is plotted against (1 / n2), where n is the integer shown in the following table.

Wavelength / nm 466.6 498.0 568.1 818.1

n 6 5 4 3

The origin of the linear relationship is due to fact that the (1 / n2) is proportional to the

ionization energy (I.E.) of an excited sodium atom after the valence electron has been

promoted to the nd shell:

where k is an arbitrary constant, and n is the principal quantum number, e.g. n =1 for the 1s shell; 2 for the 2s and 2p shell; 3 for the 3s, 3p and 3d shell and so on.

[Note for the graph, n cannot be one or two since there is no 1d or 2d shell.]

(d) What does this ionization energy of the excited nd electron tend to as the electron is removed from successively higher d shells? Circle one of the following answers in the answer booklet.

one zero infinity the constant k

Each of the atomic emission lines included in the table and graph are due to transitions where the valence electron drops from an excited d shell to the 3p shell in the neutral sodium atom.

(e) Give an expression for the energy change of these transitions in terms of I.E.(nd), and

I.E.(3p) (the ionization energy of the excited sodium atom after the valence electron has been promoted to the 3p).

(f) (i) Use the graph to determine I.E.(3p) in J (atom–1).

(ii) What is the ionization energy of the ground state of sodium in kJ mol–1?

6

3. This question is about spot cream

The drug tazarotene (sold under the trade names of Zorac® or Tazorac®) can be prescribed as a cream that can be applied to the skin to help to treat acne and certain other skin conditions. It is commonly sold as a 0.05% cream by mass.

(a) The molar mass of tazarotene is 351.46 g mol−1. Assuming that tazarotene cream has a density of 0.90 g cm−3, calculate the concentration of tazarotene in the cream in mol dm−3.

The synthesis of tazarotene is shown below. Not all of the reaction by-products are shown.

The synthesis begins with the conversion of 2-chloro-5-methylpyridine to Ester B.

(b) Draw the structure of Compound A and Ester B.

The second part of the synthesis begins with thiophenol, which is converted into Compound I by a number of steps.

(c) Draw the structures of Compounds D, E, F and I, and anions C− and G−.

7

(d) How would you classify the reaction of Compound H into Compound I?

Circle one of the following answers in the answer booklet.

Oxidation Reduction Addition Elimination Substitution

Finally, Compound I is treated with a very strong base to form anion J.

Anion J can be reacted with Compound B to form tazarotene.

(e) Suggest a structure for Anion J.

(f) How many signals would you expect to see in the 13C NMR spectrum of tazarotene?

Tazarotene is actually a pro-drug, meaning it is metabolised to its active form when inside the body.

The active form has a molar mass of 323.41 g mol−1 and two fewer signals in its 13C NMR spectrum than tazarotene.

(g) Suggest a structure for the active form of the drug.

8

4. This question is about bombardier beetles

Bombardier beetles get their name from the defence mechanism that they have when attacked, whereby they shoot a hot chemical spray at their attackers. In their abdomen they have two separate chambers, one containing an aqueous solution of hydrogen peroxide and the other an aqueous solution of an organic compound, denoted here as A.

When the beetle is attacked, fluid from both chambers is squirted into a mixing chamber which contains enzymes. One of these enzymes, catalase, catalyses the breakdown of hydrogen peroxide into oxygen and water.

(a) (i) Write a balanced equation for this reaction.

(ii) How would you classify this reaction? Circle one of the following answers in the answer booklet.

Oxidation Reduction Disproportionation Hydrolysis Dehydration

Some of the intermediates produced during the breakdown of hydrogen peroxide react with the organic compound, A, to give a product B. The overall equation for the reaction can be considered as the result of A reacting with all of the oxygen produced by the equation in part (a)(i) as in the equation below.

A + ½O2 → B + H2O

(b) Write an overall equation for the reaction of hydrogen peroxide with A to form B.

The temperature of the mixture ejected from bombardier beetles has been measured to reach the boiling point of water, considerably hotter than the beetle’s body temperature, which matches the surroundings (20 °C).

(c) (i) Calculate the amount of energy needed to heat 1 dm3 of this mixture by this amount. Assume the specific heat capacity of the mixture is the same as that of pure water, 4.18 J g−1 K−1 and that the density of the mixture is the same as that of pure water, 1.00 g cm−3.

(ii) Assuming equal volumes of the two solutions are mixed, what is the minimum concentration of the hydrogen peroxide solution in the beetle’s hydrogen peroxide chamber?

The standard enthalpy change for the overall reaction (your answer to part (b)) per mole of A is −203 kJ mol−1.

9

Compound A is a 1,2,4-trisubstituted benzene, shown on the right, where there are substituents present at the positions labelled with an X.

(d) How many possible structures are there of this 1,2,4-trisubstituted benzene when:

(i) All of the X substituents are different from each other?

(ii) Two of the X substituents are identical?

Combustion analysis of Compound A indicates that it contains only carbon, hydrogen and oxygen.

In compound A, stretching in some of the bonds contained in the X substituents contributes to characteristic peaks in the IR spectrum of this compound (shown below).

These characteristic peaks are marked I and II on the spectrum.

(e) Suggest which bond stretches are responsible for I and II.

10

In compound A, two of the three X substituents are the same (i.e. the substituents are Xa, Xa and Xb).

The 13C NMR spectrum of compound A contains seven signals.

The 1H NMR spectrum of compound A is shown below.

Peaks III and IV in the 1H NMR spectrum disappear on addition of D2O to the sample.

(f) (i) Suggest an identity for the two identical substituents (Xa and Xa).

(ii) Suggest an identity for the other substituent (Xb).

The protons on the benzene ring in compound A (H1, H2 and H3) can be assigned by analysis of their coupling constants in the 1H NMR spectrum.

The splitting that is observed in the spectrum due to coupling between two non-equivalent benzene ring protons in a 1,2-relationship to each other, is considerably larger than the splitting observed due to coupling between two non-equivalent benzene ring protons in a 1,3-relationship.

The splitting due to coupling between two non-equivalent benzene ring protons in a 1,4-relationship is generally too small to be observed.

(g) Assign one of the signals III-VIII in the 1H NMR spectrum of A to each of the protons H1, H2 and H3.

Bombardier beetles also use a simpler organic compound, C for the same purpose.

Compound C is a relative of compound A where one of the X substituents is replaced by a hydrogen atom, making a disubstituted benzene.

The 1H NMR spectrum of compound C only contains two signals.

(h) Suggest a structure for compound C and hence also a structure for compound A.

Compound C is oxidised to D in the same way that compound A is oxidised to B.

The 1H NMR spectrum of compound D only contains one signal.

(i) Suggest a structure for compound B and a structure for compound D.

11

5. This question is about fire and ice

Water is able to form a range of cage structures called clathrates which can contain many small molecules; a common example is methane hydrate (CH4)x(H2O)y. Methane hydrates potentially offer a valuable supply of methane to satisfy our demand for natural gas. Large reserves of methane hydrates have been found at the bottom of Lake Baikal in Russia.

(a) Write the balanced equation for the combustion of methane in excess oxygen.

A 100 g sample of methane hydrate, in the form known as clathrate structure II, (CH4)x(H2O)y was burnt in excess oxygen in a sealed container. After the reaction had completed and the products had cooled, 116.92 g of water was recovered from the container, and the gas, when shaken with excess limewater, gave 84.73 g of CaCO3.

(b) (i) Determine the empirical formula of the methane hydrate in structure II. The molar mass of CaCO3 is 100.09 g mol

−1.

(ii) The molar mass of the hydrate was 2835.18 g mol-1. Determine the molecular formula of the methane hydrate in structure II.

(c) 6.67 × 1011 kg of methane is estimated to be at the bottom of Lake Baikal. Determine the volume the gas would occupy if it were to escape the lake during the winter, at −19.0 °C.

Use the ideal gas equation pV = nRT where p is the pressure in Pascal, V is the volume in m3, T is the temperature in Kelvin and R is 8.31 J K−1 mol−1. [1 atm = 1.0 × 105 Pa]

The methane hydrate at the bottom of Lake Baikal is found as clathrate structure I having the formula (CH4)8(H2O)46 and a molar mass of 957.07 g mol

−1.

(d) (i) Determine the percentage by mass of methane within the methane hydrate.

(ii) Using your answer to part (d)(i) to determine the mass of methane hydrate found at the bottom of Lake Baikal.

(iii) The density of methane hydrate crystals is 0.95 g cm−3. Calculate the volume of methane hydrate crystals at the bottom of Lake Baikal.

Clathrate structure I

Clathrate structure II

12

The smallest repeating unit of methane hydrate that contains all the symmetry of the crystal is known as the unit cell. The unit cell for methane hydrate is a cube that entirely contains 8 methane molecules and 46 waters.

(e) (i) What is the mass of a single unit cell?

(ii) Using the density of methane hydrate (0.95 g cm−3) calculate the length of an edge of the cubic unit cell.

Although the methane and water molecules are not spheres, we can treat them as spheres, and an effective spherical radius can be estimated for them: 0.14 nm for water and 0.21 nm for methane.

The formula for the volume of a sphere, V, is given below.

(iii) Determine the volume of methane and the volume of water within a unit cell.

(iv) Work out the percentage of space in the crystal that is occupied with molecules.

When (CH4)8(H2O)46 is burnt in excess oxygen at 298 K, the enthalpy change of combustion is −6690.4 kJ mol−1.

(f) Use the data below to determine the enthalpy change when methane hydrate is formed from methane and water at 298 K.

Standard enthalpy changes of formation at 298 K in kJ mol−1:

CH4: −74.8, H2O: −285.8, CO2: −393.5

13

Acknowledgements & References References for questions Q1 Process for the preparation of lanthanum carbonate dehydrate P. R. Muddasani et al., United States Patent US 2012/0058200

The Rare Earth Metals and their Compounds. XII. Carbonates of Lanthanum, Neodymium and Samarium M. L. Salutsky and L. L. Quill, J. Am. Chem. Soc., 72, 3306-7, 1950. Q2 Images from Department of Chemistry, University of Cambridge Q3 Contemporary Drug Synthesis J.-J. Li et al., Wiley-Blackwell, 2004 The image is from © Shutterstock. Q4 A Biomimetic Study of the Explosive Discharge of the Bombardier Beetle N. Beheshti & A. C. McIntosh, Int. Journal of Design & Nature, 1, 1-9, 2006. Chemical Defence of a Primitive Australian Bombardier Beetle (Carabidae): Mystropomus regularis T. Eisner et al., Chemoecology, 2, 29-34,1991. Photograph from Patrick Coin (Wikipedia) The 1H NMR spectrum and IR spectrum were adapted from those available from Sigma Aldrich. Q5 Image of burning methane hydrate from: http://energy.gov/articles/new-methane-hydrate-research-investing-our-energy-future Images of crystal structures from Steve Dutch at the University of Wisconsin–Green Bay: http://www.uwgb.edu/dutchs/Petrology/Clathrate-1.HTM http://www.uwgb.edu/dutchs/Petrology/Clathrate-2.HTM Further details about clathrate structures can be found at: http://www.lsbu.ac.uk/water/clathrat2.html

The UK Chemistry Olympiad is supported by INEOS is a leading global manufacturer of petrochemicals, specialty chemicals and oil products. It comprises 19 businesses each with a major chemical company heritage. The production network spans 73 manufacturing facilities in 19 countries throughout the world. The chemicals INEOS produce enhance almost every aspect of modern life.

46

th INTERNATIONAL

CHEMISTRY OLYMPIAD

2014

UK Round One

MARK SCHEME

Although we would encourage students to always quote answers to an appropriate number of significant figures, do not penalise students for significant figure errors. Allow where a student’s answers differ slightly from the mark scheme due to the use of rounded/non-rounded data from an earlier part of the question. For answers with missing or incorrect units, penalise one mark for the first occurrence in each question and write UNIT next to it. Do not penalise for subsequent occurrences in the same question.

Question 1 2 3 4 5 Total

Marks Available

9 9 13 18 16 65

1. This question is about controlling phosphate levels

(a) (i) La2(CO3)3 + 6HCl → 2LaCl3 + 3H2O + 3CO2

State symbols not required 1

(ii) La3+ + PO4

3− → LaPO4

State symbols not required 1

(b) 2La(NO3)3 + 3Na2CO3 → La2(CO3)3 + 6NaNO3

State symbols not required 1

(c) (i)

Do not accept a molecular formula. Accept if O‒H bond not drawn out, but all other bonds must be drawn out.

1

(ii) La2O3 + 6CCl3CO2H → 2La(CCl3CO2)3 + 3H2O

State symbols not required 1

(iii) 2La(CCl3CO2)3 + 3H2O → La2(CO3)3 + 6CHCl3 + 3CO2

State symbols not required 1

(d) 2LaCl3 + 6NH4HCO3 → La2(CO3)3 + 6NH4Cl + 3CO2 + 3H2O

State symbols not required 1

(e) (i) Amount of La3+ = 1000 mg / 138.91 g mol−1 = 7.20 × 10−3 mol

Amount of La2(CO3)3·2H2O = 3.60 × 10−3 mol

Molar mass of La2(CO3)3·2H2O = (2 × 138.91 + 3 × 12.01 + 11 × 16.00 + 4 × 1.008) g mol−1

= 493.88 g mol−1

Mass of La2(CO3)3·2H2O = 493.88 g mol−1 × 3.60 × 10−3 mol

= 1780 mg (or 1.78 g)

1

(ii) Amount of La3+ = 7.20 × 10−3 mol = Amount of PO4

3−

Molar Mass of PO43− = (30.97 + 4 × 16.00) g mol−1 = 94.97 g mol−1

Mass of PO43− removed = 7.20 × 10−3 mol × 94.97 g mol−1

= 684 mg (or 0.684 g)

Allow ECF when an incorrect amount of La3+ from (e)(i) has been used correctly in this calculation.

1

Question Total 9

2. This question is about a sodium street lamp

(a) 1s2, 2s2, 2p6, 3s1 1

(b) 3s 1

(c) (i) Energy = 6.626 × 10−34 J s × 2.998 × 108 m s−1 / 589 × 10−9 m

= 3.37 × 10−19 J (atom−1) 1

(ii) Energy per mole = 3.37 × 10−19 J × 6.02 × 1023 mol−1

= 2.03 × 105 J mol−1

= 203 kJ mol−1

Allow ECF from (c)(i)

1

(d) one Zero infinity the constant k 1

(e) Energy change = I.E. (nd) − I.E. (3p)

Note this answer is negative as energy is given out. Award one mark for the expression: Energy change = I.E. (3p) − I.E. (nd)

2

(f) (i) Intercept = 0.00245 nm−1 (Allow values from 0.00243-0.0247 nm−1)

I.E. (3p) = 6.626 × 10−34 J s × 2.998 × 108 m s−1 × 2.45 × 106 m−1

= 4.87 × 10−19 J (atom−1)

1

(ii) Ionisation energy of sodium = I.E. (3p) + ΔE (3s→3p)

I.E. (3p) = 4.87 × 10−19 J (atom−1) × 6.02 × 1023 mol−1

= 293 kJ mol−1

Ionisation energy of sodium = 293 kJ mol−1 + 203 kJ mol−1

= 496 kJ mol−1

Allow ECF from (c)(ii) or (f)(i), as long as they have shown that these two quantities must be added together, and both in the units of kJ mol−1

1

Question Total 9

3. This question is about spot cream

(a) Effective density of tazarotene = 0.90 g cm−3 × 0.0005 = 0.00045 g cm−3

= 0.45 g dm−3

Concentration of tazarotene = 0.45 g dm−3 / 351.46 g mol−1

= 0.00128 M (or 1.28 mM)

1

(b)

1

1

(c)

1

1

1

1

1

The alkyne C‒H bond does not have to be explicitly drawn in (as in normal skeletal structure drawing).

1

(d) Oxidation Reduction Addition Elimination Substitution 1

(e)

1

(f) 20 signals 1

(g)

1

Question Total 13

4. This question is about bombardier beetles

(a) (i) 2H2O2 → 2H2O + O2

State symbols not required 1

(ii) Oxidation Reduction Disproportionation Hydrolysis Dehydration 1

(b) Combining 2H2O2 → 2H2O + O2 and A + ½O2 → B + H2O gives

H2O2 + A → B + 2H2O 1

(c) (i) Amount of energy = specific heat capacity ×temp. change ×mass of water

= 4.18 J g−1 K−1 × 80 K × 1000 g

= 334 kJ

1

(ii) Conc. of H2O2 in mixed solution = energy needed per litre / enthalpy change per mole of H2O2

= 334 kJ dm−3/ 203 kJ mol−1

= 1.65 mol dm−3

Therefore with equal volumes mixed, conc. of H2O2 initially must be double this value = 3.30 mol dm−3

Award one mark for the value of 1.65 mol dm−3, and one mark for the realisation of the need to double the concentration. Allow ECF from (c)(i).

2

(d) (i) 6 1

(ii) 3 1

(e) Peak I O‒H 1

Peak II C‒H 1

(f) (i) ‒OH (or hydroxyl) 1

(ii) ‒CH3 (or methyl) 1

(g) H1

H2

H3

V

VII

VI 2

All correct: two marks, two correct: one mark, one correct: half a mark

(h)

1

1

(i)

1

1

Question Total 18

5. This question is about fire and ice

(a) CH4 + 2O2 → CO2 + 2H2O

State symbols not required 1

(b) (i) Amount of CH4 = amount of CO2 = amount of CaCO3

Amount of CaCO3 = 84.73 g / 100.09 g mol−1 = 0.847 mol

Amount of H2O = 116.92 g / 18.016 g mol−1 = 6.49 mol

(CH4)x(H2O)y + 2xO2 → xCO2 + (2x+y)H2O

So x = 0.847 mol, and 2x+y = 6.49 mol

Therefore y = 6.49 − (2 × 0.847) = 4.80 mol

Expressing as integers: x = 3, y = 17

Award one mark for correct calculation of the amount of CaCO3 and H2O, one mark for correct algebraic expression or equivalent and one mark for final answer. Correct final values score full credit.

3

(ii) (CH4)3(H2O)17 Mr = 354.40 g mol−1;

therefore 2835.18 g mol−1 / 354.40 g mol−1 = 8

Molecular formula is (CH4)24(H2O)136

1

(c) Amount of CH4 = 6.67 × 1014 g / 16.04 g mol−1 = 4.16 × 1013 mol

V = nRT/p = 4.16 × 1013 mol × 8.31 J K−1 mol−1 × 254 K / 1.0 × 105 Pa

= 8.78 × 1011 m3

1

(d) (i) Mass % of methane in methane hydrate =

(8 × 16.04) g mol−1/ 957.07 g mol−1

= 13.4%

1

(ii) Mass of Baikal methane hydrate = 6.67 × 1014 g / 0.134 = 4.98 × 1015 g

Allow ECF from (d)(i) 1

(iii) Volume of methane hydrate = 4.98 × 1015 g / 0.95 g cm−3

= 5.24 × 1015 cm3 (or 5.24 × 109 m3)

Allow ECF from (d)(ii)

1

(e) (i) Unit cell mass = 957.07 g mol−1 / 6.02×1023 mol−1

= 1.59 × 10−21 g 1

(ii) Volume of unit cell = 1.59 × 10−21 g / 0.95 g cm−3 = 1.67 × 10−21 cm3

Length of unit cell edge = (1.67 × 10−21 cm3)⅓ = 1.19 × 10−7 cm

One mark for correct volume of unit cell. Allow ECF from (e)(i)

1

(iii) Volume of methane in unit cell = 8 × 4/3 × π × (0.21 × 10−9 m)3

= 3.10 × 10−28 m3

Volume of water in unit cell = 46 × 4/3 × π × (0.14 × 10−9 m)3

= 5.29 × 10−28 m3

1

(iv) Percentage of space occupied =

(3.10 × 10−28 + 5.29 × 10−28) m3 / 1.67 × 10−27 m3

= 50%

1

(f) ΔfHo (CH4)8(H2O)46 = 8ΔfH

o(CO2) + 62ΔfHo(H2O) − ΔcH

o (CH4)8(H2O)46

= (8(−393.5) + 62(−285.8) − (−6690.4)) kJ mol−1

= −14177.2 kJ mol−1

Forming methane hydrate from methane and water has the enthalpy change

8CH4 + 46H2O → (CH4)8(H2O)46

ΔrHo = (−14177.2 − 8(−74.8) − 46(−285.8)) kJ mol−1 = −432 kJ mol−1

Final answer scores full marks. One mark for a correct value for ΔfH

o (CH4)8(H2O)46,one mark for the idea of using two cycles and one mark for correct second cycle calculation. If mistake is made in calculation of ΔfH

o (CH4)8(H2O)46 but then this answer is used correctly in the second cycle this should be given two marks overall.

3

Question Total 16

Paper Total 65