NOV2005 P3

Suggested Answers and Comments N05/III/1 1 (a)(i) Rate = k[NO]2[O2]

Comments: Note that N2O2 is an intermediate and it cannot appear in the rate equation. The following is using sequential logic to derive the rate equation. Since (A) 2NO → N2O2 (fast) (B) N2O2 + O2 → 2NO2 (slow) Substituting equation (A) into (B) (since N2O2 comes from 2NO), thus, (C) (2NO) + O2 → 2NO2 Thus, it appears that the slow step constains two molecules of NO and 1 molecule of O2. Hence rate equation is rate = k[NO]2[O2]

(a)(ii) Rate constant: The rate constant is represented by the symbol ‘k’ in the rate equation in (a)(i) and it is affected by temperature and size of activation energy, and it is independent of the concentration of the reactants. Order of reaction: It is given with respect to a given reactant and it is defined as the power to which the concentration of that reactant raised to in the experimentally determined rate equation. Using the above rate equation, the order of the above reaction would be 2 + 1 = 3. Rate determining step: If the reaction progresses in a series of steps, the slowest step will be the rate determining step. Comments: Instead of explaining these terms from scratch, it is easier to explain them with the aid of a rate equation, which you have given in (a)(i).

(b)(i)

Ea = + 79 kJ mol-1

Ho = - 98 kJ mol-1

Enthalpy/ kJ mol-1

Reaction coordinate / progress of reaction

H2O2

H2O + 1/2O2

Transition state

Comments: All the necessary details in this question are illustrated by the above diagram. In addition, they are also bolded.

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(b)(ii) Activation energy is the minimum amount of energy the reactants need to possess to enable a chemical reaction to occur. Comment: Activation energy isn’t the amount of energy to “kick start” a reaction. To ‘kick start’ a reaction implies that it is the factor which initiates the reaction and then the reaction will just run by itself. Activation energy is the one of the criteria that causes a reaction to run by it self. Thus, it doesn’t “kick start” the reaction.

(b)(iii) The catalyst participates in the reaction by looking for another reaction pathway which has a lower activation barrier. This results in more number of reactants, at the same temperature, to have possessed the minimum energy for reaction to proceed. The number of effective collisions increase and hence rate increases. Comments: A catalysed reaction is faster not only because the energy barrier of the reaction is lowered. More importantly, more molecules possess that minimum energy requirement for the reaction to proceed.

(b)(iv) The rate constant will increase. Since, using the catalyst the reaction rate becomes faster, despite the concentration of reactants remains the same. Hence, using the rate equation, the rate constant increases.

In addition, with reference to the Arrhenius equation: RTEa

Aek−

= , the smaller the Ea, the larger the k value. Comments: Having the knowledge of Arrhenius equation will conveniently assist you in answering this question. If you cannot remember this equation, you can use the rate equation and logically deduce that the rate constant would increase when Ea decreases.

N05/III/2 2 (a)(i) Formula Mass of NaClO3 = 23 + 35.5 + 3(16) = 106.5

n(NaClO3) = 1 x 106/106.5 = 9389 ≈ 9390 moles Since ClO3

- ≡ 6 e- n(e-) = 6 x 9389 = 56338 ≈ 56 300 moles Q = It 56 338 x 96500 = I x (24 x 60 x 60) I = 6.29 x 104 A Comments: Students need to remember to do convert time from day to seconds. A step wise manner to calculate this answer is needed for coherence of answer.

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(a)(ii) Since current passing through the cathode is the same as the anode. n(e-) = 56 338 moles H2 ≡ 2 e- n(H2) = 0.5 x 56 338 = 28 169 ≈ 28 200 moles volume of H2 = 28 169 x 24 ≈ 6.76 x 105 dm3

Comments: The same amount of current passing through the cathode and anode. Therefore, this enables them to know the amount of electrons flowing from cathode and anode. Thus enable the calculation of the volume of H2 produced. In addition, do remember that at room temperature pressure, which is 25 oC and 1 atm, the molar volume of gas is 24 dm3. It is at 273 K and 1 atm, where the molar volume of the gas is 22.4 dm3.

(b)(i) Sr(IO3)2 (s) Sr2+(aq) + 2IO3-(aq)

Ksp = [Sr2+][IO3

-]2 Units: mol3 dm-9 Comments: Calculation of units: Units of concentration = mol dm-3 Units of Ksp = (mol dm-3)(mol dm-3)2 = (mol dm-3)(mol2 dm-6) = mol3 dm-9

The Kc expression is not suitable because it includes a concentration of Sr(IO3)2 solid, which is a constant. Thus, re-writing the expression would give the Ksp expression.

(b)(ii) Let solubility of Sr(IO3)2 be x Therefore, this implies that [Sr(IO3)2 (aq)] = x. [Sr2+] = x, [IO3

-] = 2x 1.1 x 10-9 = x(2x)2 = 4x3 => x = [Sr2+] = 6.5 x 10-9 mol dm-3 Comments: I have obtain the concentration of [Sr2+] and [IO3

-] by the following:

Sr(IO3)2 Sr2+ 2IO3

i / mol dm-3

e / mol dm-3 2x

+(s) (aq) (aq)

0 0

x

Where x is solubility of Sr(IO3)2

Note 1: All aqueous ionic compounds will fully dissociate into its component ions.

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N05/III/3 Either 3E (a) Lattice energy is the energy evolved when 1 mole of solid ionic

compound is formed from its respective component gaseous ions when the ions are infinitely apart. Comments: This question calls for the definition of lattice energy. Do note that it is for 1 mole of ionic compound formed. Do not be confused with using the following equation and answer:

−+

−+

+∝

rrqqLE

Lattice energy measures the strength of the ionic bond between opposite charge ions. The larger the charges (represented by the letter q) of the respective ions and smaller the radius (represented by the letter r) of the respective ions, the stronger the ionic bond. This is the answer when you are asked about the factors affecting LE.

(b)(i) Mg2+ (g) + O2-(g) → MgO (s) Comments: Question implies a chemical equation. Take note of state symbols as well, since it is an equation for a thermochemistry question.

(b)(ii) Mg (s) + ½O2 (g) → MgO (s) Comments: Take note of the state symbols of the reactants and products. Do remember that enthalpy change of formation is the heat change due to the formation of 1 mole of MgO.

(c)

LEMg2+ (g) + O2-(g) MgO (s)

LE = ∆Hf (MgO) - ∆Hatm (Mg) - ½ Bond Energy (O=O) – (1st + 2nd IE) – (1st + 2nd Ea) = (-602) – (+148) - ½(+496) – (+736 + (+1450)) – (-140 + (+798)) = - 3840 kJ mol-1

Mg (s) + ½O2 (g)

Mg (g) + O (g)

2nd Electron Affinity of O

Mg+ (g) + O-(g)

½ Bond Energy (O=O) ∆Hatm (Mg)

1st Electron Affinity of O

2nd Ionisation Energy of Mg

∆Hf (MgO) 1st Ionisation Energy of Mg

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Comments: Note that you are not given the ∆Hatm (O). But this is equivalent to the ½ Bond Energy (O=O). Do note that O2 exists as a double bond linking two O atoms.

(d) Ca2+ and Mg2+ has the same charge, but Ca2+ has a larger radius than Mg2+ Using the relationship below,

−+

−+

+∝

rrqqLE

Thus, the larger the cation the smaller the lattice energy. Hence, MgO as a larger lattice energy than CaO. Comments: Do note that the correct comparison is to compare the ionic radius and NOT the atomic radius.

(e) MgO is being used as a furnace lining due to its high melting point. Comments: You are to use the effect of high lattice energy of Group II oxides on their physical property. The only logical physical property that can be related to lattice energy is melting point. You shouldn’t be discussing about chemical properties, as it doesn’t relate to the lattice energy. Neither should you be talking about ease of thermal decomposition as it is not one of a relevant uses of Group II oxide.

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N05/III/3 Or 3O (a) The energy required to break one mole of a given bond into it

component gaseous atoms/molecules. Comments: It is important to stress that it is the energy which breaks 1 mole of a given bond. It is not the energy needed to convert 1 mole of compound into its component gaseous atoms.

(b)(i) C8H18 (l) + 12½O2 (g) → 8CO2 (g) + 9H2O (l) Comments: Generally: Alkanes with 1 – 4 carbons are gases in room temperature. While alkanes with 5 – 10 carbons are liquid.

(b)(ii) C8H18 + 12½O2 → 8CO2 + 9H2O Bond broken: Bonds formed: 7 C – C 16 C = O 18 C – H 18 H – O 12½ O = O ∆HC (C8H18) = 7(+350) + 18(+410) + 12½(+496) – 16(+740) – 18(+460) = (+16030) – (+20120) = - 4090 kJ mol-1 Comments: Remember that enthalpy change of combustion is the heat evolved when one mole of substance is completely burnt. Note 1: Formation of CO is due to incomplete combustion! Note 2: Bond energy for O=O is represented by this equation: O=O (g) → 2O (g). The reactants are in gaseous state, since you only want to break the covalent bond and not the IMF. Note 3: Using bond energies to calculate gives inaccurate enthalpy change because 1) they are averages and 2) the heat required to change the reactants to gaseous state is not accounted for.

(c)(i) 1) The bond energies used in the data booklet are averages. Hence, it results in the difference in the calculated value in (b)(ii) and the table in (c). 2) The ∆HC of the experiment data refer to enthalpy change of combustion at standard conditions, which implies that H2O exists as a liquid. However, the use of bond energies to calculate enthalpy change of combustion, we have taken H2O to be a gas. Thus, the energy released due to the formation of strong H-bonding in water is not accounted for. 3) The C = O bond strength in CO2 is stronger than the C = O that is quoted in the data booklet. Choose any one of the three as answer.

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Comments: 1) The bond energies in the data booklet are average. The strength of a covalent bond is also dependent on the type of atom that surrounds the molecule. Hence, we would expect the C – H bond CH4 to differ from C2H4. But this is not reflected in your data booklet. 2) To obtain the bond energy of O – H from H2O, H2O must be in gaseous state. If liquid H2O is used, additional energy would be required to overcome the hydrogen bonding between H2O molecules. 3) If the C = O bond in CO2 is stronger, it would explain for why the ∆HC in (c) is more exothermic than your calculation in (b)(ii) 4) I have left out accounting for heat required to vaporize octane because the bond energy calculation is less exothermic than the enthalpy change of combustion given by the question.

(c)(ii) The increase in ∆HC from C7H16 to C8H18 to C9H20 is regular. Approximately, the increase is by a magnitude of 654. Hence, this value represents the ∆HC of a CH2 unit.

(d)

In 1.0 dm3 of ethanol: Mass of ethanol = 0.79 x 1000 = 790g n(ethanol) = 790/(2(12) + 6(1) + 16) = 17.2 moles Heat produced by complete combustion of 1.0 dm3 of ethanol = 17.2 x 1367 ≈ 23. 5 x 103 kJ (3 S F) In 1.0 dm3 of octane: Mass of octane = 0.70 x 1000 = 700g n(Octane) = 700 / (8(12) + 18(1)) = 6.14 moles Heat produced by complete combustion of 1.0 dm3 of octane = 6.14 x 5470 ≈ 33.6 x 103 kJ (3 S F) Comments: Care is required when doing conversion of units. The units of density is given in g cm-3, whereas volume is given in dm3. Take note that the enthalpy change of combustion is the combustion of one mole of organic substance (C8H18 and CH3CH2OH)

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N05/III/4 4 (a) Elements which forms an ion or compounds with a partially-filled d-

orbital. Comments: Students may be tempted to describe the physical properties of transition elements as a means to define transition elements. Hence, to state that transition elements merely have variable oxidation states and/or colours are not suitable definitions. For example, lead has 3 different oxidation states, 0, 2 and 4, but lead is not a transition element. In addition, I2, Br2, and Cl2 all exhibit colours but neither are any of them a transition element.

(b)(i) Cr: 1s22s22p63s23p63d54s1 Comments: Despite 4s is being filled before 3d, do remember that Cr prefers the half filled d orbitals instead of having 3d44s2.

(b)(ii) Cr3+: 1s22s22p63s23p63d3 Comments: 4s electrons are removed before the 3d. Hence, 1 electron is removed from 4s and 2 are removed from 3d. Therefore, providing the above electronic configuration.

(c)(i) The following are the steps: 1. Dissolve the tablet in a 150 cm3 of HCl solution. 2. Pipette out 25.0 cm3 of the solution created in (1) into a conical flask. 3. Add one or two drops of the indicator into the conical flask. 4. Fill the burette with the standard solution of K2Cr2O7. Titrate it until the solution changes from red to blue. 5. Repeat the titration steps for unit consistent readings are obtained. 6. Calculate the average volume of K2Cr2O7 used. 7. Using this equation Cr2O7

2- + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O Calculate the number of moles of Fe2+ present in 25 cm3 of solution. 8. Calculate the number of moles of Fe2+ that is originally present in the

150 cm3 solution. Comments: You have to ensure that the solution is acidified as potassium dichromate acts as an oxidising agent in acidified medium. Remember the titration has to be repeated so as to obtain a set of consistent readings.

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(c)(ii) Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O

n(Cr2O7

2-) used = (10.50/1000) x 0.025 = 2.625 x 10-4 moles n(Fe2+) present = 2.625 x 10-4 x 6 = 1.575 x 10-4 moles n(Fe) = n(Fe2+) = 1.575 x 10-4 moles Mass of Fe present = 1.575 x 10-4 x 55.8 ≈ 0.0879g (3 S F)

N05/III/5 Either 5E (a) When concentrated H2SO4 is added to:

KCl, white fumes appears. KBr, brown fumes appears. KI, purple fumes appears. Down the group, Cl2, Br2 and I2, the reducing strength decreases. Hence, this results in I- to be most easily oxidised and Cl- the least. As a result, concentrated H2SO4 is strong enough to oxidised I- to I2. It is marginally strong enough to oxidise Br- to Br2, but HBr still remains. But HBr is white in colour, hence we still see the brown fumes. While, it is unable to oxidised Cl-, thus is only able to protonate Cl- to give HCl. KCl + H2SO4 → HCl + KHSO4 KBr + H2SO4 → HBr + KHSO4 2HBr + H2SO4 → Br2 + SO2 + 2H2O KI + H2SO4 → HI + KHSO4 8HI + H2SO4 → 4I2 + H2S + 4H2O Comments: 1) Do remember the trend of ease of oxidation. When a species is more willing to be reduced, it is more difficult to be oxidised. 2) Ease of oxidation is not the same as thermal stability. 3) Do note that in these set of equations, in order to oxidise the halide ions, they are first converted into acid halide first. 4) You need to remember the equation to show the oxidation of the hydrogen halides. YOUR DATA BOOKLET IS UNABLE TO HELP YOU WITH THIS ONE!

(b)(i) Na Br O % mass 15.2 53.0 31.8 % mass/Ar 15.2/23 = 0.66 53/79.9 = 0.66 31.8/16 =1.98 Ratio 1 1 3 Formula of salt: NaBrO3 Comments: For ionic compounds, their formula is an empirical formula. Hence, this calculation would be sufficient.

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(b)(ii) Let oxidation number of Br be x (+1) + (x) + 3(-2) = 0 x – 5 = 0 x = +5 Thus oxidation number of Br is +5. Comments: Oxidation number must have a positive or negative sign which precedes a number.

(b)(iii) 3Cl2 + NaBr + 6NaOH → NaBrO3 + 6NaCl + 3H2O Comments: Workings: Cl2 is passed into alkaline Br-. NaBrO3 is formed, hence Br- is oxidised while Cl2 is reduced. By writing the respective half equation: Equation 1: Cl2 + 2e → 2Cl- Equation 2: Br- +6OH- → BrO3

- + 6e + 3H2O (Note: Oxidation using OH- because it is in alkaline condition) Equation 2 + (3 x equation 1) = answer.

N05/III/5 Or 5O (a) The ease of the thermal decomposition of Group II nitrates decrease

down the group. This is because the down the group, the M2+ ion gets larger and hence it becomes less polarizing. Therefore, the larger NO3

- becomes less polarised by the cation. Hence, when the anion is less polarised, the ease of decomposition to form the metal oxide decreases. Comments: There are three marks to this question and there are three components to this question, 1) Describe, 2) Explain and 3) Equation. Another reason for the ease of decomposition to decrease, would be that the formation of MO is not as stable as we progress down the group as cation ion size becomes larger and lattice energy decreases. Hence, this results in the decomposition reaction to become less exothermic. Therefore, the incentive for M(NO3)2 is decompose to the corresponding MO decreases.

(b) MgCl2 (s) → Mg2+ (aq) + 2Cl- (aq) MgCl is an ionic compound. Thus, MgCl2 dissolves in water to produce the respective hydrated ions. Mg, being a relative small cation with a relatively high charge, is slightly polarising. It polarises a water molecule and causes the water to dissociate, releasing H+ ion. Therefore, partial hydrolysis occurs and hence MgCl2 dissolves in water to give a slightly acidic solution of

M(NO3)2(s) MO(s) + 2NO2(g) + ½O2(g) ∆

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pH > 6.5. The way SiCl4 reacts with water is different from how MgCl2 reacts with water. This is because SiCl4 is a covalent compound and exists as a simple discrete molecule. Si being a period 3 element has available d-orbital to accept lone pair of electrons from water to cause hydrolysis. Therefore, the hydrolysis of SiCl4 produces HCl to appear (as seen in the equation below), and since SiCl4 undergoes full hydrolysis hence creating a strongly acidic solution. SiCl4 + 2H2O → SiO2 + 4HCl Comments: It is important to link the structure of the two different chloride and account for the difference in observation when both react with water. Some common misconception is to assume that SiCl4 is a giant covalent structure (on the pretext that SiO2). This is not true. A giant covalent structure of SiO2 would not be able to interact with water, and hence if SiCl4 possess such a structure, it will not be able to undergo hydrolysis in water. Note that SiCl4 produces a more acidic solution because it undergoes full hydrolysis.

(c)(i) Ba O % mass 81.1 18.9 % mass/Ar 81.1/137 = 0.59 18.9/16 = 1.18 Ratio 1 1.18/0.59 = 2 Formula of the oxide: BaO2 Comments: Please show by means of calculation to show the formula. Since, this is an ionic compound obtaining formula by this means will be suffice.

(c)(ii) BaO2 + 2H2O → Ba(OH)2 + H2O2 A solution of Ba(OH)2 and H2O2 is obtained. This solution is acidified before KI being added. Since I- will be oxidised to I2 by the acidified solution, thus, the solution contains an oxidising agent. Therefore, the oxidising agent is H2O2.

H2O2 + 2H+ + I- → 2H2O + I2 The acidification is done to enable reduction of H2O2 which works in an acid medium. In addition, it is done to remove Ba(OH)2. Comments: Since BaO2 is added to water to give Ba(OH)2 and another compound. You should write the equation and balance the elements. In the identity of the other compound would be H2O2, which can act as either an oxidising agent or reducing agent. Since adding KI produces I2, it implies that H2O2 functions as an oxidising agent in this case.

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N2005/III/8 either

8E (a)(i) Cracking is a process where a larger hydrocarbon is broken into smaller fragments: Breaking of a larger alkane molecule into smaller alkane or alkene fragments. Comments: Please remember such definition questions. Other definition questions include “relative atomic, isotopic, molecular mass” and “strong (or weak) acid”. You might be call to define or explain jargon (chemistry terms) which you have taken for granted its meaning.

(a)(ii) Heat and catalyst such as Pt is being used. Comments:

(a)(iii) C10H22 → C2H4 + C8H18 Comments: You need to show a smaller molecule is being produced from the larger molecule. Therefore, ensure that the number of C and H is balanced.

(a)(iv) Large molecules in crude oil can be converted to more usefully smaller molecules such as ethene, which can be used to produce polymer and various other useful products. Comments: Usually students tend to forget about the purpose of a reaction, hence they will face difficulties in answering this question. Therefore, it is quite important to remember a few reactions which are applicable to industries, e.g. cracking, Haber Process, Contact Process.

(b) Electrophilc addition

C C

H

H H

H

Br Brd+ d-

slowC C

H

H H

H

Br+

Br-

fastC C

H

H H

H

Br Br

Comments: This is an electrophilic addition reaction. The electrophile is the d+ on Br-Br molecule. This d+ is generated as the Br2 molecules moves towards the double bond. Once the d+ is generated, the π electrons would be attracted to Br with the d+, therefore causing the flow of electrons pairs as denoted by the curly arrows. It is important to take note of the partial charges. They are crucial for the electrophilic addition reaction. Besides showing the carbocation as the intermediate (product of the slow step), you could also show the bromonium ion.

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(c)(i)

C C

CH3

CH3 CH3

CH3

H OH

Comments: The reagents used are similar to saying an alkene into steam and H3PO4 at 300 oC and 70 atm. Therefore, it is the formation of an alcohol. To form a diol, you would have to use cold, dilute alkaline KMnO4.

(c)(ii)

C C

H

C C

H

H H

CC

H

H

H

H

O

OH

O

OH

Comments: The reagent used for this question is that for vigorous oxidation. Thus, there is a cleavage of the C=C. Since, the alkene fragments are both in the generic form of RHC=, where R = rest of the molecule, the double bond would be cleaved to form a carboxylic acid (-COOH). Since both alkene fragment’s R is linked (we are dealing with a cyclohexene), therefore, we will only obtain one product. This is a very useful reaction to remember. Not only you have only one product formed, your final product contains the same number of carbon atoms as your reactants.

N2005/III/8 or 8O (a)(i) PBr3

Comments: Nucleophilic substitution reaction has occurred.

(a)(ii) K2Cr2O7, heat and distill. Comments: Do not use K2Cr2O7 and heat under reflux. This is because the alcohol will be fully oxidsed to give carboxylic acid. (Note: aldehyde can be oxidised to form carboxylic acids). Hence, slight warming would be good enough and distillation will help to take out the aldehyde. Note: Aldehyde has a lower boiling point than alcohol. Hence, using distillation will enable the aldehyde to be removed from an aqueous solution containing the excess alcohol and oxidising agent.

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(a)(iii) Concentrated H2SO4, 170 oC. Comments: Elimination reaction has occurred.

(b) Nucleophilic substitution

C

CH3

H

HBr

d+ d-

NCxx

C

CH3

H

H

BrNCd-d- d+

C

CH3

H

HNC

d+d- + Br x

x

transition state

Comments: Remember than in the cyanide ion, the lone pair of electron lies on the carbon atom. In addition, the charge on the ion is also on the carbon atom. Ensure that the charges are properly shown. Include the partial charges as they are required for nucleophilic substitution.

(c)(i)

C C

CH3

CH3 H

H Comments: Using ethanolic OH-, we will obtain the alkene. The usage of aqueous OH-, we will have nucleophilic substitution of Br for OH. Extra information: • Using aqueous hydroxide will only produce the alcohol. • Using ethanolic hydroxide will generally produce the alkene but you

may get the alcohol or even a nucleophilic substitution of Br with the solvent (ethanol).

The reason to why aqueous medium only produce one product is because elimination reaction requires a much higher Ea and hence doesn’t occur, we use OH- in ethanol because that is more reactive that OH- in water. Having said all that the main point of questions like (c)(i), we actually state the MAIN product. Therefore, the formation of alkene is the answer.

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(c)(ii)

C CNCH3

H

CH3 K

C CCH3

H

CH3

O

OCH2CH3

L

Comments: In the second step, ethanol and concentrated H2SO4 is added. Therefore, this is an esterification step.

(c)(iii)

C CH3CH2CHN

H

H

H

H

NHCH2CH3

M

Comments: There are two C – Br bonds. Both would undergo nucleophilic substitution with CH3CH2NH2. Since excess CH3CH3NH2 is used, it will ensure that the two Br are being substituted. In addition, excess amines would prevent multiple substitutions on N (thus prevent the formation of 2o and 3o amines). Note that amines are more nucleophilic than NH3 because of alkyl group which are electron donating. Hence, 2o and 3o would be even more nucleophilic.