2010 GCE ‘A’ Level Chemistry Paper 9647/03

Question 1: 14 marks (most popular)

Question 2: 14 marks (least popular)

Question 3: 11 marks

Question 4: 14 marks

Question 5: 10 marks

Total P3: 49 – 53 / 80 (norm)

Highlighted parts are questions where students did not do well. Refer to Cambridge report for

details on why students did not do well. They should also be the parts that the lecture should

focus on (if there is time constraint.)

1 (a) (i) Entropy of a chemical system measures the degree of disorder in a

system. It gives a measure of the extent to which particles and energy

are distributed within the system.

[1]

(ii) The mixing of Cl2 and N2 will result in an increase in the entropy of

the system as there are now more ways in which the energy can be

distributed among the molecules.

[1]

The heating of Cl2 to a higher temperature means that there are more

ways for the energy to be distributed among the molecules. Thus,

the particles will move more randomly and at higher speeds. Thus,

entropy will increase.

[1]

The chemical reaction between Cl2 and I2 to form ICl3 will result in a

decrease in entropy as there is a reduction in the number of

molecules of gas in a chemical reaction by 1 mol. There are now less

particles moving randomly and less ways to distribute the energy.

[1]

There will be an increase in entropy during photolysis as the number

of particles in gaseous state increase by 1 mol. Each molecule is

split into two chlorine radicals.

[1]

(b) (i) For reaction of hydrogen with chlorine, the reaction is explosive in

sunlight, but slow in the dark below 200 oC.

For reaction of hydrogen with bromine, the reaction is slow at 300 oC

with Pt catalyst.

The rate of reaction for H2 with Cl2 is faster than that between H2 and

Br2.

[1]

(ii) It is expected that the rate of reaction of hydrogen with fluorine will be

higher. The reactivity of the halogen with hydrogen decreases down

the Group as the oxidizing power of the halogens decreases down the

group.

[1]

(c) (i) Cl2 in liquid CCl4, uv light, r.t.p [1]

(ii) Free radical substitution

Initiation step:

Cl2 2 Cl (slow step, uv light)

Propagation step:

Cl + C2H6 C2H5 + HCl

C2H5 + Cl2 C2H5Cl + Cl

Termination step:

2 Cl Cl2

2 C2H5 CH3CH2CH2CH3

C2H5 + Cl C2H5Cl

[1]

[1]

[1]

[1]

(iii) This is because in the first propagation step, in the abstraction of

hydrogen radical from alkane to form HI, this reaction is highly

endothermic. (Bond energy of HI = 299 kJ mol–1 vs Bond energy of

HCl = 436 kJ mol–1) Therefore, it is not possible for the reaction to

occur.

[1]

(d)

A

Cl

B

Cl

Cl OR

Cl

Cl C D

Br

Br

[1]

[1]

[1]

[1]

(e) (i) CFCs are inert, non-toxic OR can be liquefied under pressure and

is easily volatilised when pressure is released

[1]

(ii) CFCs, when reaching the ozone layer, undergoes homolytic fission

between C-Cl bond under the action of uv radiation via photolysis.

This forms Cl radicals which attack the ozone via a radical

mechanism, resulting in a chain reaction of destruction of ozone

molecules.

[1]

(iii) Presence of C-H bonds in alkanes makes it flammable. Thus, alkanes

may undergo combustion at high pressure conditions.

[1]

2 (a) (i) for a weak acid, HA H A where Ka = [ ][ ]

[ ]

H A

HA

pKa = –log Ka

[1]

[1]

(ii)

pH1 HO2C CO2H

NH3+

[1]

pH3HO2C CO2

-

NH3+

OR HO2C CO2H

NH2

[1]

pH7-O2C CO2

-

NH3+

Note: the electron-donating alkyl group on –CO2H reduces the acidic

strength of –CO2H, thus it will be protonated later.

[1]

pH11-O2C CO2

-

NH2

[1]

(b) (i) Quarternary and tertiary structure of a protein will be altered.

This occur because denaturation breaks the weak bonds holding

the secondary, tertiary and quaternary structure, but not the covalent

bonds within primary structure, resulting in a more disordered

arrangement. Such action is normally done by salts, acids or

enzymes. The weak bonds that are broken are the side chain

interactions, which include van der waals’ forces of attraction,

hydrogen bonds, ionic bonds and disulphide bonds.

[1]

[1]

(ii) by metal ions such as Ca2+,

it disrupts the ionic bonds between the charged R groups of the

amino acids by forming new ionic bonds with the carboxylate anions.

[1]

(iii) by weak acid,

it disrupts the ionic bonds holding the tertiary and quaternary

structures.

The weak acid will dissociate to cause carboxylate anions or amino

groups to be protonated.

–CO2– + H+ –CO2H

–NH2 + H+ –NH3+

[1]

[1]

(iv)

2

[ ]

[ ][ ]c

gluconic acidK

GDL H O

Amount of GDL = 1.00

178= 0.005618 mol

Initial [GDL] = 0.005618

50

1000

=0.1124 mol dm–3

0.0670

(0.1124 0.0670)(55.5)

cK =0.0266 mol–1 dm3

[1]

[1]

[1]

(c) (i)

HO O

OOH

Br

BrBr

Br

Br

OH

OR

HO O

OOH

Br

BrBr

Br

Br

Br

[1]+[1]

For type

of rxn

(easy E.S

+ E.A)

(ii) HO O

OHOH

HO O

OOH

Compound F Compound G

Diadzein undergoes reduction of alkene and ketone with H2 and Ni to

form F. A secondary alcohol is formed.

F has three hydroxyl groups, and thus able to react with three moles

of sodium via redox reaction.

F dissolves in NaOH(aq) as it contains phenolic groups which can

undergoes neutralization with NaOH(aq)

F undergoes oxidation with K2Cr2O7 to form a ketone in G.

G undergoes condensation with 2,4-DNPH to form hydrozone.

[1] + [1]

for each

structure

[1] for all

chiral C

[1] for

reasons

3 (a) (i) Cathode: O2(g) + 4H+(aq) + 4e 2 H2O (l) [1]

(ii) Cathode x 3 + Anode x 2

Overall: 2 CH3OH + 3O2 2 CO2 + 4H2O

[1]

[1]

(iii) Eocell = Eo

red – EoCO2/CH3OH

1.18 = 1.23 – EoCO2/CH3OH

EoCO2/CH3OH = +0.05 V

[1]

(iv) Methanol cell is portable, easy storage and less flammable as

methanol is liquid. Thus, it need not be stored under pressure, unlike

hydrogen gas.

[1]

(b) (i) CH3CH2OH(l) + 3O2(g) 2CO2(g) + 3H2O (l) [1]

(ii) ΔG = ΔH – TΔS

= (–1367000) – 298(–140)

= –1339280 J mol–1

= –1340 kJ mol–1

[1]

(iii) Oxidation: CH3CH2OH(l) + 3H2O(l) 2CO2(g) + 12H+(aq) + 12e

cell

GE

zF

z = 12

[1]

[1]

1339280

12(96500)cellE

= +1.16V

(c) (i) ΔH = bond breaking – bond forming

= [(C-C)+5(C-H)+(C-O)+(O-H)+3(O=O)] – [4(C=O)+6(O-H)]

= [(350)+5(410)+(360)+(460)+3(496)] – [4(805)+6(460)]

= –1272 kJ mol–1

[1]

[1]

(ii) The enthalpy change of vapourisation of water and ethanol is not

considered in the calculation for (c)(i)

OR

The bond energies are of average values.

[1]

(d) (i) CxHyOH + Na CxHyO–Na+ + ½ H2

Amount of H2 produced = ½ (Amount of J used)

=10.9

24000= 0.0004542 mol

Amount of J used = 0.0009084 mol

CxHyOH + 1

4

yx

O2 xCO2 +

1

2

y

H2O

Amount of CO2 absorbed by NaOH(aq) = 109

24000= 0.004542 mol

Comparing mole ratio CxHyOH ≡ xCO2

0.0009084 : 0.004542

1 : 5

x = 5

In the combustion of the alcohol,

V(O2) used + V(CO2) produced = –54.4 cm3 (i.e. reduction in volume)

V(O2) used = 109 + 54.4 = 163.4 cm3

Amount of oxygen used =163.4

24000 =0.006808 mol

Comparing mole ratio 1

4

yx

O2 ≡ xCO2

11

20

y

O2 ≡ CO2

0.006808 : 0.004542

1.50 : 1

1.5 = 1

120

y

y = 11

[1]

[1]

(ii) J undergoes oxidation with K2Cr2O7.

J must be either primary or secondary alcohol.

J undergoes elimination to form alkene K

Reasons

[max 2]

K undergoes vigorous oxidation to result in the cleavage of the

alkene bond to form ethanoic acid and propanone.

K

CH3

C

CH3

C

CH3

H J

CH3

CCH3

H

C

CH3

H

OH

[1]+[1]

(iii) Concentrated H2SO4 , reflux at 170 °C [1]

(iv) K will not show geometrical isomerism as there are two methyl

groups attached on the same carbon involved in the C=C bond.

Note: Statements that are not acceptable:

(1) Methyl groups are on the same side of the double bond. (2) There is a carbon with two methyl groups.

[1]

4 (a) Nucleon number is the total number of protons and neutrons in a

nucleus of an atom.

[1]

(b) In an electric field, the electron will be deflected to the positive

plate, while the proton will be deflected to the negative plate.

In an electric field, the angle of deflection for an electron is larger

than that of the proton as it is lighter than the proton.

[1]

(c) (i)

NOX

XX

X

X

XX

X

XX O X

X

X

XXX

XX OO O

NO2 O3

[1]+[1]

(ii) For O3, there are 2 bond pairs and 1 lone pair. The 3 electron pairs will

point towards the corners of an equilateral triangle. The lone pair-

bond pair repulsion > bond pair-bond pair repulsion because the

lone pair is closer to the oxygen than a bond pair. Thus, bond

angle is 115° with the shape being bent.

[1]

For NO2, there are 2 bond pairs and 1 unparied electron. The 3

electron pairs will point towards the corners of an equilateral triangle.

The unpaired electron-bond pair repulsion is weaker than that of

the lone pair-bond pair repulsion in O3. Thus, bond angle is (any

value between O3 and 170°) with the shape being bent.

[1]

(iii) F, being a period 2 element, is not able to expand to accommodate

more than 8 valence electrons. To form FO2, F will need to form two

dative bonds. It would not do this as F is too electronegative for

dative bonding.

[1]

Cl, being a period 3 element, is able to use its 3d orbitals to

accommodate more than 8 valence electrons through the use of

energetically accessible 3d orbitals.

[1]

(d) (i) A sample of calcium is being held on a platinum wire that has been

repeatedly cleaned with hydrochloric acid to remove traces of previous

analytes. The sample is then burnt in a hot, non-luminous flame in the

presence of oxygen and the colour is being observed.

[1]

A brick red flame colour is observed. [1]

(ii) Oxides of Group II elements will react with water to form hydroxides.

As the Group II metal hydroxides goes down the group from

magnexium to barium, the solubility of the hydroxides increases.

This increases the extent of dissociation of the hydroxides, thus

resulting in the increase in the concentration of hydroxide ions,

and therefore the increase of the pH value from Mg to Ba.

[1]

(e) Ba(O3)2 + H2O

5

2 O2 + Ba(OH)2

[1]

(f) (i) Amount of Na2S2O3 =

150.100 0.001500

1000mol

Amount of I2 = 0.000750 mol

[1]

(ii) Amount of O3 = 0.000750 mol

Volume of O3 = 0.000750 x 22.4 = 0.0168 dm3

% O3 = 16.8

100% 3.36%500

[1]

[1]

(g) (i) For production of aldehyde, use acidified K2Cr2O7 and heat/reflux

with distillation.

[1]

For production of carboxylic acid, use acidified K2Cr2O7 and

heat/reflux

[1]

(ii)

L

From molecular formula, compound is unsaturated. To be able to react

with Br2(aq) indicates that a double bond is present, capable of

undergoing electrophilic addition.

M

Molecular formula is the same as L, indicating that it must contains a

cycloalkane in the compound as it does not react with Br2(aq),

indicating the absence of an alkene bond.

N

C

OH

CH3H

CH3

N must undergo oxidation with alkaline aqueous iodine, thus having

[1]

[1]

[1]

the structure

C

OH

CH3

H

5 (a)

In the electrochemical purification of copper

Anode: Crude copper

Cathode: Pure copper

Electrolyte used: Copper (II) sulphate

From the Data Booklet:

Ag+ + 2e ⇌ Ag +0.80V

Cu2+ + 2e ⇌ Cu +0.34V

2H2O + 2e ⇌ H2 + 2OH– +1.23 V

Zn2+ + 2e ⇌ Zn –0.76V

When an electric current is applied, both zinc and copper are oxidised

to Zn2+ and Cu2+ respectively since Eθ of Zn2+/Zn is less positive than

the Eθ of Cu2+/Cu.

Cu(s) Cu2+(aq) + 2e [including state symbols]

Zn(s) Zn2+(aq) + 2e

Silver, due to its more positive (E) reduction potential will not be

oxidised.

Silver will drop to the electrolytic bed as elemental silver.

Zn2+ and Cu2+ migrate to the cathode.

Only Cu2+ is reduced to copper due to its more favourable reduction

potential and higher concentration.

Cu2+(aq) + 2e Cu(s) [including state symbols]

[1] for

copper

[1] for

zinc

[1] for

silver

[1] for

quoting

E values

Zn2+ remains in the solution.

(b) When NH3 is first added to Cu2+ (aq):

NH3 + H2O NH4+ + OH-

Cu2+ (aq) + 2 OH- (aq) Cu(OH)2 (s) --- (1)

blue ppt

Ionic product, [Cu2+][OH-]2 > Ksp of Cu(OH)2

Blue ppt of Cu(OH)2 is formed.

When excess NH3 is added to Cu2+ (aq):

[Cu(H2O)6]2+(aq) + 4NH3(aq) [Cu(H2O)2(NH3)4]

2+(aq) + 4H2O(l) (2)

(excess)

NH3 ligands replaces the H2O ligands, forming a more stable deep

blue [Cu(NH3)4(H2O)2]2+ complex with Cu2+(aq).

[Cu2+] decreases as it is being used to form the complex

equilibrium position in (1) shifts left to increase [Cu2+]

Ionic product, [Cu2+] [OH-]2 < Ksp of Cu(OH)2

Pale blue ppt,Cu(OH)2, dissolves.

[1]

[1]

[1]

[1]

(c) O2- + NH3 NH2- + OH-

The colour of the solution is dark blue in colour.

[1]

[1]

(d) [Cu(H2O)6]2+ + 4Cl– [CuCl4]

2– (aq) + 6 H2O (l)

Ligand exchange reaction has occurred.

[Cl-] increases in concentrated hydrochloric acid, equilibrium

position will shift right to form yellow [CuCl4]2- complex.

Presences of both Cu2+ (aq) and [CuCl4]2- in the equilibrium mixture

results in a yellow-green solution.

[1]

Solution remains blue when concentrated sulfuric acid is added as the

sulfate ions are weaker ligands.

[1]

Dilution of the yellow-green solution will reduce the [Cl–], hence by

LCP, the position of the equilibrium will shift to the left, producing

the original pale blue colour.

[1]

(e) (i) 4I– + 2Cu2+ I2 + 2CuI [1]

(ii) Ecell = –0.39 V

Reaction is thus not feasible.

[1]

(iii) CuI is insoluble in aqueous solutions.

Cu2+ + e Cu+

Thus [Cu+] will be very low. Therefore, by LCP, the position of

equilibrium will shift towards the production of CuI, hence resulting

in the favouring of the forward reaction.

[1]

[1]

(f) (i) Pale blue solution will decolourised, and a brick-red precipitate

will be produced.

[1]

(ii) Positive result:

C

O

H

CC

H

C

H

C

H

HH

H

H

H

H

C

O

H

CC

H

C

C

H

H

HH

H H

HH

Negative result:

CCC

H

C

H

H

H

HH

H

O

C

H

H

H

[1]

[1]