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)
(ii)
(b)
(i)
(ii)
(c)
(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. The mixing of Cl 2 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. The heating of C l 2 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. increase . The chemical reaction between C l2 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. 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. 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 H 2 and Br2. It is expected that the rate of reaction of hydrogen with fluorine will be higher. higher. The reactivity of the halogen with hydrogen decreases down the Group as the oxidizing power of the halogens decreases down the group. light, r.t.p Cl2 in liquid CCl 4, uv light,
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
(ii)
(iii)
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 This is because in the first propagation step, in the abstraction of hydrogen radical from alkane to form H I, this reaction is highly endothermic. (Bond energy of H I = 299 kJ mol –1 vs Bond energy of HCl = 436 kJ mol –1) Therefore, it is not possible for the reaction to occur.
(d) Cl Cl
(e)
(i) (ii)
(iii) 2
(a)
(i)
Br
Cl
Cl A B OR Cl C D Br CFCs are inert, non-toxic OR can be liquefied under pressure and is easily volatilised when pressure is released 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. Presence of C-H bonds in alkanes makes it flammable. Thus, alkanes may undergo combustion at high pressure conditions. H A where Ka = [ H ][ A ]
for a weak acid, HA
[1] [1] [1]
[1]
[1]
[1] [1] [1] [1]
[1] [1]
[1] [1]
[ HA]
[1]
pKa = –log Ka (ii)
[1]
+
NH3
pH1
HO2C
CO2H +
NH3
pH3
HO2C
NH2 -
CO2
OR
HO2C
CO2H
[1]
[1]
+
NH3 -
-
O2C
CO2
pH7 Note: the electron-donating alkyl group on –CO2H reduces the acidic strength of –CO2H, thus it will be protonated later. [1]
NH2 -
O C
(b)
CO
-
2 pH11 2 (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. (ii) by metal ions such as Ca 2+, it disrupts the ionic bonds between the charged R groups of the amino acids by forming new ionic bonds with the carboxylate anions. (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 (iv) [ gluconic acid ] K c [GDL][ H 2O]
Amount of GDL = Initial [GDL] =
1.00 178
[1] [1]
[1]
[1] [1]
[1]
= 0.005618 mol
0.005618 =0.1124 mol dm –3 50
[1]
1000
K c
0.0670 (0.1124 0.0670)(55.5)
=0.0266 mol –1 dm3
[1]
(c)
(i)
Br
Br HO
O
HO
Br
O
Br Br
Br
Br
Br
Br
OH
O
O
OH
OH Br
(ii)
HO
Br
OR
O
HO
OH
3
(a)
(i) (ii) (iii)
(iv)
(b)
(i)
O OH
Compound G
Diadzein undergoes reduction of alkene and ketone with H 2 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. Cathode: O2(g) + 4H+(aq) + 4e 2 H2O (l ) Cathode x 3 + Anode x 2 Overall: 2 CH3OH + 3O2 2 CO2 + 4H2O o o o E cell = E red – E CO2/CH3OH 1.18 = 1.23 – EoCO2/CH3OH EoCO2/CH3OH = +0.05 V Methanol cell is portable, easy storage and less flammable as methanol is liquid. Thus, it need not be stored under pressure, unlike hydrogen gas. CH3CH2OH(l ) + 3O2(g) 2CO2(g) + 3H2O (l )
(ii) ΔG
(iii)
[1] + [1] for each structure
O
OH
Compound F
[1]+[1] For type of rxn (easy E.S + E.A)
= ΔH – TΔS
[1] for all chiral C [1] for reasons
[1] [1] [1]
[1] [1]
[1] [1]
= ( –1367000) – 298( –140) = –1339280 J mol –1 = –1340 kJ mol –1 Oxidation: CH3CH2OH(l ) + 3H2O(l ) 2CO2(g) + 12H+(aq) + 12e
E
cell
z = 12
G zF
[1] [1]
E (c)
(i)
(ii)
(d)
(i)
cell
1339280 12(96500)
= +1.16V
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 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. CxHyOH + Na CxHyO –Na+ + ½ H2 Amount of H 2 produced = ½ (Amount of J used) 10.9 = = 0.0004542 mol 24000 Amount of J used = 0.0009084 mol
CxHyOH + x
[1] [1] [1]
y 1
y 1 H2O O2 xCO2 + 4 2
Amount of CO2 absorbed by NaOH(aq)
=
109 24000
= 0.004542 mol
Comparing mole ratio C xHyOH ≡ xCO2 0.0009084 : 0.004542 1:5 x=5
[1]
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 cm 3 163.4 Amount of oxygen used = =0.006808 mol 24000
Comparing mole ratio x
y 1
O2 ≡ xCO2
4
y 1 1 O2 ≡ CO2 20 0.006808 : 0.004542 1.50 : 1 1.5
(ii)
= 1
y 1
20
y = 11 J undergoes oxidation with K2Cr2O7. J must be either primary or secondary alcohol. J undergoes elimination to form alkene K
[1] Reasons [max 2]
K undergoes vigorous oxidation to result in the cleavage of the alkene bond to form ethanoic acid and propanone. CH3
CH3 CH3
CH3 C
C
CH3
C
C
OH
H
CH
H H 3 K J (iii) Concentrated H2SO4 , reflux at 170 °C (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. Nucleon number is the total number of protons and neutrons in a nucleus of an atom. 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.
4 (a) (b)
(c)
(i)
O N O X
XX
(ii)
(iii)
(d)
(i)
XX
XX X X
X
X X
XX
X
O OXO X X
NO2 O3 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 pairbond 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. 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 O 3 and 170°) with the shape being bent. F, being a period 2 element, is not able to expand to accommodate more than 8 valence electrons. To form FO 2, F will need to form two dative bonds. It would not do this as F is too electronegative for dative bonding. 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. 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]+[1] [1] [1]
[1] [1]
[1]+[1]
[1]
[1]
[1]
[1]
[1]
(ii)
(e) (f)
A brick red flame colour is observed. 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.
(i)
(ii)
(i)
2
Amount of Na2S2O3 Amount of I 2 Amount of O 3 Volume of O3 % O3
(g)
5
Ba(O3)2 + H2O
O2 + Ba(OH)2 =
15
0.100 0.001500mol
1000 = 0.000750 mol = 0.000750 mol = 0.000750 x 22.4 = 0.0168 dm3 16.8 = 100% 3.36% 500
For production of aldehyde, use acidified K2Cr2O7 and heat/reflux with distillation. For production of carboxylic acid, use acidified K2Cr2O7 and heat / reflux
(ii)
[1] [1]
[1] [1]
[1] [1] [1] [1] [1]
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. [1] M Molecular formula is the same as L, indicating that it must contains a cycloalkane in the compound as it does not react with Br 2(aq), indicating the absence of an alkene bond. OH
[1]
C H
CH3
CH
3 N N must undergo oxidation with alkaline aqueous iodine, thus having
OH C
CH3
H
the structure 5 (a)
[1] for copper [1] for zinc [1] for silver
In the electrochemical purification of copper Anode: Crude copper Cathode: Pure copper Electrolyte used: Copper (II) sulphate From the Data Booklet: Ag+ + 2e Cu2+ + 2e 2H2O + 2e Zn2+ + 2e
⇌
⇌
⇌
⇌
Ag
+0.80V
Cu
+0.34V
H2 + 2OH – Zn
+1.23 V –0.76V
When an electric current is applied, both zinc and copper are oxidised to Zn2+ and Cu2+ respectively since E θ of Zn2+ /Zn is less positive than the Eθ of Cu2+ /Cu. Cu(s) Zn(s)
Cu2+(aq) + 2e [including state symbols] 2+ Zn (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 quoting E values
(b)
(c) (d)
(e)
(f)
Zn2+ remains in the solution. When NH3 is first added to Cu 2+ (aq): NH3 + H2O NH4+ + OHCu2+ (aq) + 2 OH- (aq) Cu(OH)2 (s) --- (1) blue ppt 2+ - 2 Ionic product, [Cu ][OH ] > Ksp of Cu(OH)2 Blue ppt of Cu(OH)2 is formed. When excess NH3 is added to Cu 2+ (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 [Cu 2+] Ionic product, [Cu2+] [OH-]2 < Ksp of Cu(OH)2 Pale blue ppt,Cu(OH)2, dissolves. O2- + NH3 NH2- + OHThe colour of the solution is dark blue in colour. [Cu(H2O)6]2+ + 4Cl – [CuCl4]2 – (aq) + 6 H 2O (l) Ligand exchange reaction has occurred. [Cl -] increases in concentrated hydrochloric acid, equilibrium position will shift right to form yellow [CuCl 4]2- complex. Presences of both Cu 2+ (aq) and [CuC l4]2- in the equilibrium mixture results in a yellow-green solution. Solution remains blue when concentrated sulfuric acid is added as the sulfate ions are weaker ligands. 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. (i) 4I – + 2Cu2+ I2 + 2CuI (ii) Ecell = –0.39 V Reaction is thus not feasible. (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. (i) Pale blue solution will decolourised, and a brick-red precipitate will be produced. (ii) Positive result:
[1] [1]
[1] [1]
[1] [1] [1]
[1] [1]
[1] [1] [1] [1]
[1]
H
H
H
H
H
C
C
C
C
H
H
H
H
O C H
H
H
H
H
C
C
C
C
H
H C H H
H
O
[1]
H
Negative result: H
H
H
H
O
H
C
C
C
C
C
H
H
H
H
[1] H
