Answers to Section A
Q 1(a) (i) 0.88 0.50 0.10 = 0.28 mole He (ii) (iii) (iv) (v)
1(b)(i) (ii)
Answer
Mark 1 1 1 1 1 1 1 1 1 1
Total marks
0.30 mole H2 and 0.28 mole He (both mentioned) 28 kPa 2.4 + 28 +
30 (working) OR 60.4 kPa (final answer) 2H2 + O2 H2O n = 6 to n = 2
visible light Balmer Series
6
(iii)
10 lines
4 10 marks
1
Q 2(a) (i)
Answer Heat given out when 1 mol of hydrocyanic acid is burnt
completely in excess oxygen (air) under standard conditions of 1
atm pressure and 298K. HCN(l) + 9/4 O2(g) CO2(g) + NO2(g) + H2O(l)
[Note: state symbols must be given to get mark] Expression: Hrxn =
Hf (products) - Hf (reactants) Substitution: -1628 kJ = (-393kJ) +
(-89.0kJ) + (_286kJ) - ( x + 0 ) x = -625 kJ + 1628 kJ = + 1003 kJ
Answer: Hence Hf of HCN is + 1003 kJ mol-1 Or alternative method
(algebraic / energy cycle)
Mark
Total Marks
1 1 1 2
(a) (ii)
1
1 3 1 1 2
(a) (iii)
CO2 :Linear shape NO2: bent / V-shape
(b)
At anode: 4OH- 2H2O + O2 + 4e 2H2O O2 + 4H+ + 4e OR 1
1 mol of O2 gas requires 4 F of electricity to produce 22, 400
cm3 of O2 gas is produced by 4 x 96, 500 C 140 cm3 of O2 gas is
produced by 140 x 4 x 96, 500 22, 400 = 2412.5 C 2412.5 C = 1.25 A
x t (s) t = 2412.5 ............................................
1.25 = 1930 s = OR 1.93 x 10 s3
1
1 1 (max3)
Max 10
2
Q. 3 (a)(i)
Answer A white precipitate is formed
Mark 1 1
Total Marks
(ii) [Al(H2O)6]3+ + 3OH- [Al(OH)3(H2O)3] + 3H2O OR Al3+ + 3OH-
Al(OH)3 (iii) Observation: The white precipitate dissolves in
excess ammonia Formula: [Al(OH)4](b)(i) [Fe(CN) 6]4-
1 1 4 1 1 1 1 1
-
Hexacyanoferrate(II)
[Fe(SCN)(H2O)5]2+ - Pentaaquathiocyanatoiron(III) (ii) Have
empty d orbitals / (valence orbitals) to accept lone pair electrons
from ligands Has high charge density (iii) Splitting of d orbitals
d-d electronic transitions. OR Energy is absorbed when electrons
from the lower energy d orbitals are promoted to higher energy d
orbitals
1 6 10 marks
3
Q. 4 (a) (i) 1-butene
Answer
mark
Total marks
1 1
(a) (ii) Name of mechanism : electrophilic addition
1
1 4 (b)(i) (ii) 3-methyl-2-butanol Type of isomerism: optical
isomerism Structures: 1 1
1+ 1 4
(c) (i) (ii) Epoxypropane
1 1 2 10 marks
4
Q. 5
Answer All are simple covalent molecules / simple molecular
structure Weak van der waals forces between molecules
Mark
Total marks
(a) (i)
1 1 1 +1 +1
(ii)
(iii) O atom has no d orbitals to extend its octet (iv) XeF bond
is polar because F more electronegative than Xe Since the bonds are
arranged symmetrically / symmetrical shape dipoles in each XeF bond
cancel each other / no net dipole therefore the molecule is
non-polar (b) (i) CO(g) + NO(g) + O2(g) CO2(g) + NO2(g)
1 1
1 1 1
8
(ii) Expression: Rate = k[CO]x [NO]y [O2]z Working: Exp 2 4.40 x
10-4= k [2.00 x 10-3]x [1.00 x 10-3]y [1.00 x 10-1]z Exp 1 4.40 x
10-4= k [1.00 x 10-3]x [1.00 x 10-3]y [1.00 x 10-1]z x = 0 if
working shown Working: Exp 3 3.96 x 10-3= k [2.00 x 10-3]x [3.00 x
10-3]y [1.00 x 10-1]z Exp 2 4.40 x 10-4= k [2.00 x 10-3]x [1.00 x
10-3]y [1.00 x 10-1]z y = 2 if working shown Working: Exp 2 4.40 x
10-4= k [2.00 x 10-3]x [1.00 x 10-3]y [1.00 x 10-1]z Exp 4 1.32 x
10-3= k [5.00 x 10-3]x [1.00 x 10-3]y [3.00 x 10-1]z z = 1 if
working shown Rate = k[NO]2 [O2]1 (ii i) Using exp. 1 k = Rate
[NO]2[O2] = 4.40 x 10-4 (1.00 x 10-3 )2 (1.00 x 10-1) = 4400 mol-2
dm6 s-1
1
1
1
1
1 1 8 Max 155
Q6 (a) (i)
Answer
Mark
Total marks
H2CO3 + H2O
H3O+ + HCO3
[ H 3O + ][ HCO 3 ] K1 = [ H 2CO3 ]HCO3- + H2O K2 = Ka = H3O+ +
CO322
1
[ H 3O + ][CO3 ] [ HCO3 ]K1 x K22
1
[ H 3O + ]2 [CO 3 ] = [ H 2CO 3 ](a) (ii)
1 3
[ H 3O + ][ HCO 3 ] K1 = [ H 2CO3 ][H3O+] = [HCO3-] K1 =[ H 3O +
] 2 [ H 2CO 3 ]
O [H3O+] = K1x[H 2C 3 ] 0 = ( 4.5 x1 7 )( 0.100 ) OR = 4.5 x10 8
OR = 2.12 x 10-4 mol dm-3
1
pH = - lg [H+] = - lg (2.12 x 10-4) = 3.7 (a) (iii) The
resultant mixture contains CO3 and HCO3 and is a buffer This buffer
helps to maintain the pH of human blood2-
1 2 1 1
2 (b) (i) An azeotrope is a constant boiling point mixture
whereby the 16
composition of the vapour is same as the composition of the
liquid. (b) (ii)
126 correct shape B. pt / C 100 0
vap liquid liquid
vap.
1 curves labelled axes labelled 100, 126, -67 47% - -67 1
47% HBr 0% -----> %HBr ----- > ------ 100% 100% .---<
-- %H2O -------- < ---- 0% composition 2 (b) (iii) (b) (iv) The
composition of the first distillate is pure HBr or 100% HBr The
composition of second distillate is 47%HBr and 53% H2O / or
azeotrope. A mixture of HBr and H2O shows negative deviation from
Raoults law. In pure HBr the forces are van der waals between HBr
molecules AND In pure H2O forces are hydrogen bonds between H2O
molecules When mixed together, There are ion dipole forces between
H+ and H2O molecules and between Br- ions and H2O molecules.
Ion-dipole forces stronger than hydrogen bonds 1 3 1 1 2 1
1
1
Q.7
Answers
marks
Total7
marks (a) Statement: Sodium and magnesium burn in oxygen to form
basic oxides. Equation: 4Na(s) + O2(g) 2Na2 O(g) OR 2Mg(s) + O2(g )
2MgO(g) Aluminium burns in oxygen to produce aluminium oxide, an
amphoteric oxide. 4Al(s) + 3O2(g) 2Al2O3(g) 1 1 1 1 1
Silicon, phosphorous and sulphur burn in oxygen to form acidic
oxides Si(s) + O2(g) 2SiO(g) P4(s) + 3O2(g) P4 O6(l) (any one
equation) P4(s) + 5O2(g) P4 O10(l) S(s) + O2(g) SO2(g) Chlorine
does not react with oxygen. (b) All Group 2 elements react with
water according to the following equation. [given any equation]:
M(s) + 2H2O M(OH)2 + H2 In the reaction, the Group 2 element acts
as reducing agents. Going down Group 2, the reducing power of the
element increases as the E0 values become more negative. Magnesium
reacts only with steam, Calcium reacts rapidly with hot water, but
slowly with cold water Strontium and barium react vigorously with
cold water, OR Reactivity increases down the group. (c) [Equation]:
2M(NO3)2 2MO + 4NO2 + O2 Going down Group 2, the size of the cation
increases while the charge remains the same / Charge density of
cation decreases.The polarizing power of the cations towards NO3-
ion decreases making the nitrates more stable to heat Thermal
stability increase down the group
1
1
7
1 1 1
1
1 1 1 1 8
8(a)(i)
Pb4+ + 2e = Pb2+
E = +1.69 V8
Sn4+ + 2e = Sn2+
E = +0.15 V
1 1 1 1
The +4 oxidation state of both Pb and Sn tend to be less stable
than the +2 oxidation state PbO2 is highly unstable and decomposes
readily to PbO While SnO readily combines with O2 to form SnO2 OR
PbO2 PbO + O2 SnO + O2 SnO2 Pb combines readily with Cl2 to form
the more stable PbCl2 / PbCl4 decomposes slowly at room temperature
to PbCl2 Sn combines with Cl2 to give the more stable SnCl4 OR Pb +
Cl2 PbCl2 / PbCl4 PbCl2 + Cl2 Sn + 2Cl2 SnCl4 (ii) Sn 2MnO4-(aq) +
16H+ (aq) + 5Sn2+(aq) 2Mn2+(aq) + 8H2O(l) + 5Sn4+(aq) ( states of
species not needed) Ecell = +1.52 - (+0.15) = + 1.37 V 4NH3 + 5O2
4NO + 6H2O 2NO + O2 2NO2 4NO2 + O2 + 2H2O 4HNO3 1 mole of NH3 gives
1 mole of HNO3 OR 17 g of NH3 gives 63 g of HNO3 1 tonne of NH3
gives 63/17 tonnes of HNO3 OR 3.7 tonnes (iii) Reduction 2NO N2 +
O2 OR 2NO + 2CO N2 + 2CO2
1 1 6
1 1 1 1 1 1 1 3
(b)(i)
(ii)
1 1 1 7 Total 16 Max 15
9(a)(i) + CH2=CHCH3 CH(CH3)2 19
(ii)
Catalyst : phosphoric acid / H3PO4 Oxidation / reaction with
oxygen CH3CHCH3 + O2 CH3 CH3COOH
1 1 1
2
Decomposition of hydroperoxide with dilute sulphuric acid CH3
CH3C-O-O-H
1 1
--OH + CH3COCH3 4
(b)(i)
-OH in phenol is a ring activating group ./ Lone pair electrons
on oxygen atom in OH group can be delocalized into the benzene ring
/ donated to the benzene ring Phenol more easily attacked by an
electrophile (or NO2+) compare to benzene Equation to forn
nitrobenzene Equation to form 2-nitrophenol and 4-nitrophenol
1 1 1 1 1 1 1 1 1 5 15
4
(ii)
Elimination reaction / dehydration To form but-1-ene (1-butene)
and 2- butene Any 1 equation to form 2-butene / 1-butene 2-butene
exists as pair of cis-trans isomers / shows geometrical isomers
[Draw structure of the cis and trans isomers and labelled them]
Q 10 (a)
Answer Describe test using either one of the following
reagents:
Mark 1
10
Tollens reagent / Fehlings solution / iron(III) chloride (cannot
use Br2(aq) or KMnO4) Observation: Equation: (b) Step 1 : reagent :
hydrogen cyanide / HCN Step 2: reagent : LiAlH4 in ether solvent /
NaBH4 / Na in ethanol Identity of X H CH=CHCOH CN Observation:
bubbles of nitrogen gas obtained CH=CHCH(OH)CH2NH2 + HNO2 N2 + H2O
CH=CHCH(OH)CH2OH (c) (i) CH3CCl + HO CH2CH=CH2 O OCH3 CH3CO
CH2CH=CH2 + HCl O OCH3 (ii) Products are CH3COO-Na+ and Na+O-
CH2CH=CH2
1 1 3 1 1
1 1 1 5
1
1+1 3 1 1
OCH3 (d) (i) Observation: brown colour of Br2 decolourised / and
white ppt Formula is OHCH3O Br
CH2CH(OH)CH2Br
(ii)
Observation: purple colour of MnO4- is decolourised / and
bubbles of CO2 observed Formula CH3OHO COOH
1 1 4
11
