Nov 2004 P3

N2004 P3 Nov 2001 Paper 1 Suggested solutions and Mark scheme Section A (Physical Chemistry) 1

(a) (b)

The average mass of an atom of an element compared to 1/12 the mass of a 12C atom. R.A.M. of Kr =

2.6 × 80 + 11.6 × 82 + 11.5 × 83 + 56.9 × 84 + 17.4 × 86 100

= 83.9 (1.d.p) Comments • Remember to give answer in 1 dp (c)

Let R.M.M. of A be x. 2.00 × 104 × 104 × 10-6 = (0.100 ÷ x) × 8.31 × 305 x = 121.9 Let molecular formula of A be KrFn. n = (121.9 – 83.9 ) ÷ 19.0 =2 A is KrF2. Comments: • Remember to convert to SI units when using the PV=nRT formula.

(d)(i)

F

x xx

F

Xe

F

x xx

x

x

F

Comments: • Xe can expand its octet. Hence, there will be a total of 6 electron pairs around Xe. • Thus, it is not surprising to see Xe exists in compounds. (d)(ii)

square planar (4 bp, 2 lp)

(d)(iii)

6XeF4 + 12H2O ⎯⎯→ aXeO3 + b Xe + 3O2 + c HF 3a + 6 = 12 a=2 a+b=6 b=4 c = 24

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N2004 P3 2

(a)

Temp: 298 K

V

Pressure: 1 atm

Pt electrode

1 atm H2 at 298K

salt bridge

Pt electrode

1 moldm-3 of Fe3+,

1 moldm-3 of H+

1 moldm-3 of Fe2+

(b)(i)

H2O2(aq) + 2H+(aq) + Sn2+(aq) ⎯⎯→ Sn4+(aq) + 2H2O(l) O: -1 (H2O2) to -2 (H2O) i.e. decreases by 1 Sn: +2 (Sn2+) to +4 (Sn4+) i.e. increases by 2

(b)(ii)

E°cell = +1.77 – (+0.15) = +1.62 V Since E°cell is positive, the reaction is feasible (energetically) in the left → right direction.

(c)(i)

Ag+ + e

Ag

3+

E° = +0.80 V 2+

Fe E° = +0.77 V Fe + e E°cell = +0.80 – (+0.77) = +0.03 V Since E°cell is positive, Ag+ would oxidise Fe2+ to Fe3+, while being reduced to Ag in the process. However, due to the small E°cell value, the reaction may not go to completion. Ag+ + Fe2+ ⎯⎯→ Ag + Fe3+ (c)(ii)

Cu2+ + 2e 2+

Cu +

E° = +0.34 V 3+

V + H2O E° = +0.34 V VO + 2H + e E°cell = +0.34 – (+0.34) =0V Since E°cell is O V, Cu2+ would oxidise V3+ to VO2+, while being reduced to Cu in the process. However, an equilibrium mixture is obtained. Comments: (for part (c)) • Note that this equation: ∆G = -nFE°cell • Hence, for very small positive E°cell value, you would expect the reaction to not go into completion. 3 E

(a)

A weak acid is an acid that does not dissociates completely to give H+ ions.

(b)(i)

CH3COOH Ka =

CH3COO- + H+

[CH3COO − ][H+ ] [CH3COOH]

1.74 × 10-5 =

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[H+ ]2 0.100 - [H+ ]

2

N2004 P3

1.74 × 10-5 =

[H+ ]2 (assuming 0.100 >> [H+]) 0.100

[H+] = 1.319 × 10-3 mol dm-3 pH = -lg (1.319 × 10-3) = 2.88 (b)(ii)

pH 14

7

2.88 0

10.0

Volume of NaOH /cm3

A suitable indicator to use is phenolphthalein as the equivalence point is about pH > 7, which is in the working pH of phenolphthalein. Comments: • When choosing an indicator for titration, always relate to the working pH of the indicator. • No need to calculate the specific pH at the equivalence point. You would need to show the relative pH at the equivalent point.

3 O

(c)

NH3 + HCl ⎯⎯→ NH4Cl n(NH3) produced = 14.0 × 10-3 × 0.200 = 2.80 × 10-3 mol NH2CONH2 + 2OH- ⎯⎯→ 2NH3 + CO32n(NH2CONH2) = ½ × 2.80 × 10-3 = 1.40 × 10-3 mol Percentage of nitrogen in fertiliser = [(1.40 × 10-3 × 2 × 14.0) ÷ 0.100] × 100 = 39.2 %

(a)

A buffer is a solution containing a weak acid (or base) and its soluble salt. It is able to resist change in pH when small amounts of acid or base is added, resulting in a negligible change in the pH. Comments:

• Do take note that pH of a buffer change very very very slightly when small amount of acid or base is added. However, the pH of the buffer remains relatively the same hence, we claim that buffer resists pH changes instead of saying no pH change occurs.

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N2004 P3 (b)(i)

CO2 + OH- ⎯⎯→ HCO3-

(c)

[HCO2 ] pH = pKa + lg [CO 2 ]

−

= -lg (7.90 × 10-7) + lg 20 = 7.40 (d)(i)

CO2 + Ba(OH)2 ⎯→ BaCO3 + H2O

(d)(ii)

Molar mass of BaCO3 = 137 + 12.0 + 3 × 16.0 = 197.0 g mol-1 n(BaCO3) produced = 0.600 ÷ 197.0 = 3.05 × 10-3 mol Total number of moles of CO2 and HCO3- = 3.05 × 10-3 mol [CO2(aq)] in plasma = [(3.05 × 10-3) ÷ (100 × 10-3)] × 1/21 = 1.45 × 10-3 mol dm-3

Section B (Inorganic Chemistry) 4

(a)

Higher density; higher melting point (or boiling point); higher conductivity; higher tensile strength (stronger); harder; smaller metallic/atomic radius (any 2)

(b)(i)

Fe, used in Haber process (industrial manufacture of NH3); Ni/Pt, used in the hydrogenation of alkenes; Pt, used in catalytic converters; Fe2+ (or Fe3+) salts, used in oxidation of I- by S2O32- (any 1)

(b)(ii)

1s22s22p63s23p63d34s2

(b)(iii)

+2; +3; +4; +5 (any two but +1 is not accepted) Comments: • Because of the electronic configuration of V, hence it results in the +1 oxidation number of V to be unfavourable, as V would rather lose a minimum of 2 electrons from the 4s subshell.

(b)(iv)

In [V(H2O)6]3+, the 3d subshell of V3+ are split into 2 sets of orbitals with different energies by the ligands i.e. H2O. The electrons from the lower energy d-orbitals absorb energy corresponding to orange-red light and are excited to the higher energy 3d-orbitals. The unabsorbed wavelengths correspond to the blue-green colour observed. Comments: • Mention about what causes the d subshell to split. • Mention about absorbing of a particular light colour. • Mention about the reflection of the complementary colour.

(b)(v)

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Volume of seawater processed = (14 × 3 × 10-3) ÷ (7.1 × 10-6) = 5915 dm3 = 5920 dm3 (3 sf)

4

N2004 P3

5E

(a)(i)

Sodium, Na, has high electrical conductivity/ is shiny/ is malleable, which is typical of metals. Its chloride, NaCl, has a very high melting point/ high solubility in water/ conducts electricity in the molten phase.

(a)(ii)

Phosphorus, P4, is an electrical insulator/ is dull/ has a low melting point, which is typical of non-metals. Its chlorides, PCl5 (or PCl3), has a low melting point/ reacts with water to form an acidic solution/ does not conduct electricity/ dissolves well in organic solvents. Comments (part (a)):

• Not stating the physical properties of Na and its chloride would render no marks to be awarded. (b)(i)

AlCl3 has a low melting point/ reacts with water to form an acidic solution/ does not conduct electricity/ dissolves well in organic solvents.

(b)(ii)

Silicon conducts electricity/ is shiny/ is hard.

(c)(i)

Al(OH)3 can act both as an acid as well as a base.

(c)(ii)

Al(OH)3 acting as an acid, typical of a non-metal: Al(OH)3 + OH- ⎯→ Al(OH)4Al(OH)3 acting as a base, typical of a metal: Al(OH)3 + 3H+ ⎯→ Al3+ + 3H2O

5O (a)

NaCl, MgCl2, AlCl3, SiCl4

(b)

NaCl – ionic bonding; MgCl2 – ionic bonding; AlCl3- covalent bonding; SiCl4 – covalent bonding

(c)

NaCl dissolved in water to form a neutral solution. NaCl(s) + aq ⎯→ Na+(aq) + Cl-(aq) MgCl2 dissolved in water and undergoes hydrolysis to a very slight extent to give a slightly acidic solution (pH 6.5). MgCl2(s) + 6H2O(l) ⎯→ [Mg(H2O)6]2+(aq) + 2Cl-(aq) [Mg(H2O)6]2+(aq) + H2O(l)

[Mg(H2O)5(OH)]+(aq) + H3O+(aq)

AlCl3 dissolved in water and undergoes substantive hydrolysis to give a slightly acidic solution (pH 3-4). AlCl3(s) + 6H2O(l) ⎯→ [Al(H2O)6]3+(aq) + 3Cl-(aq) [Al(H2O)6]3+(aq) + H2O(l)

[Al(H2O)5(OH)]2+(aq) + H3O+(aq)

SiCl4 undergoes full hydrolysis to give an acidic solution (pH 2) and a colourless solid. SiCl4(s) + 2H2O(l) ⎯→ SiO2(s) + 4H+(aq) + 4Cl-(aq) Comments:

• Please show the reaction of the chloride when dissolved in water first before showing any form of partially hydrolysis that is taking place due to the polarising power of the cation.

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N2004 P3

(d)(i)

Element % mass

S 31

Cl 100 – 31 = 69

31 32.1

mass ratio

69 35.5

mol ratio 1 Empirical formula is SCl2. (d)(ii)

2

Cl S Cl x

x

xx

2 bonding pairs, 2 lone-pairs shape is bent (d)(iii)

pH ≤ 3 as it undergoes full hydrolysis in water.

Section C (organic chemistry) 6

(a)

H

H H

H C

C H

H

H

C

C

C H

H

H

H

H C

H

H

H C

C

C

H

C

C

H

3

H

H H

H

C

H

H

H C

C H

H H

H H

H H

C H

C

2b H

C

H

C

C H

H C

C

H

H

H H

H

C

2a H

H

H

C

H

1

H C

C

H

H

H

H

C

H H

H

C

C

4

H C

C

H H

H

H

H H

5

1, 2a (or 2b), 3, 4 and 5 are 5 different structural isomers due to branching in the carbon chain or position of the C=C bond. 2a (cis-isomer) and 2b (trans-isomer) exhibit geometric isomerism which is a form of stereoisomerism. It arises due to restricted rotation about the C=C bond. Total number of alkenes with the formula C5H10 is 6. (b)

B is CH3CH2CH=CHCH3 (as it gave D, CH3CH2COOH and E, CH3COOH upon oxidation). C is (CH3)2C=CHCH3 (as it gave E, CH3COOH and F, CH3COCH3 upon oxidation).

(c)(i)

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Not in syllabus.

6

N2004 P3 and (ii) 7

(a)

CH2 CH2 CH3CH=CHCH2OH Compound A

CH2 CH

OH

Compound B

Br2(l), room conditions Compound A would decolourise the orange Br2 while compound B would not decolourise the Br2. OR KMnO4(aq), NaOH(aq), cold Compound A would decolourise the purple KMnO4 while compound B would not decolourise the KMnO4 (b)

CH2 CH2 CH2 CH3 CH

O

CO

CH3 CH

CH2 CH2

CH2 O

CO

Compound A Compound B Heat under reflux with NaOH(aq), cool down, add I2(aq) and NaOH(aq) and warm. Compound A would produce a yellow precipitate of CHI3 while compound B would not. Comments:

• In trying to distinguish two different esters, hydrolysis should be carried out first. Occasionally, distillation should also occur after so as to ensure the desired positive result is obtained. (c)

Cl

CHO

COCl

Compound A Compound B AgNO3(aq), room conditions Compound B would produce white fumes of HCl and a white precipitate of AgCl while compound A would not. OR AgNO3(aq), excess NH3(aq), warm Compound A would produce a silver deposit while compound B would produce a white precipitate. (d)

CONHCH3

CH3CONH

Compound A Compound B Heat under reflux with NaOH(aq), cool the mixture and acidify with followed by with HCl(aq. A white precipitate (benzoic acid) would be formed with compound A while mixture from compound B would not. OR

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N2004 P3 Heat under reflux with NaOH(aq), cool the mixture and neutralise the mixture carefully with HCl(aq), followed by addition of Br2(aq) under room conditions. Mixture from compound B would decolourise the reddish-brown Br2 while the mixture from compound A would not. Comments: • In trying to distinguish two different amides, hydrolysis should be carried out first. Occasionally, distillation should also occur after so as to ensure the desired positive result is obtained. • The second test which makes use of Br2(aq) has to be done carefully hence it is not a very suitable test. 8E

(a)(i)

G is formed from C2H5OH and reacts with alkaline aqueous iodine ⇒ G has the CH3CH(OH)- or CH3CO- group G reacts with 2,4-dinitrophenylhydrazine and Fehling's reagent ⇒ G is a aliphatic aldehyde Hence, G is CH3CHO. H reacts with sodium metal ⇒ H contains -OH group H is produced from G (CH3CHO) by reaction with HCN and trace of NaCN ⇒ nucleophilic addition of G has occurred to produce a cyanohydrin RCH(OH)CN Hence, H is CH3CH(OH)CN

(a)(ii)

I: KMnO4(aq) (or K2Cr2O7(aq)), H2SO4(aq), heat under reflux III: H2SO4(aq), heat under reflux

(b)

Contains NMR, which is not in syllabus.

8O (a)(i) (a)(ii)

Cl2, UV The reaction is a free radical substitution reaction. The first stage, inititaion, involves formation of Cl• radicals.

Cl

hν

Cl

2 Cl

The second stage, propagation involves the reaction between C6H5CH3 with the Cl• radical. H H Cl

H

+

C

C

+ HCl

H H The C6H5CH2• radical will then react with a Cl2 molecule to form C6H5CH2Cl and a Cl• radical. H H

C

+ Cl

Cl

C

Cl + Cl

H H Finally, in the termination stage, 2 radicals will combine e.g.

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N2004 P3 H

H + Cl

C

C

H

Cl

H

+

Cl

(b)(i)

C6H5CH2CN

(b)(ii)

alcoholic NaCN

(b)(iii)

nucleophilic substitution

(c)(i)

LiAlH4, dry ether, heat under reflux

(c)(ii)

Reduction

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Cl

Cl

Cl

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