P2 Answer Scheme

8/3/2019 P2 Answer Scheme

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TRIAL STPM 2006

Answer scheme:

1. (a) (i) n = PV/RT

= 5.24 x 103 Pa x 1.00 x 10-3 m3 [1]

8.31 x 301K

= 2.09 x 10-3 mol [3 sf] [1]

(ii) RMM = Mass

No of mole

= 0.270 g [1]2.09 x 10-3 mol

Relative molecular mass = 129 [3sf] [1]

(iii)No of molecule X = No of mole x Avogadros constant [1]

= 1.26 x 1021 [1]

(b) (i) Label the diagram energy level and orbital 1S, 2S 2P [1]

(ii) Electronic configuration must represent by arrow [1]

1

3d

3p

3s

2p

2s

1s


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(iii) Oxygen have very high I.E thus they do not form positive ion. They form negative ions as

they have high electron affinity. [1]

2.

IonCa2+ Mn2+ Ni2+

Number of electrons in3d shell in ion

0 5 8

[3 marks]

(i) The third I.E of M is the energy required to move the third outmost electron from agaseous ion M2+ to infinity, so that M3+ is formed [1]

(ii) The third I.E of calcium is so high because this third electron to be removed is from the

completely filled 3p orbital for the others, the third electrons removed is from the 3d orbitalwhich have similar energy level as 4s orbital. [1]

(b) (i) Weak intermolecular forces of ethane [1]

- bond between sodium ion and hydrogen ion are very strong [1]

2

3d

3p

3s

2p

2s

1s


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(ii)

Lewis structure Shape of molecule

BF3 Trigonal planar

NH+4 Positive charge at the N or

outside the square bracket

[ ]

Tetrahedral

SF6 Octahedral

3 (a) (i) P : 2Cl- Cl2 + 2e- [1]

Q : 2H2O + 2e- H2 + 2OH

- [1]

(ii) To exert pressure so that NaOH do not flow back to left compartment. [1]

(iii) Disproportionation occurs

Cl2 + 2NaOH

NaCl + NaClO + H2O [1]

(b) (i) PtCl2(H2O)2 is a complex that can exist as geometry isomers whereas [1]

CH2Cl2 has only one tetrahedral structure [1]

(ii) diaquadichloroplatinum(II) [1]

(iii)

Cl OH2 Cl OH2

Pt Pt

Cl OH2 H2O Cl

Cis isomer Trans isomer [2]4. (a) (i) CH3CH = CHCH3 ( Cis 2 butane and trans 2 butane ) [4]

(ii) Dehydrohalogenation @ elimination [1]

3

Both corre

[ I mark]

[ I mark]

[ I mark]


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(b) (i) I : Chlorine in tetrachloroethane; absence of UV light and heating [2]

II: Alcoholic of sodium hydroxide and heat. [2]

(ii) H H H H H

C + Cl Cl C Cl

H C Cl

C H C Cl H C Cl

H H H H [2]

5. (a) (i ) Na + C2H5OH C2H5ONa + H2 [1]

(ii) Sodium is rapidly oxidised by oxygen in the air to form a layer of oxide

OR sodium reacts with water vapour in air [1]

(iii) Graph: - consistent scale of both axis [1]

Correct points plotted [1]

Graph is smooth [1]

Total vol of gas/cm3

60

45

30

0

Time/ mina b

From the graph, t1/2 is constant (t1/2= a= b = 1.45 min). Therefore overall kinetics is first order.

[2]

(iv) A large excess of ethanol is used, its concentration is constant. Hence rate is independent of

ethanol and is zero order. [1]

4

I II


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(b) Q = It

= 2.21 x (28 x 60)= 3712.8 C

No. of moles of Cu deposited = 1.2263.5

= 0.0192

Amount of change required for 1 mole of Cu = 3712.8

0.0192

= 193375 C

Cu2+ + 2e CuCharge on 2 moles of e- = 193375 C

Charge on 1 mole of e- = 193375

2= 96687.5 C

F = NAeNA = 96687.5

1.60 X 10-19

= 6.04 X 1023 (3 s.f)

(6) (a) (i) 2SO2 + O2 2SO3

Initial/atm 2/3 x 3 = 2 1/3 x 3 = 1 0 [1]

Equilibrium/atm 2 1.9 = 0.1 1 1.9/2 = 0.05 1.9 [2]

At equilibrium, P SO2 = 0.1 atm ; P O2 = 0.05 atm

New total pressure = 0.1 + 0.05 + 1.9 = 2.05 atm . [1]

(ii) Kp = (PSO3 )2

(PSO2 )2 PO2 [1]

= (1.9 )2

(0.1 )2 ( 0.05 )

= 7.22 x 103 atm 1 . [1]

(b) (i) Steam distillation lowers the boiling point of a substance and hence can purify an organic

liquid which has a high boiling point and which decomposes at temperatures near its normal

boiling point

(ii) mass of 2-nitrophenol

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[1]

0 Time

(iii) mA = MA PA

mB MB PB= 139 x 6 = 0.487 . [1]

18 x 95

% of 2-nitrobenzene = 0.48 x 100 = 32.4% [1]

1.48

Mass of 2 nitrophenol present in 100g of the distillate = 32.4 g [1]

(c) PbCl2(s) Pb2+(aq) + 2Cl-(aq) .. [1]

Addition of dilute HCl .common ion effect of Cl -, equilibrium shift to LHS solubility of

PbCl2 decreases. [1]Addition of conc. HCl, formation of a soluble complex ion of PbCl 4

2- equilibrium shifts to

RHS solubility increases. [1]

PbCl2(s) + 2Cl-(aq) PbCl4

2- (aq) . [1]

7.(a) The solubility decreases down the group [1]- Solubility depends on enthalpy of solution = enthalpy of hydration- lattice energy[1]

- The lattice enthalpy is partly a function of the sum of the radii of the cationand anion. Since the size of the sulphate ion is significantly larger than the [1]cation, interionic distance remains essentially unchanged when descending

the group. Consequently, there is minimal decrease in the magnitude of the

lattice energies of the Group 2 sulphates when descending the group. [1]- However, the hydration enthalpy of the cations decreases rapidly in

magnitude as they get larger when the group is descended.

- Therefore, the solubility of the Group 2 sulphates falls with increasing

cation size because the hydration energy decreases much more than the [1]lattice energy.

7(b)(i) Na2O reacts with water to form sodium hydroxide.Na2O(s) + H2O(l) 2NaOH(aq) [1]

P4O10 dissolves in water to form phosphoric acid.

P4O10(S) + 6H2O(l) 4H3PO4(aq) [1]

SiO2 is a giant covalent molecule that is insoluble in water. The Si-O bond is very strong andcannot be broken by water. [1]

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The Mg2+ and the A13+ ions have high charge densities forming strong ionic bonds in MgO and

[1]

A12O3 These oxides cannot be decomposed by water and are said to be insoluble in water.[1]

Max 4 marks

(iii) Na2O and MgO are metallic oxides and are basic in nature. They can react with acids to form

salt and water. [1]

Na2O(s) + 2HCI(aq) 2NaCI(aq) + H2O(l)

[1]

MgO(s) + 2HCI(aq) MgCl (aq) + H2O(l)

Al2O3 is an amphoteric oxide. It shows properties of a base when it reacts with an acid. [1]

A12O3(s) + 6HCI(aq) 2AlCl3(aq) + 3H2O(l) [1]

A12O3 shows its acidic property by reacting with a base.

A1203(S) + 2NaOH(aq) 2NaA1(OH)4(aq) + 3H2O(l) [1]

SiO2 and P 4O10 are non-metallic oxides which are acidic. They can react with a base to form salt

and water. [1]

SiO2(s) + 2NaOH(aq) Na2SiO3(aq) + H2O(l)

[1]P4OI0(S) + 12NaOH(aq) 4Na3PO4(aq) + 6H2O(I)

8.

(a) (i) Resistant to corrosion / Easily recyclable. [1]

(ii) A very high current of about 10 000A is used. [1](iii) Pollution caused by discarded aluminium cans. [1]

(iv) The aluminium can is washed and then heated to about 660oC where aluminium melts.

It can then be drained into moulds for it to cold and solidify. [2]

(b) SiCl4 is easily hydrolysed by water to form silicon(IV) oxide and hydrogen chloride. [1]

SiCl4 + 4 H2O Si(OH)4 + 4 HClSiCl4 (l) + 2 H2O (l) SiO2 (s) + 4 HCl (g) [1]

The oxygen atom in the water molecule contributes its lone pair of electrons to the empty d

orbital on the silicon atom, to form a dative bond. The silicon atom has an expanded octet

and hydrogen chloride is eliminated. [1]

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CCl4 is not hydrolysed because carbon is a Period 2 element and does not have any 2d

orbitals that can be used to form a dative bond with water molecules. All the s and p orbitals

in carbon have been used for sp3 hybridisation and then, bonding. [2]

(c) (i) A small amount of nitric acid has decomposed to form nitrogen dioxide, which is a brown

coloured gas. [1]

4 HNO3 (l) 4 NO2 (g) + 2 H2O (l) + O2 (g) [1]

(ii) The ammonium salt has decomposed to form ammonia. [1]

2 NH4Cl + CaO 2 NH3 (g) + H2O + CaCl2 [1]

When reacted with concentrated hydrochloric acid, white fumes of ammonium chloride

are formed. NH3 (g) + HCl (g) NH4Cl (s) [1]

9. (a) C O H

5.173 1.724 10.34 [1]3 1 6 [1]

Empirical formula = C3H6O [1]

58n = 58

n = 1

molecular formula = C 3H6O [1]

H O

CH3CH2 C = O or CH3 C CH3 or CH3 CH (OH)2 [1]

(i) Yellow ppt. of iodoform, CHI3 and-O-C6H4-CH2CH2COO

-; oxidation [3]

(ii) O2 N CH2CH2COCH3; Electrophilic substitution nitration [2]

HO

(iii) HO C6H4 CH2CH2CH(OH)CH3; reduction[2]

(b) Phenol dissolves readily in aqueous sodium hydroxide to form a homogeneous solutionof sodium phenoxide which may be crystallized out.

Phenylmethanol shows no appreciable reaction and it remains as an immiscible liquid. [3]

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10. (a)(i) Both G and H has CH3 on carbon with OH or with C=O

[1]

G is aldehydes / has CHO group [1]H is an alcohol / has OH group [1]

G CH3CHO [1]H CH3CH(OH)CN [1]

(ii) Reaction I H+/ K2Cr2O7, distil (reject MnO4-/H+)/ [1]

Reaction II H2SO4(aq)/dilute H2SO4, heat [1]

(b) Absence of acid group OH group and carbonyl group [2]

K cyclic diester H

|

CH3 C O C = O| |

O = C O C CH3| [1]

H ---------

Max [9]

(c) (i) 2-amino-3-phenylpropanoic acid [1]

(ii) O H O H|| | || |

3 H2 NCHCOOH H2NCHC-N-CHC-N-CHCOOH

| | | |CH2 CH2 CH2 CH2| | | |

[2]

(iii) HNO2(aq)/ NaNO2+ HCl/ nitrous acid [1]Effervescence/ gas evolved for A but not for 3-phenylpropanoic acid [1]

C6H5 CH2-CH(COOH)NH2 + O=N-OH C6H5-CH2CH(OH)COOH + N2 [1]

---Max [6]

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P2 Answer Scheme

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