skema k2 kimia stpm 2011 trial sabah

8/4/2019 Skema k2 Kimia Stpm 2011 Trial Sabah

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Suggested Answers

Section A

1 (a) Number of protons & number of neutrons in the nucleus of an atom (1) 1m

(b) 1 7

8

( 1 )

O ( 1 )1m

(c)Particle Proton Neutron Electron

Mass /g 1.6725 1024 1.6748 1024 0.0009 1024

Relative charge +1 0 -1

(i) as shown in the table

(ii) 1.6725 1024+

0.0009 1024=

1.6734 1024 g

(iii) 1.6734 1024x

6.0225 1023 = 1.0078 g

(iv) Other isotopes present

1m

1m

1m

1m

(d)

(ii) increases

Explanation for Al: (1)

e removed from (3) p or

e

or orbital is higher in energy level orbetter shielded than (3)s orp electron is shielded by 3s electrons

Allow e

is further awayMark independentlyOr

1m

1m

1m

First ionisation

energy/kJ mol

1

1600

1400

1200

1000

800

600

400

200

0Na Mg Al Si P S Cl Ar

Al below Mg (1)

(1)

(1)

S below P and Cl

but above Si


2/10

Explanation for S:

e

removed from (3)p electron pair or

repulsion between paired e

(reduces energy required)Mark separatelyIf used d rather than p orbital only get (explanation marks)

1m

(10m)

2 (a) Rate = k[PH2O2]n[OH]m

By substituting the values of rate, [PH2O2] and [OH] from the results ofexperiments 1and 2:

3 = 3n

n = 1

The reaction is first orderwith respect to phosphinate ion.

1m

1m

(b) (a) Rate = k[PH2O2]n[OH]m

By substituting the values of rate, [PH2O2] and [OH] from the results ofexperiments 3and 4:

4 = 2m

m = 2

The reaction is second orderwith respect to hydroxide ion.

1m

1m

(c) The overall order of reaction is third order. 1m

(d) The rate equation is rate = k[PH2O2] [OH]2 1m

(e) By substituting the values of rate, [PH2O2] and [OH] from the results ofexperiment 1 in the rate equation

1m

1m

(f) By substituting the values ofk, [PH2O2] and [OH] into the rate equation:Rate = 4 x 0.6 x (3.0)2

= 21.6 dm3 min1

1m

1m


3/10

(10m)

3 The chlorides of Group 14 are CCl4, SiCl4, GeCl4, SnCl4, PbCl4,

(a) (i) State the type of hybridization present in the above tetrachloridesp3

1m

(ii) Name the shape of each molecule.

Tetrahedral

1m

(b) (i) The tetrachloride from the above that is totally not hydrolysed in water is

CCl4,

Water molecules cannot form dative bonds with CCl4

Because carbon atom does not have 2d orbitals.

1m

1m

(ii) PbCl4 ,has the highest boiling point.

It has the strongest dispersion forces (temporary induced dipole) because

it is the largest molecule.

1m

1m

(c) Why is carbn tetrachloride used in extinguishers?

Chemically inert (not reactive/passive) /

Its denser vapour sinks excludes oxygen and puts the fire off 1m

(d) (i) Name one of the compounds : Lead (IV) tetrachloride 1m

(ii) Why is this compound that you choose thermally unstable?

The Pb-Cl bond length is the longest and the weakest

1m

(iii) Write an equation for the decomposition.

PbCl4 (s) PbCl2(s) + Cl2 (g)

1m

(10m)

4 (a) (i) secondary alcohol 1m

(ii)H3C CH3

CHOCOCH3 + HCl (1 mark )

CH3 (1 mark )

2 m


4/10

(iii)

H3C CH3

CH

O

CH3

1m

(iv) No, because it is a secondary alcohol and not a phenol 1m

(b) (i) aqueous NaOH/KOH and boil under reflux 1m

(ii) nucleophilic substitution (SN1 or SN 2 ) 1m

(iii) NaOH in ethanol, heat the mixture 1m

(iv) elimination reaction 1m

(v) hydroxide ions act as a base to remove all the H+ ions from the

haloalkane molecule.

1m

(10m)

Section B

5 (a) (i)

(ii)

More 63Cu atoms than 65Cu atoms (1)(idea of more abundant63Cu isotope - NOT just reference to peakheights)

Electron from electron gun / high speed electron / high energy

electron

1(accept electron gun fired at) (1)[NOT bombarded with electrons]

knock electron off (Cu atom) / idea of loss of e

/ appropriateequation (1)1(Mark independently)

(iii) Identity of the ion = 63Cu2+or equivalent (1)[NOT 63.0 - penalise this error once only]

1

Explanation for the m/e value : m/e = 63/2 (=31.5) or equivalent (1)

More energy needed to remove second electron OR1

63Cu2+ statistically less likely to remove second electron(Idea that not many 63Cu2+ ions formed OR explains why few areformed e.g. more energy needed) (1)

1m

1m

1m

1m

1m

1m


5/10

If63Cu not given, can only award M2 & M3Notes on [If 65 used, lose M1 and M2]

[If mass number missing from identity but appears inexplanation,penalise M1 but allow M2 if earned]

TOTAL = 6 MARKS

(6m)

(b) (i)

(ii)

(iii)

105 (allow 104106) (1) 4 electron pairs round O (1)

tetrahedral (or 109) (1)

lone pairs repel more than bonding pairs (1)

angle < 109 (or less than tetrahedral angle) (1)

hydrogen bonding (1)

F more electronegative than H, thus will pulls electrons from H (1)

Prediction > 105 (or increases) (1) Explanation lone pair more like bonding pair

does not repel so strongly (1)

TOTAL = 9 MARKS

1m1m

1m

1m

1m

1m

1m

1m

1m

(9m)

6 (a)

(i) Diagram: fuel cell e-

beaker + 2 carbon electrodes:Anode : carbon electrode + ethanoic NaOH solutionCathode : carbon electrode with oxygen gas flowing in + NaOH aqElectrons flow from anode to cathode

1m1m1m

(ii) Function of oxygen : oxidized ethanol to ethanoic acidFunction of NaOH : as electrolyte

1m1m

(iii) At Anode : C2H5OH + H2O CH3COOH + 4e + 4H+

At Cathode: O2 + 2H2O + 4e 4OH-

Emf = + 0.4 (+ 0.2 ) = + 0.20 V

1m1m1m

(8m)

(b)

(i) State Hesss law: The overall enthalpy change is equal to the sum of theenthalpy changes for the individual steps in a reaction.

1m

(ii) N2H4(l) + O2 (g) N2(g) + 2H2O (l) 1 + 1

anode cathode

oxygen gas

carbon

electrodes

ethanoic


6/10

(iii) Enthalpy change = 4(-242.7) (-19.57 + 2 x 50.63)= -1052 kJ mol-1

The reaction is violently explosive because the reaction is veryexothermic, a lot of heat is given off.

1m1m1m1m

(7m )

7 (a) When aluminium reacts with chlorine gas, aluminium chloride, Al2Cl6, is formed 1m

(i) Draw the structural formula of aluminium chloride.

1m

(ii) 2Al(s) + 3Cl2(g) Al2 Cl 6 (s) 1m

(iii) State one important use of aluminium chloride

Catalyst in alkylation of benzene to increase the rate of reaction1m

(iv) When 1.0 g of aluminium was used, 2.3 g of aluminium chloride.was

produced. Calculate the percentage yield of the product.

2 moles Al produce 1 mole Al2 Cl 6

54 g Al produce 267 g Al2 Cl 6

1g Al produce 267/54 = 4.9 g Al2 Cl 6

% yield = 2.3/4.9 x 100 = 46.9

1m

1m

(b) The solution contains the complex ion Al(H2O)63+ and chloride ion.

The complex ion undergoes hydrolysis to produce excess of hydronium ions

making its pH


7/10

This is done by placing the aluminum as the anode in electrolysis.

4Al (s)+ 3O2 (g) 2Al2O3(s) 1m

1m

8 (a) Why are transition elements coloured?-In the presence of ligands, the 3d orbitals are split into different energy

levels /d-d splitting occurs (can illustrate).

-The electrons from a lower 3d orbitals absorb energy from light and are

excited/promoted to a higher energy level.

-however, only certain wavelengths in the visible spectrum is absorbed.

The unabsorbed wavelength corresponds to the colour of the solution of

complexes.

-the difference in energy of the d-orbitals determines the wavelength of

light absorbed and hence its colour.

What determines its colour?

Strength of the ligands, oxidation state of metals ions,

Geometry of the complexes affects splitting of d orbitals and hence the

colour of the complexes.

1m

1m

1m

1m

1m

1m

(5max)(b) (i) The addition of the SCN- displaces the H2O ligands, forming the blood-

red [Fe(SCN)(H2O)5]2+ complex ion

Since SCN- is a stronger ligand.

However when F- is added, it displaces the SCN- and H2O ligands to

form the colourless [FeF5 (H2O)]2- ion since F- is a stronger ligand.

1m

1m

1m

(3m)(ii) [Co (H2O)6]2+ + 4Cl- [CoCl 4]2- + 6H2O

Pink blue

According to Le Chateliers principle, the addition of water causes the

equilibrium to shift to the left, colour changes fro blue to pink.

1m

1m

(2m)

(c) (i) 4NH(g) +5O2(g) 4NO (g) + 6 H2O

2NO(g) + O2 2NO(g)

4NO2(g) + O2(g) + 2 H2O(l) 4HNO3(aq)

1m

1m

(ii) From the equations,

4 moles of ammonia gives 4 moles of NO4 moles of No gives 4 moles of NO2

4 moles of NO2 gives 4 moles of nitric acid

Thus, 1 mole ammonia gives 1 mole nitric acid

17 g of ammonia gives 63 g of nitric acid

Hence, 1 kg of ammonia gives 63/17 x 1 = 3.71 kg of nitric acid.

1m

1m

1m

(5m)


8/10

(15 m)

9 (a)

This process involves the hydrolysis of chlorobenzene and sodiumhydroxide at 300 C and 150atm

Cl + 2NaOH O-Na+ + NaCl + H2O

Phenol is obtained by adding acid to the products of hydrolysis.

O-Na+ + HCl OH + NaCl

1m

1m

1m

1m

(b)

The C* is the carbon atom attached to the OH group.The (+) enantiomer rotates the plane of polarization clockwise while the (-)enantiomers rotates the plane of polarization anti-clockwise.

1m1m1m

(c) (i) CH3

CH3 CH2 - CH2 C CHBr - CH2BrC C (1)

CH3 H OHBr Br

(1)

2m

(ii) CH3

CH3 CH2 - CH2 C CHCl CH3C C (1)

CH3 H ClCl H (1)

(1)

3m

(iii) CH3

CH3 CH2 - CH2 C CH(OH) CH2(OH)C - C (1)

CH3 H OH

OH OH (1) (1)

2m

(iv) CH3

CH3 CH2 - CH2 C CH = CH2C = C

CH3 H O- C-CH2CH3

1m


9/10

O (1)

(15m)


10/10

10 (a) CH3OH + CH3CH2COOH CH3CH2COOCH3 + H2O 1m

(b)(nucleophilic) addition-elimination NOT acylation 1m

Any3m

(c)(i) faster/not reversible/bigger yield/purer product/no(acid) (catalyst) required 1m

(ii) anhydride less easily hydrolysed

orreaction less violent/exothermic no (corrosive) (HCl) fumes formed

orsafer

orless toxic/dangerous expense of acid chlorideoranhydride cheaper

1m

(d) (i) C8H8O2 1m(ii)

2m

(e) (i) electrophilic addition 1m

(ii) CH3 -CH=CH-CH3 must show C=C 1m

(iii) nucleophilic substitution (1)

Max

3m

(15m)

END OF ANSWER SCHEME

skema k2 kimia stpm 2011 trial sabah

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