pahang 2008 stpm chem - p2 ans

8/8/2019 Pahang 2008 Stpm Chem - p2 Ans

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STPM Trial Examination 2008

Chemistry Paper 2

Marking scheme

Section A [40 marks]

1.( a ) Energy released when one mole of lithium chloride is formed from 1 mole of

Na+(g) and 1 mole of Cl

(g) ions under standard conditions. [2M]

( b ) Lattice energy is inversely proportional to size of cation.

Size of Li+ smaller than Na+ [2M]

( c ) Hsoln = Hhyd lattice energy [1M]

( d ) ( i )

Li+ ( g ) + Cl ( g )

LiCl ( s )

Li+ (aq ) + Cl ( aq ) [2M]

( ii ) Hsoln = [ ( 499 ) + ( 381) ] [ 848 ]

= 32 kJ mol1

[2M]

( iii ) Temperature increases [1M]

Total : 10 M


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2. (a) the rates of the forward and reverse reactions are the same 1Mthe amount of substances present does not change with time 1M

(b) (i) iron 1M

(ii) catalyst increases the rate of the forward and reverse reactionby the same amount 1M

(c) (i) NH3 + HCl NH4Cl 1M(ii) sp3 hybrid orbitals, 1M

sp3

hybrid orbitals 1M(iii) N atom has an unbonded pair of electrons to be donated to the

central atom 1M

(iv) hydrogen and nitrogen gases 1M

ammonia has dissociated to the two gases. 1M

3.( a ) Cl [1M]

( b ) Ar [1M]

( c ) ( i ) Si [1M]

( ii ) SiCl4 + 2H2O SiO2 + 4HCl [1M]

( d ) ( i ) Aluminium [1M]

( ii )

Al2O3 + 6HCl 2AlCl3 + 3H2O

Al2O3 + 2NaOH + 3H2O 2Na [Al(OH)4]4 [2M]

( e ) ( i ) hot, concentrated NaOH [1M]

( ii ) 3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O [1M]

( iii ) weedkiller ( etc...) [1M]

Total : 10M

4.( a ) ( i ) Polymers that are stretchable. [1M]

( ii ) natural rubber ( etc....) [1M]

to make gloves,...... [1M]


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H2 CO

H2O

H2 H2O

[ CH3COO ]

[ CH3COOH ]

Section B [60 marks]

5.( a ) ( i ) Pressure increased :

Position of equilibrium shifts to the left.

Backward reaction involves a reduction in volume.

Temperature increased:

Position of equilibrium shifts to the right.

Forward reaction is endothermic. [4M]

( ii )

( P ) ( P )

Kp = kPa

( P )

P = PCO = 183kPa , P = 90 kPa

Kp =90

183183x

= 3.72 x 103 kPa [4M]

( b ) ( i )

[CH3COO][ H

+]

Ka = [1M]

[CH3COOH]

( ii )

pH = pKa + lg

= lg ( 1.80 x 105 ) + lg(40.0

20.0)

= 4.44 [3M]

( iii ) [ H+

] =050.0

1000.1 14

x= 2.00 x 10

13mol dm

3

pH = lg [ H+ ] = 12.70

Therefore, change in pH = 12.70 7.00 = 5.70 [3M]

Total : 15M


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6.(a) It states that the heat evolved or absorbed in a chemical change is the same

whether the change is brought about in one stage or through intermediate stages. [1]

(b) (i) The enthalpy change when one mole of carbon monoxide is completely burnt in

oxygen under the standard conditions of 298K and 1.0 atm. [1]

CO (g) + 1/2 O2 (g) CO2 (g) [1]

(ii)

Equation 1: CO (g) + 1/2 O2 (g) CO2 (g) H

c [1]

Equation 2: 2 Fe (s) + 3/2 O2 (g ) Fe2O3(s) H

= -822kJ

Equation 3: Fe2O3(s) + 3 CO (g ) 2 Fe(s) + 3 CO2 (g) H

= -27kJ

Add equations (2) and (3), [1]

3CO (g) + 3/2 O2 (g) 3 CO2 (g) [1]

H = (-822) + (-27)

= -849 kJ [1]

H

c = -283kJmol-1 [1]

(b)

Na (s) + Cl2(g) NaCl(s) H

f [1]

Na+

(g) + Cl-

(g) NaCl(s) H = lattice energy [1]

H

f= 604 260 + H [1]

Since NaCl is ionic , its lattice energy is very negative. [1+1]

So, H

fis very exothermic [1]

The lattice energy is sufficient to overcome the enthalpy of atomisation

and ionization energy of sodium [1]

Total mark: 15M


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7(a) An ion containing a central metal ion with empty orbitals and ligands with lone pairelectrons.

Ligand forms dative bond with the central metal ion. [3M]

(b)(i) Tetraamminecopper(II) ion [1M]

Square planar [1M]

( ii ) Tetraamminedichlorochromium(III) ion [1M]

Geometrical /Cis-trans isomerism [1M]

trans cis

Cl Cl

NH3H3N H3N NH3

Cr Cr

NH3 H3N Cl

H3N

Cl NH3

[2M]

(c) Involves reactants and catalyst in different phases. [1M]

Finely-divided platinum metal and vanadium pentoxide used as catalyst [1M]

SO2(g) + O2(g) --- > SO3(g) [1M]

SO2 and O2 molecules are initially adsorped on the surface of the solid catalyst. [1M]

Bonds between atoms in the molecules are weaken. [1M]

Ea is lowered. [1M]

Total : 15 M


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8. (a) (i) The aluminium oxide ( gibbsite) layer is sandwiched betweentwo silicate layers. 1M

Four oxygen atoms are shared between one AlO6 unit and

two SiO4 units 1M

(ii) No hydrogen bonding in the 2:1 clay arrangement 1MLittle attraction between the silicon oxygen of the tetrahedral(Silicon/

oxygen ) sheets in the different layers. 1M

(b) (i) The aqueous solution of aluminium salts contains hydrated aluminiumion of [Al(H2O)6 ]

3+. 1M

The high charge density of the Al3+ ion polarizes the O-H bonds of water

1M

of water molecules and this weakens some of the bonds.

One or more of the water molecules can donate a proton to produce H3O+

1M

[Al(H2O)6]3+ + H2O [ Al(H2O)5(OH)]

2+ + H3O+ 1M

(ii) Due to high charge density ( charge/radius) ratio of Al3+ ion,the ion 1Mis able to polarise the electron cloud of an anion such as O

2-or Cl

-1M

to give a high degree of covalent character in the ionic bond. 1M

At room conditions, aluminium chloride exists in the dimeric form 1M

of Al2Cl6 with weak van der waal forces between the discrete Al2Cl6molecules. 1M

Due to the weak forces between molecules ,the solid Al2Cl6 sublime

to give Al2Cl6 vapour. [1M]

Total: 15M


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9.(a) 2-phenylmethylbutan-2-ol is a tertiary alcohol. Hence, a Grignard

reagent and a ketone is required for the reaction.

(i) Any two of these pairs.

[4M]

(ii)

[1M]

(b) (i)

[4M]


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9(a)(ii) Compound (I) , is produced. Compounds (II), (III) and (IV) do not

react with AgNO3.Conditions: Pass chlorine into methylbenzene boiling under reflux and in the

presence of sunlight.

[2M]

(iii)

[2M]

(iv) is insoluble in water, but soluble in ethanol. So ethanolic silver

nitrate is used to enable to dissolve in the reaction mixture.

[2M]

Total : 15 M


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10. (a) (i)

Element C Cl H

% by mass 66.4 28.1 5.5

Mole ratio66.4 =5.5

12

28.1 = 0.79

35.5

5.5 = 5.5

1

Simplest ratio 7 1 7

Empirical formula ofP is C7H7Cl [2M]

(12x7 + 7 + 35.5)n = 126.5

n = 1

Molecular formula is C7H7Cl [1M]

(ii) P is an aromatic compound [ ratio C:H is 1:1]

Cl atom is on side chain since it can be hydrolysed.

P is

[1M]

Hydrolysis of P produces phenyl methanol, Q

+ KOH + KCl [1M]

Controlled oxidation of Q produces benzaldehyde, R

+ (O) + H2O [1M]

Benzaldehyde, R reacts with 2,4-dinitrohydrazine to form a hydrazone,S

CH2Cl

CH2Cl CH2OH

CH2OH CHO


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+

+ H2O 1M

Further oxidation of benzaldehyde, R produces benzoic acid (T)

+ (O)

1M

(b) (i) Nucleophilic substitution: 1M

+ KOH + KBr 1M

Mechanism:

Dissociation of 2-bromo-2-methylpropane to produce a carbonium ion:

1M

Combination of the carbonium ion with OH-from KOH:

+ OH-

1M

CHO COOH

C = O

H

H2N NO2NH

NO2

NO2NH

NO2

C = N

H

C

CH3

CH3 CH3

Br

C

CH3

CH3

CH3

Br

slow

C

CH3

CH3 CH3+

+ Br -

C

CH3

CH3 CH3

OH

C

CH3

CH3 CH3+

C

CH3

CH3 CH3

OH


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(ii) The rate equation is rate = k[2-bromo-2-methylpropane] 1M

is independent of conc [OH] or [KOH] 1M

hence, concentration of KOH has no effect on rate. 1M

Total : 15 M

pahang 2008 stpm chem - p2 ans

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