Chemistry Trial Paper 2010 _Pahang Answers Paper 2

8/8/2019 Chemistry Trial Paper 2010 _Pahang Answers Paper 2

1/10


2/10

1

(b) Ka = [H+][A]/[HA] or Ka = [H3O+][A]/[HA] (1) 1

(c) (i) Increases (1)

(ii) Decreases (1)

(iii) No change (1) 3

(d) Sodium ethanoate or sodium hydroxide (1)Buffer solution (1) 2

(e) (i)

(1)

(ii) Explanation for neon

Neons electron is in a lower (2p) shell attracted more strongly to (or lessshielded from) the nucleus (1)

Explanation for magnesium

more protons electrons in same shell or similar shielding (1)

(Total 10 marks)

Ne Na Mg Al Si P S

First

ionisationenergy (1)

(1)

(1)

Element


3/10

2

QUESTION 3

3. a. (i) Mg2+ ion: 1s22s22p6 [ 1 ]

Sr2+ ion: 1s22s22p63s23p63d104s24p6 [ 1 ]

(ii) Al3+

ion is smaller than Mg2+

ion. [ 1 ]

Al3+ ion and Mg2+ ion are isoelectronic. However the nuclear charge in Al 3+ is [ 1 ]higher than that in Mg 2+ ion. Hence the size of Al 3+ ion is smaller.

(iii) An aqueous solution of beryllium salt shows acidic properties due to the

dissociation of the dehydrated Be 2+ ion. [ 1 ]

b. (i) MgCO3 (s) MgO(s) + CO 2(g) [ 1 ]

(ii) The activation energy for the reaction is high. [ 1 ]

At room temperature the particles do not have enough energy to overcome

the activation energy. [ 1 ]

(iii) As the size of the cation increases, the charge d ensity and the polarising [ 1 ]

power towards the CO 32- decreases. Hence the carbonates become more

stable. [ 1 ]

Total : [ 15 ]

QUESTION 4

4. a. (i) Concentrated nitric acid and concentrated sulphu ric acid. [ 1 ]

Temperature not exceeding 550

C [ 1 ]

C6H6 + HNO3 2HSO4--

+ NO2+

+ H3O+

[ 1 ]

(ii) Nitronium ion, NO 2+ [ 1 ]

(iii) Step 1

+ NO2+

H [ 1 ]

NO2

Step 2

H NO 2 + H+

[ 1 ]NO2

(iv) Benzene is used as a solvent for oils and fats / can be converted to

Cumene, used for making phenol. [ 1 ]

+

+


4/10


5/10

4

(Allow explanation # e.g. $ attra%

ts&

on' ingele

%

trons more strongl(

t)an

0

1

Fluorine more/very electronegative oriodine less electronegative 1orelectronegativity difference too small in HIComparison required, may be implied .

HI dipole weaker or bonding e more equally shared - wtte 1(c) NaCl is ionic (lattice) 1

(Treat atoms/mole2

ules as a2

ontra3 i2

tion4

(Accept cu5iclattice)

Diamond is macromolecular/giant covalent/giant atomic/giant molecular 1(NOTmolecular or tetra

6

e3ral)

(Ionic/van3er Waals

7

8E

7

9

)

(Many) covalent/C-C bonds need to be broken / overcome 1(NOT just weakene

3etc.)

(8 ovalentma@ 5 e inferre3 from 3 iagram)

(Treat3iagram of grap

6

ite (wit6

out one of3iamon

3)

as a contra3iction A lose

B

C

5ut allow

B

D

/B E

])

Which takes much energy orcovalent bonds are strong 1

[15]

QUESTION 6

(a) (i) enthalpy change when 1 mol of a substance (or compound) 1

is (completely) burned in oxygen (or reacted in excess oxygen) 1at 298 K and 100 kPa (or under standard conditions) 1

(ii) heat produced = mass of water Sp heat capacity xT(ormcT) 1= 1504.1864 (note if mass = 2.12 lose first 2 marks then conseq)= 40100 J or = 40.1 kJ (allow 39.9 -40.2 must have correct units) 1

moles methanol = mass/Mr= 2.12/32 (1) 1

= 0.0663

H = 40.1/0.0663 = 605 kJ (mol1) 1(allow 602 to 608 or answer in J)

(note allowconseq marking after all mistakesFut

note use ofG . H G g loses G marks

(b) (i) equilibrium shifts to left at high pressure 1because position of equilibrium moves to favour fewer moles (of gas) 1

(ii) at high temperature reaction yield is low (or at low Tyield is high) 1at low temperature reaction is slow (or at high Treaction is fast) 1therefore use a balance (or compromise) between rate and yield 1


6/10


7/10

6

2nd

portion in 2 extractions,et b be the amount of S left in the aqueous layer,

(8.00 1.10 - c)/12.50----------------------------- = 12.50

c/25.0c = 2.00 g [ 1 ]

total extraction = 4 + 2 = 6.00 g [ 1 ]______

[ 15 ]

8. a. In aqueous solution, the Al 3+ ion which has the empty valence orbitals [ 1 ]

forms coordinate bond with water molecules producing [Al(H2O)6]3+ [ 1 ]

complex ions.

The Al3+

ion has a small ionic radius and a high charge density. [ 1 ]

In the coordinated H 2O molecule, the O-H bond is weakened as the

electrons of the oxygen atom are strongly attracted towards the Al 3+ ion.

The complex ion acts as a Bronsted - owry acid by donating a proton

to an uncoordinated H 2O molecule forming the H 3O+ ion which makes [ 1 ]

the solution acidic.

[Al(H2O)6]3+

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

(aq) + H3O+(l)

[ 1 ]

(i) Al2Cl6 (s) + 12H2O(l) 2[Al(H2O)6]3+ (aq) + 6Cl- [ 1 ]

Anhydrous aluminium chloride exists as a dimer which dissolves [ 1 ]in water forming [Al(H2O)6]

3+ and Cl- ions.

(ii) 2[Al(H2O)6]3+

(aq) + 3CO32-

(aq) 2[Al(H2O)3(OH)3](s) + [ 1 ]3 CO2 (g) + 3H2O(l)

The Al3+ ion solution contains the [Al(H2O)6]3+ complex ion which

forms the solid aluminium hydroxide with a strong base such as [ 1 ]a carbonate or a sulphide. Aluminium carbonate and aluminium sulphide are not formed in [ 1 ]the reaction between the Al

3+(aq) solution with sodium carbonate

or sodium sulphide solutions.

2[Al(H2O)6]3+ (aq) + 3S2- (aq) 2 [Al(H2O)3(OH)3](s) + [ 1 ]

3H2S (g)

(iii) Aluminium chloride is a simple covalent compound. The molecules are

held by weak van der Waals forces. [ 1 ]

It has a low melting point and sublimes at 1800C. [ 1 ]

It fumes in moist air because of hydrolysis. [ 1 ]

The white fume is hydrogen chloride gas.

AlCl3 (s) + 3H2O (l) Al(OH) 3 ( s) + 3HCl (g) [ 1 ]

_______

[ 15 ]


8/10

7

9. a (i) phenylamine < ammonia < ethylamine [ 1 ]

(ii) The strength of a base depends on its ability to donate its lone pair [ 1 ]

electrons for sharing with a proton.

In phenylamine, the lone pair electron on N atom is delocalised into [ 1 ]

the ring.

The p orbital of N atom overlaps with those of the carbon atoms in

the ring. Thus it is not readily shared with protons. [ 1 ]

In ethylamine, the ethyl group has an electron donating effect. [ 1 ]

This makes the lone pair electron on N atom more avail able for

sharing with protons.

b. (i) Add acidified KMnO 4 to methylbenzene and heat. [ 1 ]

Benzoic acid is formed.

C6H5CH3 + 3[O] C 6H5COOH + H2O [ 1 ]

Add PCl5 and heat, benzoyl chloride is formed [ 1 ]

C6H5COOH + PCl5 C6H5COCl + POCl3 [ 1 ]

Add concentrated ammoniaC6H5COCl + NH3 C6H5CONH2 + HCl [ 1 ]

(ii)Chlorine gas , sunlight and boiling methylbenzene [ 1 ]

Chloromethylbenzene is formed

C6H5CH3 + Cl2 C6H5CH2Cl + HCl [ 1 ]

Add aqueous sodium hydroxide and reflux

C6H5CH2Cl + NaOH C 6H5CH2OH + NaCl [ 1 ]

To prove the presence of OH group in (ii), add PCl5 [ 1 ]

White fumes of hydrogen chloride is observedC6H5CH2OH + PCl5 C6H5CH2Cl + HCl + POCl3 [ 1 ]

_____

[ 15 ]

10. a. X is an ester ( sweet smelling liquid ) [ 1 ]

Y is phenol ( formation of white precipitate with bromine water ) [ 1 ]

Reduction of X would produce two alcohols. Y is a tertiary alcohol [ 1 ]

And Z is a primary alcohol.

Structure of X

CH3 OCH3 C C O [ 1 ]

H

Name of X: phenyl-2-methylpropanoate [ 1 ]


9/10

8

Hydrolysis of X :

CH3 O

CH3 C C O + H 2O [ 1 ]

H

(CH3)2CHCOOH + OH

(Y) [ 1 ]

Reaction of Y with bromine water

Br

OH + 3 Br2 Br OH + 3HBr [ 1 ]

Br

Reduction of X :

CH3 O

CH3 C C O + 4[H] (CH3)2CHCH2OH (Z) [ 1 ]

H

+ OH [ 1 ]

Dehydration of Z :

(CH3)2CHCH2OH (CH3)2C= CH2 + H2O [ 1 ]

b. (i) Ethyl magnesium bromide can be prepared by heating

bromoethane with magnesium metal [ 1 ]

at 350C in dry ether [ 1 ]

(ii)CH3CH2MgBr + CH3COCH3 CH3CH2C(CH3)2 + MgBr+

[ 1 ]

O-

CH3CH2C(CH3)2 + MgBr+ + H2O CH3CH2C(CH3)2 + Mg(OH)Br [ 1 ]

O-

_____

[ 15 ]


10/10

9

Chemistry Trial Paper 2010 _Pahang Answers Paper 2

Documents

Transcript of Chemistry Trial Paper 2010 _Pahang Answers Paper 2