Answers Paper 2(2)

8/3/2019 Answers Paper 2(2)

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ANSWER AND MARK SCHEME (Pahang 2010)

QUESTION 1

1. (a) (i) sum of dependent on 2nd mark or reference to orders (1)

powers in rate equation (1)allow reference to annotated rate equation

2

(b) (i) (H2) 1 (1)

eg rate trebles as [H2] trebles not rate doubles as [H2] doubles (1)

(I2

) 1 (1)

eg rate doubles as [I2] doubles (1)

mark these points independentlyallow 1 mark for orders if not clear that both are first order

4

(ii) rate = k[H2][I2] (1)

must be [H2] not [H]

allow e.c.f from (b)(i)allow mark for correct answer for (b)(ii), even if (b)(i) wrong

1

(iii) rearrange correctly

correct substitution and correct consequential answer(1)

correct units (mol1

dm3

s1

) (1)

allow e.c.f. from (b)(ii)allow e.c.f. for incorrect arrangement

2

(c) 4 106

(1)allow e.c.f. from (b)(ii) and (b)(iii)could be worked by ratio from values in table or from rate equation

1[10]


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QUESTION 2

2. (a) HA H+

(aq) + A

(aq) orHA + H2O H3O+

+ A

(1)

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)

N e N a M g A l S i P S

F i r s ti o n i s a t i o ne n e r g y ( 1 )

( 1 )

( 1 )

E l e m e n t

(1)

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(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)

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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 Al3+ is [ 1 ]

higher than that in Mg2+ ion. Hence the size of Al3+ ion is smaller.

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

dissociation of the dehydrated Be2+ ion. [ 1 ]

b. (i) MgCO3 (s) MgO(s) + CO2(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 density and the polarising [ 1 ]

power towards the CO32- decreases. Hence the carbonates become more

stable. [ 1 ]

Total : [ 15 ]

QUESTION 4

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

Temperature not exceeding 550C [ 1 ]

C6H6 + HNO3 2HSO4-- + NO2

+ + H3O+ [ 1 ]

(ii) Nitronium ion, NO2+ [ 1 ]

(iii) Step 1

+ NO2+ H [ 1 ]

NO2

Step 2

H NO2 + H+ [ 1 ]

NO2

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

Cumene, used for making phenol. [ 1 ]

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+

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b. (i) CH3CONHC6H5 [ 1 ]

OH

(ii) CH3 [ 1 ]

(iii) (CH3)2C(OH)CH2CH3 [ 1 ]

TOTAL : [15]

SECTION B

5(a) Tendency orstrength orability orpower of an atom/element/nucleus to 1

attract/withdraw electrons / e

density / bonding pair / shared pair

In a covalent bond 1(tied to M1 unless silly slip in M1)

(If molecule/ion then = CE = 0) (NOT electron(singular) for M1)Mark as 2 + 2

Increase in size ornumber of shells orincreased shielding orbonding 1electrons further from nucleus

[NOT increase in number of electrons]

Decreased attraction for (bonding) electrons 1(tied to M3)

(If ion here, lose M3 and M4) (NOT attraction ofcovalent bond)

(Ignore reference to proton number or effectivenuclear charge)

(b) Hydrogen bonding (full name) 1

Diagram shows at least one+

H and at least one

F 1(If full charges shown, M2 = 0)

3 lone pairs shown on at least one fluorine atom 1H-bond indicated, between H and a lone pair on F

H F H F

+ +

: :

:

. .

. .

(If atoms not identified, zero for diag)

(Fl for fluorine - mark to Max 2)

(Max 1 if only one HF molecule shown, orHClshown)

Dipole results from electronegativity difference orvalues quoted 1

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(difference may be inferred)

(Allow explanation e.g. F attracts bondingelectrons more strongly than H)

Fluorine more/very electronegative oriodine less electronegative 1

orelectronegativity 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/molecules as a contradiction)

(Accept cubic lattice)

Diamond is macromolecular/giant covalent/giant atomic/giant molecular 1(NOT molecular or tetrahedral)

(Ionic/van der Waals = CE = 0)

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

(Covalent may be inferred from diagram)

(Treat diagram of graphite (without one of diamond)as a contradiction lose M2 but allow M3/M4])

Which takes much energy orcovalent bonds are strong 1

[15]

QUESTION 6

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

1is (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 capacityxT(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) 1moles 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 allow conseq marking after all mistakes but

note use of 2.12 g loses 2 marks

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(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

(c) H= Hc(reactants) Hc

(products) (or correct cycle)

1

Hc

(CH3OH) = Hc

(CO) + 2 Hc(H2) H

1= (283) + (2 286) (91) (mark for previous equation or this)

= 764 (kJ mol1

) ( units not essential but lose mark if units wrong)

1(note + 764 scores 1/3)

[15]

7. (a) During the electrolysis of aqueous sodium chloride, hydrogen is liberated at thecathode

(1)While chlorine is liberated at the anode.

(1)At the cathode,4OH- 2H2O + O2 + 4e

Hence for every mole of electron that passes through the electrolyte, one mole ofNaOH will be produced.

(1)NaOH + HCl NaCl + H2O

0.25 x 15.50No of moles of HCl used in titration = ------------------- = 3.88 x 10-3 mol

(1)1000

No of moles of NaOH produced = 3.88 x 10-3 mol(1)

Quantity of charge required ,Q = 3.88 x 10-3 mol x 96500 C = 374.4 C

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Using Q = It,374.4 = I x 25 x 60I = 0.25 A/

(1)

(b) (i) The partition law states that a solute, s distributes itself between the twoImmiscible solvents, ether and water in such a way that the ration of theconcentrations in the two solvent is constant.

(1)

Concentration of S in ether--------------------------------- = 12.50Concentration of S in water

(1)

(ii) The partition law only holds when the temperature is constant and the molecular conditionin both solvents. (2)

(iii) single extractionLet a be the amount of S left in the aqueous layer,8.00 a---------- = 12.50

aa = 0.59 g (1)

1st portion in 2 extractions,Let b be the amount of S left in the aqueous layer,(8.00 b)/12.50-------------------- = 12.50

b/25.0

b = 4.00 g (1)

2nd portion in 2 extractions,Let 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 Al3+ 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 H2O molecule, the O-H bond is weakened as the

electrons of the oxygen atom are strongly attracted towards the Al3+ ion.

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

to an uncoordinated H2O molecule forming the H3O+ ion which makes [ 1 ]

the solution acidic.

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[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 whichforms 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 Al3+ (aq) solution with sodium carbonateor 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 ]

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 available forsharing with protons.

b. (i) Add acidified KMnO4 to methylbenzene and heat. [ 1 ]

Benzoic acid is formed.

C6H5CH3 + 3[O] C6H5COOH + H2O [ 1 ]

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

C6H5COOH + PCl5 C6H5COCl + POCl3 [ 1 ]

Add concentrated ammonia

C6H5COCl + NH3 C6H5CONH2 + HCl [ 1 ]

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

Chloromethylbenzene is formed

C6H5CH3 + Cl2 C6H5CH2Cl + HCl [ 1 ]

Add aqueous sodium hydroxide and reflux

C6H5CH2Cl + NaOH C6H5CH2OH + NaCl [ 1 ]

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

White fumes of hydrogen chloride is observed

C6H5CH2OH + 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 O

CH3 C C O [ 1 ]

H

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

Hydrolysis of X :

CH3 O

CH3 C C O + H2O [ 1 ]

H

(CH3)2CHCOOH + OH

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(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- _____

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[ 15 ]

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