Transcript
Page 1: STPM Trials 2009 Chemistry Answer Scheme (SMI Ipoh)

S.M.K. ST. MICHAEL IPOH STPM TRIALS 2009 CHEMISTRY PAPER 1 (ANSWERS)

1 B 26 A 2 A 27 C 3 C 28 D 4 A 29 B 5 D 30 B 6 B 31 C 7 B 32 C 8 C 33 C 9 B 34 D

10 A 35 B 11 C 36 B 12 B 37 B 13 D 38 C 14 D 39 B 15 B 40 A 16 C 41 D 17 C 42 D 18 A/C 43 B 19 C 44 A 20 C 45 B 21 A 46 B 22 C 47 D 23 D 48 D 24 B 49 A 25 C 50 A

Page 2: STPM Trials 2009 Chemistry Answer Scheme (SMI Ipoh)

SMK St. Mich:lel, IpOh

Name:

2009 Trilll ex:lIl1inatiun : pg I 19

Class:

ST. MICHAEL'S INSTITUTION, IPOH 2009 TRIAL EXAMINATION

CHEMISTRY PAPER 2 TIME: 2 y, HOU RS

Instructions to candidates : Answer ALL the questions in Section A in the spaces prov ided. All work ing MUST be shown. For numerica l answers, units MUST be quoted where they are appropriate.

Answer any FOUR questions from Section B. For this section, wr ite your answers on test pad papers provided. Begin each answer on a fresh sheet of paper and arrange your answers in numeri cal order. Tie yo ur answer sheets to this book let.

A Data Booklet is prov ided.

Section A [40 marks] Answer ALL the questions in this section.

I. (a) Arrange the elements (i) in the Third Period of the Period ic Table in the order of decreas ing atomic radius,

Na , Mg , AI • Si , P , S • CI , Ar ./

( ii) of chlorine, bromine and iodine in Group 17 of the Period ic Table in the order of increasing fi rst ionisat ion energy.

I • Be . CI,( NOT I, • Be, • CI, [2J

(b) Draw the Lewis structure and state the l?eomet -v for the SF, molecule and the 13 - ion. SF6 molecu le 13 - ion

Lewis stnlcture Lewis structure

•• •• • F • • F • • • • • ., ., [ .. " .. r •• •• : I : I

, I : • F • S • F • •

• , , • .. xx xx .. .. .. . , , . ,( • F • • F • • • • • MUST show with different dot for the e -.. ..

received ,(

GeometrY: octahedral ./ Geometry: linear -/

(c) State why (i) SF, is used as an insulating gas in high voltage e lectric appliances,

SF, is chemically inert-/

(ii) SI6 do not exist,

Due 10 the steric hindrance./ and because the $- 1 bond is weak. hence mak ing SI6

unstable ./

(iii) h - and Br3 - ions exist but.F3 - ion does not exist under normal conditions.

TIle repU lsion between the lone pair of e lectrons on adjacent fluorine atoms makes

[4J

the F 3 - unstable ./ [4]

Page 3: STPM Trials 2009 Chemistry Answer Scheme (SMI Ipoh)

SMt< St. Michael. Ipoh 2009 Trial examination: pg 2 19

2. (a) p • and Q ~ ions are isoe lectronic with Ihe ~~ X isotope.

(i) Identify P ~ and Q ~ ions.

P has 19 e, P = K, p . = K ~ ./

( ii ) Elements P and Q fo rm an ionic compound PQ. The entha lpy changes involved in the formation of thi s compound under standard conditions are as follows.

P (s) --> P (g) y, Q, (g) --> Q (g) P(g)--> p ' (g) + e Q (g) + e -->Q - (g) p (s) + y, Q, (g) --> PQ (s)

b. HO::: + 77 kJ ,,11' = + 121 kJ "H' =+ 4 19kJ "He = _ 349 kJ ,,11' = - 436 kJ

Write equation which represents the lattice energy of compound PQ and calculate the va lue of the lattice energy.

- 436 P (s) + y, Q, (g) --> PQ (s)

l +77 l+121 P (g) Q (g) LlH laC!

1+419 1-349

P , (g) + Q - (g),- ----"

Eq uation , P ' (g) + Q - (g) --> PQ(s)./ - 436 = + 77 + 4 19 + ! 21 - 349 + b.H lall ./

b.H lall = - 704 kJ ./ (with unit and negative sign)

(b) Carboxylic acids are organ ic acids obtained from plants and animals. The structura l formu lae of four carboxylic acids are given below.

CH,COOH , @-eOOH , O,N-@-eOOH Ha-@--COOH

(i)

[2]

[3]

Arrange the acids in the order of increasing acidity,

CH,COOH < @-eOOH < O,N-@-eoOH < H<>-@-eOOH ./

Oi) Which of the above acids has the strongest conjugate base?

CH)COOH ./

(iii) State the effect ofthe-N~ group on the acidity of H~OOH Presence of the -NOz group will increase the acid strength ofthis compound ./

-NO} , an electron- withdrawing group ./ helps to withdrav .. ' electrons from the

O-H bond in thc-COOH group making the O---H bond easier to break./,

OR -N02 • an electron- withdrawing group./ will disperse the negat ive charge on

the carboxylate ion, -COO ~ ./ • thus stabilises the ca rboxylate ion [5]

Page 4: STPM Trials 2009 Chemistry Answer Scheme (SMI Ipoh)

S MK St. Michael. Illoh 2009 Tri:.1 exa rnin a lion : Ilg 3 / 9

3. (a) The tetrachlorides of Group 14 elements, exam ple CC I.1 , SiCI4 • GeCI4 • SnCI4 and PbCI4 are liquids at room temperature. All the tetrachlorides, with the exception of CCI-, , are hydrolysed in aqueous solution to form acidic so lut ions.

( i) State the mo lecular shape of a ll the Group 14 tetrachlorides.

Tetmhedral ./

(ii) Write a ba lanced equation fo r the hydrolys is of SiCI4 .

S;CI, (I) + 2 H,O (I) -----> S;O, (s) + 4 He l (g) ./

(iii) Explain why CC I~ does not undergo hydrolysis.

has the e lectron ic configuration of I s22s22p2 ./

does not have empty d orbital s, cannot ex pand its va lence ./

OR cannot fonn the inrermediate coord inate bond with the water molec ule

(b) Aqueous aluminium sulphate contains [AI(H20)6]1'. Aqueous ammonia is added to aqueous a luminium sulphate unti l in excess. The reaction that occurs in limited aq ueous ammonia is as follows. [AI(H,O).)'· (aq) + 3 OH - (aq) ) AI(H,O)J(OH), (s) + 3 H,O (I)

(i) Describe one observation in the above reaction.

A white prec ipitate is formed ./

(ii) What chemical nature of the aluminium ion is shown in (a) (i) ?

Acidic in nature ./

(ii i) What happens when excess aqueous ammonia is added to the solution ? Write an equation for the reaction invol ved.

The white precipitate is soluble in excess aqlleous am moni a ./ and form a colourless

solut ion ./

) [AI(H,Oh(OH),1 - (aq) + H,O (I)'"

(tv) What chemical nature of the aluminium iOIl is shown in (a) (iii) ?

Acidic in nature ./

4. (a) Ethylamine, C1HsNHl • is a foul smelling liquid that reacts with dilute hydrochloric acid at room temperature.

( i) Write a balanced equation for the reaction between ethylamine and dilute hydrochloric acid.

C21"I,NH2 + HC1--» C2HsNH) .. CI - ./

(ii) Name the type of reaction taking place between ethylami ne and dilute hydrochloric acid .

Neutra li sation reaction . ./

(iii) State two observable changes for the reaction between elhylamine and dilute hydrochloric acid.

The fi shy smell of ethylamine disappears . ./

Temperature of mi xture rises ./ as heat of neutra li sation is li berated

[41

(6)

(4)

Page 5: STPM Trials 2009 Chemistry Answer Scheme (SMI Ipoh)

SMK St. Michel, Ipoh 2009 Trial examination : pg 4 / 9

(b) State Ihe reagents and wrile ba lanced equJlions ror the preparation o f ethylamine from the fo ll owing compounds.

(c)

ethanol

cone. in excess ./

ethanol

(ii) CH,CN

CHlCN + 2 H! -----» ./ CHj CH2NH2 ./ 180 °C

Aspirin has the following structure. o I I

©:(O--<:---CH'

C-OH

II o

O - K +

©:(e--o -K '

II o ./

Give Ihe products fonned when aspirin is heated with aqueous polassium hydroxide solution .

o II

• K . O--<:---CH,

Section B [60 marks]

[4J

[2J

5. (a) (i) The relat ive isotopic mass of an isotope is the mass of one atom of tile isotope ./ re lative to 1/ 12 the mass of one atom of carbon- I 2 ./ .

Oi)

(iii)

2B + 2 mol I

- mol x

_I = 1.278 ./ 2x 2x+6

J H,

2x + 6 = 2.556x x= 10. 79(4sl)./

IO a+ llb = 10.79 ./ a+b

--->

10 a+ II b = 10.79 0+ 10.79 b

0.21 b = 0.79a a I - =- ./ b 4

B2H6 ./

I mol

1.278 I -- rno 2x+ 6

[2J

e :l·j , ;). ;, 1:3 _I ". J1g -- I' 000

7<. ' 1 =- 1:3 -0·1, &

/

)( ~ 3/ ;/0 , ;).78

_ [JJ

- 10 ·79 ( If ~l)

[2J

Page 6: STPM Trials 2009 Chemistry Answer Scheme (SMI Ipoh)

S MK St. M ichael, IllO r. 2009 Trial examination: pg 5 / 9

(iv) Relative in tensity

4 - - - - - - - -J 2

0 L-~8--~9~~ILO--~II~·

axes ./ 2 peaks ./

m

• (b) (i) A chemical substance that can donate proton, H ., a proton donor./

CH,COOH (ag) + H,O (I) co CH,COO - (ag) + H,O' (ag) ./ NOT CH,COOH (ag) co CH,COO - (ag) + H' (ag)

(i i) Ethanoate ion ./ NOT CH~COO -, NOT sod ium ethanoate

(ii i) pH = pK a +

= - log lO (1.80 x W- 5) + log lo [ 0~9:00 ) ./ 1.00

2.00 = 4.74 - 0.0458

= 4.69(3sf) ./

NOT ( 0.900) 1.00

(iv) CH,COOH (ag) + OH - (ag) co CH,COO - (ag) + H,O (I) ./ NOT W (aq) + OH - (ag) co H,O (I)

[2J

[2J

[I J

[2J

[l J

6. (a) (i) CH,COOH (I) + CH,CH,oH (I) co CH,COOCH,CH, (I) + HOH (I) [IJ

(ii) 120.0 = 2 mol 60

- x mo l (2 - x) mol

92.0 = 2 mol 46

- x mol (2-x)mol

o + x mol x mol

4x 2 - 16x + 16 = x2

3x 2 - 16x+ 16 = 0

(3x - 4)(x - 4) = 0

4 x = -3

o + x mol x mol ./

4 2 Ethanol left = 2 - - = - mol ./

3 3 2

Mass left = 3"(46) = 30.7 g (3 sf)

(b) II, O,(g) + 2 W (ag) + 2. ---> H,O (ag) 2 H,O (I) + 2. ---> H, (g) + 2 OH - (ag) Cu" (ag) + 2. ---> Cu (s)

- (I) - (2) - (3)

At anode, NO]- can't react as reducing agent./

E' = + 1.23 V E' =- 0.83 V EO = + 0.34 V

H20 reacts as reducing agent and is oxidised to O2 ./ O2 gas is liberated H,O (ag) ---> y, 0, (g) + 2 H • (ag) + 2e ./

At cathode, E O ofCu2+ I Cu is more positive than that of H 20 I OH - ./

:;:::) Cu2" is a stronger oxidising agent than H20 Cu2" is chosen to be reduced to reddish brown copper METAL ./ Cu2• (aq) + 2e ~ Cu (s) ./

./ [4]

[6J

Page 7: STPM Trials 2009 Chemistry Answer Scheme (SMI Ipoh)

SMK St. Michael, Ipoh

(e) •• •• CI·· •• CI ••

· x X· •• AI

· X •• · X

•• CI •• •• CI ··

•• ·· CI •• X·

AI X·

•• CI ··

2009 Trial elimination : pg 6 / 9

AI : 2.8.3, Is2 2S2 2p6 3s2 3pt C1 : 2.8.7, Is2 2S2 2p6 3s2 3ps AbC I6 form s di mer with sha rin g electrons with CI atoms ./ AI has empty p orbita ls to rece ive the lone pair of e lectrons from CI atom ./ 2 coordinate bonds are formed ./

•• •• Each atom achieved the stable noble gas max diagram ./ configurat ion./ [4}

7. (a) ( i)

( ii )

The graph shows a minimum boil ing point .:::::> a maximum vapou r pressure PT > P A + Ps calculated from Raoult's ./

Positive deviat ion from Raoult' s law ./

100 "' __

T,

Boi ling point I °C 78.1

o 12 C, 96 100 Composition of ethanol

Whet! the mixture with the composition of 12 % of ethanol is di slilled, the mixture will boi l at T t • ./

The vapour produced with the composition of C2 is richer in ethanol . ./ When this vapour is condensed, the composition of the liquid obtained is the same as the compos ition of its vapour, C2 . ./

If the vaporisation and the condensation processes are repeated, the disti llate produced at 78.1 °C is an azeotrope I azeotropic mixture . ./ The res idua l liquid in the distillation flask is pure water and evenlUa lly distilled over at

12J

100 °C . ./ No ethanol will be obtained . [5]

(b) (i) White fumes of He l arc given off and white precipitate of 8e(OH)1 ./ is formed BeCl, (s) + 2 H,O (I) ---> Se(OH), (s) + 2 HCI (S) -'

Be(OHh dissolves in excess NaOH to produce a colourless solution ./ Be(OH), (s) + 2 01-1 . (aq) ---> [Be(OH), J' · (aq) -'

: .. JJ:k (i i) Ye llow precipitate of silver R-itmte./ is formed,

does not dissolve in ammonia aqueous ./ AS' (aq) + I . (aq) ---> AgI (5) -'

(c) CO2 s imple covalent molecu les./ Weak Van der Waal's forces between molecules ./

8. (a) (i) All Group 2 elements react with water to form their respective hydroxides with the liberation of hydrogen ./. M (5) + 2 H,O (I) --+ M(OHh (aq) + H, (S) -' The reactivity increases due to the increase in the reducing strength of the metals ./

(ii) Nitrates of the Group 2 elements dissoc iate to respective metal oxide when heated

2 M(NO,), (s) ---+ 2 MO (5) + 4 NO, (g) + 0 , (g) -'

" Down the Group The cationic size of M2+ 't, the cationic charge density J.. ./ The polarising power of the cation J.. ./ The thennal stability ofthe nitrates 't ./

max 13J

(3)

(2)

[3J

[4)

Page 8: STPM Trials 2009 Chemistry Answer Scheme (SMI Ipoh)

,sMK.st. Michael, Ipoh 2009 Trial uamination : pg 7 / 9

9. <a)

Homogeneous catal yst - the cata lyst is in the same physical state / phase of matter as the reactants (products are not included) ./ The oxidation state of nitrogen atom temporarily chan ge from +2 in NO to +4 in N02 and back to +2 in NO ./ NO provides an alternative reaction pathway of lower activation energy, Ea ./ The intermediate is NO] ./

(ii) Heterogeneous catalyst - (he catalyst is in a different physical stale / phase of matter from the reactants (products are not included)./

Pt, a transition metal, has empty orbitals ./ S02 and O2 molecules adsorb (adsorption) on the Pt surface, form temporary bonds with S02 and O2 molecules ./ New bonds fonned between S02 and 0 1 molecules and forms an activated complex, thus lowering the aetivat~on energy for the reaction ./

rv1'f" 3.J NaOH (aq) (~) MnO,-1 H -

C4H100 Q

decolourises Mn04~ / H ~

orange precipitate with 2,4-d initrophenylhydrazine

R gives orange precipitate with 2,4-dinitrophcnylhydrazine

I - contains carbonyl group, ---C = 0 => aldehyde or keton0

Both Q and R react with alkaline iodine to fonn a yellow precipitate

H H H I I I

Q - alcohol contains -C-CHJ, Q = CHJCH,--{:-CH, , P = CHJCH,--{:-CHJ I . OH 8'

I OH ,/

o 0 II II

I Br ,/

Q is oxidised to R - ketone, contains ;, ,,\

- C- CH)i "./ . R = C H]- CHr-C---CH3 ./ '-

CHJ CH,

I dt)' ether I CHJCH1---C- Br + Mg --::---:-:-:c::-'

reflux 35°C CHJCH,--{:-MgBr

I H

P

I H

S ,/

T reacts with PCls to liberate white fumes - T has -OH grouP2)

CHJ

I CHJCH,--{: = 0

R - ketone

CHJ

I

CH, I

+ CH,CH,--C--MgBr I

H S - Grignard reagent

CH]CHz--C-OH + MgBr · + H20

I CH(CH,)CH,CH,

dt)' ether )

[4]

[4]

max [8]

Page 9: STPM Trials 2009 Chemistry Answer Scheme (SMI Ipoh)

Cb) Ci) Phenol is a weak acid, reacts with aq ueous sodium hydroxide to produce a water­so luble ionic salt ./

@-oH + NaOH (aq) room temperature) @-o-Na++ H20'/

Pheno l does n OI react with aqueous sodium carbonate because phenol is a weaker acid than carbonic acid ./ [31

o II

(ii) Methanoic acid / formic acid, H--C-OH oxidised to carboxy lic acid.

contains a ldehyde group ./ , is

(iii)

10. Ca) 0)

o II KMo0 4 . H2S0 4 H-C-OH + [0]

Styren e undergoes addition polymeri sation ./, forms a transparent solid when exposed to air.

H H I I

n C~C

l~~ phenylethcnc styrene

poly(phenylcthenc) / poly(styrcne) / abbrev iation: PS

Relative molecular mass (C7 1-4) n = 180 ( 12 x 7 + 6) n = 180 :::::) 90 n = 180 => n = 2./ The molecular formula = Cl4HIl ./

Oi) The structure of W

H H

@J~~-@ .' Cis-trans isomers of W

The structure of X ;

H H I I

@-C~C--@

w

I~ H-C ~ c-@ .'

+ HBr ~ gas or

concentrated

H H I I

@ii*--@ H B,

y .'

[2]

[2]

[2]

Page 10: STPM Trials 2009 Chemistry Answer Scheme (SMI Ipoh)

SMK St. Michael, Ipoh 2009 T rial u.a min ation : pg 9 / 9

I ~ 2 H-C = C-@

X

I~ H-C - C- I()

I I~ H B,

major product Z./

+

+

2 HBr - > gas or

concentrated

I ~ H-C - C-@

I I Sr H

minor product Z

plane of mirror

(b) When aldehydes react with a solution of2,4-dinitrophenylhydrazine, a small molecule of H20 is eliminated ./

H I

R-C = O +

O,N

H'N-NH~-NO' room temp.

o

I O,N,

R-C = N-NH--@-NO'

(i) reagent + condition ./ , equation ./

Bc(CH,).B, + 2 KCN reflux

N = C(CH,).C = N + 2 KB,

N _ C(CH,).C = N + 4 (HJ ---t H,N(CH,).NH,

Reducing agents 1. Na I C2HsOH at room temperature 2. H, (g) / Ni at 180 "C 3. LiAI~ I dry ether, followed by H30 ~ at room temperature

H H 0 ° (ii ) n

I I c---------, II II H-N-(CH,).-N-j-H + n HO!--C-(CH,).-C-OH _________ _ J

-f1 H O O~ I I II II -(CH,).--N-C-(CH,).-c n + >

<ckchong\exam\chemistry\09u6essay2.doC>

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