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SIJIL TINGGI PERSEKOLAHAN MALAYSIA

MARK SGHEME (SKEMA PEMARKAHAN)

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SGHEME PAPER 1

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1D2B3D4A5C6A7B8B9D

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21 C22C23D24C25A.26D27D28D29c30D31 A32D33D34D35D36D37438A39C40D

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2

JADUAL SPEFIKASI ITEM (JSl)PEPERIKSAAN PERCUBAAN STPM 2OO9

NEGERI JOHOR

No. I oprc

Question No. No. ofquestionseaS Moderate Difficult

I Mattter 1,46,43 2 4

2 Electronic structure of Atoms 3,4,41 aJ

aJ Chemicalbondinq 5.6 24 Reaction kinetics 7,8 z

5 lonic Equilibria 9, 10,20,42 4

6 Chemical Equilibria 11, 12 z

7 Phase Equilibria 14 13 z

I Electrochemistry 44,45 15 16 4

9 Thermochemistry 18 17 190 Period 3 and Group 2 21 1

I Group 13 22 1

2 Group 14 23 1aJ Group 15 24 1

4 Group 17 25 1

5 d-Block elements 26 1

6 Chemistry of Carbon 27,29 31,48 4

Hydrocarbon 28,47 37 49 4

8 Haloalkanes 40 1

9 Alcohols ?A 38 z

20 Carbonyl compounds 32, 33 39 J

2l carboxvlic acids22 Carboxvlic acids derivatives 35 1

23 Amines 30 1

24 Amides25 nitriles 50

26 rmino acids27 rolymerisation 34 1

Total no. of questions 15 27 8 50

3

Question 1(a) (i)

(1)

(1)

(1)

?

I(

(1)

(1)(ii) (8X2X1/8) + (6X2Xl

(b) (i) A set of temPerature

equilibrium.

(ii)

12\=g

and pressurevalues where solid,liquid and gas exist

(1)

General sketch of graPh'

The correct axes.

The labellings.

( i i i )becauseitstr ip|epointoccursatapressureabovethenorma|atmospheric(1)

pressure.

Tamparaturs /bC

(iv) PV = nRT @ o' =#

p = 5.50 x 8.31 x (24.0+273) x 10-3 kPa

10.0 x10-3 x 44.0

= 30'9 kPa

Pt = 30.9 + 93.7 = 124'6 kPa

(1)

(1)

(1)

Total: 10 marks

Question 2

(a) Ca2*(g 1+ 2e-+ 2Cl(g) (This is the only answer for the top line)

Ca'*(g) +2e- +Clz(g)

Ca. (g) + e- + Clz (g)

Ca(g) + Clz (g) (state symbols and electrons essential)

Note:C|zto2C|canbeinanyorderbutCamustbeinsequence)Max.3 marks

(b)Enthalpyofformat ion=178+590+1145+(2x121).(2x364)-2237= - 810 kJ mol-1

(c) AH = -AH(lattice formation) + EAH(hydration)

(or cycle with state symbols' numbers or labels)

= 2237 - 1650 - (2 x 364)= '141kJ mol-1

(d) Moles of NHaCI = 2t53'5 = 0'0374

Heat absorbed = 15 x 0'0374 = 0'561

Q=mcATLT = Q/m6'= (0'561x1000)/(50x4'2) = 2'6 (oC)

(allow 2.5 to 2.1)canuse 52; ignore units, answer must be at least 2 sig figs)

Final temPerature = 2O -2'6 = 17 '4oC

lnnr*"r is for 20 - previous ans; must be < 20)

(1)(1)(1)(1)

(1)(1)

(1)

(1)

(1)

(1)

(1)

Note; may not use moles (loses first 2 marks)

so Af = (15x1000)/(50x4'2)

So answers of 71'4 and 68'7 score last 2 out of first 4'

trffo* no units for temperature' penalise wrong units)

{

Total: 10 marks

Question 3(a) 3s2 3p1

(b) (i) aluminium hydroxide

(ii) Ar(oH)4-

(c) (i) white precipitate

Effervescence

(ii)Al(HzO)o 13* + HzO +

(d) (i) AlCle + CH3CI +

(ii) as lewis acid

electron-pa ir accepto r.

(e) light

QuEstion 4

(a) (i) Ag*/silver

A!(H2O)5(OH)12. + HsO*

AlCl+- + CH:*

(1)

(1)

(1)

(1)

(1)

(1)

(1)

(1)

In AlClg ,Al has an incomplete octet and accepts a pair of electrons from the Cl

atom in CHgCI @ Alcl3 is an electron-deficient compound. and acts as an

(1)

(1)

Total = 10 marks

(ii) Carbonyl or C = O

(iii) Helps to speed up the reaction/slow at room temperature/flammable

(iv) Propanal: silver minor formed

Propanone: no change/solution remains colourless

(b) (i) optical isomerism

(iiXa)

(1)

(1)

(1)

(1)

(1)

(1)

(B)(1)

(1)

(1)

(1)

Total = {0 marks6

Hill l

Nao-( O l-c -c_N_=H\ " / r l lN^{- Ho H cH.

HHl l

Ho -(O L-l -?--TH2* *

,or- *6 i Jr.

(c) (i) nucleophilic addition

(ii) dry ether, room temperature

Question 5

(a) Lone pair donated / both electrons supplied by one atom

from N (to H.) [ignore missing charge or hydrogen atom]

d ative/dative cova lent/coord i nate bond ing

(b) Van der waals ( or Vdw) forces between methane molecules

(or VdW forces in methane)

Hydrogen bonding in.ammonia and water

Hydrogen bonds are stronger than van der Waals forces

( or VdW forces are the weakest)

Hydrogen bonds in water more extensive than ammonia because O has two lone

pairs (N one) (1)

[or Water forms more H bonds per molecule (than ammonia)]

(or H bonds in water stronger because O more electronegative than N)

(or H bonds in water stronger because O - H bond more polar than N - H)

(c) (i) The change in concentration per unit of time (1)

Both axes must be labetled to gain marks for graph. y axis conc NOz

and x axis timeCurve starts at origin and levels off.

(1)

(1)

(1)

(1)

(1)

(1)

(1)(1)

Note: tf candidates graph does not level off then second mark can be scored for a

cuwe with a continuously decreasing gradient.

Initial rate can be found by finding the gradient at t = 0 (1)

Note: Candidates may score this mark if they have shown this on their graph

(ii) 2SOz + Oz * 2SOs

NO is a catalyst

it is regenerated at the end of the reaction

provides an alternative route of lower activation energy

(1)

(1)

(1)

(1)

7

Total = 15 marks

Question 6

(a) (i) A buffer solution has the ability to resist changes in pH when a little acid or base is

added (1)

when a little acid is added, it reacts with NHg' The acid is used up' pH remains

unchanged

NHs + H*

whenal i t t lebaseisadded, i t reactswithoH- 'Thebaseisusedup'pHremains

unchanged

NHa* + OH-

(ii) POH = 14.0 - 9'0 = 5'0

pOH = PKu+ logffi

5.0 = - logl .7x10-5 + log [NHa.l / 0 '10

[NH+.] = 0.17 mol dm-3

mass of NH+CI in 450 cm3 = 0'17 x53'5 x 450 | 1000 = 4'1 S

(b) (i) acid strength increases from no' 1 to no' 3 or

down the table oras number of Cl increasedue to the electron-withdrawing effect or

electronegativity of chlorine (atoms)

stabilising the anion or weakening the O-H bond

NOT H* more available

(ii) chlorine atom is further away (from O-H) in no' 4' so has less influence

(iii) either: PH = PKa - logro [acid]= (4.9 + 2)= 3.4 (allow 3.5)

Note: [1]for correct expression & values;

[1] for correct working

(1)

(1)

(1)

(1)

(1)

(1)

(1)

(1)

(1)

(1)(1)

(1)

or Ka = 1Q -PKa = 1.259 x 10-3

[H.] = {(Ka' c) = 3'55 x 1oapH = 3.4

(1)(1)

r

Total = 15 marks

Question 7

(a) (i) lncrease (1)

(ii) Colourless solution turns to a brown solution/ black ppt(need both colours) (1)Cl2 + 2Kl -) 12 + 2 KCI (1)Clz is an oxidising agent (1)

(ii i) Add silver nitrate solution (1)' KBr forms creamy ppt (1)

Kl forms yellow ppt (1)AgNOg + KBr -+ AgBr + KNO. orAgNOg + Kl -> Agl + KNOg (1)or ionic equationsThen add (dilute or conc) ammonia (1)AgBr/ cream ppt dissolves in conc NH3 orslightly dissolves in dilute NHs (1)Agl is insoluble in dilute or conc NH3 (1)

(b) SiCIo reacts/hydrolyses, CClt doesnot (1)[This must be clearly stated and not just implied]

(lone) pair of electrons (from the oxygen atom) in a water molecule (1)cannot form a bond with/be donated to the C atomOr cannot be accepted by C atom (1)

because C has no available orbitalOR no 3d orbitals in COR C is a small atom'surrounded by Cl atomsOR Cl atoms are large and surround C atom (so attack is sterically hindered) (1)Rejects CCI+ has no d orbitals

Si has (available) 3d orbitalsRejects SiCl4 has available 3d orbitals

(1)

(

q

Question 8

(a) white precipitate formed for barium nitrate

no visible change for magnesium nitrate

) BaSOaG) + 2HNOs(aq) @HzSO+(ag)+Ba(NOs\z -

-Ba2*(aq) + SOa2-(aq) BaSO+(s)

(1)(b) ionic size of Mg2* ' Ba2*

magnitude of hydration energy of Mg2* >> Ba2* (1)

ionic size of SOa2- >> size of cations Mg2* and Ba2*

magnitude'of lattice energy for MgSOa slightly more than Basoe (1)

AHsolution = AHnvdration -AHl"tti"" (1)

ForMgSoa,AHsoution=€Xothermic;ForBaSo+,AHsotut ion=endothermic;.(1)

barium sulphate - insoluble in water ; magnesium sulphate - soluble (1)

(1)

(1)

(1)(1)

(1)(1)

I

(c) (i) (K"p=) lMgt-tt^ott1'units are mol3dm-e

(ii) Let tMg(oFl)z(?.q)l = [^M!tt] = t'

KsP=2x10-11 =4x3

x= 1.71x 10a moldm-3

' rcause of the common ion effect(iii) less t?Hb]":: ,*(aq) + 2oH-(aq1 is moved,tothe teftor the equilibrium lrag(oH)z(s) + Mg'-(aq) + zuil' (aq'

(1)

(1)

Total = 15 marks

to

Question 9

(a) (i)

t cH.cH" IcH3cH2\

| " r -

I

Ho-+ t-)-cr -> I "o-- / \ -" ' I " -H' lHHlL_l

(1)

bimolecular nucleophilic substitution

CHgCHzCHzOH18ooc

cH"cH","

Ho-c-H +Cl-

\ .H

(1)(1)

(1)(1)(1)

(ii) rate increases.atomic size of BPCI 'Hence the c-Br bsnd is weaker and is easier to break than G-Cl bond

(iii) reflux (1) - l

CHiCH2CFlzCl + NaOH(aq)(1)

conc. HzSOa (1)

CH3CH= CH2 + HCI -+ CHgCHCICHg(1)

t3 E{TrH,N-f -fi-T-f-cffiFIff{, o i{ cll2CItl per:atise {:#CINH- agein

ETITI

H.N-C-ql.i-{-C(l0i{

Ctlz0tt O H CHs prnalise palvmer'both tirnes

serylalanine or alanYlserine

fl

IEI3C-q-s-c+-ic-cooH

i l l0 +rI

Ic-0Iffl*3

CH3CH= CH2(1)

(b)

(1)

(1)

(1)(i i)

(iii)

(1)

t ( Total = 15 marks

Question 10

They'don't react with Fehling's solution - they are ketones

AandBgivesposit iveiodoformtest-theyhavethestructureCHgC=O ormethYlketone

I

Boiling point of A > B - A is less branched compared to B

A is CHgCOCHzCHzCHg

B is CHgCOCH(CHg)z

a,: CHgCHzCgCl-l2CHa

D and E gives ,6 ,positive testwith Fehlingls solution - they afe €ddehydes

D is optically active. Hence isomers of D are

NO^9H. !----'

cH3cH2cHrC+rr'ftr-Q) - NO2+ HrO

cHl. + cH3cH2cH2coo-

(1)(1)

(1)

(1)

(1)

(2)

(1)(1)

(1)(3)

Reduction of E gives a long chain compound' F

E is CHgCHzCHzCHzCHOF is CHgCHzCHzCHzCHzOH

P is copper(l) oxide (reject Cu2O)

Examples of equations:

9H. Yi*r.cHrcHrb=o . HTNNHIO)-No, -->

CH.CH"I

II

- - ' / \ ,

Hrct /CHO ,

-_--->ACH.CH2CHTCOCHT+ l, + NaOH

lL

Total = 15 marks