skema stpm johor chemistry 2011 trial (edu.joshuatly)

JA BA TA N PELAJA RAN JOHOR JA

JA' Ckao i R A N J O H O R J A

JAt a t > Z iRAN JOHOR JA

lABATAN PELAJARANJOHOR JA

JABATANPELAJARANJOHOR JA

JA BA TA N PELAJA RAN JOHOR JA

JABATAN PELAJARANJ

JABATAN PELAJARANJ

JABATANPETAJARANJ

/ JOHOR JABATANPELAJARjlNJOHOR JABATAN

PERCUBAAN STPM 2011 A UOHOh

UOlLOh ~ . . . » . ^wi i iN

J JOHOR JABA TAN PFXAJARANJOJTOR JA BA TAN

J JOHOR JA BA TA N PELAJA RA N JOHOR JA BA TA N

UOHOR JABATANPEJAJARAN JOHOR JABATAN

BA TAN PELAJA RA N JOITOR JA BA TA N

CHEMISTRY (KIMIA) BATAHPELAJARANJOHOR JABATAN ' i 7 n V ^ Z n T r T / , T ^ BATANPEJ,AJARANJOHORJABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABA TAN PElAJARAN JOHOR JABA TAN

JABATANPELAJARANJOHORJABATANPELAJARANJOllORJABATANPELAJARANJOHOR JABATAN JABA TAN PELAJARAN JOHOR JABATAN PELAJARANJOHOR JABA TAN PELAJARAN JOHOR IABA TAN

J t B n I ^ J 3 r A ^ l A N J O H O R " M A R K Q P H F M P lORJABATANPELAJARANJOHOR JABATAN JAJiATAN PELAJARAN JOHOR ^ MARK SCHEME lORJABATANPELAJARANJOHORJABATAN

^BATANPELAJARANJOHORJABATANPELAJARANJOHORJABATANPELAJARANJOHOR JABATAN

•MBATANPELAJARANJOHORJABATANPELAJARANJOHORJABATANPELAJARANJOHORJABATAN

JABATANPELAJARANJOHOR JABATANPELAJARANJOHOR JABATANPELAJARANJOHOR JABATAN

JABA7ANPELAJARAN JOHOR JABATANPELAJARAN JOHOR JABATAN PELAJAJtAN JOHOR lABATAN

JABATAN PELAJARAN JOHOR

S I J I L T I N G G I PERSEKOLAHAN MALAYSIA

STPM %'-/l http://edM.joShuatly.GOmJ

CONFIDENTIAL*

Question Details Mark Awarded

Ka)(i) PV = nRT 1 500 x 103 x 20.0 x 10"3 = (1.5 + x + 1.2) X 8.31 x298 1 X= 1.3 1 Or

X = 500 X IO3X 2Q.Q X iO'3

8.31 x 298

= 1.3

1(a){ii) , g I'5, x500.— 188 kPa 1.5 + 1.3 + 1.2

1

1{b)(i) N : 1 s22s22p3 1

1{b){ii) >M "

H o l f N"* N* 0H OX O t

H H 1

1(c)(i) Relative molecular mass of propanorie is higher, therefore stronger Van der Waa!s forces of attractions exist between molecules compared to the ether.

1

1

1(c)(ii) Ether: sp3 1

Propanone : sp3, sp2 1

http://edu/. jo$hMatly.GOmJ

Questiori Details Mark Awarded

2(a) (i) The standard electrode potential is the potential difference between the standard hydrogen efectrode and other electrode systems with the concentration of1.0 M at 1 atm I under standard conditions

1

t

2(a) (ii) S2O8 2" + 2 e ^ 2S0/ -

2l ^ I2 + 2e ^J

> 1

S2O82- + 2T ^ 2 S 0 / ' + I2 1

2(a) (iii) Pt(s) / l " (aq), I2 {aq) // S2O0 2" (aq) , SO4

2' (aq) / Pt{s) 1

2{a) (iv) E0c e l l = +2.01 V - (+0.54 V ) = +1.47 V 1

2 (b) (i) P: solid Q: solid &liquid R: liquid S: liquid & gas T : Gas 4-5 correct:2m

2-3 correct:1m

2 (b) (ii) Density of solid W is lower than liquid W. 1

2 (b) (iii) The melting point decrease with increase in pressure 1

http://edu/. jo$hMatly.GOmJ

Question Details Mark Awarded

3 (a) (i) ' ^ f l P f ~ " 1

3 (a) (ii) R.M.Mof Pbl2 = 461,

concentration of Pbl2 in water =0.46 g dm"3 1

The concentration of the Iead ion = 0.46/461

= 9.978X 10"1 moi dm'3

=10.0X10"4 mol drn"3 1

3 (a) (iii) The concentration ofthe iodide ion = 2 X [ Pb2+]

= 2 X 9.978 X 10^

= 1.996 X 10"3 moldnT3

1

= 2.0X10"3 mol dm'3 1

3 (a) (iii) K,p = [Pb 2 t ] [ l f

=[ 9.978X 10^ ][ 1.996 X 10"3]2

= 3.975 X 10"9 mol 3dm"9

= 4.0 X 10"9 mol 3dm"9

1

3{b) {i) Nitrogen monoxide gas, NO 1 3 {b) <ii) 3Cu(s) + 8HN03(aq) ^3Cu(N03)2(aq) + 2NO (g) + 4H2Ow 1

3 (b) (iii) Brown gas : nitrogen dioxide ,NO2 2NO<g, + 0 2 ( g ) ~ 2N02(g) Colourless Brown

gas gas

1 1

http://edu/.jo$buatly.GOm/

1

Question ___ Details Mark Awarded

4a(i) Acid-base reaction Amine functional group

1 1

<ii) Functional group : Ester 1

Typeofreaction : hydrolysis 1

Structural Formula:

CH3

N: / \

CH2 CH2

C f t ^ ^ H 2

C

Z " Z ^ A COO- Na+ + C2H5OH <0> t

4(b) (i) Reflux 1

OD Nucleophilic substitution 1

(iii) Step 1 - magnesium and dry ether 1

Step 2 - CH3CHO followed by di!ute acid 1

%f) Sodium / PCI5 / SOCl2 / PCI3 1

_ .

http://edu/.jo$hMatly.GOm/

Question 5(a)

5(b)

Detai!s H2C03(aq) * HCCV{aq) + hT(aq)

In the presence of acid, HCO3-(aq) + H f ( a q ) * = t H2C03(aq)

Carbonic acid being unstable, decompose to CO2 and H2O In the presence of alkaii, H2CQ3(ag) + OH"(aq) -4 • HC03"(aq) + H2O(I)

Mark Awarded

pH

i IO 1} X M .Wi JS ^ <5 S9 V&hfme<tfKCI ntliit(lfcm'

Hydrolysis of NH4* ions in water, thus is acidic

NH4+ + H2O ^=dfe- NH3 + H3O+ pH<7

^ ^ _ ,

Equivalence point A ^ ^

_ , / -

Shape -1m

Label - 1m

End point < 7 -1m

5(c) (i)

(ii)

(iii)

Ig k = Ig A -2.3037? I T

^ * L = J k J j L * L k2 2.303^7½ T1

Substituting k, = 1.63 x 10"3, k2 = 4.75 x 10"4 , T2 = 293 Ea= 9.10 x 104 J or 91.0kJ

unsubstituted

T1 = 303 substituted

Rate = ^CO(CH2COOH)2]

Time = 0.693

l.63xlO"3

= 425 s

http://edu/.jo$buatly.GOm/

Question Details ~ Mark Awarded

6(a) y* Ch <g) + m > ei(g)

AH4 -364

T v

Cl"(aq) < "381 cr(g)

AH4 = +121+(- 364) +{ -381) = 121- 364 -381 = -624 kJ

AH4 very exothermic, Cl?is a very strong oxidising agent, itself reduced easily

4 correct: 2 m 3 correct: 1 m

1

1

1 1

6(b)(i)

, . ' _ rT^ -̂

Critical temperature of CH4 < NH3 < H2O

- CH4 has weak Van der Waals forces of attraction between molecules,

- NH3 and steam has strong H-bonding

- H-bonding in water is stronger than in ammonia

- the stronger the intermolecular forces of attraction , the more energy is required to overcome them, thus the critical temperature is higher.

1

1

1

1

1

08 -Energy is required to overcome the intermolecular forces of attraction,

- energy is drawn from the surroundings, temperature drop

-ideal gas has negligible intermolecular forces of attraction

- expansion of ideal gas does not require energy to overcome the intermolecular forces of attraction

1

1

1

1

http:ffedu/.jo$hMatly.com//

Question Details Mark Awarded

7(a)(i) AH1 =. Atomisation enthalpy of magnesium

AH2 = First ionisation energy of magnesium

AH7 = Lattice energy of magnesium oxide

I

I

I

(ii) AHf =AH1 +AH2 +AH3 +AH4 +AH5 +AH6 + AH7

= +150+{+ 736)+{ + 1450)+(+ 950) +(-3889)

= - 603 kJ/mol

The charge of Mg 2+ ion and Ba 2+ ion are the same but the size of the Mg + ion is smaller compared to the Baz+ ion

I

I

I

1

(iii)

Lattice energy = GUQ_ r+ +r.

I

Sop Lattice energy of MgO > BaO I

b{i) - CFCs can cause depletion of ozone layer - because they are unreactive

1

initiation step: CCI2F2 - w > »CClF2 + *CI 1

Propagation step : • Cl + O3 —~>CIO* + O2

CIO- + O > • Cl + O2 1

Net reaction : 203 -—~> 302 1

(ii) Hydrofluorocarbon, HFC 1

_ ^ ^ _

Will not release damaging chlorine free radicals into the atmosphere OR C-H bor.ds will break down before they reach stratosphere

1

http://edu/. jo$buatly.GOm/

Question Details Mark Awarded

8(a)(i) With cold NaOHfaq)

-Chlorine undegoes disproportionation reaction in cold NaOH{aq) to form NaCI and NaCIO. 0 -1 +1 Cl2 + 2NaOH ^ NaCI + NaCIO + H2O

1

1

-Chlorine oxidised to NaCIO and reduced to NaCI.The NaCl and NaCIO produced are sodium salts and have no chlorine smell.

1

With hot NaOHfaq) -Chlorine undegoes disproportionation reaction in hot NaOH{aq) to form NaCI and NaCIO3. O -1 +5 3CI2 + 6NaOH ^ 5NaCI + NaCIO3 + 3H20

I

1

Chlorine oxidised to NaClO3 and reduced to NaCI.The NaCl and NaCIO3 produced are sodium salts and have no chlorine smell.

1

(max 5m)

8(a)(ii) - Fluorine is highly electronegative element.HF has the highest h.p because hydrogen bonding exist between HF molecules. 1

- Hydrogen bonding does not exist in HCI,HBr and Hl because Cl j Br and I atoms are not highly electronegative. 1

- b.p increases from HC! to Hl becausethe molecular size increases from HCI to HI. 1

-The larger the molecular size, the stronger the Van der Waals forces of attraction between the molecules and the higher the b.p.

1

http://edu/.jo$hMatly.GOmJ

8(b)(i) Formuia of X is [Cr (H2O)4 Cl2]+. Cf" .2H20 ~ Geometric (or cis-trans) isomerism

8(b)(ii)

1 1

1

1 trans-tetraaquadichlorochromium(lll) ion

cis-tetraaquadichlorochromium{ll[) ion

1

1

I

Question Details Mark Awarded

9(a) Optical isomerism occurs in compounds with the same structural formula but differ in their effect on plane-polarised light.

~ r ~

- Optical isomers contain a chira! carbon, 1

- Are nonsuperimposabie or mirror images to one another, 1 - Rotate the plane of plane-polarised light in opposite direction but to the same degree. 8 9 OH 1

CH3 — C — CH2CH3

t H

(max 3 marks)

(b) Since X gives white fumes with PCI5, X is an a!cohol.

X gives a give yellow precipitates with a!ka!ine iodine shows that X has structure . CH3CH(OH)R

Hence X is 1-phenylethanol. CHXH(OH)CgH i

Equations :

CH3CH(OH)C6H5 + PCI5 > CH3CH(Cl)C6H5 +

X HCl + POCl3

White fumes

CH3CH(OH)C6H5 + *I2 + 6Na OH" — * CHI3 + v yellow precipitate A

C6H5COONa + 5NaI + 5H20

I

!

I

I

1

Excess H5SO4 CH3CH(OH)C6H5 • CH2-CH(C6H5) +

X Y H2O

1

X undergo dehydration to form Y, phenylethene, j

CPb=CHC6H5

1

http://edM/.j0shu&dy.G0m/ (7)

_____

(c) - A and B are aromatic compounds, highly unsaturated^

- A i s a pheno!

- B contains -OH group and the structure

OH I

CH3-C-I H

- 4-ethylaniline reacts with nitrous acid to produce A

- bromobenzene reacts with Mg/dry ether to produce a

Grignard reagent The Grignard reagent then reacts with

ethanal to produce a 2° alcohol, B OH

OH I I CHCH3 A: v s: 6 CH2CH3

1

1

1

1

1

http://edu/.jo$hMatly.GOm/

I

Question Details Mark Awarded

10(a)(i) < ^0^CH 2 CH 2 NH 2 t HCI HP < ^ 5^CH 2CH 2NH 3+CI~ 1

< ^ 3 ^ C H 2 C O N H 2 + HCI + H2O ^ < ^ O ^ C H , C O O H + NH4CI 1

(N) Reagents used - sodium hydroxide \ Conditions - reflux J Observations -

1

R - no visible reaction 1

S - gas with a characteristic smell, NH3, given off. 1

10(b) Step I - KCN / ethanol, reflux

Step l l -HCJ, reflux

Step III - PCI5 or SOCI2

A : CH3CH2CH2CN

8 : O

^ O y - O-C-CH2CH2CH3

1

1

1

1

1

~^m~~

< S ^ " > W$ <^°°" 1

C2H5OH / Conc H2SO4, reflux

< S >

*

-COOC2Hs

Step 1 - oxidation of methylbenzene 1

Reagent - KMnO4 and H2SO4 Condition - reflux

1

Step 2 - esterification 1

Reagent - ethanoi and a concentrated H2SO4 condition- reflux I

1

http://edu/.jo$hMatly.GOmJ