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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
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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
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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
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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
_ .
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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
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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
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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
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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
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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
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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
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