Trial Spm Mark Scheme p2_smkbsp 2013

7/29/2019 Trial Spm Mark Scheme p2_smkbsp 2013

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SULIT

4541/2

Kimia

Kertas 1

Kertas 2

Ogos/Sept

2013

Skema Pemarkahan

SEKOLAH MEN. KEB. BANDAR SERI PUTRA, KAJANG

__________________________________________________________________________________________

PEPERIKSAAN PERCUBAAN

SIJIL PELAJARAN MALAYSIA 2013

SKEMA PEMARKAHAN

KIMIA

Kertas 1 and Kertas 2

Kertas soalan ini mengandungi 12 halaman bercetak

4541/2 [Lihat sebelah


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2

SKEMA PERMARKAHAN

KIMIA 4541/1

No

soalanJawapan

No

soalanJawapan

No

soalanJawapan

No

soalanJawapan

No

soalanJawapan

1 B 11 B 21 D 31 C 41 D

2 D 12 A 22 C 32 D 42 B

3 D 13 D 23 B 33 C 43 C

4 A 14 D 24 A 34 A 44 A

5 C 15 A 25 D 35 A 45 C

6 B 16 A 26 D 36 C 46 D

7 C 17 D 27 A 37 A 47 D

8 A 18 C 28 B 38 B 48 D

9 A 19 C 29 B 39 D 49 A

10 A 20 C 30 C 40 D 50 D


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PAPER 2

Section A

1 (a) Acetamide / CH3CONH2 1

(b) Sublimation 1(c) (i) 1. Temperature at which liquid changes into a solid

2. at a particular pressure 1

(ii) 82oC 1(iii) 1. Heat loss to the surrounding

2. balanced by heat energy liberated when particles attract one

Another // when liquid change to solid

1

1 2

(d) (i)

proton

neutron

electron

All correct 1(ii) 4 + 3 = 7 1

(e) [Able to draw a diagram that shows the following information]

1. Diagram of combustion tube containing copper oxide and isclamp using retort stand

2. Dry hydrogen is flow into the combustion tube, excess of

hydrogen is flow out and copper oxide is heated3. Label hydrogen, copper oxide and heat

2

10


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2 (a) Silicon / Chlorine / Argon 1

(b)(i) 2.8.8 1

(ii) it has achieved the stable octet electron arrangement 1

(c) Used as semiconductor to make diodes/transistor 1

(d) (i) Atomic radius/size decreases 1(ii) -The number of proton increases across the period from sodium to

argon hence the positive nuclear charge also increases

-thus, stronger attraction force between the nucleus and the electrons

in the first three occupied shells- causing the electron to be pulled closer to the nucleus ,therefore the

atomic radius decreases

1

1

1

(e)(i) 2 Na + Cl2 2 NaCl 1

(ii) Ionic bond 1

10


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3 (a) Chemical energy to electrical energy 1

(b)(i) Zinc 1

(ii) Zink is more electropositive than iron. 1

(c) Green to colourless 1

(d) Allow ions moving through 1

(e) From zinc to iron electrode 1

(f) Zinc

Zinc donate electron to iron.

1

1

(g) Zn Zn+

+ 2e 1

(h) No of mol = 0.056 / 56

No of atoms = 0.056/ 56 x 6.02 x 1023

1

10

4 (a) 1

(b) C3H7OH + 5O2 3CO2 + 4H2O 1

(c) (i) Pleasant smell / fruity smell 1

(ii) Ethanoic acid 1

(d)(i) Oxidation 1

(ii) Orange to green 1

(e) (i) 1

(ii) 1,2 dibromo propane 1

(iii) [Able to draw and labeled the diagram correctly ]1. Functional diagram

2. Label the diagram

1

1

10

5 (a) (i) Stanum / tin


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(ii) The stanum atoms disrupt the orderly arrangement of atoms in copper.This prevents the layers of pure metal (copper) atoms from sliding on one

another easily, making bronze harder than copper.(iii)

Pure copper:

//

Bronze:

`

(b) (i) Oleum

(ii) 2SO2 + O2 2SO3(iii) Moles of sulphur = 48 / 32 = 1.5

Moles of SO2 = moles of sulphur = 1.5

Volume of SO2 = 1.5 x 24 dm3 = 36 dm3

TOTAL

5

(c)

[Labeled of axes are correct]

[The curves are correct] 11

2

10

6 (a) (i) Butanol 1 11

I

II

Volume of hydrogen gas / cm3

Time / min

copper

Stanum


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(ii) The heat given off when 1 mol of butanol burn completely in

excess oxygen is 2679 kJ.

1

(b) (i) (1) Number of mole = 0.37 mol //0.005 mol

74(2) Heat released = 0.005 x 2679/ 13.395 kJ/13.395 x 10

3J

1

1

(ii) (1) Substitution/Replacement

13.395 x 103

= 500 x 4.2 x

/ = 13.395 x 103500 x 4.2

(2) = 6.379/6.40C

1

1

(c) Draw and label of the energy axis in the form of arrow with

two level of energyFormula/Name the reactants and products for exothermic

reaction

Energy C4H9OH + 6O2

H = -2679 kJmol-1

4CO2 + 5H2O

1+1

(d) (i) Any value between: -2016 kJ mol-

H - 2036 kJ mol-

1

(ii) As the number of carbon atoms increases, there are morecarbon dioxide and water molecules are produce

Thus more bonds are formed

1+1

SECTION B


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7 (a) (i) 1. The brown colour of bromine water becomes colourless.

2. Ethene has undergone an addition reaction.

(ii) 1. The product is 1,2-dibromoethane

2.

C C

1

1

1

1 4

(b) (i) 1. % mass of hydrogen = 10082.75 = 17.252. Moles ratio of C : H = 82.75/12 : 17.25/1

3. = 2 : 5

4. The empirical formula of Z is C2H5

(ii) 1. Let the molecular formula be (C2H5)n

Therefore, (12x2 + 1x5)n = 58

n = 2

2. Hence, Z is C4H10

11

1

1

1

1

1 7

(c)

Property Unvulcanised rubber Vulcanised rubber

(i) Elasticity Less elastic.Reason: the rubber

molecules can slide

over each other easilywhen it is stretched.

More elastic.Reason: the presence of

sulphur cross-links can pull

back the rubber moleculesto their original positions

after it is stretched.

(ii) Strength and

hardness

Soft and less strong.

Reason: the rubbermolecules can slideover each other easily

when it is stretched.

Stronger and harder.

Reason: the presence ofsulphur cross-links makesliding of rubber molecules

more difficult.

(iii) Resistance

towards

oxidation

Easily oxidised by

oxygen.

Reason: the number of

carbon-carbon doublebonds, C=C per

molecule is high.

More resistance to

oxidation.

Reason: the number of

carbon-carbon doublebonds, C=C per molecule

is low.

11+1

1

1+1

1

1+1 9

Total 20

Br Br

H

H H

H


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No. Rubric Mark8 (a) (i) Atom Ris located in Group 17, Period 3 1

1

5

(ii) Electron arrangement of atom R is 2.8.7.It is located in Group 17 because it has seven valenceelectron.It is in Period 3 because it has three shells filled with

electron

1

11

(b) (i) Atoms P and R form covalent bond.

To achieve the stable electron arrangement,atom P needs 4 electrons while atom R needs one electron.Thus, atom P shares 4 pairs of electrons with 4 atoms ofR,forming a molecule with the formula PR4 // diagram

1

11

1

1

11

(ii) Atom Q and atom R form ionic bond.Atom Q has the electron arrangement 2.8.1. and atom R

has the electron arrangement 2.8.7.To achieve a stable (octet )electron arrangement,atom Q donates 1 electron to form a positive ion//

equationQ Q+ + e

Atom R receives an electron to form ion R-//equationandachieve a stable octetelectron arrangement.

R + e R-

Ion Q+

and ion R-

are pulled together by the strongelectrostatic forces to form a compound with the formula

QR// diagram

1

1

1

11

1

RR

R

R

+


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(c) The ionic compound/ (b)(ii) dissolves in waterwhile the covalent compound / (b)(i)does not dissolve in water.Water is a polar solvent that can cause the ionic compound todissociate into ions.

Covalent compounds are non-polar and can only dissolve inorganic solvents.

OR

The melting point of the ionic compound/ (b)(ii) is higher thanthat of the covalent compound/ (b)(i) .

This is because in ionic compounds ions are held by strongelectrostatic forces.

High energy is needed to overcome these forces.In covalent compounds, molecules are held by weakintermolecular forces.

Only a little energy is required to overcome the attractive forces.

OR

The ionic compound/(b)(ii) conducts electricity in the molten oraqueous statewhereas the covalent compound/(b)(i) does not conductelectricity.This is because in the molten or aqueous state, ionic compoundsconsist of freely moving ions.

Covalent compounds are made up of molecules only

11

1

1

1

1

1

11

1

1

1

1

4

max4/5

4

Total 20

SECTION C


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9 (a) (i) Based on the equation, iron(II) ion is oxidised toiron(III) ionSo, iron(II) ion acts as a reducing agent

11

(ii) Based on the equation, iron(II) ion is reduced toironSo, iron(II) ion acts as an oxidising agent

1

1 4

(b) eMgMg 22

Oxidation number of magnesium increases from 0 to +2, Somagnesium undergoes oxidation

CueCu

22

oxidation number of copper decreases from +2 to 0, so copper(II)ion undergoes reduction

1

1

1

1 4

(c) (i) Reduction is a reaction that involves gain of electron.Oxidation is a reaction that involves loss of electron.

11 2

(ii) At the negative terminal:

Iron(II) ion releases one / loses one electron andis oxidised to iron(III) ion.Fe2+ Fe3+ + e

The green coloured solution of iron(II) sulphate turns brown.The electron flows from the negative terminal// carbon

immersed in iron(II) sulphate solution to the positiveterminal// carbon immersed in bromine water.

At the positive terminal:

Bromine accepts electron andis reduced to bromide ions, Br

-

Br2 + 2e 2Br-

The brown coloured bromine water turns colourless.The deflection of the galvanometer needle shows that there is

a flow of current

111

11

11111 10

Total 20

10 (a) Able to state


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(i) Sulphuric acid

Mg + H2SO4 MgSO4 + H2

(ii) Able to draw an energy profile diagram that shows the following five

information:1. label of energy on vertical axis

2. The position of the energy level of the reactant is higher than the

energy level of the product3. Correct position of Ea4. Correct position of Ea

5. Correct position of H

If energy level diagram given, not energy profile diagram, award pt1 and pt2only)

1

1

1

1

11

1

Able to give explanation by stating the following information:

1. Reaction is exothermic2. The reactants contain moare energy than the products3. Heat given nout during bond formation is greater than heat

absorbed during bond breaking4. H is the energy difference between the reactants and products5. Activation energy, Ea must be overcome in order for the reaction

to take place

6. The use of catalyst reduces the activation energy(must correspond to the energy profile diagram: negativecatalyst increases the activation energy)

7. the use of catalyst increases the frequency of effective collisionbetween H

+ion and magnesium atom

(Neagative catalyst reduces the frequency of effective collisionbetween H

+

and magnesium atom)any fivepoints:

Note: If endothermic reaction: Pts 1,2,3 lost

5

(b) (i) Able to calculate the average rate of reaction

1. volume divided by time

2. Correct answer with the unitAvarage rate of rfeactiuon = 400 cm

3

200 s

= 2 cm3

s-1

1+

1

(ii) Able to explain the difference in the rate of reaction by stating thefollowing

Information

1. The rate of reaction for Experiment II is higher/greater thanExperiment I

r: slower/faster

2. This is because sulphuric acid in Experiment II isdiprotic/dibasic acid

1

1

1


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3. Hydrochloric acid in Experiment I is monoprotic/monobasicacid(Basicity of acid in Experiment II is higher than acid in

Experiment I gives pts 2 and 3)

4. Diprotic acid has higher concentration of H+ ion //monoproticacid has lower concentration of H

+ion

a: more/higher number/less H+

5. The frequency of collision between H+ ion and magnesium inExperiment II is higher than in Experiment I// The frequency ofcollision between H

+ion and magnesium in Experiment I is

lower than in Experiment II

r: between particles

6. The frequency of effective collision in Experiment II is

higher/greater than in Experiment I// The frequency of effective

collision in Experiment I is lower/smaller than in Experiment II

1

1

1

6

END OF MARKING SCHEME

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Trial Spm Mark Scheme p2_smkbsp 2013