Transcript
Page 1: Pecutan Akhir Kimia Spm 2015

PANITIA KIMIASEKOLAH MENENGAH SAINS TAPAH

JALAN PAHANG, 35000 TAPAH,PERAK DARUL RIDZUAN

PECUTAN AKHIR SPM 2015

Page 2: Pecutan Akhir Kimia Spm 2015

RATE OF REACTION• A group of students carried out three experiments to investigate the factors affecting the rate of reaction between

hydrochloric acid and zinc. Table shows the results of the experiments.

i) Calculate the average rate of reaction for Experiment I and Experiment II. [2 marks]

ii) On the same axis , sketch the graph for the three sets of experiments for the liberation of

40 cm³ of hydrogen gas [3 marks]

iii) Write the ionic equation for the reaction between zinc and hydrochloric acid. [2 marks]

iv) Based on Table above, compare the rate of reaction between Experiment I and Experiment II

Experiment II and Experiment III

Explain the difference in the rate of reaction based on the Collision Theory. [10 marks]

Experiment Reactants Time taken to collect 40 cm of hydrogen gas (s)

I 50 cm of 1.0 mol dm¯³ hydrochloric acid + zinc granule + a few drops of copper(II) sulphate solution 90

II 50 cm of 1.0 mol dm¯³ hydrochloric acid + zinc granule 150III 50 cm of 0.5 mol dm¯³ hydrochloric acid + zinc granule 270

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ANSWERi) Average rate of reaction experiment I = 40/90 = 0.44 cm³ s¯¹

Average rate of reaction experiment II = 40/150 = 0.27 cm³ s¯¹

ii)

iii) Zn + 2H⁺ Zn²⁺ + H₂

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iv) Experiment I and II

1. Rate of reaction of Experiment I is higher than Experiment II

2. Copper(II) sulphate solution acts as catalyst

3. Which provide an alternative path with lower activation energy

4. More colliding particles able to achieve the activation energy

5. The frequency of effective collision between zinc atom and hydrogen ions is higher in Experiment I than in Experiment II

Experiment II and III

6. Rate of reaction in experiment II is higher than Experiment III

7. The concentration of hydrochloric acid in experiment II is higher than in Experiment III

8. The number of hydrogen ions per unit volume is higher in Experiment II

9. The frequency of collision between zinc atom and hydrogen ions is higher in Experiment II than Experiment III

10. The frequency of effective collision between zinc atom and hydrogen ions is higher in Experiment II than in Experiment III

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MANUFACTURED OF SUBSTANCES IN INDUSTRY

Table shows three substance, example and their components respectively

i) Name substance P and suggest how the strength of concrete can be increased to be used as pillars of

building. [2 marks]

ii) Polymer R is formed through polymerization process.

Write the chemical equation to produce the polymer R and give a name for the polymer. [3 marks]

iii) Q is one type of alloy. Compare and explain the hardness between alloy Q and its pure metal. [5 marks]

Substance Example ComponentP Reinforced concrete cement, sand, small pebbles and steel

Alloy Q copper and zincPolymer R ethene

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ANSWERi) Composite material - Adding steel rod into the concrete

ii) nC2H4 [C2H4 ]

Polyethene

iii) - Alloy Q/brass is harder than its pure metal/copper

- the presence of zinc atom in alloy Q disrupts the orderly arrangement of copper atom

- These make the atomic layers of atoms harder to slide over on another

- in pure metal/copper the atoms are arranged packed closely and in orderly manner.

- this allow the layers of atoms are easily to slide one another

n

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CHEMICAL FOR CONSUMERS

Diagram shows the apparatus used to investigate the cleaning action of cleaning agent X and Y to remove oily stain from the cloth.

• Based on Diagram compare and explain the effectiveness of cleaning action between experiment I and II. Identify the cleaning agent X and Y.

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ANSWER- Cleaning agent Y in experiment II is more effective than cleaning agent X- Cleaning agent Y do not form scum in hard water therefore it can remove oily stain from

the cloth- Cleaning agent X in experiment I is not effective in hard water because hard water

contain high calcium ion and magnesium ion- These ions will react with cleaning agent X to formed an insoluble precipitate/scum

- The formation of scum will reduces the number of cleaning agent A

• Cleaning agent X is soap• Cleaning agent Y is detergent

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CHEMICAL BOND

By using the element in Diagram above, explain the formation of compound between element R and Q , also formation of compound between element L and T. The two compounds should have different bond types.

[10 marks]

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ANSWER R and Q

1. Electron arrangement of atom R is 2.8.2

2. Electron arrangement of atom Q is 2.6.

3. Atom R donates 2 electrons to achieve stable octet electron arrangement to form R²⁺ ion

4. Atom Q receive 2 electrons to achieve stable octet electron arrangement to form Q²¯ ion

5. R²⁺ ion and Q²¯ ion are attracted by a strong electrostatic force to form ionic bond

L and T

1. Electron arrangement of atom L is 2.4

2. Electron arrangement of atom T is 2.8.7

3. Atom L contribute 4 electron and atom T contribute 1 electron for sharing

4. To achieve stable octet electron arrangement

5. 1 atom L share 4 pairs of electron with 4 atom T to form covalent bond

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ELECTROCHEMISTRYDiagram shows two type of cells.

Compare and contrast cell X and cell Y in terms of:

• Type of cell

• The energy change

• The terminals of the cells

• Ions presence in the electrolyte

• Observation

• Half equation for both electrodes

• Name of the processes occurred at the positive terminal of each cell [ 10 marks]

Cell X Cell Y

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ANSWER Cell X Cell YType of cell Electrolytic cell Voltaic cellThe energy change Electrical energy to chemical energy Chemical energy to electrical energy

The terminal of the cell Positive terminal / anode: CopperNegative terminal / cathode: copper

Positive terminal / cathode: copperNegative terminal / anode: aluminium

Ions present in the electrolyteCu 2+, H+

SO4 2- , OH-

Observation Anode: Electrode become thinnerCathode: Brown solid is deposited//thicker

Negative terminal/Aluminium plate: Electrode become thinnerPositive terminal/Copper plate:Brown solid is deposited//copper plate become thicker

Half equation for both electrodes

Anode:Cu Cu 2+ + 2eCathode:Cu 2+ +2e Cu

Al plate/- terminal:Al Al 3+ + 3eCu plate//+ terminal:Cu 2+ +2e Cu

Name of the process occurred at both electrodes/terminal

Anode/Al plate: OxidationCathode/Copper plate//negative terminal: Reduction

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EMPIRICAL FORMULAMetal X is more reactive than hydrogen.

[Relative atomic mass: O = 16 ; X = 24 , ionic formula : X²⁺ ]

Describe a laboratory experiment to determine the empirical formula of oxide X. Your answer should consist of the following: Procedure of the experiment Calculation involved

[ 10 marks ]

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ANSWER• Procedure:

1. A crucible and its lid are weighed

2. 10 cm of cleaned X ribbon is coiled loosely and placed in the crucible.

3. The crucible with its lid and content are weighed again.

4. The crucible is heated strongly without its lid.

5. Using a pair of tongs, the lid is lifted at intervals.

6. When the burning is completed, the lid is removed and the crucible is heated strongly for 2 minutes.

7. The crucible is allowed to cool to room temperature.

8. The crucible and its lid and content are weighed again

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ANSWER• Calculation:

Mass of crucible + lid = a g

Mass of crucible + lid + magnesium = b g

Mass of crucible + lid + X oxide = c g

Empirical formula is XpOq

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REDOXDiagram 10.2 shows the reactions involving Fe2+ ion and Fe3+ ion

i) State the suitable examples for reagent A and reagent B.

ii) Explain the oxidation and reduction in terms of electron transfer in Reaction I.

iii) Write the half equation involved in Reaction II and state the observation occurred. [7 marks]

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ANSWERi) A : Bromine/chlorine water // acidified KMnO4 solution

B : Zinc

ii) Fe2+ undergoes oxidation reaction because Fe2+ releases electron.

Bromine water undergoes reduction because Bromine water gains electron

iii) Fe3+ + e → Fe2+

Zn → Zn2+ + 2e

Observation : Brown solution turns to green


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