peka tingkatan 5

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Mark Scheme Chemistry PEKA Form 5

Topic :Rate of Reaction Experiment No : ........1.2........

Aim To investigate the effect of concentration on the rate of reaction.

Problem statement How does the concentration of a reactant affect the rate of reaction? [ K1PP1(i) - Able to write the aim or problem statement correctly]

Hypothesis The more concentrated the sodium thiosulphate solution, the higher the rate of reaction.

Variables Manipulated variable : Concentration of sodium thiosulphate solution

Responding variable : Time taken for the cross `X` to disappear from the sight.Controlled variable : Concentration and volume of dilute sulphuric acid , temperature

of the solution

[ K1PP1(ii) - Able to write the hypothesis or variables correctly]

Materials 0.2 mol dm-3 145 cm3 sodium thiosulphate solution, 1.0 mol dm-3 25 cm3 sulphuric acid, distilled water 80 cm3, white paper marked `X` at the centre

Apparatus 50 cm3 measuring cylinder, 10 cm3 measuring cylinder, 150 cm3 conical flask, and stop watch

[ K1PP1(iii) - Able to list all the materials and apparatus correctly]

Procedure 1. 50 cm3 of 2.0 mol dm-3 sodium thiosulphate is measured using measuring cylinder and poured into a conical flask.2. A piece of paper marked `X` is placed under the conical flask.3. 5 cm3 of 1.0 mol dm-3 sulphuric acid is measured and poured into the conical flask.4. Stopwatch is started immediately.5. The mixture in the conical flask is swirled and placed it on the paper marked `X` .6. Mark `X` is observed vertically from the top through the solution.7. Time,t is recorded for the mark `X` to disappear from sight.8. The experiment is repeated four more times using different volumes of 0.2 mol dm-3 sodium thiosulphate solution to be diluted with different volumes of distilled water as shown in Table 1.


[ K1PP1(iv) - Able to write the procedure correctly]

Data Sample Answer:

Set Volume of 0.2 mol dm-3 sodium thiosulphate solution/ cm3

Volume of distilled water/ cm3

Volume of 1.0 mol dm-3 sulphuric acid/ cm3

Total volume of reacting mixture/ cm3

I 50 0 5 55II 40 10 5 55II 30 20 5 55IV 20 30 5 55V 10 40 5 55


[ K1PP1(v) – Able to tabulate the data ]

[ K3PP1 – Able to construct a table with row consists of headings and units correctly ]

[ K3PP2 - Able to write a reading of the Time and 1/Time with correct and consistent decimal places]

[ K3PP3 - Able to write all the reading of the Time and 1/Time with correct and consistent decimal places]

Set Volume of 0.2 mol dm-3 sodium thiosulphate solution/ cm3

Volume of distilled water/ cm3

Volume of 1.0 mol dm-3 sulphuric acid/ cm3

Time taken/s

1 / Time

1/s

I 50 0 5 16 0.063II 40 10 5 20 0.050II 30 20 5 27 0.037IV 20 30 5 41 0.024V 10 40 5 83 0.012


Communication Sample Answer: Concentration against time

Concentration/mol dm-3

Time/s

Concentration against 1/time Concentration/mol dm-3

1/Time


s-1

[K4PP1- Able to draw 2 graphs; concentration of sodium thiosulphate solution against time,t and concentration of sodium thiosulphate solution against 1/t correctly]

Interpreting data 1. Concentration of sodium thiosulphate solution in the reacting mixture.

2. Based on the graphs,when the concentration of sodium thiosulphate is higher,the time taken for the mark `X` to disappear from sight is shorter.

3. The rate of reaction directly proportional to the concentration of sodium thiosulphate solution used.

[K4PP2 – Able to1. calculate the concentration of sodium thiosulphate solution in the

reacting mixture in each sets using the formula M1V1=M2V2 2. determine the values of the reciprocal of time for sets I to V3. deduce from the graphs4. state the relationship between the rate of reaction and the

concentration of sodium thiosulphate solution correctly]

Conclusion The hypothesis is accepted.

Set Concentration of sodium thiosulphate/ mol dm-3

I0.2X50 = 0.18 55

II0.2X40 = 0.15 55

III0.2X30 = 0.11 55

IV0.2X20 = 0.07 55

V0.2X10 = 0.04 55


[K4PP3- Able to state the hypothesis is accepted or rejected]

END

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