mark scheme peka chem 1.2- f5
Post on 27-Nov-2014
610 Views
Preview:
TRANSCRIPT
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
top related