PEKA Chemistry form 5 chapter 1 (To study the effect of temperature on the rate of reaction.)

Aim                             : To study the effect of temperature on the rate of reaction.

Problem statement      : How does the temperature of sodium thiosulphate, Na2S2O3, solution affect the rate of reaction?

Hypothesis                  : The rate of reaction will increase if the temperature of sodium thiosulphate, Na2S2O3, solution increases.

Variables                     :-

Manipulated variable  : Temperature of sodium thiosulphate, Na2S2O3, solution. (reaction)

                                    Responding variable   : Rate of reaction.

Controlled variables   : Volume and concentration of sulphuric acid, H2SO4, volume and concentration of sodium thiosulphate, Na2S2O3, solution.

Operational definition : The time taken for the mark “X” to disappear from sight in different temperature when the sodium thiosulphate, Na2S2O3, solution react with sulphuric acid, H2SO4

The rate of reaction is measured by how fast the 'X' is hidden by the yellow precipitate.

Apparatus                    : 50cm3 measuring cylinder, 100cm3 conical flask, 10cm3 measuring cylinder, stopwatch, thermometer, Bunsen burner, wire gauze, tripod stand

Materials                     : 0.2 mol dm-3 sodium thiosulphate, Na2S2O3, solution, 0.1mol dm-3 sulphuric acid, H2SO4, white paper marked “X” at the centre

Procedure                    :-

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1.      An “X” is marked using a pencil on the centre of a piece of white paper.

2.      50cm3 of 0.2 mol dm-3 sodium thiosulphate, Na2S2O3, solution is measured with a 50cm3 measuring cylinder and is poured into a conical flask.

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3.      Using a 10cm3 measuring cylinder, 5cm3 of 0.1mol dm-3 sulphuric acid, H2SO4 is measured.

4.      The sodium thiosulphate, Na2S2O3, solution is heated with a Bunsen burner and quickly placed on the “X” mark on the piece of white paper when it reaches 30c.

5.      Sulphuric acid, H2SO4 is immediately and carefully poured into the conical flask containing 50cm3 of 30c of sodium thiosulphate, Na2S2O3, solution. At the same time, the stopwatch is started.

6.      The conical flask is shaken slowly throughout the experiment on the white paper.

7.      The mark “X” is observed vertically from the top part of the conical flask through the solution.

8.      The stopwatch is stopped quickly as the mark “X” on the white paper is no longer visible.

9.      The time taken, for the mark “X” to disappear from sight is recorded.

10.  Steps 1 to 9 are repeated four more times using different temperature of 0.2 mol dm-3 sodium thiosulphate, Na2S2O3, solution at 35c, 40c, 45c and 50c. Whereas the other conditions remain unchanged.

11.  The results are recorded in a table.

Data and observations            :-

Temperature of sodium thiosulphate, Na2S2O3, solution/ c	30	35	40	45	50
Time/ s	9.8	8.7	8	6	5.2
1/time/ s-1

Interpreting data         :-

1.      Based on the results, two graphs are plotted.

(a)    The graph of the temperature of sodium thiosulphate, Na2S2O3, solution against the time taken for the mark “X” to disappear from sight.

(b)   The graph of the temperature of sodium thiosulphate, Na2S2O3, solution against 1/time.

2.      Based on the graph plotted, determine the relationships between the temperature of sodium thiosulphate, Na2S2O3, solution with  

(a)    The time taken

-          As the temperature of sodium thiosulphate, Na2S203, solution decreases, a longer time is needed for mark ‘X’ to disappear from sight.  Therefore, as the temperature becomes lower, the rate of reaction also decreases.

(b)   1/time

-          The temperature rises, the value of 1/time or rate of reaction also increases.

3.      Deduce the effect of the temperature a reactant on the rate of reaction.

-          The relationship between the rate of reaction and the temperature of sodium thiosulphate, Na2S2O3, solution is, when the temperature of a reactant increases, the rate of reaction increases.

Discussion                   :-

           

1.      Write down the chemical equation for the reaction between sodium thiosulphate, Na2S2O3, solution and dilute sulphuric acid, H2SO4.

-          S₂O_3(aq)   +   2H_(aq)   →   S_(s)   +   SO_2(g)   +   H₂O_(l)

2.      Name the yellow precipitate which is formed throughout the experiment?

-          The yellow precipitate formed is sulphur.

3.      Why are the volume and concentration of both the sodium thiosulphate, Na2S2O3, solution and sulphuric acid, H2SO4 used remain constant for each set of the experiment?

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Conclusion                  : Hypothesis is accepted. The rate of reaction will increase if the temperature of sodium thiosulphate, Na2S2O3, solution increases.