Determining Activation Energy of a Reaction

Determining Activation Energy of a Reaction

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Determining Activation Energy of a Reaction

Objective

To determine the activation energy for the reduction of

peroxodisulphate(VI) ions. S2O82-, by iodide ions I-, using a 'clock'

reaction.

Principle

The equation for reduction of S2O82- by I- is:

S2O82- + 2I- â†' 2SO42- + I2

The formation of iodine is 'monitored' by small & known amount of

thiosulphate ions, S2O32-:

2S2O32- + I2 â†' S4O62- + 2I-

Once the reactants are mixed, the stop-watch is started. At the time

when all of the thiosulphate is reacted, any free iodine produced will

turn starch solution (added before) into dark-blue, the time is then

recorded. The amount of thiosulphate added monitors the time in which

starch turns blue and the reaction rate is directly proportional to 1

÷ time taken for starch changes to dark blue.

By plotting a graph of log10(1/t) against 1/T (T=absolute

temperature), activation energy (Ea) can be found.

Chemicals

0.020M K2S2O8, 0.50M KI, 0.010M Na2S2O3, 0.2% starch solution

Apparatus

400 cm3 beaker, boiling tubes x2, pipettes (10ml), thermometer x2,

water bath, stop-watch

Procedure

1.> Half-fill the beaker with hot water at temperature between 49 -

51℃ (as water bath).

2.> Pipette 10 cm3 of 0.020M K2S2O8 solution into the first boiling

tube, place a thermometer in this solution and keep this in the water

bath.

3.> Pipette 5 cm3 of both 0.50M KI and 0.010M Na2S2O3 and 2.5 cm3 of

starch solution into the second boiling tube. Place another

thermometer in

this solution and stand it in the water bath.

4.> When the temperatures of the two solutions are equal and constant

(to within ± 1℃), pour the contents of the second boiling tube into

the first,

shake to mix, and start the stop-watch.

5.> When the blue colour of the starch-iodine complex appears, stop

the stop-watch and record the time.

6.> Repeat the experiment at temperatures at 50℃, 45℃, 40℃, 35℃, 30℃.

7.> Plot a graph of log10(1/t) against 1/T to calculate the Ea.

Data Analysis

Temperature /℃

50

45

40

35

30

Temperature (T) /K

323

318

313

308

303

Time (t) /sec

52

62

95

135

182

log10 (1/t)

- 1.72

- 1.79

- 1.98

- 2.13

- 2.26

1/T /K-1

3.10x10-3

3.14x10-3

3.19x10-3

3.25x10-3

3.30x10-3

â-- Plot log10(1/t) against 1/T:

[IMAGE]

Let the rate equation be: Rate = k[S2O82-]a [I-]b

From Arrhenius equation, we have k = Ae-Ea/RT, where A is Arrhenius

constant.

By taking logarithm of above equation, we have log10k = log10A - Ea÷(2.303RT)

[1]

∵ All concentration terms have been kept constant

âˆ' Rate only varies as k (rate constant)

∵ Rate is directly proportional to 1/t & varies as k in the rate

equation

âˆ' We can substitute k by 1/t in [1]:

log10(1/t) = log10A - Ea÷(2.303RT), which is the equation of the

previous graph

From the graph, the slope is,

-Ea ÷ (2.303R) = [-1.75 - (-1.98)] ÷ [(3.12 - 3.20)x10-3] = - 2875

âˆ'