Essay About 0.50M Ki And Turn Starch Solution

Determining Activation Energy of a Reaction
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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- ↠2SO42- + I2
The formation of iodine is monitored by small & known amount of
thiosulphate ions, S2O32-:
2S2O32- + I2 ↠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 /℃
Temperature (T) /K
Time (t) /sec
log10 (1/t)
– 1.72
– 1.79
– 1.98
– 2.13
– 2.26
1/T /K-1
3.10×10-3
3.14×10-3
3.19×10-3
3.25×10-3
3.30×10-3
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)
∵ All concentration terms have been kept constant
∠Rate only varies as k (rate constant)
∵ Rate is directly proportional to 1/t & varies as k in the rate
equation
∠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,