Experiment# Kinetics of an Iodine Clock Reaction 1: A color change will occur in Kinetics of an Iodine Clock Reaction that signifies the end of a trial. What chemical change occurs that causes the color change? 2: The same amount of thiosulfate ion is added to each trail to delay the onset of the color change. Calculate the initial molarity of S2O3 2- in each trail given that each trail has 1.00mL of 0.010M Na2S2O3 and the total volume of 7.00mL. Solution In Iodine clock reaction, all the colourless reactants react to form iodine which is identified by starch indicatin by the formation of a blue-black colour after a some period of time. The reactant are hydrogen peroxide which is identified using dilute sulphuric acid, small quantities of starch and sodium thiosulphate. Hydrogen peroxide reacts with potassium iodide to form iodine. H 2 O 2(aq) + 2I - (aq) + 2H + (aq) ------------> 2H 2 O (l) + I 2(aq) The iodine formed reacts with sodium thiosulphate to form iodide ion I 2 (aq) + Na 2 S 2 O 3(aq) ---------> 2NaI (aq) + S 4 O 6 2- (aq) When all the sodium thioulphate is consumed then the left over iodine reacts with iodide to form triodide ion which reacts with starch indicator to form a blue-black colour complex. I 2 + I - ----> I 3 - + starch -------> blue-black colour The only source of thiosulphate ions is sodium thiosulphate so, number of moles of thiosulphate = number of moles of sodium thiosulphate Vol of Na 2 S 2 O 3 in lts = 0.001 lt Molarity of Na 2 S 2 O 3 = 0.01 M No. of moles of sodium thiosulphate = Molarity (mol/lt) of Na 2 S 2 O 3 X Vol of Na 2 S 2 O 3 in lts = = 0.01 M x 0.001 lts = 10 -5 moles of Na 2 S 2 O 3 = 10 -5 moles of S 2 O 3 2- . Total volume in lts = 0.007 lts Molarity of thiosulphate initially = No. of moles of S 2 O 3 2- ./ total Vol in lts = 10 -5 moles of S 2 O 3 2- ./ 0.007 lts = 1.428 x 10 -3 moles/lt or M of S 2 O 3 2- . Initial molarity of S 2 O 3 2- .= 1.43 x 10 -3 M .

1. Experiment# Kinetics of an Iodine Clock Reaction
1: A color change will occur in Kinetics of an Iodine Clock Reaction that signifies the end of a
trial. What chemical change occurs that causes the color change?
2: The same amount of thiosulfate ion is added to each trail to delay the onset of the color
change. Calculate the initial molarity of S2O3 2- in each trail given that each trail has 1.00mL of
0.010M Na2S2O3 and the total volume of 7.00mL.
Solution
In Iodine clock reaction, all the colourless reactants react to form iodine which is identified by
starch indicatin by the formation of a blue-black colour after a some period of time.
The reactant are hydrogen peroxide which is identified using dilute sulphuric acid, small
quantities of starch and sodium thiosulphate.
Hydrogen peroxide reacts with potassium iodide to form iodine.
H 2 O 2(aq) + 2I -
(aq) + 2H +
(aq) ------------> 2H 2 O (l) + I 2(aq)
The iodine formed reacts with sodium thiosulphate to form iodide ion
I 2 (aq) + Na 2 S 2 O 3(aq) ---------> 2NaI (aq) + S 4 O 6
2-
(aq)
When all the sodium thioulphate is consumed then the left over iodine reacts with iodide to form
triodide ion which reacts with starch indicator to form a blue-black colour complex.
I 2 + I -
----> I 3
-
+ starch -------> blue-black colour
The only source of thiosulphate ions is sodium thiosulphate so,
number of moles of thiosulphate = number of moles of sodium thiosulphate
Vol of Na 2 S 2 O 3 in lts = 0.001 lt
Molarity of Na 2 S 2 O 3 = 0.01 M

2. No. of moles of sodium thiosulphate = Molarity (mol/lt) of Na 2 S 2 O 3 X Vol of Na 2 S 2 O 3 in lts
=
= 0.01 M x 0.001 lts = 10 -5
moles of Na 2 S 2 O 3 = 10 -5
moles of S 2 O 3
2-
.
Total volume in lts = 0.007 lts
Molarity of thiosulphate initially = No. of moles of S 2 O 3
2-
./ total Vol in lts
= 10 -5
moles of S 2 O 3
2-
./ 0.007 lts = 1.428 x 10 -3
moles/lt or M of S 2 O 3
2-
.
Initial molarity of S 2 O 3
2-
.= 1.43 x 10 -3
M