Iodine clock reaction lab 2I - (aq) + S 2 O 8 2- (aq)à I 2 (aq) + 2SO 4 2- (aq)
ID: 916885 • Letter: I
Question
Iodine clock reaction lab
2I-(aq) + S2O82-(aq)à I2(aq) + 2SO42- (aq)
Rate=k[I-]x[S2O82-]y
I need to calculate the initcial concentrations of I- and S2O8^2- then use that to calculate the reaction rate M/sec and the k value.
Run Number
3% Starch
0.012 Na2S2O3 mL
0.20 M KI mL
0.20 M KNO3 mL
0.20 M (NH4)2SO4 mL
0.20 M (NH4)2S2O8 mL
Total= 2 mL
Time sec
1
2 drops
0.200
0.800
0.200
0.400
0.400
2.00
35.75
2
2 drops
0.200
0.400
0.600
0.400
0.400
2.00
109.72
3
2 drops
0.200
0.200
0.800
0.400
0.400
2.00
218.65
4
2 drops
0.200
0.400
0.600
0.00
0.800
2.00
33.31
5
2 drops
0.200
0.400
0.600
0.600
0.200
2.00
219.18
Run Number
3% Starch
0.012 Na2S2O3 mL
0.20 M KI mL
0.20 M KNO3 mL
0.20 M (NH4)2SO4 mL
0.20 M (NH4)2S2O8 mL
Total= 2 mL
Time sec
1
2 drops
0.200
0.800
0.200
0.400
0.400
2.00
35.75
2
2 drops
0.200
0.400
0.600
0.400
0.400
2.00
109.72
3
2 drops
0.200
0.200
0.800
0.400
0.400
2.00
218.65
4
2 drops
0.200
0.400
0.600
0.00
0.800
2.00
33.31
5
2 drops
0.200
0.400
0.600
0.600
0.200
2.00
219.18
Explanation / Answer
Calculation initial conc I- and S2O8^2
Run no
Conc of I- ( moles)
Conc of S2O82- (moles)
1
0.00016
0.00008
2
0.00008
0.00008
3
0.00004
0.00008
4
0.00008
0.00016
5
0.00008
0.00004
Concentration=molarity*volume
Example) Conc of I- ( mol/L) for Run1
=0.20M * 0.800ml=0.20 mol/L * 0.800/1000 L=0.00016 moles
Conc of S2O82- (moles)=0.20 M*0.400ml=0.20 mol/L * 0.400/1000 L=0.00008 moles
2I-(aq) + S2O82- (aq) à I2(aq) + 2SO42- (aq)
Rate=k[I-]x[S2O82-]y,
calculation of initial rate
the end point of the reaction is measured in terms of appearance of blue-coloured I2-starch complex when S2O32-(thiosulphate )runs out.As S2O32- concentration is fixed ,so the reaction is measured only until the entire S2O32- is used up.
2I-(aq) + S2O82- (aq) à I2(aq) + 2SO42- (aq) (first reaction)
2S2O32-+I2=S4O62- +2I-….(second reaction)
[I2] (generated in first reaction)=[I2] consumed in second reaction=1/2*[S2O3]2-=[S2O82-]
But as [S2O3]2- is fixed , so we can write change in [S2O82-] = 1/2*[S2O3]2-
Rate=d[I2]/dt =1/2*[S2O3]2-/t
Where t=time for appearance of blue color
[S2O3]2-=0.012M*0.200 ml=0.012M*0.200 /1000 ml=2.4*10^-6 moles
run1 ,
Initial rate=1/2*[S2O3]2-/t=1/2(2.4*10^-6 moles)/35.75s=0.033*10^-6 moles/s =0.033*10^-6 moles/s /total volume=0.033*10^-6 moles/s/ 2.0 ml *1000ml/L=0.0165*10^-3 M/s
Run2,
Initial rate=1/2*[S2O3]2-/t=1/2(2.4*10^-6 moles)/109.72 s=0.0109*10^-6 moles/s =0.0109*10^-6 moles/s /total volume=0.0109*10^-6 moles/s/ 2.0 ml *1000ml/L=0.0054*10^-3 M/s
Similarly initial rate for run 3,4,5 can be calculated.
Order of reaction calculation
order of rxn x and y
for run 1,
0.0165 *10^-3 M/s = k (0.00016mol /2.0 *10^-3 L))^x (0.00008 mol/2.0 *10^-3L)^y ………(1)
Where total volume=2.00ml=2.00*10^-3 L
for run 2,
0.0054*10^-3 M/s = k (0.00008mol/2.0 *10^-3 L)^x (0.00008 mol/2.0 *10^-3 L)^y……………..(2)
divide equation (1) by (2)
0.0165*10^-3 /0.0054*10^-3 = k[ (0.00016)^x (0.00008)^y ]/[ k (0.00008)^x (0.00008)^y]
Or,3.055=(0.00016/0.00008)^x * (0.00008/0.00008)^y
Or,3.055=(2)^x (1)^y………………….(3)
Or, 3.055=(2)^x
Or, log 3.055=x log 2
X= log 3.055/ log 2=0.48/0.30=1.4=1 (approx)
So substitute x=0 in eqn (3) to get y,
3.04=(2)^1 (1)^y
Or, 3.045/2= (1)^y
Or, , 1= (1)^y
Y=1(approx)
Rate law=
Rate=k[I-][S2O82-] x=1.y=1
calculation of K
run1,
Initial rate= 0.0165*10^-3 M/s =k (0.00016 mol/2.0 *10^-3L) (0.00008 mol/2.0 *10^-3 L)……….(4)
Solve (4)
K=5.15*10^-3s-1
Run2,
Initial rate = 0.0054*10^-3 M/s= k (0.00008 mol/2.0*10^-3L) (0.00008 mol/2.0*10^-3 L)……….(5)
Solve,
K=3.37*10^-3s-1
Run no
Conc of I- ( moles)
Conc of S2O82- (moles)
1
0.00016
0.00008
2
0.00008
0.00008
3
0.00004
0.00008
4
0.00008
0.00016
5
0.00008
0.00004
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