Question

Iodine clock reaction lab

2I^-(aq) + S₂O₈^2-(aq)à I₂(aq) + 2SO₄^2- (aq)

Rate=k[I^-]^x[S₂O₈^2-]^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

Answer 1

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) + S₂O₈^2- (aq) à I₂(aq) + 2SO₄^2- (aq)

Rate=k[I^-]^x[S₂O₈^2-]^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) + S₂O₈^2- (aq) à I₂(aq) + 2SO₄^2- (aq) (first reaction)

2S2O32-+I2=S4O62- +2I-….(second reaction)

[I2] (generated in first reaction)=[I2] consumed in second reaction=1/2*[S2O3]2-=∆[S₂O₈^2-]

But as [S2O3]2- is fixed , so we can write change in [S₂O₈^{2-] = 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^-][S₂O₈^2-]                          ^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