Beaker A and beaker B are added together. 1) Calculate the initial concentration

Question

Beaker A and beaker B are added together.

1) Calculate the initial concentrations of KI and K2S2O8 after mixing A and B for each run.

Calculate the values of the exponents by using appropriate combinations of the results of runs I to V. Round off the values for each run to the nearest integer.

a) Exponent " m " from the results of runs I, I I and III, by the initial rates method.

b) Exponent " n " from results of Runs I, I and V, by the initial Rates method.

3) For this reaction what is the rate law? What is the overall order and what are the units for its rates constant?

Explanation / Answer

Kinetic Study of the Reaction of KI with K2S2O8

Beaker A and beaker B are added together.

S2O8(-2) + 2I- -> 2SO4(-2) + I2
2S2O3(-2) + I2 -> S4O6(-2) + 2I-

rate [S2O8(-2)] = rate [I2] = (rate [S2O3(-2)])/2

1) Calculate the initial concentrations of KI and K2S2O8 after mixing A and B

for each run.

Concentrations in mol/L
I
[KI] = 0.2*20/(20+10+20) = 0.08
[K2S2O8]=0.1*20/(20+10+20)= 0.04

II
[KI] = 0.2*10/(10+10+10+20) = 0.04
[K2S2O8]=0.1*20/(10+10+10+20)= 0.04

III
[KI] = 0.2*5/(5+10+15+20) = 0.02
[K2S2O8]=0.1*20/(5+10+15+20)= 0.04

IV
[KI] = 0.2*20/(20+10+10+10) = 0.08
[K2S2O8]=0.1*10/(20+10+10+10)= 0.02

V
[KI] = 0.2*20/(20+10+5+5+10) = 0.08
[K2S2O8]=0.1*5/(20+10+5+5+10)= 0.01

2) Calculate the values of the exponents by using appropriate combinations of

the results of runs I to V. Round off the values for each run to the nearest

integer.

Iodine is a self-indicator. Net iodine produced can be detected by its color

change to orange or yellow.

a) Exponent " m " from the results of runs I, II and III, by the initial rates

method.

Initial concentration, [S2O3(-2)] = 10*.005/50=0.001 mol/L
rate [S2O8(-2)] = (rate [S2O3(-2)])/2 = 0.001/(2*t)
I
time = 45.56s
rate [S2O8(-2)] = 0.001/(2*45.56) = 1.1E-5 mol/(L*s)

II
time = 98.88s
rate [S2O8(-2)] = 0.001/(2*98.88) = 0.505E-5 mol/(L*s)

III
time = 165.91s
rate [S2O8(-2)] = 0.001/(2*165.91) = 0.3E-5 mol/(L*s)

rateI/rateII = 2 = [S2O8(-2)/S2O8(-2)]^m = 2^m

m =1

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