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The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl– + I– OI– + Cl– . This rapid reaction gives the following rate data:

(a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when [OCl–] = 2.0 × 10-3 M and [I–] = 5.0 × 10–4 M.

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Jean Keeling
Jean KeelingLv2
13 May 2020

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Related questions

The iodine ion reacts with hypochlorite ion (the activeingredient in chlorine bleaches) in the following way: OCl- + I- ----> OI- +Cl-. This rapid reaction gives the following ratedata:
[OCl-],M I-,M Rate, L/s
1.6 x10-3 1.6x10-3 1.75 x 104
7.6 x10-3 1.6x10-3 2.72 x 104
1.6 x10-3 9.6x10-3 2.72 x 105
A) write the rate law for this reaction
B) What is the overall order of the reaction
C) What is the numerical value of the rate constant and what are the units?
D) Calculate the rate of this reaction when the initial concentrations of OCl- and I- are 7.6 x10-3 M and 9.6 x 10-3M respectively.

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
OCl−+I−→OI−+Cl−.
This rapid reaction gives the following rate data:

[OCl−](M) [I]−(M) Rate (M/s)
1.5×10−3 1.5×10−3 1.36×10−4
3.0×10−3 1.5×10−3 2.72×10−4
1.5×10−3 3.0×10−3 2.72×10−4

Write the rate law for this reaction.

Calculate the rate constant with proper units.

Calculate the rate when [OCl−]= 2.5×10−3 M and [I−]= 5.1×10−4 M .

The iodide ion reacts with the hypochlorite ion (active ingredient in chorine bleaches) in the following way: OCl + I → OI + Cl. This reaction was studied with the following results.

(a) What is the correct rate law expression for this reaction?

(b) Calculate the rate constant with proper units.

Exp #

[OCl] (M)

[I] (M)

Initial Rate (M/s)

1

1.5 x 10-3

1.5 x 10-3

1.35 x 10-4

2

3.0 x 10-3

1.5 x 10-3

2.73 x 10-4

3

1.5 x 10-3

3.0 x 10-3

2.71 x 10-4

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