A clock reaction is run at 20 ºC with several different mixtures of iodide, sodi

Question

A clock reaction is run at 20 ºC with several different mixtures of iodide, sodium bromate and acid, to form iodine. Thiosulfate is used to react with the iodine formed initially. Starch indicator is added to form a blue color when all the thiosulfate has been used up and the iodine concentration begins to rise. The following sets of mixtures are used.

                                    Initial concentrations in reaction mixtures
Run number [I-]      [BrO3-]      [H+]
1 0.002          0.008       0.02
2 0.002       0.016            0.02
3 0.004         0.008         0.02
4 0.002        0.008             0.04              

[S2O32- ] is equal to 0.0001M in each reaction mixture.
The rate law is :
  
Rate = k [ I-]a [ BrO3-]b [H+]c


Reaction time in run number 1 30
Reaction time in run number 2 30
Reaction time in run number 3 15
Reaction time in run number 4 30

Calculate the following: a, b and c in the rate law

k in the rate law

Explanation / Answer

Run1 and Run 2 have the same rate. Thus b=0

Rate ratio Run3/Run1 = 2/1. Thus a=1

Rate ratio Run4/Run1= 1 . Thus c=0.

During the reaction time, [S2O32- ] = - 0.0001 M

The   rate law is

Rate = - [S2O32- ] /dt = k[I-]

Using data from Run1:

0.0001 M / 30 s = k · 0.002 M

k = 1.67x10-3 s-1

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