For the reaction show below determine the rate law. H 2 SO 3 (aq) + 6I - (aq) +

Question

For the reaction show below determine the rate law.

H2SO3(aq) + 6I-(aq) + 4H+(aq) S(s) + 2I3- + 3H2O (l)

In the presence of a little starch, this solution turns from colorless to blue when I3- is formed.

The following data were obtained:

Calculate the rate law for this reaction:

rate = k [H2SO3]  [H+]  [I-]  

[H2SO3] [H+] [I-] Initial Rate (mol/L s) 1.0 x 10-2 2.0 x 10-1 2.0 x 10-1 1.66 x 10-7 3.0 x 10-2 2.0 x 10-1 2.0 x 10-1 4.98 x 10-7 1.0 x 10-2 1.0 x 10-1 2.0 x 10-1 8.30 x 10-8 1.0 x 10-2 2.0 x 10-1 4.0 x 10-1 13.2 x 10-7

Explanation / Answer

ANSWER:

rate = k [H2SO3]1 [H+]1 [I-] 3

Explaination: To solve such a question we have to determine the effect of concentration change on rate of reaction.

Whwn concentration of H2SO3 is increased by three times keeping the concentration of other reactants constants, from 1.0 x 10-2 (reading 1 of the given table) to 3.0 x 10-2 (reading 2of the given table) the rate of reaction also increases by three times, from 1.66 x 10-7 (reading 1) to 4.98 x 10-7 (reading 2)

[4.98 x 10-7/1.66 x 10-7 = 3]

Hence order with respect to H2SO3 is 1.

Simillarly when concentration of H+ is halved (from 2.0 x 10-1 reading 1 to 1.0 x 10-1 reading 3) the rate of reaction is also halved ( from 1.66 x 10-7 in reading 1 to 8.30 x 10-8 in reaing 3 of the table). therefore order with respect to H+ is 1.

Simillaly when the concentration of I-is doubled ( Consider reading 1 and reading 4) the rate of reaction increases by eight fold. Hence order with respect to I- is 3. (HINT: (2)3 = 8)

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