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:

[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

Calculate the rate law for this reaction:

rate = k [H₂SO₃]  [H⁺]  [I^-]  

Answer 1

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)