Question

The experimentally determined rate law for the reaction 

\[\ce{H2O2 + 2H+ + 2I- -> I2 + 2H2O}\]

is found to be 

Rate = k [H₂O₂][I⁻]

Postulate a mechanism for the reaction if OI⁻ ions have been detected as intermediate during the progress of the reaction.

Answer 1

Given reaction is \[\ce{H2O2 + 2H+ + 2I- -> I2 + 2H2O}\]

Rate = k [H₂O₂][I⁻]

It is also observed that OI⁻ (hypoiodite ion) is formed as an intermediate during the reaction.

To be consistent with the observed rate law and the formation of OI⁻, the following mechanism is proposed:

Slow rate-determining step:

\[\ce{H2O2 + I− −> H2O + OI−}\]

This step determines the rate of reaction and is consistent with the rate law, as it involves both H₂O₂ and I−.

Step 2 (fast):

\[\ce{OI− + H+ −> HOI}\]

Step 3 (fast):

\[\ce{HOI + H+ + I− −> I2 + H2O}\]

Overall Reaction (adding all steps):

\[\ce{H2O2 + 2H+ + 2I− −> I2 + 2H2O}\]

Thus, this mechanism is consistent with:

The observed rate law: Rate = k[H₂O₂][I⁻]

The formation of OI⁻ as an intermediate

The correct overall balanced reaction.