The mechanism for the reaction 2 H2O2(aq) > 2 H2O(l) + O2(g) in the presence of I-(aq) is proposed to be: Step 1 (slow): H2O2 + I- > H2O + OI- Step 2 (fast): H2O2 + OI > H2O + O2 + I- What is the rate law for the overall reaction?

rate: K' [H₂O₂]

Explanation:

Every chemical reaction occurs with a determined mechanism. The slow step of the reaction is the rate determining.

According to the reaction:

H₂O₂(aq) → 2 H₂O(l) + O₂(g)

The proposed mechanism (And its rate laws) in the presence of I⁻ is:

Step 1: H₂O₂ + I⁻ → H₂O + OI⁻ (slow) rate₁: K₁ [H₂O₂] [I⁻]

Step 2: H₂O₂ + OI⁻ → H₂O + O₂ (fast) rate₂: K₂ [H₂O₂] [OI⁻]

As the step 1, is the rate-determining, the rate law for the overall reaction is:

rate: K [H₂O₂] [I⁻]

Now, as I⁻ is a catalyst and its concentration over the reaction is constant, it is possible to write:

rate: K' [H₂O₂]

Where K' = K [I⁻]

a polar bond is formed when electrons are shared unevenly between atoms

2b2t

explanation:

2b2t