Question

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?

Answer 1

Answer:

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⁻]

Answer 2

The rate law for the total reaction is K' [H₂O₂]

Calculation of rate law:

A chemical reaction should be the process where one or more substances are known as reactants and are transformed into one or more different substances called products.

As per the reaction

H₂O₂(aq) → 2 H₂O(l) + O₂(g)

Here

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

Now

the rate law for the overall reaction is:

rate: K [H₂O₂] [I⁻]

Since as I⁻ am a catalyst and its concentration should be over the reaction is the same,

Hence, The rate law for the total reaction is K' [H₂O₂]

Learn more about the reaction here: https://brainly.com/question/24710640