Which represents the overall (net) reaction of this mechanism: step 1: H2O2 + I- → H2O + OI- step 2: H2O2 + OI- → H2O + O2 + I- Note that I- acts as a catalyst in this reaction. A. 2H2O2 → 2H2O + O2

B. H2O2 → H2 + O2

C. H2O2 + I- → OI- + H2O

D. H2O2 + OI- → H2O + O2 + I-

A. 2H₂O₂  → 2H₂O + O₂

Explanation:

1. H₂O₂ + I⁻ → H₂O + OI⁻2. H₂O₂ + OI⁻ → H₂O + O₂ + I⁻

If we make a net sum of both reactions, we're left with:

H₂O₂ + I⁻ + H₂O₂ + OI⁻ → H₂O + OI⁻ + H₂O + O₂ + I⁻

Grouping species:

2H₂O₂ + OI⁻ + I⁻ → 2H₂O + OI⁻ + O₂ + I⁻

There is OI⁻ at both sides, so it is eliminated -same goes for the catalyst, I⁻-.

2H₂O₂  → 2H₂O + O₂

Thus the answer is option A.

Answerδh > 0explanationc₆h₆  (s)  → c₆h₆ (l)if you are changing it from a solid to a liquid, you are pretty much melting the solid. this requires heat energy. therefore, the reaction is endothermic (takes in energy); the system enthalpy increases as the reaction proceeds: δh > 0.