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AysviL [449]
3 years ago
5

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- S

tep 2 (fast): H2O2 + OI- -----> H2O + O2 + I- What is the molecularity of the rate determining step?
Chemistry
1 answer:
andrezito [222]3 years ago
6 0

Answer:

Molecularity of the rate determining step = 2

Explanation:

Step 1 (slow): H₂O₂ + I⁻ -----> H₂O + OI⁻

Step 2 (fast): H₂O₂ + OI⁻ -----> H₂O + O₂ + I⁻

The rate determining step in a reaction mechanism is also considered as slowest step.

Slowest step is also considered its highest activation energy in energy profile diagram.

In this case intermediate  (IO⁻) is formed.

Step 1 considered as a slowest step.

So,  Rate = K [H₂O₂][I⁻]

  Molecularity = 2

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1.5 % is the percent yield in the reaction.

Explanation:

Given that:

original mass of the sample used in reaction = 1.897 grams

product formed after decomposition = 1.071 grams

The reaction for the decomposition:

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It says that 1 mole of Mg(HCO3)2  yielded 1 mole of  MgCO2  on decomposition

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putting the values in the equation:

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