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The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanisms for this reaction are: Mechanism A (1) H2(g)+I2(g)−→k12HI(g) (one-step reaction) Mechanism B (1) I2(g)⥫⥬=k−1k12I(g) (fast, equilibrium) (2) H2(g)+2I(g)−→k22HI(g) (slow) Mechanism C (1) I2(g)⥫⥬=k−1k12I(g) (fast, equilibrium) (2) I(g)+H2(g)−→k2HI(g)+H(g) (slow) (3) H(g)+I(g)−→k3HI(g) (fast) Which of these mechanisms are consistent with the observed rate law? mechanism C mechanism B mechanism A In 1967, J. H. Sullivan showed that this reaction was dramatically catalyzed by light when the energy of the light was sufficient to break the I−I bond in an I2 molecule. Which mechanism or mechanisms are consistent with both the rate law and this additional observation? mechanism A mechanism B mechanism C

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Answer:

The only mechanism consistent with observed rate law is mechanism A.

Mechanism A is the only consistent with the observation of J. H. Sullivan

Explanation:

In a reaction, rate law is determined by slow step of the reaction.

In mechanism A, rate law is:

rate = k₁ [H₂][I₂]

In mechanism B, rate law is:

rate = k₂ [H₂][I]²

In mechanism C, rate law is:

rate = k₂ [H₂][I]

Thus, the only mechanism consistent with observed rate law is mechanism A

Now, catalyst works in the slow-step of reaction, is the reaction is catalyzed when The I-I bond breaks, the mechanism consistent with the observation is the mechanism where the slow-step of reaction involves this bond break, again, mechanism A is the only consistent with the observation of J. H. Sullivan.

The mechanism that gives the rate of the reaction be similar to the experimentally determined rate has been mechanism A.

J. H. Sullivan observation has been the constituent of Mechanism A.

The reactions can be processed in a one-step fast reaction or the two-step slow reaction. The slow reaction in the two-step reaction has been used for the determination of the rate constant according to the rate law.

The experimentally determined rate law has been:

Rate = [tex]\rm k[H_2][I_2][/tex]

The rate law for the proposed mechanisms has been:

Mechanism A:

It is a one-step reaction. The rate law can be given by;

Rate = [tex]\rm k[H_2][I_2][/tex]

Mechanism B:

It is a two-step reaction. The rate can be determined with the slow reaction. It can be given by:

Rate = [tex]\rm k[H_2][I]^2[/tex]

Mechanism C:

It has been a three-step reaction. The rate can be determined with the slow reaction. It can be given by:

Rate = [tex]\rm k[H_2][I][/tex]

Thus the mechanism that gives the rate of the reaction be similar to the experimentally determined rate has been mechanism A.

With the observation of J. H. Sullivan, the catalyst act as a slow reacting step. The catalyst activity in the breaking of I-I bond, Since there has been I-I bond in the mechanism A, the J. H. Sullivan observation has been the constituent of Mechanism A.

For more information about the rate law, refer to the link:

https://brainly.com/question/4222261

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