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WE DO: The mechanism shown here is proposed for the reaction:

2A → 2B + C


2A k1↔ B + D (Fast, equilibrium)

B + D k-1→ E + C (Slow)

E → 2B (Fast)


Propose a rate law consistent with the mechanism.

Please show and explain everything.

WE DO The mechanism shown here is proposed for the reaction2A 2B C2A k1 B D Fast equilibriumB D k1 E C SlowE 2B FastPropose a rate law consistent with the mecha class=

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Based on the mechanism shown above, the correct rate law that is consistent with the sample reaction is [tex]B+D \rightarrow ^{k_{-1}} E+C[/tex].

What is the rate law?

Rate law can be defined as a chemical equation that is used to relate the initial (forward) chemical reaction rate with respect to the concentrations of both the chemical reactants and constant parameters.

Mathematically, the rate law is given by this formula:

[tex]R = K[A]^x[B]^y[C]^z[/tex]

Where:

  • k is the rate constant.
  • A is the concentration of reactant A.
  • B is the concentration of reactant B.
  • C is the concentration of reactant C.
  • x, y, and z are the order of the reaction.

Based on the mechanism shown above, the correct rate law that is consistent with the sample reaction is as follows:

                                       [tex]B+D \rightarrow ^{k_{-1}} E+C[/tex]

Read more on rate law here: brainly.com/question/24749252