The overall order of an elementary step directly corresponds to its molecularity

Question

The overall order of an elementary step directly corresponds to its molecularity. Both steps in this example are second order because they are each bimolecular. Furthermore, the rate law can be determined directly from the number of each type of molecule in an elementary step. For example, the rate law for step 1 is

rate=k[NO2]2

The exponent "2" is used because the reaction involves two NO2molecules. The rate law for step 2 is

rate=k[N2O4]1[CO]1=k[N2O4][CO]

because the reaction involves only one molecule of each reactant the exponents are ommitted.

Analyzing a new reaction

Consider the following elementary steps that make up the mechanism of a certain reaction:

3X?E+F

E+2M?F+N

Part A

What is the overall reaction?

Express your answer as a chemical equation.

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Part B

Which species is a reaction intermediate?

Which species is a reaction intermediate?

Part C

What is the rate law for step 1 of this reaction?

Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][C]3 type k*[A]*[C]^3.

Part D

What is the rate law for step 2 of this reaction?

Express your answer in standard MasteringChemistry notation. For example, if the rate law is k[A][C]3 type k*[A]*[C]^3.

Explanation / Answer

Part A

3X--------------> E+F

E+2M ---------------> F+N

Additing the above two equation, we get 3X + E + 2M ------------> E + F + F + N

So, overall reaction is 3X + 2M ------------> 2F +N

Part B

Species E is a reaction intermediate, because E is neither a reactant nor a prodyct. So the answer is b. E

Part C

Rate law for step 1

3X--------------> E+F

Rate law = k[X]3

Part D

Rate law for step 2

E+2M ---------------> F+N

rate law = k [E][M]2

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