Order and rate law of a reaction The overall order of an elementary step directl

Question

Order and rate law of a reaction

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

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.

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

The exponent "2" is used because the reaction involves two NO2 molecules. The rate law for step 2 is

rate=k[NO3]1[CO]1=k[NO3][CO]

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

Analyzing a new reaction

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

3XE+F

E+MF+N

Explanation / Answer

for the elementary steps,

3X --> E + F

E + M --> F + N

The overall reaction becomes,

3X + M ---> 2F + N

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