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 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 ommitted.

Express your answer as a chemical equation.

Part B

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

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

Part C

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

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

Explanation / Answer

So according to Kinetics, when the slow step is not the 1st step in an overall reaction, we must take in account the intermediates that are consumed in the slow step of the reaction. Here's the example:

1. NO2 + NO2 -> NO3 + NO slow step
2. NO3 + CO -> NO2 + CO2 fast step

A.rate law: k2[NO2]^2

B. but if the speeds are reversed, then it's this:

1. NO2 + NO2 -> NO3 + NO fast step
2. NO3 + CO -> NO2 + CO2 slow step

rate law NOT: k2[NO3][CO]

but : k2k1/k-1 [NO3][CO] k-1 = reverse reaction of step 1

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.