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What is the rate law for this reaction?  Is the experimental rate law reasonable given...

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sj-14 | Student, College Freshman | eNoter

Posted February 20, 2012 at 8:36 AM via web

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What is the rate law for this reaction?  Is the experimental rate law reasonable given the proposed mechanism?

If the reaction NO2 (g) + CO  (g) ----> CO2 (g) + NO (g) occurs by a one step process, what would be the expected rate kaw for this reaction?

b.) The actual rate law is Rate = k[NO2]^2. Could the reaction actually occur by a one step collision between NO2 and CO. Explain your answer.

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mlsiasebs | College Teacher | (Level 1) Associate Educator

Posted February 21, 2012 at 9:36 AM (Answer #1)

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If the reaction occurs in one step, then the given reaction can be considered an elementary reaction.

NO2 (g) + CO  (g) ----> CO2 (g) + NO (g)

For an elementary reaction, we can determine the the rate law from the reactants and their coefficients.  Treating this as an elementary reaction means that the rate law would be

rate = k[NO2][CO]

However, based on experimental data, the rate law is found to be

rate = k[NO2]^2

which indicates that the mechanism is not a one-step process as proposed but rather that the rate determining step occurs from the collison of two NO2 molecules.  Since [CO] is not found in the experimental rate law, it is not involved in the rate determining step and must enter the reaction in another step eliminating the possibility of a one-step mechanism.

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