Suppose the decomposition of nitrogen dioxide proceeds by the following mechanism:step elementary reaction rate constant1 NO2(g) → NO(g) + O(g) k12 O(g) + NO2(g) → O2(g) + NO(g) k2Suppose also k1≪k2. That is, the first step is much slower than the second.Write the balanced chemical equation for the overall chemical reaction: Write the experimentally-observable rate law for the overall chemical reaction. =ratek Note: your answer should not contain the concentrations of any intermediates.Express the rate constant k for the overall chemical reaction in terms of k1, k2, and (if necessary) the rate constants k-1 and k-2 for the reverse of the two elementary reactions in the mechanism. =k
Question
Suppose the decomposition of nitrogen dioxide proceeds by the following mechanism:step elementary reaction rate constant1 NO2(g) → NO(g) + O(g) k12 O(g) + NO2(g) → O2(g) + NO(g) k2Suppose also k1≪k2. That is, the first step is much slower than the second.Write the balanced chemical equation for the overall chemical reaction: Write the experimentally-observable rate law for the overall chemical reaction. =ratek Note: your answer should not contain the concentrations of any intermediates.Express the rate constant k for the overall chemical reaction in terms of k1, k2, and (if necessary) the rate constants k-1 and k-2 for the reverse of the two elementary reactions in the mechanism. =k
Solution
The overall balanced chemical equation for the reaction is obtained by adding the two elementary reactions:
NO2(g) → NO(g) + O(g) O(g) + NO2(g) → O2(g) + NO(g)
This gives:
2NO2(g) → 2NO(g) + O2(g)
The rate law for the overall reaction is determined by the slowest (rate-determining) step, which is the first step in this case. Since the first step is a unimolecular reaction involving only NO2, the rate law is first order in NO2:
rate = k[NO2]
where k is the rate constant for the overall reaction and [NO2] is the concentration of NO2.
Since the first step is much slower than the second, the rate constant for the overall reaction is approximately equal to the rate constant for the first step, k1. Therefore, we have:
k ≈ k1
Note that the rate constants for the reverse reactions, k-1 and k-2, are not needed in this case because the reaction is assumed to proceed predominantly in the forward direction.
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