Suppose the reaction between nitric oxide and oxygen proceeds by the following mechanism:step elementary reaction rate constant1 NO(g) + O2(g) → NO2(g) + O(g) k12 NO(g) + O(g) → NO2(g) k2Suppose also k1≫k2. That is, the first step is much faster 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 reaction between nitric oxide and oxygen proceeds by the following mechanism:step elementary reaction rate constant1 NO(g) + O2(g) → NO2(g) + O(g) k12 NO(g) + O(g) → NO2(g) k2Suppose also k1≫k2. That is, the first step is much faster 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 balanced chemical equation for the overall chemical reaction is:
2NO(g) + O2(g) → 2NO2(g)
The experimentally-observable rate law for the overall chemical reaction is:
rate = k[NO]^2[O2]
Since the first step is much faster than the second, it is in quasi-equilibrium. Therefore, we can express the concentration of the intermediate O in terms of the reactants and rate constants:
[O] = k1[NO][O2] / k-1
Substituting this into the rate law gives:
rate = k2[NO] * (k1[NO][O2] / k-1) = (k1*k2/k-1)[NO]^2[O2]
So, the rate constant k for the overall chemical reaction in terms of k1, k2, and k-1 is:
k = k1*k2/k-1
Note: The rate constants for the reverse reactions are typically denoted with a negative sign in front of the number (e.g., k-1, k-2). In this case, only k-1 is necessary because the first step is in quasi-equilibrium.
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