Question

Nitramide, NO2NH2, decomposes slowely in aqueous slution according to the following reaction: NO2NH2 (aq) ---> N2O...

Nitramide, NO2NH2, decomposes slowely in aqueous slution according to the following reaction:

NO2NH2 (aq) ---> N2O (g) + H2O

the reaction follows the rate law: Rate= k(NO2NH2)/(H3O+) *Omit H2O from the rate law that you determine from the Mechanisms*

(a) Which of the following mechanisms is the most appropriate for the interpretation of this rate law? Justify your answer.

Mechamism 1

NO2NH2 ---> N2O + H2O  rate constant = k1

Mechanism 2

NO2NH2 + H3O+ <---> NO2NH3+ + H2O Fast rate constant= k2 (forward) and k-2 (reverse)

NO2NH3+ ---> N2O + H3O+   Slow rate constant= k3   

Mechanism 3

NO2NH2 +  H2O  <---> NO2NH- + H3O+   Fast rate constant= k4 (forward) and k-4 (reverse)

NO2NH- ---> N2O + OH- Slow rate constant= k5

H3O+ + OH- ---> 2H2O Fast rate constant= k6

(b) Define k based on your mechanism

(c) Is there a Catalyst in the reaction? If so, what is it?

(d) is there an intermediate(s) in the reaction? If so, what is it?

Homework Answers

Answer #1

(a) Based on the given rate law, the most appropriate mechanism would be,

Mechanism 3

In mechanism 3 the rate of H3O+ change is inversely related to NO2NH2 as required in the given rate law.

(b) the rate constant k for the reaction

From the slow step the rate equation would be,

rate = k5[NO2NH-]

Here NO2NH- is an intermediate

From equilibrium equation (first equation) and applying steady state assumption for the intermediate,

rate of formation of interemdiate = rate of consumption of intermediate

k4[NO2NH2] = k-4[NO2NH-][H3O+]

[NO2NH-] = (k4/k-4)([NO2NH2]/[H3O+])

Feeding in the rate equation,

rate = (k5k4/k-4)([NO2NH2]/[H3O+])

Thus, the rate constant k = k4.k5/k-4

(c) The catalysts in the reaction is H+.

(d) Intermediate in the reaction is NO2NH-

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