Question

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?

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-