Nitramide, NO 2 NH2, decomposes slowely in aqueous slution according to the foll

ID: 960690 • Letter: N

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

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

(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

(d) Intermediate in the reaction is NO2NH-

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.