The decomposition of nitramide, O2NNH2, in water has the chemical equation and rate law O2NNH2(aq)⟶N2O(g)+H2O(l)rate=k[O2NNH2][H+] A proposed mechanism for this reaction is O2NNH2(aq)⥫⥬=k−1k1O2NNH−(aq)+H+(aq)(fast equilibrium) O2NNH−(aq)−→k2N2O(g)+OH−(aq)(slow) H+(aq)+OH−(aq)−→k3H2O(l)(fast) What is the relationship between the observed value of k and the rate constants for the individual steps of the mechanism?

The rate law is defined as the molar concentration of reactants raised to the power of their stoichiometric coefficients. The law states the dependency of chemical reactions on reactants.

In the given decomposition of nitramide in water, it can be stated that the third step of the reaction is fast, and therefore, does not contribute to the rate law.

The rate-determining step of a reaction is a slow step.

The decomposition reaction can be written as:

O₂NNH₂ [tex]\rightarrow[/tex] O₂NNH⁻ + H⁺ (fast equilibrium)

The value of K is:

[tex]\text{K}_{\text {eq}}&=\dfrac{K_1}{K_{-1}}&=\dfrac{\text[{O}_2\text{NNH}^-][\text{H}^+]}{\text[{O}_2\text{NNH}_{-2}]}[/tex]

[tex]\begin{aligned}\text{Rate}=\text{K}_2\frac{\text{K}_1[\text{O}_2\text{NNH}_2]}{\text{K}_{-1}[\text{H}_2]}\end{aligned}[/tex]

The second step of the reaction:

O₂NNH₂ [tex]\rightarrow[/tex] O₂N + OH⁻

The rate can be given as:

Rate = K₂ [O₂NNH⁻]

Substituting equation 1, we get:

[tex]\begin{aligned} \text[{O}_2\text{NNH}^-] &= \dfrac{\text{K}_1[\text{O}_2\text{NNH}_2]}{\text{K}_{-1}[\text{H}^+]}\end{aligned}[/tex]

Now, writing the equation in the rate constant, we get:

[tex]\text {K}&=\dfrac{K_2K_1}{K_{-1}}[/tex]

Thus, it can be concluded that the third step is very fast and does not contribute to the rate law.

To know more about rate law, refer to the following link:

https://brainly.com/question/4222261