Answer :
Answer:
[tex]\large \boxed{{\text{rate} = k\dfrac{[\text{ O$_{2}$NNH$_{2}$]}}{[\text{H}^{+}]}} }[/tex]
Explanation:
O₂NNH₂⟶ N₂O + H₂O
I think you are asking us to derive the rate law from the proposed mechanism.
A. The mechanism
[tex]\rm O_{2}NNH_{2}\xrightarrow[k_{-1}]{k_{1}} O_{2}NNH^{-} + H^{+} \,(fast \, equilibrium)\\\rm O_{2}NNH^{-} \xrightarrow{k_{2}} N_{2}O + \text{OH}^{-} \,(slow)\\\rm H^{+} +\text{OH}^{-}\xrightarrow{k_{3}} H_{2}O\, (fast)[/tex]
B. The rate expressions
[tex]-\dfrac{\text{d[O$_{2}$NNH$_{2}$]} }{\text{d}t} = k_{1}[\text{ O$_{2}$NNH$_{2}$]} - k_{-1}[\text{O$_{2}$NNH$^{-}$]}[\text{H}^{+}]\\\\-\dfrac{\text{d[O$_{2}$NNH$^{-}$]}}{\text{d}t} = k_{2}[\text{ O$_{2}$NNH$^{-}$}]\\\\[/tex]
The first step is an equilibrium, so the rates of the forward and reverse reactions are equal. The equilibrium is only slowly leaking away O₂NNH⁻ to form product.
[tex]k_{1}[\text{ O$_{2}$NNH$_{2}$]} = k_{-1}[\text{O$_{2}$NNH$^{-}$]}[\text{H}^{+}][/tex]
[tex][\text{O$_{2}$NNH$^{-}$]} = \dfrac{k_{1}}{k_{-1}}\times \dfrac{[\text{ O$_{2}$NNH$_{2}$]}}{[\text{H}^{+}]}[/tex]
C. The rate law
Substitute into the rate law for the slow step.
[tex]-\dfrac{\text{d[O$_{2}$NNH$^{-}$]}}{\text{d}t} = \dfrac{k_{1}k_{2}}{k_{-1}}\times \dfrac{[\text{ O$_{2}$NNH$_{2}$]}}{[\text{H}^{+}]}\\\\\text{rate} = k \dfrac{[\text{ O$_{2}$NNH$_{2}$]}}{[\text{H}^{+}]}\\\\\text{The rate law for the reaction is $\large \boxed{\text{rate} = k\dfrac{[\text{ O$_{2}$NNH$_{2}$]}}{[\text{H}^{+}]} }$}[/tex]