Question

The decomposition of nitramide, O 2 NNH 2 , in water has the given chemical equation and rate law. O 2 NNH 2 ( aq ) ⟶ N 2 O ( g ) + H 2 O ( l ) rate = k [ O 2 NNH 2 ] [ H + ] A proposed mechanism for this reaction is:________. ( 1 ) O 2 NNH 2 ( aq ) k 1 ⇌ k − 1 O 2 NNH − ( aq ) + H + ( aq ) (fast equilibrium) ( 2 ) O 2 NNH − ( aq ) k 2 −→ N 2 O ( g ) + OH − ( aq ) (slow) ( 3 ) H + ( aq ) + OH − ( aq ) k 3 −→ H 2 O ( l ) (fast)

Answer 1

Answer:

$\large \boxed{{\text{rate} = k\dfrac{[\text{ O _{2} NNH _{2} ]}}{[\text{H}^{+}]}} }$

Explanation:

O₂NNH₂⟶ N₂O + H₂O  

I think you are asking us to derive the rate law from the proposed mechanism.

A. The mechanism

$\rm O_{2}NNH_{2}\xrightarrow[k_{-1}]{k_{1}} O_{2}NNH^{-} + H^{+} \,(fast \, equilibrium)\\\rm O_{2}NNH^{-} \xrightarrow{k_{2}} N_{2}O + \text{OH}^{-} \,(slow)\\\rm H^{+} +\text{OH}^{-}\xrightarrow{k_{3}} H_{2}O\, (fast)$

B. The rate expressions

$-\dfrac{\text{d[O _{2} NNH _{2} ]} }{\text{d}t} = k_{1}[\text{ O _{2} NNH _{2} ]} - k_{-1}[\text{O _{2} NNH ^{-} ]}[\text{H}^{+}]\\\\-\dfrac{\text{d[O _{2} NNH ^{-} ]}}{\text{d}t} = k_{2}[\text{ O _{2} NNH ^{-} }]$

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.

$k_{1}[\text{ O _{2} NNH _{2} ]} = k_{-1}[\text{O _{2} NNH ^{-} ]}[\text{H}^{+}]$

$[\text{O _{2} NNH ^{-} ]} = \dfrac{k_{1}}{k_{-1}}\times \dfrac{[\text{ O _{2} NNH _{2} ]}}{[\text{H}^{+}]}$

C. The rate law

Substitute into the rate law for the slow step.

$-\dfrac{\text{d[O _{2} NNH ^{-} ]}}{\text{d}t} = \dfrac{k_{1}k_{2}}{k_{-1}}\times \dfrac{[\text{ O _{2} NNH _{2} ]}}{[\text{H}^{+}]}\\\\\text{rate} = k \dfrac{[\text{ O _{2} NNH _{2} ]}}{[\text{H}^{+}]}\\\\\text{The rate law for the reaction is \large \boxed{\text{rate} = k\dfrac{[\text{ O _{2} NNH _{2} ]}}{[\text{H}^{+}]} } }$