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

`\large \boxed{{\text{rate} = k\frac{[\text{ O$_(2)$NNH$_(2)$]}}{[\text{H}^(+)]}} }`

Step-by-step 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

`-\frac{\text{d[O$_(2)$NNH$_(2)$]} }{\text{d}t} = k_(1)[\text{ O$_(2)$NNH$_(2)$]} - k_(-1)[\text{O$_(2)$NNH$^(-)$]}[\text{H}^(+)]\\\\-\frac{\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$^(-)$]} = (k_(1))/(k_(-1))* \frac{[\text{ O$_(2)$NNH$_(2)$]}}{[\text{H}^(+)]}`

C. The rate law

Substitute into the rate law for the slow step.

`-\frac{\text{d[O$_(2)$NNH$^(-)$]}}{\text{d}t} = (k_(1)k_(2))/(k_(-1))* \frac{[\text{ O$_(2)$NNH$_(2)$]}}{[\text{H}^(+)]}\\\\\text{rate} = k \frac{[\text{ O$_(2)$NNH$_(2)$]}}{[\text{H}^(+)]}\\\\\text{The rate law for the reaction is $\large \boxed{\text{rate} = k\frac{[\text{ O$_(2)$NNH$_(2)$]}}{[\text{H}^(+)]} }$}`