# Using the steady state approximation in the decomposition of nitramide to determine rate law

$$\ce{O2NNH2 (aq) ->[\mathit{k}_1] O2NNH- (aq) + H+ (aq)}$$
$$\ce{O2NNH- (aq) + H+ (aq) ->[\mathit{k}_{-1}] O2NNH2 (aq)}$$
$$\ce{O2NNH- (aq) ->[\mathit{k}_2] N2O(g) + OH- (aq)}$$
$$\ce{H+ (aq) + OH- (aq) ->[\mathit{k}_3] H2O(l)}$$

We assumed the first two steps were fast and gave a rapid pre-equilibrium, while the third step that generated $$\ce{N2O}$$ was slow. The last step is very fast.

(i) For this question, use the steady state approximation for [$$\ce{O2NNH-}$$], assuming it is a reactive intermediate, to derive an expression for the rate law.

The solution given is $$\frac{\mathrm d [\ce{N2O}]}{\mathrm d t} = k_\mathrm{obs}\frac{\ce{[O2NNH2]}}{\ce{[H+]}}$$

My attempt at the solution is as follows:

$$\frac{\mathrm d[\ce{N2O}]}{\mathrm dt} = k_2\ce{[O2NNH-]}$$

\begin{align} \frac{\mathrm d[\ce{O2NNH-}]}{\mathrm dt} = k_1[\ce{O2NNH2}] - &k_{-1}[\ce{O2NNH-}][\ce{H+}] - k_2[\ce{O2NNH-}] = 0 \\ \ce{[O2NNH-]} &= \frac{k_1[\ce{O2NNH2}]}{(k_{-1}\ce{[H+]} + k_2)} \\ \frac{\mathrm d[\ce{N2O}]}{\mathrm dt} &= \frac{k_2k_1[\ce{O2NNH2}]}{(k_{-1}\ce{[H+]} + k_2)} \end{align}
• You must next assume $k_2$ is small wrt $k_{-1}[H^+]$ to get the answer given. Apr 13 at 7:37