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

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:

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

Using the steady state approximation,

\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}

I have done this so far, but this doesn’t seem right to me. I’ve been looking over at it for seemingly ages but I can’t see where I am going wrong or if I have misunderstood something about the steady state approximation. Any insight would be greatly appreciated.

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    $\begingroup$ math.meta.stackexchange.com/questions/5020/… This should help for future reference.. I'll fix the text up this time. Take care that you format the text so it is easier for people with text readers. $\endgroup$ Apr 13 at 7:08
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    $\begingroup$ You must next assume $k_2$ is small wrt $k_{-1}[H^+]$ to get the answer given. $\endgroup$
    – porphyrin
    Apr 13 at 7:37
  • $\begingroup$ How have you come up with your equation for the production of N2O, the one just under "Using the steady state approximation". The last term, −k2[O2NNH−], is telling me that the rate of production of N2O decreases if there is more O2NNH−, do you think this can be right? Also apply the steady state approximation to the fast equation - the whole point is that the quick ones reach steady state compared to the slow one. $\endgroup$
    – Ian Bush
    Apr 13 at 7:37
  • $\begingroup$ That should be for the rate of [O2NNH-], apologies someone edited this and I don’t know how to change it $\endgroup$ Apr 13 at 7:45
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    $\begingroup$ @MarinaCalder - ah, OK, that makes sense now. I've submitted an edit correcting, hopefully it will get accepted soon. In that case the earlier comment by porphyrin has the answer for you $\endgroup$
    – Ian Bush
    Apr 13 at 7:51

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