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The reaction of tert-butyl bromide with azide ion in aqueous solution is proposed to proceed by the following mechanism: $$ \begin{align} \ce{(CH3)3CBr(aq) &<=>[k$_1$][$k_{-1}$] (CH3)3C+(aq) + Br–(aq)}\\ \ce{(CH3)3C+(aq) + N3–(aq) &->[$k_2$] (CH3)3CN3(aq)} \end{align} $$ $k_i$ is the rate constant.

Assuming that $\ce{(CH3)3C+(aq)}$ achieves a steady-state concentration, but making no further assumptions about the relative magnitudes of the three rate constants, what is the rate law for this reaction?

The answer is

$$\mathrm{Rate} = \frac{k_1k_2[\ce{(CH3)3CBr}][\ce{N3–}]}{k_{–1}[\ce{Br–}] + k_2[\ce{N3–}]}$$

But how would I obtain that answer using multistep reaction mechanisms and not differential equations?

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closed as off-topic by airhuff, Todd Minehardt, M.A.R., Jon Custer, Mithoron Dec 26 '17 at 14:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

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    $\begingroup$ Welcome to chemistry.SE! If you have any questions about the policies of our community, please ‎visit the help center. Also, this is a homework question. We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. $\endgroup$ – airhuff Dec 26 '17 at 4:13
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Since $(CH_3)_3C^+(aq)$ achieves a steady-state concentration, the equations when directly added give the overall reaction whose rate constant is the product of the rate constants of individual reactions. Hence the rate law comes out to be as it is written.

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