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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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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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