The reaction $\mathrm{A}+2 \mathrm{~B}+\mathrm{C} ightarrow \mathrm{D}$ occurs by the following mechanism…

The reaction $\mathrm{A}+2 \mathrm{~B}+\mathrm{C} ightarrow \mathrm{D}$ occurs by the following mechanism


The rate law for this reaction is
  1. $r=k[\mathrm{C}]$
  2. $r=k[\mathrm{~A}][\mathrm{B}]^{2}[\mathrm{C}]$
  3. $r=k[\mathrm{D}]$
  4. $r=k[\mathrm{~A}][\mathrm{B}][\mathrm{C}]$

Solution

From the slow step, we write rate $=k_{3}[\mathrm{E}][\mathrm{C}]$
From the first step, we get $\quad K_{\mathrm{eq}}=\frac{k_{1}}{k_{2}}=\frac{[\mathrm{E}]}{[\mathrm{A}][\mathrm{B}]} \Rightarrow[\mathrm{E}]=K_{\mathrm{eq}}[\mathrm{A}][\mathrm{B}]$
Hence, $\quad$ rate $=k_{3} K_{\mathrm{eq}}[\mathrm{A}][\mathrm{B}][\mathrm{C}]=k[\mathrm{~A}][\mathrm{B}][\mathrm{C}]$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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