The reaction $2 \mathrm{NO}+\mathrm{Br}_{2} \rightarrow 2 \mathrm{NOBr}$

takes places through the mechanism given below:

$\mathrm{NO}+\mathrm{Br}_{2} \Leftrightarrow \mathrm{NOBr}_{2}$ (fast)

$\mathrm{NOBr}_{2}+\mathrm{NO} \rightarrow 2 \mathrm{NOBr}$ (slow)

The overall order of the reaction is ___________.

<p>The overall order of a reaction is determined by the slow (rate-determining) step. </p> <p>Here, the slow step is: $ \text{NOBr}_2 + \text{NO} \rightarrow 2 \text{NOBr} $</p> <p>This is a second-order reaction: first-order with respect to $\text{NOBr}_2$ and first-order with respect to NO.</p> <p>However, $\text{NOBr}_2$ is not a reactant in the overall reaction. It&#39;s an intermediate. So, we need to express it in terms of the initial reactants.</p> <p>From the first (fast) step, we get: $ \text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2 $</p> <p>In the steady state approximation, the rate of formation of $\text{NOBr}_2$ equals the rate of its consumption. We can write this as:</p> <p>$ k_1[\text{NO}][\text{Br}_2] = k_{-1}[\text{NOBr}_2] + k_2[\text{NOBr}_2][\text{NO}] $</p> <p>Since the second reaction (slow step) is much slower than the first one, $k_2[\text{NOBr}_2][\text{NO}]$ term is negligible in comparison to $k_{-1}[\text{NOBr}_2]$. </p> <p>So, we can simplify to:<br/><br/> $ k_1[\text{NO}][\text{Br}_2] \approx k_{-1}[\text{NOBr}_2] $</p> <p>Solving for $[\text{NOBr}_2]$, we get: $ [\text{NOBr}_2] \approx \frac{k_1}{k_{-1}} [\text{NO}][\text{Br}_2] $</p> <p>Substitute $[\text{NOBr}_2]$ into the rate equation for the slow step:<br/><br/> $ \text{Rate} = k_2[\text{NO}][\text{NOBr}_2] $<br/><br/> $ \text{Rate} = k_2[\text{NO}] \left( \frac{k_1}{k_{-1}} [\text{NO}][\text{Br}_2] \right) $</p> <p>So the overall reaction order is 3 (2 with respect to NO and 1 with respect to $\text{Br}_2$).</p>