122. Write the overall reaction and the rate law from the following elementary reactions:

                $$\begin{gathered} \mathrm{NO}+{\mathrm{Br}}_2\leftrightharpoons {\mathrm{NOBr}}_2 \left(\mathrm{fast}\right) \\ {\mathrm{NOBr}}_2+\mathrm{NO} \to 2\mathrm{NOBr} \left(\mathrm{slow}\right) \end{gathered}$$

What is the order of the reaction?

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Elementary step NOBr₂ + NO → 2NOBr is slow, thus it is rate determining step. Its rate is given by
           $\mathrm{rate} = \mathrm{k}\left[{\mathrm{NOBr}}_2\right] \left[\mathrm{NO}\right]$         ...(i)

For elementary reversible step

          $\mathrm{NO}+\mathrm{Br} \rightleftharpoons {\mathrm{NOBr}}_2$

the equilibrium constant k is given by

                 $\mathrm{k} = \frac{\left[{\mathrm{NOBr}}_2\right]}{\left[\mathrm{NO}\right] \left[{\mathrm{Br}}_2\right]}$

or     $\left[\mathrm{NOBr}\right] = \mathrm{k}\left[\mathrm{NO}\right] \left[{\mathrm{Br}}_2\right]$            ...(ii)

Putting the expression for [NOBr₂] from equation (ii) in to equation (i) we shall get the rate law equation as

$$\begin{gathered} \mathrm{Rate} = \mathrm{k}\text{'}\mathrm{k}\left[\mathrm{NO}\right]\left[{\mathrm{Br}}_2\right]\left[\mathrm{NO}\right] \\ = \mathrm{K}{\left[\mathrm{NO}\right]}^2 \left[{\mathrm{Br}}_2\right] \end{gathered}$$

where k = k' k = rate constant of the overall reaction. From the rate equation it is formed that the order with respect to NO is 2 with respect to ${\mathrm{Br}}_2$ the order is 1 and overall order of reaction is

2 + 1 = 3.