The reaction 2 NO + Br 2 2 NOBr takes place through the mechanism given below: NO + Br 2 NOBr 2 ( fast ) NOBr 2 + NO 2 NOBr ( slow ) The overall order of the reaction is ________ . [2023]

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Solution

Q.
The reaction $2 NO + {Br}_{2} \to 2 NOBr$

takes place through the mechanism given below:

$NO + {Br}_{2} ⇌ {NOBr}_{2} (fast)$

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

The overall order of the reaction is ________ . [2023]

Ans.
(3)

$2 NO + {Br}_{2} ⟶ 2 NOBr$

(i) $NO + {Br}_{2} ⇌_{k_{2}}^{k_{1}} {NOBr}_{2} Fast$

(ii) ${NOBr}_{2} + NO \to_{k_{3}}^{slow} 2 NOBr r = k_{3} [{NOBr}_{2}] [NO]$ ...(i)

$K_{eq} = \frac{k_{1}}{k_{2}} = \frac{[{NOBr}_{2}]}{[NO] [{Br}_{2}]} \Rightarrow [{NOBr}_{2}] = \frac{k_{1} \times [NO] [{Br}_{2}]}{k_{2}}$

Now put $[{NOBr}_{2}]$ in equation (i) we get,

$r = \frac{k_{1} k_{3}}{k_{2}} {[NO]}^{2} [{Br}_{2}]$ $r = k {[NO]}^{2} [{Br}_{2}]$

So, overall order of reaction $= 2 + 1 = 3 .$

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