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

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

What is the order of the reaction?

Solution

For the given reaction, the overall reaction will be:

NO + {Br}_{2} ⇋ {NOBr}_{2} (fast) {NOBr}_{2} + NO \to 2 NOBr (slow)

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2 NO + {Br}_{2} \to 2 NOBr (Overall reaction)

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Elementary step NOBr₂ + NO → 2NOBr is slow, thus it is rate determining step. Its rate is given by

rate = k [{NOBr}_{2}] [NO]

...(i)

For elementary reversible step

NO + Br ⇌ {NOBr}_{2}

the equilibrium constant k is given by

k = \frac{[{NOBr}_{2}]}{[NO] [{Br}_{2}]}

[NOBr] = k [NO] [{Br}_{2}]

...(ii)

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

Rate = k' k [NO] [{Br}_{2}] [NO] = K {[NO]}^{2} [{Br}_{2}]

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

{Br}_{2}

the order is 1 and overall order of reaction is

2 + 1 = 3.

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