For the chemical reaction- 2O3- 3O2 proceed as follows O3- O2 - O-fast O - O2- 2O2- slowThe rate law expression will be-

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Introduction to Oxidation and Reduction

For the chemi...

Question

For the chemical reaction, $2 O_{3} \to 3 O_{2}$ proceed as follows
$O_{3} ⇌ O_{2} + O . . . . . (f a s t)$
$O + O_{2} \to 2 O_{2} . . . . . (s l o w)$
The rate law expression will be:

r = k^{'} [O_{3}]^{2}

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r = k^{'} [O_{3}]^{2} [O_{2}]^{- 1}

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r = k^{'} [O_{3}] [O_{2}]

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Solution

The correct option is A $r = k^{'} [O_{3}]^{2} [O_{2}]^{- 1}$
As slowest step is the rate determining step.
Hence, from slow reaction, $r = K [O] [O_{3}] . . . . . (i)$
From fast reaction, $K_{C} = \frac{[O_{2}] [O]}{[O_{3}]}$
$[O] = \frac{K_{C} . [O_{3}]}{[O_{2}]}$ , i.e.
$r = \frac{K_{C} [O_{3}]}{[O_{2}]} . [O_{3}]$
$\Rightarrow K_{C} . [O_{3}]^{2} [O_{2}]^{- 1}$
$∴ R a t e (r) = K^{'} [O_{3}]^{2} \cdot [O_{2}]^{- 1} (K^{'} = K_{C} \cdot K)$ .

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2 O_{3} \to 3 O_{2}

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The rate law expression should be:

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For the chemical reaction, 2O3→3O2 proceed as followsO3⇌O2+O.....(fast)O+O2→2O2.....(slow)The rate law expression will be:

The correct option is A r=k′[O3]2[O2]−1As slowest step is the rate determining step.Hence, from slow reaction, r=K[O][O3].....(i)From fast reaction, KC=[O2][O][O3][O]=KC.[O3][O2], i.e.r=KC[O3][O2].[O3]⇒KC.[O3]2[O2]−1∴Rate(r)=K′[O3]2⋅[O2]−1(K′=KC⋅K).

For the chemical reaction, $2 O_{3} \to 3 O_{2}$ proceed as follows
$O_{3} ⇌ O_{2} + O . . . . . (f a s t)$
$O + O_{2} \to 2 O_{2} . . . . . (s l o w)$
The rate law expression will be: