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Answer

this reaction followed by two steps with [O] intermediate.

O3 --->O2 + O  (reversible step ) .....(i)

O3 +O --> 2O2 ( slow step).......(ii)

slow step is rate determining step, therefore

rate = k [O3][O].......(iii)

from eq (i)

Keq = [O2][O]/[O3]

[O] = Keq[O3]/[O2] ....(iv)

put [O] from (iv) in (iii)

rate = k' [O3]^2 [O2]^-1

Answer

To be remembered, this is a multi step reaction, so slow step is rate determining step & no intermediates in rate reaction. 

Step 1 (fast, reversible)  O3 ------> O2 + (O)

Step 2 (slow)  O3 + (O) --------> O2 + O2

So, r = k [O3][(O)]

Here, nacent oxygen (O), is an intermediate so, from step -1 k1 is a equilibrium constant. So, k1 = [O2]*[(O)] / [O3]

So, [(O)] = k1 * [O3]/[O2]

So, r = k3*k1 * [O3]^2/ [O2] = k' [O3]^2 / [O2] = k' [O3]*[O2]^-1

Answer

all four options given are wrong. 

right ans will be

-1/3[dO2]/dt=K[O3]^2[O2]^1

 

Answer
RHS of answers are not correct so can't solve

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