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The chemical reaction 2O(3) overset(k(1)...

The chemical reaction `2O_(3) overset(k_(1))rarr 3O_(2)` proceeds as follows :
`O_(3) overset(k_(eq))hArr O_(2)+O` (fast)
`O + O_(3) overset(k)rarr 2O_(2)` (slow)

What should be the rate law expression ?

A

`r=K[O_(3)]^(2)`

B

`r=K[O_(3)]^(2)[O_(2)]^(-1)`

C

`r=K[O_(3)][O_(2)]`

D

Unpredictable

Text Solution

Verified by Experts

The correct Answer is:
b
For slow step, rate `=K[O][O_(3)] …(i)`
Also for equilibrium `K_(c )=([O][O_(2)])/([O_(3)]) …(ii)`
`:. [O]=(K_(c )[O_(3)])/([O_(2)]) …(iii)`
By eqs. (iii) and (i), the intermediate `[O]` is eliminated as rate, `(K.K_(c )[O_(3)][O_(3)])/([O_(2)])=K'[O_(3)]^(2)[O_(2)]^(-1)`
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