1.

The chemical reaction, 2O_(3) rarr 3O_(2) proceeds as follows Step1 : O_(3) hArr O_(2)+O""…(fast) Step 2: O + O_(3) rarr 2O_(2)""…(slow) The rate law expression should be

Answer»

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

Solution :Acc.to slowest STEP
`r = k[O][O_(3)]""`….(i)
as `'O'` INTERMEDIATE its CONC. is calculated as
`K = ([O_(2)][O])/([O_(3)]), ""[O] = (K[O_(3)])/([O_(2)])`
Subtitute in(i)
`r = (kK[O_(3)][O_(3)])/([O_(2)]) rArr r = K'[O_(3)]^(2)[O_(2)]^(-1)`


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