The reaction ; `2O_3 -> 3O_2`, is assigned the followig mechanism :
(I) `O_3` ⇌ `O_2+O`
(II) `O_3 +O overset{mbox{(slow)}}{rightarrow} 2O_2` The rate law of the reaction will therefore be

The reaction ; `2O_3 -> 3O_2`, is assigned the followig mechanism :
(I) `O_3` ⇌ `O_2+O`

(II) `O_3 +O overset{mbox{(slow)}}{rightarrow} 2O_2` The rate law of the reaction will therefore be

`r alpha [O_3]^2[O_2]`

`r alpha [O_3]^2[O_2]^-1`

`r alpha [O_3]`

`r alpha [O_3][O_2]^-2`

Solution

0 votes

Best answer

Step II, being r.d.s. Rate of overall reaction = Rate of Step II

= `K_(II)[O_3 ][O]` Putting the value of [O] from the equilibrium of step I, we get