The reaction 2O_(3)(g)rarr3O_(2)(g) is believed to occur according to the mechanism given below. Step -I : O_(3)(g)underset(k_(-1))overset(k_(1))hArrO_(2)(g)+O(g) (fast) Step- II : O_(3)(g)+O(g)overset(k_(2))rarr2O_(2)(g) (slow) Write the rate law for the reaction . Determine the order of the reaction with respect toO_(3)(g) and O_(2)(g).

The reaction 2O_(3)(g)rarr3O_(2)(g) is believed to occur according to

1.	The reaction 2O_(3)(g)rarr3O_(2)(g) is believed to occur according to the mechanism given below. Step -I : O_(3)(g)underset(k_(-1))overset(k_(1))hArrO_(2)(g)+O(g) (fast) Step- II : O_(3)(g)+O(g)overset(k_(2))rarr2O_(2)(g) (slow) Write the rate law for the reaction . Determine the order of the reaction with respect toO_(3)(g) and O_(2)(g).
Answer» Solution :The slow step of reaction is its rate-determiningstep, becauserate of this step determinesthe overall rate of the reaction. So, the rate of the slow step will be the rate of the reaction. Thus, rate `=k_(2)[O_(3)][O]""...[1]` The rate equation contains concentration term of reaction INTERMEDIATE O, which does not appear in the overall balanced equation of the reaction. But the rate law of a reaction shouldnot appear in theoverall balanced equation. So, we REQUIRE to eliminate [O] from equation (1). This can be done by making use of the equilibrium CONDITION of step - 1.For this equilibrium, we can write equilibrium constant EXPRESSION as `k_(1) = ([O_(2)][O])/([O_(3)])` This gives`[O]=k_(1)[O_(3)]//[O_(2)]` Substituting the expression for [O] into equation (1)gives rate `k_(2)K_(1) = ([O_(3)]^(2))/([O_(2)])=k[O_(3)]^(2)[O_(2)]^(-1)[k=k_(2)K_(1)` = constant] This equation represents the rate law for the GIVEN reaction . It appears from the equation that the reaction has an order 2 with respect to`O_(3)and -1` with respect to`O_(2)`.