The rate law expresison is given for a typical reaction, `n_(1)A + n_(2) B rarrP` as `r = k[A]^(n)[B]^(n2)`. The reaction completes only in one step and `A` and `B` are present in the solution. If the reaction occurs in more than one step, then the rate law is expressed by consdering the slowest step, i.e., for `S_(N)l` reaction `r = k[RX]`. If the eraction occurs in more than one step and the rates of the steps involved are comparable, then steady state approximation is conisdered, i.e., the rate of formation of intermediate is always equal to the rate of decompoistion of the intermediate. Conisder the reaction:
`[[I_(2) underset(k_(1))overset(k_(2))hArr2I("rapid equilibrium")],[H_(2)+2I overset(k_(3))rarr 2HI("slow")]]`
Which of the following expresison is correct?

Options

A. `(d[HI])/(dt) = (k_(3)k_(1))/(k_(2))[H_(2)][I_(2)]`
B. `(d[HI])/(dt) = (k_(1)k_(3))/(k_(2))2[H_(2)][I_(2)]`
C. `(d[HI])/(dt) = k_(2)[H_(2)][I_(2)]^(2)`
D. None of these

Answer

A

Explanation

`I_(2) underset(k_(2))overset(k_(1))hArr 2I ("rapid")`
`H_(2) +2I overset(k_(3))hArr 2HI ("slow")`
`r = k_(3)[H_(2)][I]^(2)`
`= k_(3)[H_(2)](k_(1))/(k_(2))[I_(2)]`, `[(k_(1))/(k_(2)) = ([I]^(2))/([I_(2)])]`
`= k_(3)(k_(1))/(k_(2))[H_(2)][I_(2)]`