Mechanism of a reaction is given below:
`X_(2) rarr 2X, k_(1) = 10^(5) s^(-1)` (forward)
`k_(2) = 10^(5)s^(-1)` (backward)
`X + Z rarr XZ, k_(3) = 10^(-4) M^(-1) s^(-1)`
Write the rate law and calculate the order of overall reaction.

Explanation

`X_(2) hArr 2X, k_(1) = 10^(5)=k_(2)`
(This is equilibrium step)
`X + Z rarr XZ, k_(3) = 10^(-4)` (Value of `k_(3)` is very less as compared to `k_(1)` and `k_(2)`).
So it slow step and is the rate-determining step
`:.` Rate `= k_(3)[X][Z]` ...(i)
`k_(eq) = (k_(2))/(k_(1)) = (10^(5))/(10^(5)) = 1`
`:. 1 = ([X]^(2))/([X_(2)])`, `[X] = [X_(2)]^(1//2)`
substittute the value of `[X]` in Eq. (i)
Rate `= k_(3)[X_(2)]^(1//2) [Z]`
`:.` Order of reaction `= (1)/(2) + 1 = (3)/(2)`