Mechanism of a hypothetical reaction
`X_(2) + Y_(2) rarr 2XY` is given below:
(i) `X_(2) rarr X + X` (fast)
(ii) `X+Y_(2) hArr XY+Y` (slow)
(iii) `X + Y rarr XY` (fast)
The overall order of the reaction will be :

Options

A. 2
B. 0
C. 1.5
D. 1

Answer

C

Explanation

According to the rete controlling step (slow step), we have
Rate `= K [X] [Y_(2)]`
Since `X` is not the reactant in the main reaction, it must be excluded from the rate law. For this propose, we use the fast step (i):
`K_(eq) ([X] [X])/([X_(2)]) = ([X]^(2))/([X_(2)])`
`:. [X]^(2) = K_(eq) [X_(2)]`
`[X] = sqrt(K_(eq)) [X_(2)]^(1//2)`
Substituting the value of `[X]` from (ii), in (i), we get
Rate `= K sqrt(K_(eq)) [X_(2)]^(1//2) [Y_(2)]`
`K' [X_(2)]^(1//2) [Y_(2)]`
Hence the overall order of the reaction will be `1//2 + 1 3//2 = 1.5`