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. 1
B. 2
C. 0
D. 1.5

Answer

D

Explanation

The solution of this question is given by assuming step (i) to be reversible which is not given in question
Overall rate = Rate of slowest step (ii)
`=k[X][Y_(2)]`......(1)
k=rate constant of step (ii)
Assuming step (i) to be reversible, its equilibrium constant,
`k_(eq)=([X]^(2))/([X_(2)]) implies [X]=k_(eq)^(1/2)[X_(2)]^((1)/(2))`.....(2)
Put (2) in (1)
Rate `=kk_(eq)^((1)/(2))[X_(2)]^((1)/(2))[Y_(2)]`
Overall order `=(1)/(2)+1=(3)/(2)`