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

According to steady state approximation concentration of intermediate remains constant most of the time so rate of reaction with respect to intermediate is taken to by zero.
`(i) X_(2) to X+X` (fast)
(ii) `X+Y_(2) hArrXY+Y` (slow)
(iii) `X+Y to XY` (fast)
This is fact that `X+Y_(2)hArrXY+Y` cannot be slow. So mechanism of this reaction may be
`(i) X_(2)hArrX+X` (fast)
(ii) `X+Y_(2) to XY+Y` (slow)
(iii) `X+Y to XY` (fast)
So step (ii) is rate determining step
`r=k[X][Y_(2)]` .............(i)
X can be replaced by fast equilibrium given in step (i)
`K=[(X^(2)])/([X_(2)])`
`[X]^(2)=K[X_(2)]`
`[X]=[K[X_(2)]]^((1)/(2))` ...........(ii)
If value of equation (ii) is placed in equation (i) then
`r=k[Y_(2)][K[X_(2)]]^((1)/(2))`
so order of reaction is `1.5`.