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Mechanism of hypothetical reaction X(2...

Mechanism of hypothetical reaction `X_(2)+Y_(2)rarr2XY` is given below:
(i) `X_(2)rarrX+X` (fast)
(ii) `X+Y_(2)hArrXY +Y` (slow)
(iii) `X+YrarrXY ` (fast)
The overall order of the reaction will be -

A

2

B

0

C

`1.5`

D

1

Text Solution

Verified by Experts

The correct Answer is:
C

The question is solved by considering the step (i) to be in equilibrium (This is not given in the question ).
According to the law of mass action , `r=k[X][Y_(2)],`
where k = rate constant of step (ii).
From step (i) we get , `k_(eq)=([X]^(2))/([X_(2)])` , where
`k_(eq)` = equilibrium constant of step (i)
`therefore[X]^(2)=k_(eq)[X_(2)]" or, "[X]=sqrt(k_(eq))xx[X_(2)]^(1//2)`
The expression of r becomes
`r=ksqrt(k_(eq))[X_(2)]^(1//2)[Y_(2)] " or, "r=k'[X_(2)]^(1//2)[Y_(2)]`
`therefore` The overall order of the reaction = 1 + 0.5 = 1.5
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