Question

A hypothetical reaction `A_(2) + B_(2) rarr 2AB` follows the mechanism as given below:
`A_(2) hArr A+A ("fast")`
`A+B_(2) rarr AB+B` (slow)
`A+B rarr AB` (fast)
The order of the overall reaction is
A. `2`
B. `1`
C. `1.5`
D. `0`

Answer 1

Correct Answer - C
Since `(A)` is the intermediate reactive species whose concentration is determined form equilibrium step.
Slow step is:
`A + B_(2) rarr AB + B (slow)`
`r = k[A][B_(2)]` …(i)
form equilibrium step
`A_(2) hArr A+A ("fast")`
`k_(eq) = ([A]^(2))/([A_(2)])`
`:. [A] = (k_(eq)[A_(2)])^(1//2)`
Substitute the value of `[A]` in Eq. (i),
`r = k.k_(eq)^(1//2) [A_(2)]^(1//2)[B_(2)]`
Thus, order of reaction `= (1)/(2)+1=1(1)/(2)`

Answer 2

Correct Answer - C
Since `(A)` is the intermediate reactive species whose concentration is determined form equilibrium step.
Slow step is:
`A + B_(2) rarr AB + B (slow)`
`r = k[A][B_(2)]` …(i)
form equilibrium step
`A_(2) hArr A+A ("fast")`
`k_(eq) = ([A]^(2))/([A_(2)])`
`:. [A] = (k_(eq)[A_(2)])^(1//2)`
Substitute the value of `[A]` in Eq. (i),
`r = k.k_(eq)^(1//2) [A_(2)]^(1//2)[B_(2)]`
Thus, order of reaction `= (1)/(2)+1=1(1)/(2)`