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

Options

A. zero
B. 1
C. `1(1)/(2)`
D. 2

Answer

C

Explanation

As rate depends upon the slow step,
Rate `= k[A][B_(2)]`
for the equilibrium, `iff 2A , K_(c) = ([A]^(2))/([A_(2)])`
or `[A]=sqrt(K_(c)[A_(2)])`
` therefore` Rate `=k K_(c)^(1//2) [A_(2)]^(1//2)[B_(2)] = k'[A_(2)]^(1//2)[B_(2)]`
Hence, order `= (1)/(2) + 1 = 1""(1)/(2).`