For the reaction `2H_(2)+O_(2)rarr 2H_(2)O`, the rate law expression is , `r = k[H_(2)]^(n)`. When the concentration of `H_(2)` is doubled, the rate of reaction found to be quadrupled. The value of n is

Options

A. 0
B. 1
C. 2
D. 3

Answer

C

Explanation

`r=k[H_(2)]^(n)" "....(i)`
On doubling the concentration of `H_(2)`,
`4r=k[2H_(2)]^(n)" "....(ii)`
On dividing Eq. (ii) by Eq. (i), we get,
`4=(2)^(n)`
`(2)^(2)=(2)^(n)`
`n=2`