For a reaction `(d x)/(d t) = K[H^(+)]^(n)`. If `pH` of reaction medium changes from two to one rate becomes `100` times of value at `pH = 2`, The order of reaction is

Options

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

Answer

B

Explanation

pH = 2, `r_(1)=k xx (10^(-2))^(n)` {`:. [H^(+)]=10^(-pH)`}
pH = 1, `r_(2)=k xx (10^(-1))^(n)`
Given `r_(2) = 100r_(1)`
`implies ((10^(-1))/(10^(-2)))^(n)=100`
`10^(n)=100`
`:.` n = 2