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

`r_(b) = K[H^(+)]^(n)`
pH =2, `therefore [H^(+)] = 10^(-2)`
`r_(0) = K[10^(-2)]^(n)` at `pH=1, [H^(+)]=10^(-1)`
`r_(1) = K[10^(-1)]^(n), r_(1)/r_(0) = 100 = [10]^(n) , therefore n=2`