The rate of a certain reaction is given by , rate `=K [H^(+) ]^(n)` . The rate increases 100 times when the pH changes from 3 to 1 . The order (n) of the reaction is ______.

Options

A. 3
B. 0
C. 1
D. 1.5

Answer

C

Explanation

Given,rate `k[H^(+)]^n
`initial pH=
`3so[H^(+)]=1xx10^(-3)`,initial rate=
`r_(1)`Final pH=
`1so[H^](+)=1xx10^(-1)`final rate
`=r_(2)=100r_(1)`
On substituting values, we get
`r_(1)=k[1xx10^(-3)]^(n)
and" " r_(2)=100r_(1)=k[10^(-1)]^(n)" "...(ii)`On dividing Eq. (i),we get
`(r_(1))/(100r_(1))=[(1xx10^(-3))/(1 xx10^(-1))]^(n)`
`1/100=[(10^(-2))/1]^(n)implies[(1)/(100)]^(1)=[(1)/(100)]^(n)`
`therefore" "n=1`Thus, reaction is of first order.