The rate law for the reaction,
`2Cl_(2)Orarr2Cl_(2)+O_(2)`
at `200^(@)C` is found to be : rate = `k[Cl_(2)O]^(2)`
(a) How would the rate change if `[Cl_(2)O]` is reduced to one- third of its original value
(b) How should the `[Cl_(2)O]` be chabged in order to double the rate
(c) How would the rate change is `[Cl_(2)O]` is raised to threefold of its original value ?

Explanation

(a) Rate equation for the reaction,
`r=[k[Cl_(2)O]^(2)`
Let the new rate be `r^('), so,r^(')=k[(Cl_(2)O)/(3)]=(1)/(9)r`
(b) In order to have the =2r, let the concentration of `Cl_(2)O` be x.
So, `2r=kx^(2)" "....(i)`
We know, that, `r=k[Cl_(2)O]^(2)....(ii)`
Dividing eq. (i) by eq. (ii),
`" "(2r)/(r)=(kx^(2))/(k[Cl_(2)O]^(2))`
`or " " 2=(x^(2))/([Cl_(2)O]^(2))`
`or " " x^(2)=2[Cl_(2)O]^(2)`
`or " " x=sqrt(2)[Cl_(2)O]`
(c) New rate `=k[3Cl_(2)O]^(2)=9k[Cl_(2)O]^(2)=9r`