At `193^(@)C` , the rate law for the reaction `2Cl_(2)O to 2Cl_(2) + O_(2)` is rate `= k[Cl_(2)O]^(2)` .
(a) How the rate changes if `[Cl_(2)O]` is raised to therefold of the original ?
(b) How should `[Cl_(2)O]` be changed in order to order to double the rate ?

Explanation

Original rate is given as, rate `=k[Cl_(2)O]^(2)`
(a) Rate with threefold concentration of `Cl_(2)O` is k `[3Cl_(2)O]^(2)`. This is nine times to the original rate.
(b) With `[Cl_(2)O]` the rate is `r=k[Cl_(2)O)^(2)`. Requirement is that .r. should be doubled at concentration of, `Cl_(2)O` as `x`.
`(2r)/(r)=(kx^(2))/(k[Cl_(1)O]^(2))("or")x=sqrt(2)[Cl_(2)O]`
The concentration of `Cl_(2)O` should be increased by `1.414` times.