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 threefold 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 mine 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_(2)O]^(2)) (or) x = sqrt(2)[Cl_(2)O]`
The concentration of `Cl_(2)O` should be increased by 1.414 times.