The experiment rate law for the reaction `S_(2)O_(8)^(2-)(aq) + 2I^(ɵ)(aq) rarr 2SO_(4)^(2-) (aq) + I_(2)(aq)` is `k[S_(2)O_(8)^(2-)] [I^(ɵ)]`. How would the rate change if
(a) Concentration of `S_(2)O_(8)^(2-)` is halved.
(b) Concentration of `S_(2)O_(8)^(2-)` and `I^(ɵ)` are halved.

Explanation

(a) `r_(1) = k[S_(2)O_(8)^(2-)][I^(ɵ)]`
`r_(2) = k[(S_(2)O_(8)^(2-))/(2)][I^(ɵ)]`
`(r_(2))/(r_(1)) = (1)/(2) rArr r_(2) = (1)/(2)r_(1)`
(b) `r_(2) = k[(S_(2)O_(8)^(2-))/(2)][(I^(ɵ))/(2)]`
`(r_(2))/(r_(1)) = (1)/(4)`