In acidic medium, the rate of reaction between `BrO_(3)^(ɵ)` and `Br^(ɵ)` is given by the expression
`(-d[BrO_(3)^(ɵ)])/(dt) =k[BrO_(3)^(ɵ)][Br^(ɵ)][H^(o+)]^(2)`

Options

A. Rate constant of overall reaction is `4 sec^(-1)`
B. Rate of reaction is independent of the concentration of acid
C. The changes in `pH` of the solution will not affect the rate
D. Doubling the concentration of `H^(+)` ions will increase the reaction rate by `4` times.

Answer

D

Explanation

`-(d[BrO_(3)^(-)])/(d t) = k[BrO_(3)^(-)][Br^(-)][H^(+)]^(2)`
So, on doubling the concentration of `H^(+)` ion will increase the reaction rate by `4` times.