Question

In acidic medium the rate of reaction between $(Br{O}_3{)}^{-}andB{r}^{-}$ ion is given by the expression
$\frac{d\left[Br{O}_3^{-}\right]}{dt}=k\left[Br{O}_3^{-}\right]\left[B{r}^{-}\right]{\left[H^{+}\right]}^2$ it means:

• A. rate constant of overall reaction is $4se{c}^{-1}$
• B. rate of reaction is independent of the conc. of acid
• C. the change in pH of the solution will not attect the rate
• D. doubling the conc. of $H^{+}$ ions will increase the reaction rate by $4$ times

Answer 1

The correct option is D doubling the conc. of $H^{+}$ ions will increase the reaction rate by $4$ times

$\frac{d\left[Br{O}_3^{-}\right]}{dt}=k\left[Br{O}_3^{-}\right]\left[B{r}^{-}\right]{\left[H^{+}\right]}^2$
as reaction is second order w.r.t. hydrogen ion so
the change in pH of the solution will affect the rate and doubling the conc. of $H^{+}$ ions will increase the reaction rate by 4 times.