Question

In acidic medium, the rate of reaction between \[[BrO_3^ - ]\]and $[B{r^ - }]$ ions is given by the expression $ - \dfrac{{d[BrO_3^ - ]}}{{dt}} = k[BrO_3^ - ][B{r^ - }]{[{H^ + }]^2}$
It means:
 (i) Rate constant of the reaction depends upon the concentration of ${H^ + }$ions
 (ii) Rate of reaction is independent of the concentration of acid added
 (iii) The change in $pH$ of the solution will affect the rate of reaction
 (iv) Doubling the concentration of ${H^ + }$ ions will increase the reaction rate by $4$ times.
A.Only (ii)
B.Only (iii)
C.Only (i) and (ii)
D.Only (iii) and (iv)

Answer 1

Hint: The rate constant does not depend upon the initial concentration of the reactants. The rate of a reaction depends upon the concentration of the reactants. $pH$ is the measurement of hydrogen ion concentration so, if [${H^ + }$] changes pH will also change.

Complete step by step answer:
The rate constant of a reaction is equal to the rate of reaction when the concentration of each of the reactants is unity. The rate constant of a reaction has a definite value at a particular temperature. Its value increases with increase in temperature. It does not depend upon the initial concentrations of reactants. Therefore the rate constant of the reaction between \[[BrO_3^ - ]\] and $[B{r^ - }]$ does not depend upon the concentration of ${H^ + }$ ions.
The rate of the reaction rate is the rate of change of concentration of a reactant or a product with respect to time. The reaction rate at a particular instant depends upon the concentration of reactants at that instant.
The rate equation of the reaction of reaction between \[[BrO_3^ - ]\] and $[B{r^ - }]$ is given as:
Rate $ = $$ - \dfrac{{d[BrO_3^ - ]}}{{dt}} = k[BrO_3^ - ][B{r^ - }]{[{H^ + }]^2}$
So, the rate of the reaction is dependent on the concentration of acid [${H^ + }$] added.
We know that, $pH = - \log [{H^ + }]$, it means that $pH$ depends on the concentration of ${H^ + }$ ions and the rate of the reaction depends on the concentration of ${H^ + }$ions. Therefore, the change in $pH$ of the solution will affect the rate of reaction.
Rate$ = $$ - \dfrac{{d[BrO_3^ - ]}}{{dt}} = k[BrO_3^ - ][B{r^ - }]{[{H^ + }]^2}$
On doubling the concentration of ${H^ + }$ ions, we get
Rate$ = $$ - \dfrac{{d[BrO_3^ - ]}}{{dt}} = k[BrO_3^ - ][B{r^ - }]{[2{H^ + }]^2} = 4k[BrO_3^ - ][B{r^ - }]{[{H^ + }]^2}$
Therefore, doubling the concentration of ${H^ + }$ ions will increase the reaction’s rate by $4$ times.

Hence, option (D) is the correct answer.

Note:
Rate constant also known as specific reaction rate is a measure of the rate of a reaction, higher the value of rate constant greater is the rate of the reaction. It is a constant of proportionality and is equal to the rate of reaction when the molar concentration of each of the reactants is unity. A particular reaction has a definite value of rate constant at a particular temperature. The value of rate constant for a particular reaction increases with an increase in temperature. Its units depend upon the overall order of the reaction.