Respuesta :
For a general reaction,
[tex]A+B\rightarrow C[/tex]
General expression for rate law will be:
[tex]r=k[A]^{a}[B]^{b}[/tex]
Here, r is rate of the reaction, k is rate constant, a is order with respect to reactant A and b is order with respect to reactant B.
The reaction is first order with respect to [tex]BrO_{3}^{-}[/tex], second order with respect to [tex]Br^{-}[/tex] and zero order with respect to [tex]H^{+}[/tex].
According to above information, expression for rate law will be:
[tex]r=k[BrO_{3}^{-}]^{1}[Br^{-}]^{2}[H^{+}]^{0}[/tex]
Or,
[tex]r=k[BrO_{3}^{-}][Br^{-}]^{2}[/tex] ...... (1)
- When concentration of [tex]BrO_{3}^{-}[/tex] get doubled, rate of the reaction becomes,
[tex]r^{'}=2k[BrO_{3}^{-}][Br^{-}]^{2}[/tex] ...... (2)
Dividing (2) by (1)
[tex]\frac{r^{'}}{r}=\frac{2k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=2[/tex]
Or,
[tex]r^{'}=2r[/tex]
Thus, rate of the reaction also get doubled.
- When the concentration of [tex]Br^{-}[/tex] is halved, the rate of reaction becomes
[tex]r^{"}=k[BrO_{3}^{-}]([Br^{-}]/2)^{2}[/tex]
Or,
[tex]r^{"}=1/4k[BrO_{3}^{-}][Br^{-}]^{2}[/tex] ...... (3)
Dividing (3) by (1)
[tex]\frac{r^{"}}{r}=\frac{1/4k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=\frac{1}{4}[/tex]
Or,
[tex]r^{"}=\frac{r}{4}[/tex]
Thus, rate of reaction becomes 1/4th of the initial rate.
- When the concentration of [tex]H^{+}[/tex] is tripled:
Since, the rate expression does not have concentration of [tex]H^{+}[/tex], it is independent of it. Thus, any change in the concentration will not affect the rate of reaction and rate of reaction remains the same as in equation (1).
