Question

Is first order in bro3⎻ , second order in br⎻, and zero order in h+. By what factor will the reaction rate change if the concentration of bro3⎻ is doubled, the concentration of br⎻ is halved, and the concentration of h+ is tripled ?

Answer 1

For a general reaction,

$A+B\rightarrow C$

General expression for rate law  will be:

$r=k[A]^{a}[B]^{b}$

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 $BrO_{3}^{-}$, second order with respect to $Br^{-}$ and zero order with respect to $H^{+}$.

According to above information, expression for rate law will be:

$r=k[BrO_{3}^{-}]^{1}[Br^{-}]^{2}[H^{+}]^{0}$

Or,

$r=k[BrO_{3}^{-}][Br^{-}]^{2}$ ...... (1)

• When concentration of $BrO_{3}^{-}$ get doubled, rate of the reaction becomes,

$r^{'}=2k[BrO_{3}^{-}][Br^{-}]^{2}$ ...... (2)

Dividing (2) by (1)

$\frac{r^{'}}{r}=\frac{2k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=2$

Or,

$r^{'}=2r$

Thus, rate of the reaction also get doubled.

• When the concentration of $Br^{-}$ is halved, the rate of reaction becomes

$r^{"}=k[BrO_{3}^{-}]([Br^{-}]/2)^{2}$

Or,

$r^{"}=1/4k[BrO_{3}^{-}][Br^{-}]^{2}$ ...... (3)

Dividing (3) by (1)

$\frac{r^{"}}{r}=\frac{1/4k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=\frac{1}{4}$

Or,

$r^{"}=\frac{r}{4}$

Thus, rate of reaction becomes 1/4th of the initial rate.

• When the concentration of $H^{+}$ is tripled:

Since, the rate expression does not have concentration of $H^{+}$, 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).