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)

`(r^('))/(r)=(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^(`

Or,

`r^(` ...... (3)

Dividing (3) by (1)

`(r^(`

Or,

`r^(`

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).