Question

Question #7d5f5

Answer 1

Here's how you can prove this.

Explanation:

You need to prove that the half-life of a `n^"th"` order reaction, `t_"1/2"`, is proportional to the inverse of the initial concentration of the reactant, `A_0`, raised to the power `n-1`.

`color(blue)(t_"1/2" prop 1/A_0^(n-1))`

So, for a general `n^"th"` order reaction

`A -> "products"`

that is `n^"th"` order in `A`, you can say that the rate of reaction will depend on the concentration of `A` raised to the `n^"th"` power

`color(blue)("rate" = -(d[A])/dt = k * [A]^n)`

Rearrange this equation to get

`-(d[A])/([A]^n) = k * dt`

Integrate both sides of the equation to get

`-int 1/([A]^n) * d[A] = k * int dt`

`-int [A]^(-n) * d[A] = k * t + c`

`-([A]^((-n + 1)))/((-n+1)) = k * t + c`

You can get rid of that minus sign by writing

`-n+1 = -(n-1)`

This will get you

`1/(n-1) * [A]^(-(n-1)) = k * t + c`

`1/(n-1) * 1/[A]^(n-1) = k * t + c " " " "color(purple)((1))`

To get rid of that constant `c`, use the fact that at `t=0`, the concentration of the reactant is equal to its initial value, `A_0`

`color(blue)("For t = 0" implies [A] = [A_0])`

At `t=0`, the equation `color(purple)((1))` becomes

`1/(n-1) * 1/[A_0]^(n-1) = k * 0 + c`

Isolate the constant to get

`c = 1/(n-1) * 1/[A_0]^(n-1)`

Plug this into equation `color(purple)((1))` to get

`1/(n-1) * 1/[A]^(n-1) = k * t + overbrace(1/(n-1) * 1/[A_0]^(n-1))^(color(red)(=c))`

Rearrange this equation to get

`1/(n-1) * (1/[A]^(n-1) - 1/[A_0]^(n-1)) = k * t" " " "color(purple)((2))`

Now, the half-life of a chemical reaction is the time needed for the concentration of the reactant to be reduced to half of its initial value

`color(blue)(t = t_"1/2" implies [A] = 1/2 * [A_0])`

Plug this into equation `color(purple)((2))` to get - keep in mind that

`(1/(A_0/2))^n = (2/A_0)^n = 2^n/A_0^n`

`1/(n-1) * (2^(n-1)/[A_0]^(n-1) - 1/[A_o]^(n-1)) = k * t_"1/2"`

Rearrange to get

`1/(n-1) * (2^(n-1) - 1)/[A_0]^(n-1) = k * t_"1/2"`

This is equivalent to

`t_"1/2" = overbrace(1/k * (2^(n-1) - 1)/(n-1))^(color(red)("constant")) * 1/[A_0]^(n-1)`

Therefore,

`color(blue)(t_"1/2" prop 1/[A_0]^(n-1)) color(white)(x)color(green)(sqrt())`