# Question #7d5f5

##### 1 Answer

#### Answer:

Here's how you can prove this.

#### Explanation:

You need to prove that the half-life of a *proportional* to the inverse of the initial concentration of the reactant,

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

So, for a general

#A -> "products"#

that is

#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

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

At

#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

#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

#(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())#