Question #7d5f5
1 Answer
Here's how you can prove this.
Explanation:
You need to prove that the half-life of a
#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())#