# Question 7d5f5

Dec 20, 2015

Here's how you can prove this.

#### Explanation:

You need to prove that the half-life of a ${n}^{\text{th}}$ order reaction, ${t}_{\text{1/2}}$, is proportional to the inverse of the initial concentration of the reactant, ${A}_{0}$, raised to the power $n - 1$.

$\textcolor{b l u e}{{t}_{\text{1/2}} \propto \frac{1}{A} _ {0}^{n - 1}}$

So, for a general ${n}^{\text{th}}$ order reaction

$A \to \text{products}$

that is ${n}^{\text{th}}$ order in $A$, you can say that the rate of reaction will depend on the concentration of $A$ raised to the ${n}^{\text{th}}$ power

$\textcolor{b l u e}{\text{rate} = - \frac{d \left[A\right]}{\mathrm{dt}} = k \cdot {\left[A\right]}^{n}}$

Rearrange this equation to get

$- \frac{d \left[A\right]}{{\left[A\right]}^{n}} = k \cdot \mathrm{dt}$

Integrate both sides of the equation to get

$- \int \frac{1}{{\left[A\right]}^{n}} \cdot d \left[A\right] = k \cdot \int \mathrm{dt}$

$- \int {\left[A\right]}^{- n} \cdot d \left[A\right] = k \cdot t + c$

$- \frac{{\left[A\right]}^{\left(- n + 1\right)}}{\left(- n + 1\right)} = k \cdot t + c$

You can get rid of that minus sign by writing

$- n + 1 = - \left(n - 1\right)$

This will get you

$\frac{1}{n - 1} \cdot {\left[A\right]}^{- \left(n - 1\right)} = k \cdot t + c$

$\frac{1}{n - 1} \cdot \frac{1}{A} ^ \left(n - 1\right) = k \cdot t + c \text{ " " } \textcolor{p u r p \le}{\left(1\right)}$

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, ${A}_{0}$

$\textcolor{b l u e}{\text{For t = 0} \implies \left[A\right] = \left[{A}_{0}\right]}$

At $t = 0$, the equation $\textcolor{p u r p \le}{\left(1\right)}$ becomes

$\frac{1}{n - 1} \cdot \frac{1}{{A}_{0}} ^ \left(n - 1\right) = k \cdot 0 + c$

Isolate the constant to get

$c = \frac{1}{n - 1} \cdot \frac{1}{{A}_{0}} ^ \left(n - 1\right)$

Plug this into equation $\textcolor{p u r p \le}{\left(1\right)}$ to get

$\frac{1}{n - 1} \cdot \frac{1}{A} ^ \left(n - 1\right) = k \cdot t + {\overbrace{\frac{1}{n - 1} \cdot \frac{1}{{A}_{0}} ^ \left(n - 1\right)}}^{\textcolor{red}{= c}}$

Rearrange this equation to get

$\frac{1}{n - 1} \cdot \left(\frac{1}{A} ^ \left(n - 1\right) - \frac{1}{{A}_{0}} ^ \left(n - 1\right)\right) = k \cdot t \text{ " " } \textcolor{p u r p \le}{\left(2\right)}$

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

$\textcolor{b l u e}{t = {t}_{\text{1/2}} \implies \left[A\right] = \frac{1}{2} \cdot \left[{A}_{0}\right]}$

Plug this into equation $\textcolor{p u r p \le}{\left(2\right)}$ to get - keep in mind that

${\left(\frac{1}{{A}_{0} / 2}\right)}^{n} = {\left(\frac{2}{A} _ 0\right)}^{n} = {2}^{n} / {A}_{0}^{n}$

$\frac{1}{n - 1} \cdot \left({2}^{n - 1} / {\left[{A}_{0}\right]}^{n - 1} - \frac{1}{{A}_{o}} ^ \left(n - 1\right)\right) = k \cdot {t}_{\text{1/2}}$

Rearrange to get

$\frac{1}{n - 1} \cdot \frac{{2}^{n - 1} - 1}{{A}_{0}} ^ \left(n - 1\right) = k \cdot {t}_{\text{1/2}}$

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,

$\textcolor{b l u e}{{t}_{\text{1/2}} \propto \frac{1}{{A}_{0}} ^ \left(n - 1\right)} \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$