Question

What will be the expression of $t_{1/2}$ half -life when reaction follows rate law, $\text{rate}=k[\text{concentration}{]}^2$
Here,
$k$ is the rate constant
$a$ is the initial concentration of the reactant

• A. $\frac{1}{k{a}^2}$
• B. $\frac{3}{2k{a}^2}$
• C. $\frac{1}{ka}$
• D. $k{a}^2$

Answer 1

The correct option is C $\frac{1}{ka}$

The expression for integrated rate law of $n^{th}$ order reaction is
$\left[\frac{1}{(a-x{)}^{n-1}}\right]-\left[\frac{1}{a^{n-1}}\right]=(n-1)kt.....eqn\left(1\right)$

$A\to \text{Product(s)}$
$time=0atime=\left(t\right)a-x$
At $t=t_{1/2},$
$x=\frac{a}{2}\text{(reaction is half completed)}$

Substituting the value of X in equation (1) gives,

$\left[\frac{1}{(a-\frac{a}{2}{)}^{n-1}}\right]-\left[\frac{1}{a^{n-1}}\right]=(n-1)k{t}_{1/2}$

$[\frac{2^{n-1}}{a^{n-1}}-\frac{1}{a^{n-1}}]=(n-1)k{t}_{1/2}$

On rearranging we get,
$t_{1/2}=\frac{1}{k(n-1)}\left[\frac{2^{n-1}-1}{a^{n-1}}\right]......eqn\left(2\right)$

In general,
$Rate\left(R\right)=k[A{]}^n$

Given,
$\text{Rate}=k[\text{A}{]}^2$
So, order = 2

Substituting $n=2$ in equation (2) gives,

$t_{1/2}=\frac{1}{k(2-1)}\left[\frac{2^{2-1}-1}{a^{2-1}}\right]$

$t_{1/2}=\frac{1}{k}\left[\frac{2-1}{a}\right]$

$t_{1/2}=\frac{1}{k}\left[\frac{2-1}{a}\right]$

$t_{1/2}=\frac{1}{ka}$