Question

Choose the correct option:
Here,
$t_{1/2}$ is the half life period of the reaction

• A. For a first order reaction,$t_{1/2}$ is directly proportional to initial concentration
• B. For a zero order reaction ,$t_{1/2}$ is inversely proportional to initial concentration
• C. For a second order reaction, $t_{1/2}$ is inversely proportional to initial concentration
• D. All of the above

Answer 1

The correct option is C For a second order reaction, $t_{1/2}$ is inversely proportional to initial concentration

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

$t_{1/2}\propto \frac{1}{a^{n-1}}$
a : initial concentration of reactant

For zero order reaction, n=0
$t_{1/2}\propto \frac{1}{a^{n-1}}$
$t_{1/2}\propto a$
Thus, half life period of zero order reaction is directly proportional to initial concentration of reactant.

For first order reaction, n=1
$t_{1/2}\propto \frac{1}{a^{n-1}}$
$t_{1/2}\propto \frac{0.693}{k}$
Thus, half life period of first order reaction is independent of initial concentration of reactant.

For second order reaction, n=2
$t_{1/2}\propto \frac{1}{a^{n-1}}$

$t_{1/2}\propto \frac{1}{a}$
Thus, half life period of second order reaction is inversely proportional to initial concentration of reactant.