Question

For a given reaction
$A\to \text{Products}$, rate of reaction is given by
$r=k[A{]}^n$
For a constant temperature throughout, choose the correct option(s):

• A. For $n=0$, half life decreases as the concentration of reactant decreases
• B. For $n=1$, half life independent of the concentration of reactant.
• C. For $n=2$, half life increases as the concentration of reactant decreases.
• D. All of the above.

Answer 1

The correct option is D All of the above.

Method 1:
Graphical method
For $n=0$


From above graph it can be seen that half life decreases as the concentration of reactant decreases.
For $n=1$


From above graph it can be seen that half-life is unaffected by change in concentration.

For $n=2$


From above graph it can be seen that half life increases as the concentration of reactant decreases.


Method 2:
By generalized equation:
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,
$t_{1/2}=\frac{1}{k(n-1)}\left[\frac{2^{n-1}-1}{a^{n-1}}\right]......eqn\left(2\right)$
Form equation 2, at constant temperature it is evident that:

$t_{1/2}\propto a^{1-n}$

When $n=0$,
$t_{1/2}\propto a$
For $n=0$, half life decreases as the concentration of reactant decreases.
When $n=1$,
$t_{1/2}\propto a^0$
For $n=1$, half life is independent of concentration and is a constant value for a constant temperature.

When $n=2$,
$t_{1/2}\propto a^{-1}$
For $n=2$, half life increases as the concentration of reactant decreases.