Question

Show that the half-life period of a first-order reaction is independent of the initial concentration of the reactant.

Answer 1

Hint: The half-life period of a reaction is defined as that period in which the concentration of reactant is reduced to half of its initial concentration. It is denoted as ${t_{1/2}}$.

Complete answer: Let’s discuss the derivation for the half-life period of the first-order reaction:
Let us consider a general reaction for first order:
$A \to {\text{Product}}$
Suppose $[{A_o}]$ is the initial concentration
$[A]$ is the final concentration.
The rate of reaction is given as:
$ = - \dfrac{{d[A]}}{{dt}}$ (1)
According to rate law:
$Rate{\text{ }}of{\text{ }}reaction = k{[A]^1}$ (2)
Where k is rate constant
Comparing the above two equation we get:
$ - \dfrac{{d[A]}}{{dt}} = k[A]$
$ \Rightarrow - \dfrac{{d[A]}}{{[A]}} = k \times dt$ (3)
Taking integration on both sides the equation (3) will be
$\int { - \frac{{d[A]}}{{dt}}} = $$\int { - \frac{{d[A]}}{{[A]}}} = \int {kdt} $ (4)
As we know,
$\int { - \frac{{d[A]}}{{[A]}}} {\text{ }}\;{\text{ }} = {\text{ }}\;{\text{ }} - \ln [A]$
And $\int {dt} = t$
On solving equation (4) will be
$ - \ln [A] = kt + C$ (5)
Where C is integration constant.
Now we will find the value of C
When $t = 0$, $[A]$=$[{A_o}]$
Put above value in equation (5) we get,
${\text{ }}\;{\text{ }} - \ln [{A_o}] = k \times 0 + C \Rightarrow {\text{ }}\;{\text{ }} - \ln [{A_o}] = C{\text{ }}\;{\text{ }}$ Now put the value of C in equation (5)

\[
  {\text{ }}\;{\text{ }}\;{\text{ }} - \ln [A] = kt - \ln [{A_o}]{\text{ }}\;{\text{ }} \Rightarrow - \ln [A] + \ln [{A_o}] = kt{\text{ }}\;{\text{ }} \\
  \;{\text{ since }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\ln A - \ln B = \frac{{\ln A}}{{\ln B}}{\text{ }}\;{\text{ }} \\
  {\text{Hence }}\;{\text{ }}\;{\text{ }}\frac{{\ln [{A_o}]}}{{\ln [A]}} = kt{\text{ }}\;{\text{ }} \Rightarrow \frac{1}{k} \times \frac{{\ln [{A_o}]}}{{\ln [A]}} = t{\text{ }}\;{\text{ }} \\ \]
Hence, during the half-life period $[A]$=$[\frac{{{A_o}}}{2}]$ and \[\]
Putting value in the above equation we get
\[{\text{ }}\;{\text{ }}\;{\text{ }}{t_{1/2}} = \frac{1}{k} \times \frac{{\ln [{A_o}]}}{{\ln [\frac{{{A_0}}}{2}]}}{\text{ }}\;{\text{ }} \Rightarrow {t_{1/2}} = \frac{1}{k} \times \frac{{\ln [{A_o}] \times 2}}{{\ln [{A_o}]}}{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }} \Rightarrow {t_{1/2}} = \frac{1}{k} \times \ln 2\]
Since we can’t calculate the ln, so it is changed to log by multiplying 2.303

Log2=0.3010
Hence
${\text{ }}\;{\text{ }}\;{\text{ }}{t_{1/2}} = \frac{1}{k} \times 2.303 \times 0.3010{\text{ }}\;{\text{ }}\;{\text{ }} \Rightarrow {t_{1/2}} = \frac{{0.693}}{k}$ Hence proved.

Note: In general, for a reaction of ${n^{th}}$ order,
             \[{\text{ }}\;{\text{ }}\;{\text{ }}{t_{1/2}} \propto {[A]^{1 - n}}{\text{ }}or{\text{ }}\;{\text{ }}\;{\text{ }}{t_{1/2}} \propto \frac{1}{{{{[A]}^{1 - n}}}}\]