Question

For a first order reaction with rate constant $k$ and initial concentration $a$, the half-life period is given by:
(This question has multiple correct options)
A. $\dfrac{{\ln 2}}{k}$
B. $\dfrac{1}{{ka}}$
C. $\dfrac{{0.693}}{k}$
D. $\dfrac{{2.303}}{k}\log 2$

Answer 1

First order reaction: It is a chemical reaction in which the rate of the reaction depends on the concentration of one reactant only. The other reactants in the first order reaction will be of zero order. The rate law for first order reaction is ${\rm{rate}} = k\left[ A \right]$, where $k$ is the rate constant of the reaction.

Complete answer:
Differential rate law: It is the method to describe rate in terms of the change in concentration of reactants over a specific period of time. These rate laws are very helpful in determining the mechanism of overall reaction.
Integrated rate law: It is the method to describe rate in terms of initial and final concentration of reactants in a chemical reaction after a particular amount of time.
For first order reaction, the differential rate law is as follows:
$- \dfrac{{d[A]}}{{dt}} = k[A]$
Therefore, integrated rate law for first order reaction can be written as follows:
$\int\limits_{{{[A]}_o}}^{{{[A]}_t}} {\dfrac{{d[A]}}{{[A]}}} = - k\int\limits_0^t {dt}$
Where, ${[A]_o}$ is the initial concentration of reactant and ${[A]_t}$ is the final concentration of the reactant.
$\Rightarrow \left[ {\ln [A]} \right]_{{{[A]}_o}}^{{{[A]}_t}} = - kt$
$\Rightarrow \ln {[A]_t} - \ln {[A]_o} = - kt$
$\Rightarrow \ln \dfrac{{{{[A]}_o}}}{{{{[A]}_t}}} = kt\,\, - - (i)$
Converting natural logarithmic function into log function with base $10$.
$\Rightarrow 2.303\log \dfrac{{{{[A]}_o}}}{{{{[A]}_t}}} = kt\,\, - - (ii)$
Half-life period: It is the time required by a reactant to reach $50\%$ of its initial concentration.
Therefore, at $t = {t_{\dfrac{1}{2}}}$, the relation between initial concentration and final concentration is ${[A]_t} = \dfrac{{{{[A]}_o}}}{2}$
According to the given conditions, ${[A]_o} = a$. So, ${[A]_t} = \dfrac{a}{2}$
Substituting values at half-life conditions in equation $(i)$.
$\ln \dfrac{a}{{\dfrac{a}{2}}} = k{t_{\dfrac{1}{2}}}$
$\Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{k}\,\, - - (ii)$
$\Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}\; - - (iii)$
Substituting values at half-life conditions in equation $(ii)$
$2.303\log \dfrac{a}{{\dfrac{a}{2}}} = k{t_{\dfrac{1}{2}}}\,\,$
$\Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k}\log 2\, - - (iv)$

Hence according to equations $(ii)$, $(iii)$ and $(iv)$, options (A), (C) and (D) are correct.

Note:
Most of the radioactive decay reactions are of first order reaction. Hence the study of first order kinetics is very important due to its wide use in various chemical reactions. It also guides how to increase or decrease the rate of reaction over a certain set of conditions.