Question

Mathematical expression for ${{\text{t}}_{{\text{1/4}}}}$i.e., when ${{\text{(1/4)}}^{{\text{th}}}}$ reaction is over following first order kinetics can be given by:
A. ${{\text{t}}_{{\text{1/4}}}} = \,\dfrac{{2.303}}{k}\log 4$
B. ${{\text{t}}_{{\text{1/4}}}} = \,\dfrac{{2.303}}{k}\log \dfrac{1}{4}$
C. ${{\text{t}}_{{\text{1/4}}}} = \,\dfrac{{2.303}}{k}\log 2$
D. ${{\text{t}}_{{\text{1/4}}}} = \,\dfrac{{2.303}}{k}\log \dfrac{4}{3}$

Answer 1

The first-order reaction is the reaction in which the rate of reaction is directly proportional to the concentration of the reactant. At one/ fourth life time means at ${{\text{t}}_{{\text{1/4}}}}$time $1/4$ of the initial concentration of the reactant is consumed.

Formula used: $k\,\, = \,\dfrac{{2.303}}{t}\log \dfrac{{{A_ \circ }}}{{{A_x}}}$

Complete answer:
As we known the first-order rate constant formula is,
${\text{k}}\,\,{\text{ = }}\,\dfrac{{\text{1}}}{{\text{t}}}{\text{ln}}\dfrac{{{{\text{A}}_{\text{o}}}}}{{{{\text{A}}_{\text{x}}}}}$
We can multiply the expression from $2.303$ to convert ln into log.
${\text{k}}\,\,{\text{ = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}{\text{log}}\dfrac{{{{\text{A}}_{\text{o}}}}}{{{{\text{A}}_{\text{x}}}}}$
where,
$k$ is the first-order rate constant. Unit of first order rate constant is ${\text{tim}}{{\text{e}}^{ - 1}}$.
$t$ is the time.
${{\text{A}}_{\text{o}}}$ is the initial concentration of the reactant.
${{\text{A}}_{\text{x}}}$ is the concentration of the reactant left at time $t$.
The initial concentration of the reactant is not given, so we can assume that the initial concentration of the reactant is $1$.
At time ${{\text{t}}_{{\text{1/4}}}}$the concentration of the reactant consumed is $1/4$.
So, the left concentration of reactant will be Initial-consumed
So,
${{\text{A}}_{\text{x}}} = \,1 - \dfrac{1}{4}$
${{\text{A}}_{\text{x}}}\,{\text{ = }}\,\dfrac{{\text{3}}}{{\text{4}}}$ $\,$
We can use the first-order rate constant formula to determine the rate constant as follows:
On substituting $3/4$ for ${{\text{A}}_{\text{x}}}$and $1$ for ${{\text{A}}_{\text{o}}}$.
$\Rightarrow k\,\, = \,\dfrac{{2.303}}{{{t_{1/4}}}}\log \dfrac{1}{{3/4}}$
$\Rightarrow k\,\, = \,\dfrac{{2.303}}{{{t_{1/4}}}}\log \dfrac{4}{3}$
Rearrange the above expression for ${{\text{t}}_{{\text{1/4}}}}$ as follows:
$\Rightarrow{{\text{t}}_{{\text{1/4}}}}\,\,{\text{ = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{\text{k}}}{\text{log}}\dfrac{{\text{4}}}{{\text{3}}}$
So, the above formula represents the $1/4$ life of a reaction.

**Therefore, option (D) ${{\text{t}}_{{\text{1/4}}}}\,\,{\text{ = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{\text{k}}}{\text{log}}\dfrac{{\text{4}}}{{\text{3}}}$ is correct.

Note:**
The ${{\text{A}}_{\text{x}}}$ shows the concentration of the reactant left. If the initial concentration is not given then we can use $1$ or $100$ if the concentrations are given in percent for initial concentration of reactant. The formula includes the concentration of the reactant left, so subtract the concentration of product from an initial concentration of reactant to get the concentration of the reactant left. Just like ${{\text{t}}_{{\text{1/4}}}}$, ${{\text{t}}_{{\text{1/2}}}}$represents the time when half of the reactant is consumed or we can say half of the reactant remains left. ${{\text{t}}_{{\text{1/2}}}}$ is the half-life of the reaction.