Answer
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Hint: The first-order reaction is the reaction in which the rate of reaction is directly proportional to the concentration of the reactant. ${\dfrac{3}{4}}^{th}$ life means the concentration of the reactant at time $t$ is $\dfrac{3}{4}$ of the initial concentration of the reactant.
Complete step by step answer:
As we known the first-order rate constant formula is,
$\Rightarrow k\,\, = \,\dfrac{1}{t}\ln \dfrac{{{A_ \circ }}}{{{A_x}}}$
We can multiply the expression from $2.303$ to convert ‘ln into log’.
$\Rightarrow k\,\, = \,\dfrac{{2.303}}{t}\log \dfrac{{{A_ \circ }}}{{{A_x}}}$
Where, k is the first-order rate constant. Unit of first order rate constant is ${\text{tim}}{{\text{e}}^{ -1}}$.
$\Rightarrow$ ‘t’ is the time.
$\Rightarrow {A_ \circ }$ is the initial concentration of the reactant.
$\Rightarrow {A_x}$ is the concentration of the reactant left at time ‘t’.
The concentration of reactant is not given so, we can assume that the initial concentration of the reactant is ‘1’.
At time ‘t’ the concentration of the reactant left is $\dfrac{1}{4}$.
We can use the first-order rate constant formula to determine the rate constant as follows:
On substituting ‘1’ for ${A_ \circ }$ and $\dfrac{1}{4}$ for ${A_x}$.
$\Rightarrow k\,\, = \,\dfrac{{2.303}}{t}\log \dfrac{1}{{1/4}}$
$\Rightarrow k\,\, = \,\dfrac{{2.303}}{t}\log 4$
Rearrange the above expression for tas follows:
$\Rightarrow t\,\, = \,\dfrac{{2.303}}{k}\log 4$
So, the above formula represents the $\dfrac{3}{4}$ life of a reaction.
Thus, Option C is correct.
Note: The ${A_x}$ 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.
Complete step by step answer:
As we known the first-order rate constant formula is,
$\Rightarrow k\,\, = \,\dfrac{1}{t}\ln \dfrac{{{A_ \circ }}}{{{A_x}}}$
We can multiply the expression from $2.303$ to convert ‘ln into log’.
$\Rightarrow k\,\, = \,\dfrac{{2.303}}{t}\log \dfrac{{{A_ \circ }}}{{{A_x}}}$
Where, k is the first-order rate constant. Unit of first order rate constant is ${\text{tim}}{{\text{e}}^{ -1}}$.
$\Rightarrow$ ‘t’ is the time.
$\Rightarrow {A_ \circ }$ is the initial concentration of the reactant.
$\Rightarrow {A_x}$ is the concentration of the reactant left at time ‘t’.
The concentration of reactant is not given so, we can assume that the initial concentration of the reactant is ‘1’.
At time ‘t’ the concentration of the reactant left is $\dfrac{1}{4}$.
We can use the first-order rate constant formula to determine the rate constant as follows:
On substituting ‘1’ for ${A_ \circ }$ and $\dfrac{1}{4}$ for ${A_x}$.
$\Rightarrow k\,\, = \,\dfrac{{2.303}}{t}\log \dfrac{1}{{1/4}}$
$\Rightarrow k\,\, = \,\dfrac{{2.303}}{t}\log 4$
Rearrange the above expression for tas follows:
$\Rightarrow t\,\, = \,\dfrac{{2.303}}{k}\log 4$
So, the above formula represents the $\dfrac{3}{4}$ life of a reaction.
Thus, Option C is correct.
Note: The ${A_x}$ 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.
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