Answer
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Hint:A chemical reaction in which the rate of the reaction is proportional to the concentration of the reacting substance is known as a first order reaction. To solve this we must know the expression for the rate constant of a first order reaction. at half-life period only half of the reactant remains.
Complete solution:
The time taken for the reactant species to reduce to half of its initial concentration is known as the half-life period of the reaction. At the half-life, $50\% $ of the reaction is completed.
We know that the equation for the rate constant of a first order reaction is,
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}$
Where $k$ is the rate constant of a first order reaction,
$t$ is time,
${\left[ a \right]^0}$ is the initial concentration of the reactant,
$\left[ a \right]$ is the final concentration of the reactant.
At the half-life period of the reaction, only half of the reactant remains and half of the reactant has been converted to the product. Thus,
$\left[ a \right] = \dfrac{{{{\left[ a \right]}^0}}}{2}$ and $t = {t_{{\text{1/2}}}}$
Thus,
$k{t_{{\text{1/2}}}} = \ln \dfrac{{{{\left[ a \right]}^0}}}{{{{\left[ a \right]}^0}{\text{/2}}}}$
$k{t_{{\text{1/2}}}} = \ln 2$
${t_{{\text{1/2}}}} = \dfrac{{0.693}}{k}$
Thus, the expression for half-life of a first order reaction is ${t_{{\text{1/2}}}} = \dfrac{{0.693}}{k}$.
At the half-life period of the reaction, the concentration of the reactant is equal to the concentration of the product.
The expression for the half-life period of a first order reaction does not contain a concentration term. Thus, the half-life period for such reactions is independent of the initial concentration of the reactants.
Note: The unit of rate constant for first order reaction is ${\text{se}}{{\text{c}}^{ - 1}}$ or ${\text{mi}}{{\text{n}}^{ - 1}}$. The units do not contain concentration terms. Thus, we can say that the rate constant of a first order reaction is independent of the concentration of the reactant.
Complete solution:
The time taken for the reactant species to reduce to half of its initial concentration is known as the half-life period of the reaction. At the half-life, $50\% $ of the reaction is completed.
We know that the equation for the rate constant of a first order reaction is,
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}$
Where $k$ is the rate constant of a first order reaction,
$t$ is time,
${\left[ a \right]^0}$ is the initial concentration of the reactant,
$\left[ a \right]$ is the final concentration of the reactant.
At the half-life period of the reaction, only half of the reactant remains and half of the reactant has been converted to the product. Thus,
$\left[ a \right] = \dfrac{{{{\left[ a \right]}^0}}}{2}$ and $t = {t_{{\text{1/2}}}}$
Thus,
$k{t_{{\text{1/2}}}} = \ln \dfrac{{{{\left[ a \right]}^0}}}{{{{\left[ a \right]}^0}{\text{/2}}}}$
$k{t_{{\text{1/2}}}} = \ln 2$
${t_{{\text{1/2}}}} = \dfrac{{0.693}}{k}$
Thus, the expression for half-life of a first order reaction is ${t_{{\text{1/2}}}} = \dfrac{{0.693}}{k}$.
At the half-life period of the reaction, the concentration of the reactant is equal to the concentration of the product.
The expression for the half-life period of a first order reaction does not contain a concentration term. Thus, the half-life period for such reactions is independent of the initial concentration of the reactants.
Note: The unit of rate constant for first order reaction is ${\text{se}}{{\text{c}}^{ - 1}}$ or ${\text{mi}}{{\text{n}}^{ - 1}}$. The units do not contain concentration terms. Thus, we can say that the rate constant of a first order reaction is independent of the concentration of the reactant.
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