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A first-order reaction $({t_{\dfrac{1}{2}}} = 1day)$ completes $x\% $ after 4 half-lives. Value of $x$ is:
A. $87.5$
B. $75$
C. $93.75$
D. $50$

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Last updated date: 23rd Jul 2024
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Answer
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Hint: The half lifetime for 4 half-lives will be equal to 4 times the half-life of 1 half-life i.e. 1 day and substituting ${t_{\dfrac{1}{2}}}^\prime = 4(1day) = 4days$ in integrated rate equation for the first-order reaction, we can find the value of $x$ but first of all, we need to find the value of the rate constant $k$. For a first-order reaction, $k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{a}{{a - x}}$
Here, $a$ is the initial concentration
$x$ is the amount of reactant reacting in time t

Complete step by step answer:
Now, we know that the half-life formula for the first-order reaction is,
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$
Here, $k$ is rate constant
For 1 half-life,
${t_{\dfrac{1}{2}}} = 1day$ (Given)
$\therefore $For 4 half-lives,
${t_{\dfrac{1}{2}}}^\prime = 4\left( {{t_{\dfrac{1}{2}}}} \right)$
$ = 4(1)$
${t_{\dfrac{1}{2}}}^\prime = 4days$ ……. (Equation number 1)
Now,
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$
So, $k = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$
$ \Rightarrow k = \dfrac{{0.693}}{1}$
$k = 0.693da{y^{ - 1}}$ ….. (Equation number 2)
For $t = {t_{\dfrac{1}{2}}}$,
${t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k}{\log _{10}}\dfrac{a}{{a - x}}$
And for $t = {t_{\dfrac{1}{2}}}^\prime $,
${t_{\dfrac{1}{2}}}^\prime = \dfrac{{2.303}}{k}{\log _{10}}\dfrac{{100}}{{100 - x}}$ …… (Equation number 3)
As the reaction completes $x\% $ after 4 half-lives. (Given)
But ${t_{\dfrac{1}{2}}}^\prime = 4days$ …..(From Equation number 1)
And $k = 0.693da{y^{ - 1}}$ …..(From Equation number 2)
Substituting these values in Equation number 3,
$4 = \dfrac{{2.303}}{{0.693}}{\log _{10}}\dfrac{{100}}{{100 - x}}$
$ \Rightarrow 4 = 3.3232\left[ {{{\log }_{10}}\left( {100} \right) - {{\log }_{10}}\left( {100 - x} \right)} \right]$
$ \Rightarrow 4 = 3.3232\left[ {2 - {{\log }_{10}}\left( {100 - x} \right)} \right]$
$ \Rightarrow \dfrac{4}{{3.3232}} = 2 - {\log _{10}}\left( {100 - x} \right)$
$ \Rightarrow 1.2036 = 2 - {\log _{10}}\left( {100 - x} \right)$
$ \Rightarrow 1.2036 - 2 = - {\log _{10}}\left( {100 - x} \right)$
$ \Rightarrow - 0.7964 = - {\log _{10}}\left( {100 - x} \right)$
$ \Rightarrow 0.7964 = {\log _{10}}\left( {100 - x} \right)$
Now, converting logarithmic form to exponential form,
${10^{0.7964}} = 100 - x$
Taking log on both the sides,
$\log \left( {{{10}^{0.7964}}} \right) = \log \left( {100 - x} \right)$
$0.7964 = {\log _{10}}\left( {100 - x} \right)$
Now taking antilog on both sides,
$A.L.\left( {0.7964} \right) = A.L.\left[ {{{\log }_{10}}\left( {100 - x} \right)} \right]$
$6.2574 = 100 - x$
$ \Rightarrow 6.2574 - 100 = x$
$ \Rightarrow - 93.75 = - x$
$x = 93.75$
Hence, the value of $x$ is $93.75$

So, the correct answer is Option C.

Additional Information:
The concept of half-life was discovered by Ernest Rutherford in 1907. It relates to the time required by radioactive substances for disintegration or the formation of a new substance. It is also known as the half-life period.

Note: The rate constant of a second-order reaction is $k = \dfrac{1}{{2at}}\log \dfrac{x}{{a - x}}$. The first-order rate constant depends on the concentration of only 1 among the reactants. We could see that the half-life of the first order is independent of the reactant concentration. The unit of the first-order reaction is ${\sec ^{ - 1}}$. The unit of half-life is seconds.