Question

A first order reaction is half-completed in $45$ minutes. How long does it need for $99.9\% $ of the reaction to be completed?
(a) $20$ Hours
(b) $10$ Hours
(c) $7\dfrac{3}{{10}}$ Hours
(d) $5$ Hours

Answer 1

Hint: We have to know what is a first order reaction, what formulas should be used to solve these questions, what we understand by the question which says that the reaction is half completed and how we use these statements to solve these questions.

Complete step by step solution:
For first order reaction, The relation of rate of reaction to the time and initial concentration and final concentration are as follows:
$k = \dfrac{{\ln \left[ {\dfrac{{{{\left[ A \right]}_O}}}{A}} \right]}}{t}$
Where, $k$ is rate of reaction, $t$ is time, ${[A]_o}$ is initial concentration and $A$ is final concentration left.
For half-life, (Half-life: The time at which reaction is half completed i.e, $A = \dfrac{{{{\left[ A \right]}_O}}}{2}$ )
${t_{\dfrac{1}{2}}} = \dfrac{{\ln (2)}}{k} = \dfrac{{0.693}}{k}$
Where, $k$ rate of reaction and ${t_{\dfrac{1}{2}}}$ is half-life time.
Complete step-by-step answer:
Given data in the question:
Half-life time, ${t_{\dfrac{1}{2}}} = 45\min $
To find out the time required to complete the reaction, we have to find the $k$ first.
By using half-life time formula,
$k = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{0.693}}{{45}} = 0.0154{\min ^{ - 1}}$
So, $k = 0.0154{\min ^{ - 1}}$
In the question asked, the time required to complete $99.9\% $ reaction,
So, the final concentration left,
Final concentration left= Initial concentration – the reaction completed
$A = {[A]_O} - 0.999{[A]_O} = 0.001{[A]_O}$
So, the final concentration will be $A = 0.001{[A]_O}$ .
Now, putting the value of $k$ and $A$ in the rate equation ( $k = \dfrac{{\ln \left[ {\dfrac{{{{\left[ A \right]}_O}}}{A}} \right]}}{t}$ )
$t = \dfrac{{\ln \left[ {\dfrac{{{{\left[ A \right]}_O}}}{{0.001{{[A]}_o}}}} \right]}}{k} = \dfrac{{\ln [1000]}}{{0.0154}}$
Now, converting $\ln $ to $\log $ for easy calculations, $\ln = 2.303\log $
$ \Rightarrow t = \dfrac{{2.303\log [1000]}}{{0.0154}}$
 From the properties of $\log $ , $\log (1000) = 3\log (10)$ and $\log (10) = 1$ ,
$ \Rightarrow t = \dfrac{{2.303 \times 3 \times \log [10]}}{{0.0154}} = \dfrac{{2.303 \times 3}}{{0.0154}}$
Now, solving the equation,
We get,
$ \Rightarrow t = \dfrac{{2.303 \times 3}}{{0.0154}}\min = 448.64\min $
Now, convert the time in hours because in the options of the question, the time is in hours.
So, to convert the minutes in hours, we have to divide time with $60$ ,
$ \Rightarrow t = \dfrac{{448.64}}{{60}} \approx 7\dfrac{3}{{10}}hours$
Hence, the time will be approximately equal to $t \approx 7\dfrac{3}{{10}}hours$

Hence, the correct option is (c) $7\dfrac{3}{{10}}$ Hours.


Note:When the half-life time of any reaction or the time at which half of the reaction completed is given in the question, think once about the half-life formula and try to find out the rate of reaction and be careful what is given the percentage of reaction completed or left.