Question

For a first order reaction, calculate the ratio between the time taken to complete one half of the reaction and the time taken to complete one third of the reaction.

Answer 1

Hint: We can calculate the ratio between the time taken to complete one half of the reaction and the time taken to complete one third of the reaction using the rate constant, initial amount of the reactant and amount that is remaining after the process of decay.

Complete step by step answer:
We can write the integrated rate law expression for the first order reaction as,
$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)$
Where,
t represents the time taken
$a$ represents the initial amount of the reactant
$a - x$ represents the amount that is left after the process of decay
Let, ${t_{\dfrac{1}{2}}}$ is the time taken for half of the reaction and amount of concentration left after ${t_{\dfrac{1}{2}}}$ is half of initial concentration.
So $a - x = \dfrac{a}{2}$
As we know,
$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)$
Let us now rearrange the above equation to arrive the expression for ${t_{\dfrac{1}{2}}}$
${t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k}\log \dfrac{a}{{a - x}}$
Let us now substitute the value of $a - x$ in the expression for ${t_{\dfrac{1}{2}}}$.
$ \Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k}\log \dfrac{{\dfrac{a}{1}}}{{\dfrac{a}{2}}}$
$ \Rightarrow{t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k}\log 2$
Substituting the value log 2 we get,
$ \Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{2.303}}{k} \times 0.301$
Multiplying the values we get,
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k} \to \left( 1 \right)$
The time taken for one third of the reaction is given as ${t_{\dfrac{1}{3}}}$ and the concentration that is remained after ${t_{\dfrac{1}{3}}}$ is 2/3 of initial concentration.
So, $a - x = \dfrac{{2a}}{3}$
Let us now rearrange the above equation to arrive the expression for ${t_{\dfrac{1}{3}}}$
${t_{\dfrac{1}{3}}} = \dfrac{{2.303}}{k}\log \dfrac{a}{{a - x}}$
Let us now substitute the value of $a - x$ in the expression for ${t_{\dfrac{1}{3}}}$.
$ \Rightarrow {t_{\dfrac{3}{2}}} = \dfrac{{2.303}}{k}\log \dfrac{{\dfrac{a}{1}}}{{\dfrac{{2a}}{3}}}$
$ \Rightarrow{t_{\dfrac{3}{2}}} = \dfrac{{2.303}}{k}\log \dfrac{3}{2}$
$ \Rightarrow {t_{\dfrac{3}{2}}} = \dfrac{{0.405}}{k} \to \left( 2 \right)$
Let us first divide first expression by second expression to get the time taken.
$ \Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k} \to \left( 1 \right)$
$ \Rightarrow {t_{\dfrac{3}{2}}} = \dfrac{{0.405}}{k} \to \left( 2 \right)$
$ \Rightarrow t = \dfrac{{\dfrac{{0.693}}{k}}}{{\dfrac{{0.405}}{k}}}$
$t = 1.71$
The ratio between the time taken to complete one half of the reaction and the time taken to complete one third of the reaction is $1.71$.

Note:
As we know that the rate constant of a first order reaction contains a time unit and not concentration unit. From this we can understand that for a first order reaction, the numerical value of k is independent of the unit in which the concentration is expressed. Even if we change the concentration unit, the numerical value of k for a first order reaction would not change.