Question

For first order reaction, time required for $99.0\% $ completion is $x$ times for the first time required for the completion of $90\% $ of the reaction $x$ is:

Answer 1

Hint: To find the time required for the completion of a first order reaction to a certain degree, we have to write the expression for the rate constant of a first order reaction. Because the rate constant of a reaction does not change with time or concentration of the reactants.

Complete step by step answer:
The expression for the rate constant of a first order reaction is as follows,
$K = \dfrac{{2.303}}{t}\log \left[ {\dfrac{a}{{a - x}}} \right]$
Where, $K = $ rate constant of the reaction
$t = $ time taken by the reaction
$a = $ concentration of reactant initially
$a - x = $ concentration of reactant left after time ‘t’
$x = $ amount of reactant consumed after time ‘t’
After completion of $99.0\% $ of the reaction, the amount of reactant consumed is $99.0\% $ of a:
$K = \dfrac{{2.303}}{{{t_{99.0\% }}}}\log \left[ {\dfrac{a}{{a - 0.99a}}} \right]$
$ \Rightarrow K = \dfrac{{2.303}}{{{t_{99.0\% }}}}\log \left[ {\dfrac{a}{{0.01a}}} \right]$
$ \Rightarrow K = \dfrac{{2.303}}{{{t_{99.0\% }}}}\log \left[ {100} \right]$
$ \Rightarrow K = \dfrac{{2.303}}{{{t_{99.0\% }}}} \times 2$ …(As $\log (100) = 2$)
$ \Rightarrow {t_{99.0\% }} = \dfrac{{2.303}}{K} \times 2$ …$(1)$
After completion of $90.0\% $ of the reaction, the amount of reactant consumed is $90.0\% $ of a:
$K = \dfrac{{2.303}}{{{t_{90.0\% }}}}\log \left[ {\dfrac{a}{{a - 0.9a}}} \right]$
$ \Rightarrow K = \dfrac{{2.303}}{{{t_{90.0\% }}}}\log \left[ {\dfrac{a}{{0.1a}}} \right]$
$ \Rightarrow K = \dfrac{{2.303}}{{{t_{90.0\% }}}}\log \left[ {10} \right]$
$ \Rightarrow K = \dfrac{{2.303}}{{{t_{90.0\% }}}}$ …(As $\log (10) = 1$)
$ \Rightarrow {t_{90.0\% }} = \dfrac{{2.303}}{K}$ …$(2)$
It is given in the question that the time required for the completion of $99.0\% $ reaction is $x$ times the time required for the completion of $90\% $ reaction.
So, ${t_{99.0\% }} = x \times {t_{90.0\% }}$
Putting equation $(1)$ and $(2)$ in the above equation:
$ \Rightarrow \dfrac{{2.303}}{K} \times 2 = x \times \dfrac{{2.303}}{K}$
$ \Rightarrow x = 2$
Therefore, the time required for the completion of $99.0\% $ reaction is $2$ times the time required for the completion of $90\% $ reaction.

Note:
First order reactions are the reactions in which the rate of reaction is directly proportional to the concentration of one of the reactants or partially dependent on all the reactants such that the rate of reaction varies linearly with their concentrations as a whole. And it should be kept in mind that for a reaction the concentration may change with time but rate constant is always constant.