Question

For the first order reaction, the half-life is 14 sec, the time required for the initial concentration to reduce to 1/8 of its value is:
(a) \[{{(14)}^{3}}\] sec
(b) 28 sec
(c) 42 sec
(d) 1.75 sec

Answer 1

Hint:The half-life of a reaction is defined as the time required to reduce the concentration of the reactant to half of its initial value. It is demoted by the symbol \[{{t}_{{\scriptstyle{}^{1}/{}_{2}}}}\]. The half-life of a reaction depends on the order of reaction.

Complete step by step solution:
The sum of the power of the concentration terms on which the rate of reaction actually depends, as observed experimentally, is called the order of reaction. In a first order reaction rate varies as the first power of the concentration of the reactant, i.e. the rate increases as the number of times as the concentration of reactant is increased. For a first order reaction, the half-life is given by the equation:
\[{{t}_{{\scriptstyle{}^{1}/{}_{2}}}}=\dfrac{\ln \,2}{k}\] where, k = rate constant or velocity constant.
For the given problem, the half-life of reactant (\[{{t}_{{\scriptstyle{}^{1}/{}_{2}}}}\]) is 14 sec. This means that it takes 14 seconds to reduce the initial concentration to half.
\[{{t}_{{\scriptstyle{}^{1}/{}_{2}}}}=\dfrac{\ln \,2}{k}=14\,\sec \]
On rearranging, we get:
\[k=\dfrac{\ln \,2}{14}\]
For first order reactions, \[k=\dfrac{1}{t}\ln \dfrac{{{a}_{0}}}{{{a}_{t}}}\]
We have been given in the problem that \[{{a}_{t}}=\dfrac{{{a}_{0}}}{8}\]
Substituting the value, we get
\[k=\dfrac{1}{t}\ln \dfrac{{{a}_{0}}}{{}^{{{a}_{0}}}/{}_{8}}=\dfrac{1}{t}\ln \,8\]
\[\dfrac{\ln \,2}{14}=\dfrac{1}{t}\ln \,{{(2)}^{3}}\]
\[\dfrac{\ln \,2}{14}=\dfrac{3}{t}\ln \,(2)\]
\[\dfrac{1}{14}=\dfrac{3}{t}\]
\[t=42\sec \]
So, option (c) is correct.

Note: We can also compute it intuitively, it will take another 14 seconds to reduce the concentration to \[\dfrac{1}{4}\]and will take further 14 seconds to reach \[\dfrac{1}{8}\] of its initial concentration.

So, the total time taken will be, 14+14+14 = 42 seconds.