Question

Chlorine-34 has a half-life of only 1.53 sec. How long does it take for 99.99% of a $^{34}Cl$ sample to decay?
A.t=20.34 sec
B. t= 30.34 sec
C. t=40.34 sec
D. None of these

Answer 1

Hint: Try to find out the amount of chlorine left, and after that substitute the data given in the question in the rate equation. On solving the equation, the value of ‘t’ can be obtained.

Complete step by step solution:
In order to answer our question, we need to learn about kinetics of a reaction and study about the half lives. When a chemical reaction occurs, then the concentration of reactants decrease and that of the product increase. Now, the total time of the reaction to occur is the time taken by the reactants to achieve a concentration of 0 or for the maximum formation of the products. Similar to the total time, a half-life for a reaction is also defined which helps us to understand the kinetics of that particular reaction. It is defined as the time during which the concentration of the reactants is reduced to half of the initial concentration or it is the time required for the completion of half of the reaction. It is denoted by ${{t}_{1/2}}$. Now, it is to be noted that for a zero order reaction, the expression of half life is :${{t}_{1/2}}=\dfrac{0.693}{k}$, where k is the rate constant and the expression for half life of a first order reaction is:
\[k=\dfrac{2.303}{{{t}_{1/2}}}\log \dfrac{[{{N}_{0}}]}{[N]}\]
Here, $[{{N}_{0}}],[N]$ refers to the initial and final concentrations. Now, let us come to our question. Let the initial amount of the chlorine be $100{{N}_{0}}$, so the amount which is left after getting decomposed is 100-99.99=$0.01{{N}_{0}}$ By substituting the data given in the question into the formula, we can write:
\[\begin{align}
& k=\dfrac{2.303}{t}\log \dfrac{{{N}_{0}}}{N} \\
& \Rightarrow \dfrac{2.303}{t}\log \dfrac{100}{0.01}=\dfrac{0.693}{1.53} \\
& \Rightarrow t=\dfrac{0.693}{1.53\times 2.303}\times \dfrac{1}{\log (\dfrac{100}{0.01})} \\
\end{align}\]

On solving, we get the value of t as 20.34 seconds. So, we get our correct answer as option A.

Note: It is to be noted that the expression for the half time can be taken out from the general rate equation, it is just that the final concentration has to be substituted as $\dfrac{[{{N}_{0}}]}{2}$.So,${{t}_{1/2}}=\dfrac{2.303}{k}\log \dfrac{{{R}_{0}}}{(\dfrac{{{R}_{0}}}{2})}=\dfrac{2.303}{k}\log 2=\dfrac{0.693}{k}$, this is the half life expression for a zero order reaction.