Question

In a first order reaction the amount of reactant decayed in three half lives would be, given the initial amount is $a$ :
A. $\dfrac{{7a}}{8}$
B. $\dfrac{a}{8}$
C. $\dfrac{a}{6}$
D. $\dfrac{{5a}}{6}$

Answer 1

Hint: We will find the amount of reactant left after n half lives. The given half lives here are 3 and hence we can find the amount of reactant left after 3 half lives. The difference between initial amount and the amount left will give us the amount of reactant decayed in three half lives.


Complete solution:
Given to us, the initial amount of reactant is $a$ and that the reaction is first order.
For a first order reaction, the amount of reactant left after $n$ half lives is given as ${\left( {\dfrac{1}{2}} \right)^n}$ times the initial concentration.
Here, $n = 3$ so the amount of reactant left after three half lives is ${\left( {\dfrac{1}{2}} \right)^3} \times a = \left( {\dfrac{a}{8}} \right)$
The difference between the initial amount and the amount left is $a - \dfrac{a}{8} = \dfrac{{7a}}{8}$ .

Therefore, the amount of reactant decayed in three half lives is $\dfrac{{7a}}{8}$ i.e. option A.

Additional information: The time that is required for a particular amount of a reactant to decrease or decay to half of its initial amount is known as half-life. If two reactions have the same order, then the slower reaction will have a longer half-life and the faster reaction will have a shorter half-life.

Note:One might mistakenly finish the solution at $\dfrac{a}{8}$ which is only the amount left after three half-lives and not the amount decayed. It is important to remove the amount left from the initial amount to find the amount of reactant decayed.