Question

Two substances are present such that $[{A_0}] = 4[{B_0}]$ and half-life of A is 5 min and that of B is 15 min. If they start decaying at the same time following first-order kinetics how much time later will it take if the concentration of both of them would be the same?
A. 15 min
B. 10 min
C. 5 min
D. 12 min

Answer 1

Hint: The rate of a reaction depends on the concentration of the reactants and the products that participate in the reaction process. The disappearance of the reactant and the increase in the concentration of the product formed have a direct relation with the rate of reaction. Rate of reaction can be calculated by knowing the change in the concentration of the reactants or product in unit time.

Complete step by step answer:
The half-life of a substance is the time in which it reduces to half of its original mass. Each half-life period will reduce the substance to half of its original mass.
So analysing the above question we have,
Amount of A left in ${n_1}$ halves $ = {(\dfrac{1}{2})^{{n_1}}}[{A_0}]$
and similarly, we can calculate,
Amount of A left in ${n_2}$ halves $ = {(\dfrac{1}{2})^{n2}}[{B_0}]$
To at the end according to the rate law and kinetics we have
$\dfrac{{[{A_o}]}}{{{2^{n1}}}} = \dfrac{{[{B_0}]}}{{{2^{{n^2}}}}}$
Which will give
$\dfrac{4}{{{2^{n1}}}} = \dfrac{1}{{{2^{n2}}}}$ since we have $[{A_0}] = 4[{B_0}]$
So upon simplification we get
$\dfrac{{{2^{n1}}}}{{{2^{n2}}}} = 4$
${2^{n1 - n2}}$ $ = {(2)^2}$
${n_1}{n_2} = 2$
$ \Rightarrow {n_2} = ({n_1} - 2)$
Also, we can say that $t = {n_1}{t_{1/2(A)}};t = {n_2}{t_{1/2(B)}}$
Let the concentration of both become equal after a given time t, so we have
$\dfrac{{{n_1} \times {t_{1/2}}_{(A)}}}{{{n_2} \times {t_{1/2(B)}}}} = 1\dfrac{{{n_1} \times 5}}{{{n_2} \times 15}} = 1$
$ \Rightarrow \dfrac{{{n_1}}}{{{n_2}}} = 3$
So from the equation given above, we get,
$ \Rightarrow {n_1} = 3,{n_2} = 1$
$t = 3 \times 5 = 15\min $

So, the correct answer is Option A.

Note: The rate of the reaction can be represented in terms of the decrease in the concentration of the reactant as $ - \dfrac{{\Delta [A]}}{{\Delta t}} = - \dfrac{{\Delta [B]}}{{\Delta t}}$ Here the negative sign indicates the decrease in the concentration.
The rate can also be expressed in terms of product as $Rate = + \dfrac{{\Delta [Product]}}{{\Delta t}}$
Here $\Delta t$ represents the change in time and $\Delta [A]$ represents the change in concentration of A in $\Delta t$ time.
This approach can be used for the calculation of rate when the time interval is given in place of the rate constant.