Question

Two substances A and B are present such that $\left[A_0\right]=4\left[B_0\right]$ 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 take if the concentration of both of them would be same?

• A. 15 min
• B. 10 min
• C. 5 min
• D. 12 min

Answer 1

Answer: (A)

15 min

Amount of A left in $n_1$ halves = ${\left(\frac{1}{2}\right)}^{n_1}\left[A_0\right]$
Amount of A left in $n_2$ halves = ${\left(\frac{1}{2}\right)}^{n_2}\left[B_0\right]$
At the end, according to the question
$\frac{\left[A_0\right]}{2^{n^1}}=\frac{\left[B_0\right]}{2^{n^2}}\Rightarrow \frac{4}{2^{n_1}}=\frac{1}{2^{n_2}},\left(\right[A_0]=4[B_0\left]\right)$
$\frac{2^{n_1}}{2^{n_2}}=4\Rightarrow 2^{n_1-n_2}=(2{)}^2\Rightarrow n_1{n}_2=2$
$\therefore n_2=(n_1-2)$
Also t = $n_1{t}_{1/2\left(A\right)};t=n_2{t}_{1/2\left(B\right)}$
(Let concentration of both become equal after time t)
$\therefore \frac{n_1\times t_{1/2\left(A\right)}}{n_2\times t_{1/2\left(B\right)}}=1\Rightarrow \frac{n_1\times 5}{n_2\times 15}=1\Rightarrow \frac{n_1}{n_2}=3$ ...(II)
From equations (i) and (ii), we get
$n_1=3,n_2=1$
t = 3 $\times$ 5 = 15 min