• # question_answer Two substances A and B are present such that$[{{A}_{0}}]=4[{{B}_{0}}]$ and half-life of A is 5 minute and that of B is 15 min. If they start decaying at the same time following 1st order kinetics, how much time later will the concentration of both of them would be same? A) 15 min     B) 10 min C) 5 min                D) 12 min

 Amount of A left in ${{n}_{1}}$ halves $={{\left( \frac{1}{2} \right)}^{{{n}_{1}}}}[{{A}_{0}}]$ Amount of B in ${{n}_{2}}$ halves $={{\left( \frac{1}{2} \right)}^{{{n}_{2}}}}[{{A}_{0}}]$ At the end halves, $\frac{[{{A}_{0}}]}{{{2}^{ni}}}=\frac{[{{B}_{0}}]}{{{2}^{{{n}_{2}}}}}$ $\Rightarrow$   $\frac{4}{{{2}^{{{n}_{1}}}}}=\frac{1}{{{2}^{{{n}_{2}}}}}$ As ${{A}_{0}}=4{{B}_{0}}$ $\therefore$      $\frac{{{2}^{{{n}_{1}}}}}{{{2}^{_{2}}}}=4$ $\Rightarrow$   ${{2}^{{{n}_{1}}-{{n}_{2}}}}={{(2)}^{2}}$ $\therefore$      ${{n}_{1}}-{{n}_{2}}=2$ so ${{n}_{2}}={{n}_{1}}-2$         (i) Also,     $t={{n}_{1}}\times {{t}_{1/2}}(A)$ $t={{n}_{1}}\times {{t}_{1/2}}(A)$ $t={{n}_{2}}\times {{t}_{1/2}}(B)$
 Let conc. of both become equal after time (t) $\therefore$      $\frac{{{n}_{1}}\times {{t}_{1/2}}(A)}{{{n}_{2}}\times {{t}_{1/2}}(B)}=1$ $\Rightarrow$   $\frac{{{n}_{1}}\times 5}{{{n}_{2}}\times 5}=1$ $\frac{{{n}_{1}}}{{{n}_{2}}}=3$                                           (ii) From equation (i) and (ii), ${{n}_{1}}=3$ and ${{n}_{2}}=1$ $t=3\times 5=15\,\min$