Two substances A(t(1/2) = 5 m i n) and B(t(1/2) = 15 m i n) follow first order kinetics, are taken in such a way that initially [A]=4[B] . The time after which the concentration of both substances will be equal is:

Question

Two substances A(t(1/2) = 5 m i n) and B(t(1/2) = 15 m i n) follow first order kinetics, are taken in such a way that initially [A]=4[B] . The time after which the concentration of both substances will be equal is: (A) 5 min.  (B) 10 min. (C) 15 min. (D) 20 min.

Tardigrade Admin · Accepted Answer

Q. Two substances $A (t_{\frac{1}{2}} = 5 min)$ and $B (t_{\frac{1}{2}} = 15 min)$ follow first order kinetics, are taken in such a way that initially $[A] = 4 [B]$ . The time after which the concentration of both substances will be equal is:

$5 min .$

$10 min .$

$15 min .$

$20 min .$

Q. Two substances A(t21​​=5min) and B(t21​​=15min) follow first order kinetics, are taken in such a way that initially [A]=4[B] . The time after which the concentration of both substances will be equal is:

5min.

10min.

15min.

20min.

Q. Two substances $A (t_{\frac{1}{2}} = 5 min)$ and $B (t_{\frac{1}{2}} = 15 min)$ follow first order kinetics, are taken in such a way that initially $[A] = 4 [B]$ . The time after which the concentration of both substances will be equal is:

$5 min .$

$10 min .$

$15 min .$

$20 min .$