Two substances A and B are present such that [A_(0)]=4[B_(0)] and half-life of A is 5 minutes and that of B is 15 minutes . If they start decaying at the same time following first order kinetics how much time later will the concentration of both of them would be same ?

Two substances A and B are present such that [A_(0)]=4[B_(0)] and half

1.	Two substances A and B are present such that [A_(0)]=4[B_(0)] and half-life of A is 5 minutes and that of B is 15 minutes . If they start decaying at the same time following first order kinetics how much time later will the concentration of both of them would be same ?
Answer» `15` minutes `10` minutes `5` minutes `12`minutes Solution :AMOUNT of A left in `n_(1)` half-lives `=((1)/(2))^(n_(1))[A_(0)]` Amount of B left `n_(2)` half-lives `=((1)/(2))^(n_(2))[B_(0)]` At the END, `([A_(0)])/(2^(n_(1)))=([B_(0)])/(2^(n_(2)))` `implies(4)/(2^(n_(1)))=(1)/(2^(n_(2))){[A_(0)]=4[B_(0)]}` `:.(2^(n_(1)))/(2^(n_(2)))=4implies2^(n_(1)-n_(2))=(2)^(2)` `:.n_(1)-n_(2)=2` `:.n_(2)=(n_(1)-2)` Also, `t=n_(1)xxt_(1//2(B))t=n_(2)xxt_(1//2)(B))` (LET, concentration of both become equal after time `t`) `:.(n_(1)xxt_(1//2(A)))/(n_(2)xxt_(1//2(B)))=1implies(n_(1)xx5)/(n_(2)xx15)=1implies(n_(1))/(n_(2))=3` From Eqs. `(1)` and `(2)`, we get `n_(1)=3`, `n_(2)=1` `t=3xx5=15` minutes