JEE MainChemistryChemical Kinetics
Two different first order reactions, A products and B products , have half-lives in the ratio t_ 1/2, A : t_ 1/2, B = 2:1 . If t_A is the time taken for the concentration of A to reduce to 1 16 of its initial value, and t_B is the time taken for the concentration of B to reduce to 1 8 of its initial value, then the ratio t_A t_B will be :
Options
- A4 3
- B2 3
- C3 8
- D8 3
Correct answer
D. 8 3
Step-by-step solution
For a first order reaction, the time required to reduce the concentration to ( 1 2 )^n of its initial value is n t_ 1/2 . For reaction A , the concentration reduces to 1 16 = ( 1 2 )^4 . Therefore, t_A = 4 t_ 1/2, A . For reaction B , the concentration reduces to 1 8 = ( 1 2 )^3 . Therefore, t_B = 3 t_ 1/2, B . The ratio of the times is: t_A t_B = 4 t_ 1/2, A 3 t_ 1/2, B Given that t_ 1/2, A t_ 1/2, B = 2 1 , we substitute this into the equation: t_A t_B = 4 3 2 = 8 3 Answer: 8 3