A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as :
`k_(1) =1.26xx10^(-4)s^(-1), k_(2) =3.8xx10^(-5) s^(-1)`
The percentage distribution of B and C are :

Options

A. 75% B and 25% C
B. 80% B and 20% C
C. 60% B and 40% C
D. 76.83% B and 23.17% C

Answer

D

Explanation

`% B =(k_(1))/(k_(1)+k_(2))xx100=(12.6xx10^(-5))/((12.6+3.8)10^(-5))xx100=76.83%` Therefore `% C =100-76.83=23.17%`