A follow parallel path of first-order reactions giving `B` and `C` as
If the initial concentration of `A` is `0.25 M`, calculate the concentration of `C` after 5 hr of reaction.

Explanation

`k_(av)=k_(1)+k_(2)==(1.5xx10^(-5)+5xx10^(-6))s^(-1)=(15xx10^(-6)+5xx10^(-6))s^(-1)=20xx10^(-6)s^(-1)`
Also, `2.303log""([A]_(0))/([A]_(t))=kt" or "log""([A]_(0))/([A]_(t))=(kt)/(2.303)`
`:." "log""(0.25)/([A]_(t))=(20xx10^(-6)xx5xx60xx60)/(2.303)" or "(0.25)/([A]_(t))=1.433:.[A]_(t)=(0.25)/(1.433)=0.1744" M"`
`:.[A]" decomposed"=[A]_(0)-[A]_(t)=0.25-0.1744=0.0756" M"`
Concentration of C formed `=(k_(2))/(k_(1)+k_(2))xx[A_(0)]_("decomposed")xx(2)/(5)(because5" moles of A give 2 moles of C")`
`=(5xx10^(-6))/(20xx10^(-6))xx0.0756xx(2)/(5)=7.56xx10^(-3)"M"`