The half-life for a first order reaction is `5xx10^(6)` s. What percentage of the initial reactant will react in 2 hours ?

Explanation

Given `t_(1//2)=5 xx10^(-6)sec.`
`k=(0.693)/(1_(1//2))=(0.693)/(5xx10^(-6))=0.1386xx10^(6) sec ^(-1)`
` k=0.1386xx10^(6) sec ^(-1) =1.386xx10^(5) sec ^(-1)`
Here ` a=100, t =2 hrs. =2 xx 60xx60 sec, (a-x)?`
`k=(2.303)/(t )log ""(a)/(a-x)`
`1.386xx10^(5) =(2.303)/(2xx60xx60)log ""(a)/(a-x)`
`log ""(a)/(a-x)=(1.386xx10^(5) xx2xx60xx60)/(2.303)`
`=(9979.2xx10^(5))/(2.303)=4333.1xx10^(5)`
`log ""(a)/(a-x)=9.2712`
`a-x =(a)/(9.2712)=(100)/(9.2712)=10.786`
`100-x =10.786`
`x=100-10.786=89.214%`