The half-life period of a `1^(st)` order reaction is 60 minutes. What percentage will be left over after 240 minutes ?

Options

A. 0.0625
B. 0.0425
C. 0.05
D. 0.06

Answer

a

Explanation

`:t_(1//2)=(0.693)/(K)Rightarrow(0.693)/(t_(1//2))=Krightarrow(0.693)/(60)=K`
`K=0.01155min^(-1)`
`K= (2.303)/(t)log ((a)/(a-x))`
Let the initial amount (a) 100
`0.01155min^(-1)= (2.303)/(240min)log((100)/(a-x))`
`(0.01155min^(-1)xx40min)/(2.303)=log ((100)/(a-x))`
`1.204=log100-(a-x)`
`1.204=2-log(a-x)`
`log(a-x)=2-1.204`
log(a-x)=0.796
(a-x)=6.25%