Sucrose decomposes in acid solution into glucose and fructose according
to the first order rate law, with `t_(1//2) = 3.00` hours. What fraction of sample
of sucrose remains after 8 hours

Answer

`0.158M`

Explanation

Decomposition of sucrose is of first order reaction
`:.k=(2.303)/(t)log``(a)/(a-x)=(2.303)/(t)log``([A]_(0))/([A])`
Fraction of sucrose remaining after `8hr =([A])/([A])`
`:.k=(2.303)/(t_(1//2))=(0.693)/(3hr)=0.231hr^(-1)`
Hence, `0.231hr^(-1)=(2.303)/(8hr)lot``([A]_(0))/([A])`
`or log``([A]_(0))/([A])=0.8024`
`or ([A]_(0))/([A])=Antilog (0.8024)=6.345`
`or ([A])/([A]_(0))=(1)/(6.345)=0.158M`
Second method
Let `x%` of the sucrose decomposes. Thuen use the relation `:`
`(t_(1//2))/(t_(x%))=(0.3)/(log((100)/(100-x)))`
`:.log((100)/(100-x))=(t(x%)xx0.3)/(t_(1//2))=(8hrxx0.3)/(3hr)=0.8`
`:. (100)/(100-x)=(a)/(a-x)=Antilog(0.8)=6.345`