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

Answer

0.158

Explanation

Decomposition of sucrose is first order reaction,
`k=2.303/t log [A_(0)]/[A]`……..(i)
For time `t_(1//2) = 3.0` hr
`k=0.693/t_(1//2) = 0.693/(3 hr) = 2.231 hr^(-1)`
Substituting `k=0.231 hr^(-1)` in (i) we get
`2.231 = (2.303)/8 log ([A_(0)])/([A])`
`[A_(0)]/[A]` = antilog 0.8024 = 6.345
`therefore` Fraction of sample remains
`=[A]/[A_(0)] = 1/(6.345) = 0.158`