Question

Sucrose decomposes in acid solution into glucose and fructose according to first order rate law with $t_{1/2}=3Hrs$. What fraction of the sample of sucrose remains after $8$ hours.

Answer 1

Half-life of a reaction is the time required for the reaction to be half completed.

Amount of substance left unreacted after one half life= $\frac{[A{]}_0}{2}$; $[A{]}_0$= initial amount of substance at $t=0$

Amount of substance left after $2$ half lives= $\frac{[A{]}_0}{2^2}$

In general,

amount of substance left after $n$ half lives= $\frac{[A{]}_0}{2^n}$

Now, given half life= $t_{1/2}=3$ hrs

Time after which amount left is asked= $8$ hours

No. of half lives=$\frac{8}{3}$

Amount of sucrose left= $\frac{[A{]}_0}{2^{8/3}}$

Fraction of sucrose left= $\frac{1}{2^{2/3}}=\frac{1}{6.35}=0.16$