Question

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

Answer 1

Given: $t_{1/2}=3$hours, $t=8$hours
As source decomposes according to first order rate law,
$k=\frac{2.303}{t}log\frac{[A{]}_0}{[A{]}_t}$
Using the formula $k=\frac{0.693}{t_{1/2}}$
$\therefore k=\frac{0.693}{t_{1/2}}=\frac{0.693}{3}=0.231h{r}^{-1}$
From formula, $k=\frac{2.303}{t}log\frac{[A{]}_0}{[A{]}_t}$
$\therefore 0.231=\frac{2.303}{8}log\frac{[A{]}_0}{[A{]}_t}$
$\therefore log\frac{[A{]}_0}{[A{]}_t}=0.8024$
Taking antilog on both sides,
$\therefore \frac{[A{]}_0}{[A{]}_t}=$ Antilog $\left(0.8024\right)=6.345$
$\therefore \frac{[A{]}_0}{[A{]}_t}=\frac{1}{6.345}=0.158$.
Hence, the fraction of the sample of sucrose that remains after $8$ hours is $0.158$.