Question

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with $t_{\frac{1}{2}}=3.00h$. What fraction of sample of sucrose remains after 8h?

Answer 1

For 1st order reactions
$k=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{\left(3.0h\right)}t=\frac{2.303}{k}log\frac{[A{]}_0}{\left[A\right]}$
or $log\frac{[A{]}_0}{\left[A\right]}=\frac{k\times t}{2.303}$
$log\frac{[A{]}_0}{\left[A\right]}=\frac{0.693}{3h}\times \frac{\left(8h\right)}{2.303}=0.8024\frac{[A{]}_0}{\left[A\right]}=Antilog0.8024=6.345[A{]}_0=1M;[A]=\frac{[A{]}_0}{6.345}=\frac{1M}{6.345}=0.1576M$
After 8 h sucrose left = 0.1576M