Question

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

• A. $1.023M$
• B. $0.8725M$
• C. $0.023M$
• D. $0.1576M$

Answer 1

The correct option is D $0.1576M$

For first order reactions,

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{\left(3.0\right)}$

$t=\frac{2.303}{k}log\frac{\left[A_0\right]}{\left[A\right]}$

Thus, $log\frac{\left[A_0\right]}{\left[A\right]}=\frac{k\times t}{2.303}=\frac{0.693}{3}\times \frac{8}{2.303}=0.8024$

$\frac{[A{]}_0}{\left[A\right]}=$ Antilog $0.8024=6.345$

$[A{]}_0=1$M;

$\left[A\right]=\frac{1}{6.345}=0.1576M$.