Question

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

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

Answer 1

Answer: (D)

For first order reactions, $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{(3.0)}$
$t=\frac{2.303}{K}\log \frac{[A_0]}{[A]}$
Thus, $\log \frac{[A_0]}{A}=\frac{kxt}{2.303}=\frac{0.693}{3}\times \frac{8}{2.303}=0.804$
$\frac{{[A]}_0}{[A]}=\text{Antilog}\text{0}\text{.8024}\text{=}6.345$
${[A]}_0=1M;$
$[A]=\frac{1}{6.345}=0.1576M$