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}}=1\,M; $
$ [A]=\frac{1}{6.345}=0.1576\,M $