Question

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with \(t_{\frac 12} = 3.00\  hours\). What fraction of sample of sucrose remains after \(8 \ hours\)?

Answer 1

For a first order reaction,

\(k = \frac {2.303}{t} log \ \frac {[R]_o}{[R]}\)

It is given that, \(t_{\frac 12} = 3.00\  hours\)

\(k = \frac {0.693}{t_{1/2}}\)

Therefore, \(k = \frac {0.693}{3} h^{-1}\)

\(k = 0.231\  h^{-1}\)

Then, \(0.231 \ h^{-1} = \frac {2.303}{t} log \ \frac {[R]_o}{[R]}\)

⇒ \(log\ \frac { [R]_o}{[R]} =\frac { 0.231 h^{-1} \times 8 h}{2.303}\)

⇒ \(\frac {[R]_o}{[R]}\) = \(antilog \ (0.8024)\)

⇒ \(\frac {[R]_o}{[R]}\) = \(6.3445\)

⇒ \(\frac {[R]_o}{[R]}\) = \(0.1576  \ (approx)\)

⇒ \(\frac {[R]_o}{[R]}\) = \(0.158\)

Hence, the fraction of sample of sucrose that remains after 8 hours is \(0.158\).