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 $0.158$.