Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law,
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
For a first order reaction,
$k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$
It is given that, t1/2 = 3.00 hours
Therefore, $k=\frac{0.693}{t_{1 / 2}}$
$=\frac{0.693}{3} \mathrm{~h}^{-1}$
= 0.231 h−1
Then, $0.231 \mathrm{~h}^{-1}=\frac{2.303}{8 \mathrm{~h}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$
$\Rightarrow \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}=\frac{0.231 \mathrm{~h}^{-1} \times 8 \mathrm{~h}}{2.303}$
$\Rightarrow \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}=\operatorname{antilog}(0.8024)$
$\Rightarrow \frac{[R]_{0}}{[R]}=6.3445$
$\Rightarrow \frac{[\mathrm{R}]}{[\mathrm{R}]_{0}}=0.1576$ (approx)
$=0.158$
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.
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