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

M manish

For first order reaction,

$k = \frac{2.303}{t}\log\frac{[R]_{0}}{[R]}$

given that half life = 3 hrs ($t_{1/2}$)

Therefore k = 0.693/half-life
= 0.231 per hour

Now,

$\\0.231 = \frac{2.303}{8}\log\frac{[R]_{0}}{[R]}\\ \log\frac{[R]_{0}}{[R]} = 0.231\times \frac{8}{2.203}$
= antilog (0.8024)
= 6.3445

$[R]_{0}/[R] = 6.3445$

$[R]/[R]_{0} = 0.157$(approx)

Therefore fraction of sample of sucrose remains after 8 hrs is 0.157

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