Sucrose decomposes in acid solution into glucose and fuctose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Solution

We have given that

t_{1 / 2} = 3.00 hours . For the first order reaction, t_{1 / 2} = \frac{0.693}{K} or K = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{3} = 0.231 {hr}^{- 1} Now K = \frac{2.303}{t} \log \frac{a}{a - x} 0.231 = \frac{2.303}{8} \log \frac{a}{a - x} \frac{0.231 \times 8}{2.303} = \log \frac{a}{a - x} 0.8024 = \log \frac{a}{a - x} or \frac{a}{a - x} = Antilog 0.8024 \frac{a}{a - x} = 6.345 or fraction left = \frac{a - x}{a} = \frac{1}{6.345} = 0.157 M .

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