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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?
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30/01/2022 5:17 pm
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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?
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30/01/2022 5:24 pm
First order reaction,
k = \(\frac{2.303}{t}\)log \(\frac{[R]_0}{[R]}\)
It is given that, t_{1/2} = 3.00 hours
k = \(\frac{0.693}{t_{1/2}}\)
Therefore, \(\frac{0.693}{3}h^{1}\)
= 0.231 h^{1}
Then, 0.231 h^{1}
= \(\frac{2.303}{8 \ h}\)log \(\frac{[R]_0}{[R]}\)
⇒ log \(\frac{[R]_0}{[R]}\) = \(\frac{0.231 h^{1} \times 8 h}{2.303}\)
⇒ \(\frac{[R]_0}{[R]}\) = antilog (0.8024)
⇒ \(\frac{[R]_0}{[R]}\) = 6.3445
⇒ \(\frac{[R]_0}{[R]}\) = 0.1576 (approx)
= 0.158
Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.
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