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

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 ?

Answers (1)

best_answer

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

Posted by

manish

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