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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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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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First order reaction,

k = $$\frac{2.303}{t}$$log $$\frac{[R]_0}{[R]}$$

It is given that, t1/2 = 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.

This post was modified 5 months ago by admin
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