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Q.

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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a

2.518 M

b

3.182 M

c

0.158 M

d

1.158 M

answer is C.

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Detailed Solution

For a first order reaction,
$k = 2.303 / tlog [R]_{0} / [R]$
It is given that, $t_{1 / 2} = 3.00$ hours
Therefore, $k = 0.693 / t_{1 / 2}$
$\begin{aligned} = 0.693 / 3 h^{- 1} \\ = 0.231 h^{- 1} \end{aligned}$
Then, $0.231 h^{- 1} = 2.303 / 8 hlog [R]_{0} / [R]$

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