#### Question

Sucrose decomposes in acid solution to give glucose and fructose according to the first order rate law. The half life of the reaction is 3 hours. Calculate fraction of sucrose which will remain after 8 hours.

#### Solution

For a first order reaction

`k = 2.303/t log ([R]_0/[[R]])`

it is given that t_{1/2} = 3 hours

therefore,

`k= 0.693/t_(1/2)`

k = 0.693/3 h^{-1}

k = 0.231 h^{-1}

`therefore 0.231=2.303/(8h) log([R]_0/[[R]])`

`therefore log([R]_0/[[R]]) = (0.231h^-1 times 8h)/2.303`

`therefore [R]_0/[[R]] = antilog(0.8024)`

`therefore [R]_0/[[R]] = 6.3445`

`therefore [[R]]/[R]_0 ~~ 0.1576`

`therefore [[R]]/[R]_0 = 0.158`

Hence the fraction of the sample of sucrose that remains after 8 hours is 0.158.

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Solution Sucrose decomposes in acid solution to give glucose and fructose according to the first order rate law. The half life of the reaction is 3 hours. Calculate fraction of sucrose which will remain after 8 hours Concept: Integrated Rate Equations - Half-life of a Reaction.