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.
For a first order reaction
`k = 2.303/t log ([R]_0/[[R]])`
it is given that t1/2 = 3 hours
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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