Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with `t_(1//2)=3.00hr.` What fraction of sample of sucrose remains after `8 hr` ?

Sucrose decomposes in acid solution into glucose and fructose accordin

1.	Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with `t_(1//2)=3.00hr.` What fraction of sample of sucrose remains after `8 hr` ?
Answer» For a first order reaction , `k = (2.303)/(t) "log" ([R]_(0))/([R])` It is given , `t_(1//2) = 3.00` hours `k = (0.693)/(t_((1)/(2)))` Therefore , = `(0.693)/(3) h^(-1)` = `0.231 h^(-1)` Then , `0.231 h^(-1) = (2.303)/(8h) "log" ([R]_(0))/([R])` `implies "log" ([R]_(0))/([R]) = (0.231 h^(-1) xx 8 h)/(2.303)` `implies ([R]_(0))/([R])`= antilog (`0.8024`) `implies ([R]_(0))/([R]) = 6.3445` `implies ([R])/([R]_(0)) = 0.1576` (approx) =`0.158` Hence , the fraction of sample of sucrose that remains after 8 hours is `0.158`

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Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with `t_(1//2)=3.00hr.` What fraction of sample of sucrose remains after `8 hr` ?

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