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` ?
Text Solution
Verified by Experts
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`
Topper's Solved these Questions
CHEMICAL KINETICS
NCERT|Exercise Exercise|1 Videos
CHEMICAL KINETICS
NCERT|Exercise SOLVED EXAMPLE|1 Videos
BIOMOLECULES
NCERT|Exercise Exercise|33 Videos
CHEMISTRY IN EVERYDAY LIFE
NCERT|Exercise Exercise|32 Videos
Similar Questions
Explore conceptually related problems
Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25^(@)C . After 9h , the fraction of sucrose remaining if f. The value of log_(10)((1)/(f)) is _________ xx 10^(-2) . (Rounded off to the nearest integer ) [Assume , ln 10 = 2.303, ln 2= 0.693 ]
Mean life of a radioactive sample is t_0 . What fraction of sample remains left after time t_0ln_2 ?
A first order reaction has half life of 14.5 hrs. What percentage of the reactant will remain after 24 hrs ?
A substance having a half-life period of 30 min decomposes according to firt order rate law. a) What fraction of this will be decomposed and what will remian behind after 1.5 hr? b) How long will it take to be 60% decomposed, If the molar concentration is just doubled?
