Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?


t1/2 = 3.00 hours


We know, t1/2 = 0.693/k


k = 0.693/3


k = 0.231hrs-1


We know, time


Where, k- rate constant


[R]° -Initial concentration


[R]-Concentration at time ‘t’


Thus, substituting the values,


log([R]0/[R]) = 0.8


log([R]/[R]0) = -0.8


[R]/[R]0 = 0.158


Hence, 0.158 fraction of sucrose remains.


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