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 9 h , the fraction of sucrose remaining is f . The value of

Question

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 $9 h,$ the fraction of sucrose remaining is $f .$ The value of $\log_{10} (\frac{1}{f})$ is ________ $\times 10^{- 2} .$ (Rounded off to the nearest integer)

[Assume: In $10 = 2.303, \ln 2 = 0.693$ ]

TestHub · Accepted Answer

C 12 H 22 O 11 + H 2 O → t 1 / 2 = 10 3 hr order C 6 H 12 O 6 Glucose + C 6 H 12 O 6 Fructose t = 0 a = A 0 t = 9 hr a - x = A t from I order kinetic: k × t 2 . 303 = log A 0 A t ⇒ ln 2 × 9 10 3 × 2 . 303 = log 1 f ⇒ 0 . 693 × 9 × 3 23 . 03 = log 1 f ⇒ log 1 f = 0 . 81246 = 81 . 24 × 10 - 2 ⇒ x = 81

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