JEE Main202124 Feb 2021Evening ShiftChemistryChemical KineticsActual
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 1 f is ________ × 10 - 2 . (Rounded off to the nearest integer) [Assume: In 10 = 2 . 303 , ln 2 = 0 . 693 ]
Correct answer
0
Step-by-step solution
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