Sucrose decomposes in acid solution into | Class 12 Chemistry Chapter Chemical Kinetics, Chemical Kinetics NCERT Solutions

Question:

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?

Answer:

For a first order reaction,

k = 2.303/t  Log  [R]º / [R]

It is given that, t1/2 = 3.00 hours

Therefore, k = 0.693 / t1/2

= 0.693 / 3  h-1

= 0.231 h - 1

Then, 0.231 h - 1 = 2.303 / 8h  Log  [R]º / [R]

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.


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Comments

  • Rohit
  • 2019-07-24 06:08:31

Antilog is very simple just split values Which comes then it becomes easy


  • Sanyam khattar
  • 2018-08-21 00:43:56

U will have to learn log table from 1 to 10 as log2=0.3010 log 3=0.4771 and so on so as the log of 2 is 0.3010 so antilog for 0.3010 will be 2 do this with all the no. Upto 10


  • Yashashwini
  • 2018-05-31 15:58:48

How will we know the antilog value?


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 25: Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law,....