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?
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.
In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
A/ mol L - 1 |
0.20 | 0.20 | 0.40 |
B/ mol L - 1 |
0.30 | 0.10 | 0.05 |
r0/ mol L - 1 s - 1 |
5.07 × 10 - 5 |
5.07 × 10 - 5 |
1.43 × 10 - 4 |
What is the order of the reaction with respect to A and B?
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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,....
Comments
Antilog is very simple just split values Which comes then it becomes easy
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
How will we know the antilog value?