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Decomposition of H2O2 was studied by titration against KMnO4 solution. It was found that 0.4 mol of H2O2 was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr, the order of reaction must be
- a)3
- b)1
- c)0
- d)2
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Decomposition of H2O2was studied by titration against KMnO4solution. I...
Decomposition of nH2O2 is 1st order.
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Decomposition of H2O2was studied by titration against KMnO4solution. I...
Solution:
The given reaction is the decomposition of H2O2, which can be represented as follows:
2H2O2 → 2H2O + O2
The reaction is being studied by titration against KMnO4 solution. KMnO4 is used as an oxidizing agent here, which oxidizes H2O2 to H2O and O2. The reaction between H2O2 and KMnO4 can be represented as follows:
5H2O2 + 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 8H2O + 5O2
In this reaction, KMnO4 is reduced to MnSO4, and H2O2 is oxidized to O2.
Given:
0.4 mol of H2O2 is reduced to 0.2 mol in 20 min
0.4 mol of H2O2 is reduced to 0.1 mol in 40 min
0.4 mol of H2O2 is reduced to 0.05 mol in 60 min
To determine the order of the reaction, we need to find the rate constant (k) of the reaction.
Rate of reaction = k[H2O2]n
where [H2O2] is the concentration of H2O2 and n is the order of the reaction.
Let's calculate the rate constant for the given data:
For the first reaction:
Rate of reaction = (0.4 - 0.2) / (20 × 60) = 0.0001667 mol L-1 min-1
[H2O2] = 0.2 mol/L
k = rate of reaction / [H2O2]n
0.0001667 = k(0.2)n
For the second reaction:
Rate of reaction = (0.4 - 0.1) / (40 × 60) = 0.0000833 mol L-1 min-1
[H2O2] = 0.1 mol/L
k = rate of reaction / [H2O2]n
0.0000833 = k(0.1)n
For the third reaction:
Rate of reaction = (0.4 - 0.05) / (60 × 60) = 0.0000556 mol L-1 min-1
[H2O2] = 0.05 mol/L
k = rate of reaction / [H2O2]n
0.0000556 = k(0.05)n
Dividing equation 2 by equation 1, we get:
0.0001667 / 0.0000833 = (0.2)n / (0.1)n
2 = 2n
n = 1
Dividing equation 3 by equation 2, we get:
0.0000833 / 0.0000556 = (0.1)n / (0.05)n
1.5 = 2n
n = 0.585
Since n is not a whole number, we can conclude that the reaction is a first-order reaction. Therefore, the correct answer is option B.
The given reaction is the decomposition of H2O2, which can be represented as follows:
2H2O2 → 2H2O + O2
The reaction is being studied by titration against KMnO4 solution. KMnO4 is used as an oxidizing agent here, which oxidizes H2O2 to H2O and O2. The reaction between H2O2 and KMnO4 can be represented as follows:
5H2O2 + 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 8H2O + 5O2
In this reaction, KMnO4 is reduced to MnSO4, and H2O2 is oxidized to O2.
Given:
0.4 mol of H2O2 is reduced to 0.2 mol in 20 min
0.4 mol of H2O2 is reduced to 0.1 mol in 40 min
0.4 mol of H2O2 is reduced to 0.05 mol in 60 min
To determine the order of the reaction, we need to find the rate constant (k) of the reaction.
Rate of reaction = k[H2O2]n
where [H2O2] is the concentration of H2O2 and n is the order of the reaction.
Let's calculate the rate constant for the given data:
For the first reaction:
Rate of reaction = (0.4 - 0.2) / (20 × 60) = 0.0001667 mol L-1 min-1
[H2O2] = 0.2 mol/L
k = rate of reaction / [H2O2]n
0.0001667 = k(0.2)n
For the second reaction:
Rate of reaction = (0.4 - 0.1) / (40 × 60) = 0.0000833 mol L-1 min-1
[H2O2] = 0.1 mol/L
k = rate of reaction / [H2O2]n
0.0000833 = k(0.1)n
For the third reaction:
Rate of reaction = (0.4 - 0.05) / (60 × 60) = 0.0000556 mol L-1 min-1
[H2O2] = 0.05 mol/L
k = rate of reaction / [H2O2]n
0.0000556 = k(0.05)n
Dividing equation 2 by equation 1, we get:
0.0001667 / 0.0000833 = (0.2)n / (0.1)n
2 = 2n
n = 1
Dividing equation 3 by equation 2, we get:
0.0000833 / 0.0000556 = (0.1)n / (0.05)n
1.5 = 2n
n = 0.585
Since n is not a whole number, we can conclude that the reaction is a first-order reaction. Therefore, the correct answer is option B.
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Decomposition of H2O2was studied by titration against KMnO4solution. I...
Correct option is 1
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Decomposition of H2O2was studied by titration against KMnO4solution. It was found that 0.4 mol of H2O2was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr,the order of reaction must bea)3b)1c)0d)2Correct answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Decomposition of H2O2was studied by titration against KMnO4solution. It was found that 0.4 mol of H2O2was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr,the order of reaction must bea)3b)1c)0d)2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Decomposition of H2O2was studied by titration against KMnO4solution. It was found that 0.4 mol of H2O2was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr,the order of reaction must bea)3b)1c)0d)2Correct answer is option 'B'. Can you explain this answer?.
Decomposition of H2O2was studied by titration against KMnO4solution. It was found that 0.4 mol of H2O2was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr,the order of reaction must bea)3b)1c)0d)2Correct answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Decomposition of H2O2was studied by titration against KMnO4solution. It was found that 0.4 mol of H2O2was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr,the order of reaction must bea)3b)1c)0d)2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Decomposition of H2O2was studied by titration against KMnO4solution. It was found that 0.4 mol of H2O2was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr,the order of reaction must bea)3b)1c)0d)2Correct answer is option 'B'. Can you explain this answer?.
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Decomposition of H2O2was studied by titration against KMnO4solution. It was found that 0.4 mol of H2O2was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr,the order of reaction must bea)3b)1c)0d)2Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Decomposition of H2O2was studied by titration against KMnO4solution. It was found that 0.4 mol of H2O2was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr,the order of reaction must bea)3b)1c)0d)2Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Decomposition of H2O2was studied by titration against KMnO4solution. It was found that 0.4 mol of H2O2was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr,the order of reaction must bea)3b)1c)0d)2Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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