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Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:
- a)2.66 L min–1 at STP
- b)1.34 × 10–2 mol min–1
- c)6.96 × 10–2 mol min–1
- d)6.93 × 10–4 mol min–1
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Decomposition of H2O2 follows a first order reaction. In fifty minute...
For a first order reaction
Given a = 0.5 , (a - x) = 0.125, t = 50 min
∴ 
= 2.78 x 10-2min-1
r = k[H2O2] = 2.78 x 10-2 x 0.05
= 1.386 x 10-3 mol min-1
Now,
= 
= 6.93 x 10-4 mol min-1
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Decomposition of H2O2 follows a first order reaction. In fifty minute...
Decomposition of H2O2
The decomposition of hydrogen peroxide (H2O2) follows a first-order reaction. This means that the rate of decomposition is directly proportional to the concentration of H2O2.
Given Information
In fifty minutes, the concentration of H2O2 decreases from 0.5 to 0.125 M.
Calculating the Rate Constant
Since the reaction follows a first-order kinetics, we can use the integrated rate equation:
ln([A]t/[A]0) = -kt
Where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
Using the given information, we can substitute the values into the equation:
ln(0.125/0.5) = -k(50)
Simplifying the equation gives:
-1.09861 = -50k
Solving for k:
k = 0.0219722 min^-1
Calculating the Rate of Formation of O2
The rate of formation of O2 is equal to half the rate of decomposition of H2O2, as the balanced chemical equation for the decomposition of H2O2 is:
2H2O2 -> 2H2O + O2
At the point when the concentration of H2O2 reaches 0.05 M, we can calculate the rate of formation of O2 using the rate constant (k) we obtained earlier.
Using the first-order rate equation:
k = ln([A]0/[A]t)
Substituting the values:
0.0219722 = ln(0.5/0.05)
Simplifying the equation gives:
0.0219722 = ln(10)
Taking the anti-logarithm of both sides:
10^(0.0219722) = 1.051
Multiplying this value by 2 (as the rate of formation of O2 is half the rate of decomposition of H2O2) gives:
2 * 1.051 = 2.102
Therefore, the rate of formation of O2 when the concentration of H2O2 reaches 0.05 M is approximately 2.102 L min^-1 at STP. This corresponds to option 'D': 6.93 × 10^(-4) mol min^(-1).
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Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:a)2.66 L min–1 at STPb)1.34 × 10–2 mol min–1c)6.96 × 10–2 mol min–1 d)6.93 × 10–4 mol min–1Correct answer is option 'D'. Can you explain this answer?
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Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:a)2.66 L min–1 at STPb)1.34 × 10–2 mol min–1c)6.96 × 10–2 mol min–1 d)6.93 × 10–4 mol min–1Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:a)2.66 L min–1 at STPb)1.34 × 10–2 mol min–1c)6.96 × 10–2 mol min–1 d)6.93 × 10–4 mol min–1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:a)2.66 L min–1 at STPb)1.34 × 10–2 mol min–1c)6.96 × 10–2 mol min–1 d)6.93 × 10–4 mol min–1Correct answer is option 'D'. Can you explain this answer?.
Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:a)2.66 L min–1 at STPb)1.34 × 10–2 mol min–1c)6.96 × 10–2 mol min–1 d)6.93 × 10–4 mol min–1Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:a)2.66 L min–1 at STPb)1.34 × 10–2 mol min–1c)6.96 × 10–2 mol min–1 d)6.93 × 10–4 mol min–1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:a)2.66 L min–1 at STPb)1.34 × 10–2 mol min–1c)6.96 × 10–2 mol min–1 d)6.93 × 10–4 mol min–1Correct answer is option 'D'. Can you explain this answer?.
Solutions for Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:a)2.66 L min–1 at STPb)1.34 × 10–2 mol min–1c)6.96 × 10–2 mol min–1 d)6.93 × 10–4 mol min–1Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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