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The decomposition of NO2 at 400 K proceeds at a of rate of 5.4 x 10 -5 mol L-1 s-1 when [NO2] = 0.01 mol-1
2 NO2(g) → 2NO(g ) + O2(g).
Q. Rate constant of the reaction will be 
  • a)
    0.54 L mol-1 s-1
  • b)
    0.54 x 10-3 L mol-1 s-1
  • c)
    5.4 s-1
  • d)
    0.54s-1
Correct answer is option 'A'. Can you explain this answer?
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The decomposition of NO2 at 400 K proceeds at a of rate of 5.4 x 10 -5...

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The decomposition of NO2 at 400 K proceeds at a of rate of 5.4 x 10 -5...
Understanding the Reaction Rate
The given reaction is:
2 NO2(g) → 2 NO(g) + O2(g)
The rate of the reaction is expressed as:
Rate = - (1/2) d[NO2]/dt
Given:
- Rate = 5.4 x 10^-5 mol L^-1 s^-1
- [NO2] = 0.01 mol L^-1
Rate Law Expression
For this reaction, the rate law can be expressed as:
Rate = k [NO2]^n
Where:
- k = rate constant
- n = order of the reaction
Assuming the reaction is first-order with respect to NO2 (n = 1):
Calculating the Rate Constant (k)
1. Substitute the given values into the rate law:
- Rate = k [NO2]
- 5.4 x 10^-5 mol L^-1 s^-1 = k (0.01 mol L^-1)
2. Rearranging the equation to solve for k:
- k = (5.4 x 10^-5 mol L^-1 s^-1) / (0.01 mol L^-1)
- k = 5.4 x 10^-3 s^-1
3. Converting to appropriate units:
- Since k = 5.4 x 10^-3 s^-1 can also be expressed as:
- k = 0.54 x 10^-2 s^-1 = 0.54 L mol^-1 s^-1 (taking into account the units of rate and concentration)
Conclusion
Thus, the rate constant k is 0.54 L mol^-1 s^-1. The correct answer is option 'A'.
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The decomposition of NO2 at 400 K proceeds at a of rate of 5.4 x 10 -5 mol L-1 s-1 when [NO2] = 0.01 mol-12 NO2(g) → 2NO(g ) + O2(g).Q. Rate constant of the reaction will bea)0.54 L mol-1 s-1b)0.54 x 10-3 L mol-1 s-1c)5.4 s-1d)0.54s-1Correct answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The decomposition of NO2 at 400 K proceeds at a of rate of 5.4 x 10 -5 mol L-1 s-1 when [NO2] = 0.01 mol-12 NO2(g) → 2NO(g ) + O2(g).Q. Rate constant of the reaction will bea)0.54 L mol-1 s-1b)0.54 x 10-3 L mol-1 s-1c)5.4 s-1d)0.54s-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The decomposition of NO2 at 400 K proceeds at a of rate of 5.4 x 10 -5 mol L-1 s-1 when [NO2] = 0.01 mol-12 NO2(g) → 2NO(g ) + O2(g).Q. Rate constant of the reaction will bea)0.54 L mol-1 s-1b)0.54 x 10-3 L mol-1 s-1c)5.4 s-1d)0.54s-1Correct answer is option 'A'. Can you explain this answer?.
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