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Chemical Kinetics and Nuclear Chemistry question

2023 · 29 Jan · Shift 2 · Q23
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Chemical Kinetics and Nuclear Chemistry question

2023 · 29 Jan · Shift 2 · Q23

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
For conversion of compound A →\to→ B, the rate constant of the reaction was found to be 4.6×10−5 L mol−1 s−1\mathrm{4.6\times10^{-5}~L~mol^{-1}~s^{-1}}4.6×10−5 L mol−1 s−1. The order of the reaction is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 2

  1. Identify the unit of the rate constant

    Given: k=4.6×10−5  L mol−1s−1k = 4.6\times 10^{-5}\;\text{L mol}^{-1}\text{s}^{-1}k=4.6×10−5L mol−1s−1

  2. Use the relation between units of kkk and order of reaction

    For an nnn-th order reaction, Rate=k[A]n\text{Rate} = k[A]^nRate=k[A]n

    Since rate has units of concentration per time, [Rate]=mol L−1s−1[\text{Rate}] = \text{mol L}^{-1}\text{s}^{-1}[Rate]=mol L−1s−1

    Therefore, [k]=mol L−1s−1(mol L−1)n[k] = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^n}[k]=(mol L−1)nmol L−1s−1​

    [k]=Ln−1mol1−ns−1[k] = \text{L}^{n-1}\text{mol}^{1-n}\text{s}^{-1}[k]=Ln−1mol1−ns−1

  3. Compare with the given unit

    Given unit: L mol−1s−1\text{L mol}^{-1}\text{s}^{-1}L mol−1s−1

    Match with: Ln−1mol1−ns−1\text{L}^{n-1}\text{mol}^{1-n}\text{s}^{-1}Ln−1mol1−ns−1

    Comparing powers: n−1=1⇒n=2n-1 = 1 \Rightarrow n=2n−1=1⇒n=2

    Also, 1−n=−1⇒n=21-n = -1 \Rightarrow n=21−n=−1⇒n=2

  4. Conclusion

    The reaction is of second order.

    2\boxed{2}2​

  5. Comparison with stored correct answer

    Stored correct answer = 222

    This matches the derived answer.

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