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Chemical Kinetics and Nuclear Chemistry question

2021 · 27 Jul · Shift 1 · Q3
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Chemical Kinetics and Nuclear Chemistry question

2021 · 27 Jul · Shift 1 · Q3

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
For a reaction of order n, the unit of the rate constant is :
  1. A
    mol1 −-− n L1 −-− n s
  2. B
    mol1 −-− n L2n s −-− 1
  3. C
    mol1 −-− n Ln −-− 1 s −-− 1
  4. D
    mol1 −-− n L1 −-− n s −-− 1
View written solutionFree

Correct answer: C

  1. Write the rate law

For an nthn^{\text{th}}nth order reaction, Rate=k[C]n\text{Rate} = k[C]^nRate=k[C]n where [C][C][C] is concentration.

  1. Write units of each term

  • Unit of rate =concentration/time= \text{concentration}/\text{time}=concentration/time [Rate]=mol L−1s−1[\text{Rate}] = \text{mol L}^{-1}\text{s}^{-1}[Rate]=mol L−1s−1

  • Unit of concentration: [C]=mol L−1[C] = \text{mol L}^{-1}[C]=mol L−1

So, [C]n=(mol L−1)n=molnL−n[C]^n = (\text{mol L}^{-1})^n = \text{mol}^n\text{L}^{-n}[C]n=(mol L−1)n=molnL−n

  1. Find unit of rate constant kkk

From k=Rate[C]nk = \frac{\text{Rate}}{[C]^n}k=[C]nRate​

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

Simplifying, [k]=mol1−nLn−1s−1[k] = \text{mol}^{1-n}\text{L}^{n-1}\text{s}^{-1}[k]=mol1−nLn−1s−1

  1. Match with the given options

Option C is: mol1−nLn−1s−1\text{mol}^{1-n}\text{L}^{n-1}\text{s}^{-1}mol1−nLn−1s−1

So, Option C is correct.

  1. Final answer

mol1−nLn−1s−1\boxed{\text{mol}^{1-n}\text{L}^{n-1}\text{s}^{-1}}mol1−nLn−1s−1​ which corresponds to Option C.

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