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Chemical Kinetics and Nuclear Chemistry question

2023 · 10 Apr · Shift 2 · Q16
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Chemical Kinetics and Nuclear Chemistry question

2023 · 10 Apr · Shift 2 · Q16

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
The number of incorrect statement/s from the following is ‾\underline{\hspace{2cm}}​ A. The successive half lives of zero order reactions decreases with time. B. A substance appearing as reactant in the chemical equation may not affect the rate of reaction C. Order and molecularity of a chemical reaction can be a fractional number D. The rate constant units of zero and second order reaction are mol L−1 s−1\mathrm{mol} ~\mathrm{L}^{-1} \mathrm{~s}^{-1}mol L−1 s−1 and mol−1 L s−1\mathrm{mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}mol−1 L s−1 respectively
Numerical answer
View written solutionFree

Correct answer: 1

  1. We check each statement one by one.
  1. Statement A: The successive half lives of zero order reactions decreases with time.

For a zero order reaction,

[A]=[A]0−kt[A]=[A]_0-kt[A]=[A]0​−kt

Half-life is

t1/2=[A]02kt_{1/2} = \frac{[A]_0}{2k}t1/2​=2k[A]0​​

Now the next half-life (from [A]0/2[A]_0/2[A]0​/2 to [A]0/4[A]_0/4[A]0​/4) is

t1/2′=[A]0/4kt'_{1/2} = \frac{[A]_0/4}{k}t1/2′​=k[A]0​/4​

which is smaller than the first half-life. Thus, successive half-lives keep decreasing.

So, A is correct.

  1. Statement B: A substance appearing as reactant in the chemical equation may not affect the rate of reaction.

Yes, this is possible when the reaction is complex and the reactant does not appear in the rate-determining step or when the experimentally observed rate law does not depend on that reactant.

So, B is correct.

  1. Statement C: Order and molecularity of a chemical reaction can be a fractional number.
  • Order can be fractional.
  • Molecularity cannot be fractional; it is always a whole number for an elementary reaction.

Hence the statement says both can be fractional, which is wrong.

So, C is incorrect.

  1. Statement D: The rate constant units of zero and second order reaction are mol L−1 s−1\mathrm{mol\,L^{-1}\,s^{-1}}molL−1s−1 and mol−1 L s−1\mathrm{mol^{-1}\,L\,s^{-1}}mol−1Ls−1 respectively.

For an nnnth order reaction,

[k]=(concentration)1−n(time)−1[k] = (\text{concentration})^{1-n}(\text{time})^{-1}[k]=(concentration)1−n(time)−1
  • For zero order:
[k]=(concentration)1−0(time)−1=mol L−1 s−1[k] = (\text{concentration})^{1-0}(\text{time})^{-1} = \mathrm{mol\,L^{-1}\,s^{-1}}[k]=(concentration)1−0(time)−1=molL−1s−1
  • For second order:
[k]=(concentration)−1(time)−1=L mol−1 s−1[k] = (\text{concentration})^{-1}(\text{time})^{-1} = \mathrm{L\,mol^{-1}\,s^{-1}}[k]=(concentration)−1(time)−1=Lmol−1s−1

which is the same as mol−1 L s−1\mathrm{mol^{-1}\,L\,s^{-1}}mol−1Ls−1.

So, D is correct.

  1. Therefore, only one statement is incorrect:
1\boxed{1}1​
  1. Comparison with stored correct answer:

Stored correct answer = 111

Our derived answer = 111

So they match.

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