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Chemical Kinetics and Nuclear Chemistry question

2023 · 25 Jan · Shift 2 · Q16
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Chemical Kinetics and Nuclear Chemistry question

2023 · 25 Jan · Shift 2 · Q16

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
A first order reaction has the rate constant, k=4.6×10−3 s−1\mathrm{k=4.6\times10^{-3}~s^{-1}}k=4.6×10−3 s−1. The number of correct statement/s from the following is/are ‾\underline{\hspace{2cm}}​ Given : log⁡3=0.48\mathrm{\log3=0.48}log3=0.48 A. Reaction completes in 1000 s. B. The reaction has a half-life of 500 s. C. The time required for 10% completion is 25 times the time required for 90% completion. D. The degree of dissociation is equal to (1−e−kt\mathrm{1-e^{-kt}}1−e−kt) E. The rate and the rate constant have the same unit.
Numerical answer
View written solutionFree

Correct answer: 1

  1. Given

For a first-order reaction, k=4.6×10−3 s−1k = 4.6\times 10^{-3}\ \text{s}^{-1}k=4.6×10−3 s−1

We check each statement one by one.

  1. Statement A: Reaction completes in 1000 s

For a first-order reaction, the reactant concentration decreases exponentially: [A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}[A]t​=[A]0​e−kt

Such a reaction never becomes exactly zero in finite time, so it never truly completes in 1000 s.

Hence, A is false.

  1. Statement B: The reaction has a half-life of 500 s

For a first-order reaction, t1/2=0.693kt_{1/2} = \frac{0.693}{k}t1/2​=k0.693​

Substitute k=4.6×10−3 s−1k = 4.6\times10^{-3}\ \text{s}^{-1}k=4.6×10−3 s−1: t1/2=0.6934.6×10−3t_{1/2} = \frac{0.693}{4.6\times10^{-3}}t1/2​=4.6×10−30.693​ t1/2≈150.65 st_{1/2} \approx 150.65\ \text{s}t1/2​≈150.65 s

This is not 500 s.

Hence, B is false.

  1. Statement C: The time required for 10% completion is 25 times the time required for 90% completion

For a first-order reaction, t=2.303klog⁡aa−xt = \frac{2.303}{k}\log\frac{a}{a-x}t=k2.303​loga−xa​

For 10% completion:

x=0.1ax = 0.1ax=0.1a, so t10%=2.303klog⁡a0.9a=2.303klog⁡109t_{10\%} = \frac{2.303}{k}\log\frac{a}{0.9a} = \frac{2.303}{k}\log\frac{10}{9}t10%​=k2.303​log0.9aa​=k2.303​log910​

For 90% completion:

x=0.9ax = 0.9ax=0.9a, so t90%=2.303klog⁡a0.1a=2.303klog⁡10=2.303kt_{90\%} = \frac{2.303}{k}\log\frac{a}{0.1a} = \frac{2.303}{k}\log 10 = \frac{2.303}{k}t90%​=k2.303​log0.1aa​=k2.303​log10=k2.303​

Now, t10%t90%=log⁡109\frac{t_{10\%}}{t_{90\%}} = \log\frac{10}{9}t90%​t10%​​=log910​

Since log⁡(10/9)\log(10/9)log(10/9) is much less than 1, certainly it is not 25.

Actually, t90%t_{90\%}t90%​ is much greater than t10%t_{10\%}t10%​.

Hence, C is false.

  1. Statement D: The degree of dissociation is equal to (1−e−kt)(1-e^{-kt})(1−e−kt)

For a first-order decomposition/dissociation process, [A]t=[A]0e−kt[A]_t = [A]_0 e^{-kt}[A]t​=[A]0​e−kt

Fraction remaining undissociated: [A]t[A]0=e−kt\frac{[A]_t}{[A]_0} = e^{-kt}[A]0​[A]t​​=e−kt

Therefore, degree of dissociation (fraction reacted) is α=1−e−kt\alpha = 1 - e^{-kt}α=1−e−kt

Hence, D is true.

  1. Statement E: The rate and the rate constant have the same unit

For a first-order reaction:

  • Rate has unit of concentration/time, e.g. mol L−1s−1\text{mol L}^{-1}\text{s}^{-1}mol L−1s−1
  • Rate constant has unit of s−1\text{s}^{-1}s−1

These are not the same.

Hence, E is false.

  1. Count the correct statements
  • A: False
  • B: False
  • C: False
  • D: True
  • E: False

So, the number of correct statements is 111

  1. Comparison with stored answer

Stored correct answer = 111

Our derived answer = 111

So the answer agrees with the stored correct answer.

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