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Chemical Kinetics and Nuclear Chemistry question

2018 · 15 Apr · Shift 2 · Q14
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Chemical Kinetics and Nuclear Chemistry question

2018 · 15 Apr · Shift 2 · Q14

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
For a first order reaction, A →\to→ P, t1/2 (half-life) is 10 days The time required for 14{1 \over 4}41​ th conversion of A (in days) is : (ln 2 = 0.693, ln 3 = 1.1)
  1. A
    5
  2. B
    3.2
  3. C
    4.1
  4. D
    2.5
View written solutionFree

Correct answer: C

  1. For a first-order reaction,

t1/2=0.693kt_{1/2} = \frac{0.693}{k}t1/2​=k0.693​

Given:

t1/2=10 dayst_{1/2} = 10\text{ days}t1/2​=10 days

So,

k=0.69310=0.0693 day−1k = \frac{0.693}{10} = 0.0693\ \text{day}^{-1}k=100.693​=0.0693 day−1

  1. "14{1 \over 4}41​th conversion" means 25%25\%25% of AAA has reacted. Thus, 75%75\%75% of AAA remains.

So,

[A]t[A]0=34\frac{[A]_t}{[A]_0} = \frac{3}{4}[A]0​[A]t​​=43​

  1. For a first-order reaction,

t=1kln⁡[A]0[A]tt = \frac{1}{k} \ln \frac{[A]_0}{[A]_t}t=k1​ln[A]t​[A]0​​

Substitute:

t=10.0693ln⁡13/4=10.0693ln⁡43t = \frac{1}{0.0693} \ln \frac{1}{3/4} = \frac{1}{0.0693} \ln \frac{4}{3}t=0.06931​ln3/41​=0.06931​ln34​

Now,

ln⁡43=ln⁡4−ln⁡3=2ln⁡2−ln⁡3\ln \frac{4}{3} = \ln 4 - \ln 3 = 2\ln 2 - \ln 3ln34​=ln4−ln3=2ln2−ln3

=2(0.693)−1.1=1.386−1.1=0.286= 2(0.693) - 1.1 = 1.386 - 1.1 = 0.286=2(0.693)−1.1=1.386−1.1=0.286

Therefore,

t=0.2860.0693≈4.13 dayst = \frac{0.286}{0.0693} \approx 4.13\text{ days}t=0.06930.286​≈4.13 days

  1. Hence, the required time is

4.1 days\boxed{4.1\text{ days}}4.1 days​

So, the correct option is C.

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