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Chemical Kinetics and Nuclear Chemistry question

2004 · Shift 0 · Q41
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Chemical Kinetics and Nuclear Chemistry question

2004 · Shift 0 · Q41

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is
  1. A
    30 minutes
  2. B
    60 minutes
  3. C
    7.5 minutes
  4. D
    15 minutes
View written solutionFree

Correct answer: A

  1. Use the property of a first-order reaction

For a first-order reaction, t=2.303klog⁡[A]0[A]tt = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}t=k2.303​log[A]t​[A]0​​

A very important result is that the half-life is constant: t1/2=0.693kt_{1/2} = \frac{0.693}{k}t1/2​=k0.693​

  1. Find the half-life from the given data

The concentration decreases from 0.8 M0.8\,\text{M}0.8M to 0.4 M0.4\,\text{M}0.4M in 151515 minutes.

Since 0.40.40.4 is exactly half of 0.80.80.8, this time is the half-life.

So, t1/2=15 minutest_{1/2} = 15\text{ minutes}t1/2​=15 minutes

  1. Analyze the required change

We need the time for concentration to change from 0.1 M0.1\,\text{M}0.1M to 0.025 M0.025\,\text{M}0.025M.

Now, 0.1→0.05→0.0250.1 \to 0.05 \to 0.0250.1→0.05→0.025

This is two successive halvings.

  • First halving: 0.1→0.050.1 \to 0.050.1→0.05 takes 151515 min
  • Second halving: 0.05→0.0250.05 \to 0.0250.05→0.025 takes 151515 min

Therefore, total time: t=15+15=30 minutest = 15 + 15 = 30\text{ minutes}t=15+15=30 minutes

  1. Option check
  • A: 303030 minutes ✅
  • B: 606060 minutes ❌
  • C: 7.57.57.5 minutes ❌
  • D: 151515 minutes ❌

Hence, the correct answer is A.

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