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Chemical Kinetics and Nuclear Chemistry question

2025 · 4 Apr · Shift 2 · Q9
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Chemical Kinetics and Nuclear Chemistry question

2025 · 4 Apr · Shift 2 · Q9

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
Half life of zero order reaction A→\mathrm{A} \rightarrowA→ product is 1 hour, when initial concentration of reactant is 2.0 mol L−12.0 \mathrm{~mol} \mathrm{~L}{ }^{-1}2.0 mol L−1. The time required to decrease concentration of A from 0.50 to 0.25 mol L−10.25 \mathrm{~mol} \mathrm{~L}^{-1}0.25 mol L−1 is :
  1. A
    0.5 hour
  2. B
    15 min
  3. C
    60 min
  4. D
    4 hour
View written solutionFree

Correct answer: B

  1. Use the integrated rate law for a zero-order reaction

For a zero-order reaction,

[A]t=[A]0−kt[A]_t = [A]_0 - kt[A]t​=[A]0​−kt

and the half-life is

t1/2=[A]02kt_{1/2} = \frac{[A]_0}{2k}t1/2​=2k[A]0​​
  1. Find the rate constant using the given half-life

Given:

  • Initial concentration, [A]0=2.0 mol L−1[A]_0 = 2.0\,\text{mol L}^{-1}[A]0​=2.0mol L−1
  • Half-life, t1/2=1 hourt_{1/2} = 1\,\text{hour}t1/2​=1hour

So,

1=2.02k1 = \frac{2.0}{2k}1=2k2.0​ 1=1k1 = \frac{1}{k}1=k1​ k=1 mol L−1hour−1k = 1\,\text{mol L}^{-1}\text{hour}^{-1}k=1mol L−1hour−1
  1. Calculate time for concentration to decrease from 0.500.500.50 to 0.25 mol L−10.25\,\text{mol L}^{-1}0.25mol L−1

For zero-order kinetics,

t=[A]1−[A]2kt = \frac{[A]_1 - [A]_2}{k}t=k[A]1​−[A]2​​

Here,

  • [A]1=0.50 mol L−1[A]_1 = 0.50\,\text{mol L}^{-1}[A]1​=0.50mol L−1
  • [A]2=0.25 mol L−1[A]_2 = 0.25\,\text{mol L}^{-1}[A]2​=0.25mol L−1
  • k=1 mol L−1hour−1k = 1\,\text{mol L}^{-1}\text{hour}^{-1}k=1mol L−1hour−1

Thus,

t=0.50−0.251=0.25 hourt = \frac{0.50 - 0.25}{1} = 0.25\,\text{hour}t=10.50−0.25​=0.25hour

Convert to minutes:

0.25×60=15 min0.25 \times 60 = 15\,\text{min}0.25×60=15min
  1. Check options
  • A: 0.50.50.5 hour = 303030 min ❌
  • B: 151515 min ✅
  • C: 606060 min ❌
  • D: 444 hour ❌

Therefore, the correct option is B.

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