JEE Main · 2025 · Shift-IImediumCK-141

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are t1t_1 and t2t_2 (s), respectively. The ratio t1t2\tfrac{t_1}{t_2} will be:

Options
  1. a

    43\tfrac{4}{3}

  2. b

    32\tfrac{3}{2}

  3. c

    34\tfrac{3}{4}

  4. d

    23\tfrac{2}{3}

    ✓
Correct Answerd

23\tfrac{2}{3}

Detailed Solution

Step 1: Apply first order integrated rate law

For a first order reaction: t=1kln⁡[A]0[A]t = \tfrac{1}{k} \ln \tfrac{[A]_0}{[A]}

Step 2: Calculate t1t_1 (concentration falls to 14\tfrac{1}{4} of initial)

t1=1kln⁡A0A0/4=1kln⁡4=2ln⁡2kt_1 = \tfrac{1}{k} \ln \tfrac{A_0}{A_0/4} = \tfrac{1}{k} \ln 4 = \tfrac{2 \ln 2}{k}

Step 3: Calculate t2t_2 (concentration falls to 18\tfrac{1}{8} of initial)

t2=1kln⁡A0A0/8=1kln⁡8=3ln⁡2kt_2 = \tfrac{1}{k} \ln \tfrac{A_0}{A_0/8} = \tfrac{1}{k} \ln 8 = \tfrac{3 \ln 2}{k}

Step 4: Find the ratio

t1t2=2ln⁡23ln⁡2=23\tfrac{t_1}{t_2} = \tfrac{2 \ln 2}{3 \ln 2} = \tfrac{2}{3}

Key Points to Remember:

  • For first order: t1/2=ln⁡2kt_{1/2} = \tfrac{\ln 2}{k}
  • Time to reach 14\tfrac{1}{4} of initial = 2×t1/22 \times t_{1/2}
  • Time to reach 18\tfrac{1}{8} of initial = 3×t1/23 \times t_{1/2}

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