JEE Main · 2020 · Shift-IhardCK-023

A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is: (Use ln⁡2=0.693\ln 2 = 0.693)

Options
  1. a

    180

  2. b

    900

    ✓
  3. c

    300

  4. d

    120

Correct Answerb

900

Detailed Solution

Let initial concentration of both = C0C_0.

At time tt: [A]=C0e−kAt,kA=ln⁡2300[\mathrm{A}] = C_0 e^{-k_A t},\quad k_A = \frac{\ln 2}{300} [B]=C0e−kBt,kB=ln⁡2180[\mathrm{B}] = C_0 e^{-k_B t},\quad k_B = \frac{\ln 2}{180}

Condition: [A]=4[B][\mathrm{A}] = 4[\mathrm{B}]: C0e−kAt=4C0e−kBtC_0 e^{-k_A t} = 4 C_0 e^{-k_B t} e(kB−kA)t=4e^{(k_B - k_A)t} = 4 (kB−kA)t=ln⁡4=2ln⁡2(k_B - k_A)t = \ln 4 = 2\ln 2

kB−kA=ln⁡2(1180−1300)=ln⁡2×300−180180×300=ln⁡2×12054000=ln⁡2450k_B - k_A = \ln 2\left(\frac{1}{180} - \frac{1}{300}\right) = \ln 2 \times \frac{300 - 180}{180 \times 300} = \ln 2 \times \frac{120}{54000} = \frac{\ln 2}{450}

t=2ln⁡2ln⁡2/450=2×450=900 st = \frac{2\ln 2}{\ln 2/450} = 2 \times 450 = 900\,\mathrm{s}

Answer: Option (2) — 900 s

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