JEE Main · 2019 · Shift-ImediumCK-120

The following results were obtained during kinetic studies of the reaction: {2A + B product} | Experiment |…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

The following results were obtained during kinetic studies of the reaction: 2A+B→product\mathrm{2A + B \rightarrow product}

| Experiment | [A] (mol L−1)[\mathrm{A}]\,(\mathrm{mol\,L^{-1}}) | [B] (mol L−1)[\mathrm{B}]\,(\mathrm{mol\,L^{-1}}) | Initial rate of reaction  (mol L−1 min−1)\,(\mathrm{mol\,L^{-1}\,min^{-1}}) | |:---:|:---:|:---:|:---:| | I | 0.10 | 0.20 | 6.93×10−36.93 \times 10^{-3} | | II | 0.10 | 0.25 | 6.93×10−36.93 \times 10^{-3} | | III | 0.20 | 0.30 | 1.386×10−21.386 \times 10^{-2} |

The time (in minutes) required to consume half of A is

Options
  1. a

    100

  2. b

    10

    ✓
  3. c

    5

  4. d

    1

Correct Answerb

10

Detailed Solution

Order w.r.t. B: Exps I & II: [A] same, [B] changes (0.20 to 0.25), rate unchanged → order w.r.t. B = 0.

Order w.r.t. A: Exps I & III: [A] doubles (0.10 to 0.20), rate doubles (6.93×10−36.93 \times 10^{-3} to 1.386×10−21.386 \times 10^{-2}) → order w.r.t. A = 1.

Rate law: r=k[A]r = k[\mathrm{A}]

From Exp I: k=6.93×10−30.10=6.93×10−2 min−1k = \frac{6.93 \times 10^{-3}}{0.10} = 6.93 \times 10^{-2}\,\mathrm{min^{-1}}

t1/2=ln⁡2k=0.6936.93×10−2=10 mint_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{6.93 \times 10^{-2}} = 10\,\mathrm{min}

The rate is the rate of reaction, not rate of disappearance of A. Rate of disappearance of A =2×= 2 \times rate of reaction =2×6.93×10−3= 2 \times 6.93 \times 10^{-3}. So keff=2×6.93×10−3/0.10=0.1386k_{eff} = 2 \times 6.93 \times 10^{-3}/0.10 = 0.1386... Answer: Option (2)

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