JEE Main · 2020 · Shift-IImediumCK-004

The results given in the below table were obtained during kinetic studies of the following reaction: {2A + B C + D} |…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

The results given in the below table were obtained during kinetic studies of the following reaction: 2A+B→C+D\mathrm{2A + B \rightarrow C + D} | Experiment | [A] / mol L−1[\mathrm{A}]\,/\,\mathrm{mol\,L^{-1}} | [B] / mol L−1[\mathrm{B}]\,/\,\mathrm{mol\,L^{-1}} | Initial rate mol L−1 min−1\mathrm{mol\,L^{-1}\,min^{-1}} | |:---:|:---:|:---:|:---:| | I | 0.1 | 0.1 | 6.00×10−36.00 \times 10^{-3} | | II | 0.1 | 0.2 | 2.40×10−22.40 \times 10^{-2} | | III | 0.2 | 0.1 | 1.20×10−21.20 \times 10^{-2} | | IV | X | 0.2 | 7.20×10−27.20 \times 10^{-2} | | V | 0.3 | Y | 2.88×10−12.88 \times 10^{-1} |

X and Y in the given table are respectively:

Options
  1. a

    0.4, 0.40.4,\,0.4

  2. b

    0.4, 0.30.4,\,0.3

  3. c

    0.3, 0.40.3,\,0.4

    ✓
  4. d

    0.3, 0.30.3,\,0.3

Correct Answerc

0.3, 0.40.3,\,0.4

Detailed Solution

Step 1 — Determine orders:

From Exp I & II (A constant): Rate doubles when B doubles → order w.r.t. B = 2 (rate quadruples: 2.40/0.6=42.40/0.6 = 4, so nB=2n_B = 2)

From Exp I & III (B constant): Rate doubles when A doubles → order w.r.t. A = 1

Rate law: r=k[A][B]2r = k[\mathrm{A}][\mathrm{B}]^2

From Exp I: k=6.00×10−30.1×(0.1)2=6.00 L2 mol−2 min−1k = \frac{6.00 \times 10^{-3}}{0.1 \times (0.1)^2} = 6.00\,\mathrm{L^2\,mol^{-2}\,min^{-1}}

Step 2 — Find X (Exp IV): 7.20×10−2=6.00×X×(0.2)2=6.00×X×0.047.20 \times 10^{-2} = 6.00 \times X \times (0.2)^2 = 6.00 \times X \times 0.04 X=7.20×10−20.24=0.3X = \frac{7.20 \times 10^{-2}}{0.24} = 0.3

Step 3 — Find Y (Exp V): 2.88×10−1=6.00×0.3×Y22.88 \times 10^{-1} = 6.00 \times 0.3 \times Y^2 Y2=0.2881.8=0.16⇒Y=0.4Y^2 = \frac{0.288}{1.8} = 0.16 \Rightarrow Y = 0.4

Answer: Option (3) — 0.3, 0.4

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