JEE Main · 2019 · Shift-IImediumCK-085

For an elementary chemical reaction, A2 k-1k1 2A, the expression for d[A]dt is:

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

For an elementary chemical reaction, A2⇌k1k−12A\mathrm{A_2} \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{2A}, the expression for d[A]dt\frac{d[\mathrm{A}]}{dt} is:

Options
  1. a

    2k1[A2]−2k−1[A]22k_1[\mathrm{A_2}] - 2k_{-1}[\mathrm{A}]^2

    ✓
  2. b

    2k1[A2]−k−1[A]22k_1[\mathrm{A_2}] - k_{-1}[\mathrm{A}]^2

  3. c

    k1[A2]+k−1[A]2k_1[\mathrm{A_2}] + k_{-1}[\mathrm{A}]^2

  4. d

    k1[A2]−k−1[A]2k_1[\mathrm{A_2}] - k_{-1}[\mathrm{A}]^2

Correct Answera

2k1[A2]−2k−1[A]22k_1[\mathrm{A_2}] - 2k_{-1}[\mathrm{A}]^2

Detailed Solution

For the forward reaction A2→2A\mathrm{A_2 \rightarrow 2A}: rate of formation of A =2k1[A2]= 2k_1[\mathrm{A_2}]

For the reverse reaction 2A→A2\mathrm{2A \rightarrow A_2}: rate of consumption of A =2k−1[A]2= 2k_{-1}[\mathrm{A}]^2

Net rate: d[A]dt=2k1[A2]−2k−1[A]2\frac{d[\mathrm{A}]}{dt} = 2k_1[\mathrm{A_2}] - 2k_{-1}[\mathrm{A}]^2

Answer: Option (1)

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