2A + B → A_2B For the above reaction, the rate law equation is, rate…

The correct answer is Option (2) → \(3.89 × 10^{-9}\, mol\, L^{-1}\, s^{-1}\)

The rate at any moment is found by putting the concentrations at that moment into the rate law. When [A] has fallen, [B] has fallen too, in the ratio of the stoichiometric coefficients: in \( \ce{2A + B -> A2B} \), one mole of B is used for every two moles of A.

The amount of A used up:

    \( \displaystyle 0.10 - 0.06 = 0.04\ \mathrm{mol\,L^{-1}} \)

The amount of B used up is half of this:

    \( \displaystyle \frac{0.04}{2} = 0.02\ \mathrm{mol\,L^{-1}} \)

    \( \displaystyle [\mathrm{B}] = 0.20 - 0.02 = 0.18\ \mathrm{mol\,L^{-1}} \)

Now the rate:

    \( \displaystyle \text{rate} = k[\mathrm{A}][\mathrm{B}]^{2} \)

    \( \displaystyle \text{rate} = (2.0 \times 10^{-6}) \times (0.06) \times (0.18)^{2} \)

    \( \displaystyle \text{rate} = 2.0 \times 10^{-6} \times 0.06 \times 0.0324 = 3.89 \times 10^{-9}\ \mathrm{mol\,L^{-1}\,s^{-1}} \)

So the rate is 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹.

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Exercises – “… 3.2 For the reaction: 2A + B → A₂B the rate = k[A][B]² with k = 2.0 × 10^–6 mol^–2 L² s^–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^–1, [B] = 0.2 mol L^–1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^–1.”