$2A + B → A_2B$

For the above reaction, the rate law equation is, rate = $k[A][B]^2$ with $k = 2.0 × 10^{-6}\, mol^{-2}\, L^2s^{-1}$. If the initial concentrations of A and B were $0.1\, mol\, L^{-1}$ and $0.2\, mol\, L^{-1}$, respectively, the rate of reaction after [A] is reduced to $0.06\, mol\, L^{-1}$ will be

$3.89 × 10^{-9}\, mol\, L^{-1}\, s^{-1}$

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

Reaction:

$2A + B \to A_2B$

Rate law:

$\text{rate} = k[A][B]^2$

Initial concentrations:

$[A]_0 = 0.10 \text{ mol L}^{-1}$

$[B]_0 = 0.20 \text{ mol L}^{-1}$

When $[A]$ becomes $0.06 \text{ mol L}^{-1}$, the amount of $A$ consumed is:

$A \text{ consumed} = 0.10 - 0.06 = \mathbf{0.04 \text{ mol L}^{-1}}$

From the stoichiometry:

$2A + B \to A_2B$

If $2$ moles of $A$ react, $1$ mole of $B$ reacts.

Thus $B$ consumed:

$B \text{ consumed} = 0.04 / 2$

$B \text{ consumed} = \mathbf{0.02 \text{ mol L}^{-1}}$

New concentration of $B$:

$[B] = 0.20 - 0.02$

$[B] = \mathbf{0.18 \text{ mol L}^{-1}}$

Now calculate the rate:

$\text{rate} = k[A][B]^2$

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

$(0.18)^2 = \mathbf{0.0324}$

$\text{rate} = 2.0 \times 10^{-6} \times 0.06 \times 0.0324$

$\text{rate} = \mathbf{3.888 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1}}$

$\text{rate} \approx \mathbf{3.89 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1}}$