$2A + B → A_2B$

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

Calculate the rate of reaction using the rate law:

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

Given:

• $k = 2.0 \times 10^{-6} \, \text{mol}^{-2} \text{L}^2 \text{s}^{-1}$
• Initial $[A]_0 = 0.1$ M, $[B]_0 = 0.2$ M
• $[A]=0.06$ M

Step 1: Express [B] in terms of [A]

For the reaction:

$2A+B→A_2​B$

• Stoichiometry: 2 moles of A react with 1 mole of B
• Change in [A]: Δ[A] = 0.1 − 0.06 = 0.04 M
• Corresponding change in [B]: Δ[B] = (0.04)/2 = 0.02 M

$[B] = 0.2 - 0.02 = 0.18 \, \text{M}$

Step 2: Calculate rate

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

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

$0.18^2 = 0.0324$

$0.0324×0.06=0.001944$

$\text{rate} = 2.0 \times 10^{-6} \times 0.001944 = 3.888 \times 10^{-9} \, \text{mol L}^{-1}\text{s}^{-1}$