For a reaction A + B → C; rate = k[A][B]2, on doubling the concentrat…

The correct answer is Option (2) → Half

In a rate law, each concentration is raised to the power of its order. So the effect of changing a concentration depends on that power: if a concentration is multiplied by x, the rate is multiplied by x raised to its order.

Here, Rate = k[A][B]²: the reaction is first order in A and second order in B.

New concentrations: [A]′ = 2[A] and [B]′ = [B]/2.

    \( \displaystyle \text{Rate}^{\prime}=k\,(2[\mathrm{A}])\left(\frac{[\mathrm{B}]}{2}\right)^{2} \)

    \( \displaystyle \text{Rate}^{\prime}=k\times2\times\frac{1}{4}\,[\mathrm{A}][\mathrm{B}]^{2} \)

    \( \displaystyle \text{Rate}^{\prime}=\frac{1}{2}\,k[\mathrm{A}][\mathrm{B}]^{2}=\frac{1}{2}\,\text{Rate} \)

So the initial rate becomes half.

Shortcut: Multiply the factors directly: doubling A gives 2¹ = 2, halving B gives (1/2)² = 1/4, and 2 × 1/4 = 1/2.