A reaction has rate law $\text{rate} = k[A][B]^2$. If $[A]$ is DOUBLED and $[B]$ is HALVED, the new rate compared to the original rate is

1. Let the original rate be $r_0 = k[A]_0[B]_0^2$. 2. New concentrations: $[A]' = 2[A]_0$ and $[B]' = (1/2)[B]_0$. 3. Substitute: $r' = k(2[A]_0)((1/2)[B]_0)^2 = k\cdot 2\cdot [A]_0 \cdot (1/4)[B]_0^2 = (1/2)\,k[A]_0[B]_0^2$. 4. Compare to original: $r' = (1/2)\,r_0$. The new rate is HALF the original. 5. Logic check: doubling $A$ multiplies rate by 2; halving $B$ multiplies rate by $(1/2)^2 = 1/4$. Net factor $= 2 \cdot (1/4) = 1/2$. ✓ 6. Distractor A forgets the squared dependence on B. Option C forgets the doubling of A. Option D would require the two effects to exactly cancel (they don't because B is squared and A is linear). _Source: NCERT Class 12 Chemistry Part 1, Ch 3, §3.2.3 (Order from rate law, Examples 3.2-3.4), p. 6–9._