For a reaction, the rate = k[A]2[B], when the initial concentration of A is doubled, the initial rate will
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → increase to 4 times
In a rate law, each concentration is raised to its order with respect to that reactant. When the concentration of one reactant is multiplied by a factor, the rate is multiplied by that factor raised to its order; the other concentrations stay the same.
Here the reaction is second order in A and first order in B:
\( \displaystyle r_1=k[A]^2[B] \)
\( \displaystyle r_2=k(2[A])^2[B]=4\,k[A]^2[B] \)
\( \displaystyle \frac{r_2}{r_1}=2^2=4 \)
So the initial rate will increase to 4 times.