You are given a rate law equation Rate = K [A]^2 [B] . By how many times will the rate increase or decrease for the reaction if i) concentration of A is doubled while that of B remain constant. [1]…

i) Effect of doubling concentration of $\mathrm{A}$ while keeping $\mathrm{B}$ constant

Given rate law: $$\text{Rate}=K[A{]}^2[B]$$

Initial rate (R₁): $$R_1=K[A{]}^2[B]$$

New rate (R₂) when [A] is doubled: $$R_2=K(2[A]{)}^2[B]=K\cdot 4[A{]}^2[B]=4\cdot K[A{]}^2[B]$$

Comparison: $$\frac{R_2}{R_1}=\frac{4\cdot K[A{]}^2[B]}{K[A{]}^2[B]}=4$$

Conclusion: The rate increases by 4 times.

---

ii) Effect of doubling concentration of $\mathrm{B}$ while keeping $\mathrm{A}$ constant

Initial rate (R₁): $$R_1=K[A{]}^2[B]$$

New rate (R₃) when [B] is doubled: $$R_3=K[A{]}^2(2[B])=2\cdot K[A{]}^2[B]$$

Comparison: $$\frac{R_3}{R_1}=\frac{2\cdot K[A{]}^2[B]}{K[A{]}^2[B]}=2$$

Conclusion: The rate increases by 2 times.