When the initial concentration of a particular reactant is artificially increased by a factor of , t — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

Let the rate law be $\text{Rate} = k[A]^n$. If the concentration is multiplied by $4$, the new rate is $\text{Rate}_{new} = k(4[A])^n = 4^n \times k[A]^n = 4^n \times \text{Rate}$. We are given that the rate increases by a factor of $8$, so $4^n = 8$. Expressing both bases as powers of $2$ gives $(2^2)^n = 2^3 \Rightarrow 2^{2n} = 2^3 \Rightarrow 2n = 3 \Rightarrow n = 1.5$.