A dilute aqueous reaction has the rate law $\text{rate}=k[A{]}^{3/2}[B{]}^2$. If the concentration of $A$ increases from $0.01\mathrm{M}$ to $0.04\mathrm{M}$ while $[B]$ remains $0.02\mathrm{M}$, by what factor does the instantaneous rate change compared to the original rate?

Step-by-step solution

1. Identify the given rate law.
2. Express the new rate as $k(0.04{)}^{3/2}(0.02{)}^2$.
3. Express the original rate as $k(0.01{)}^{3/2}(0.02{)}^2$.
4. Take the ratio of the new rate to the original rate: $(0.04/0.01{)}^{3/2}=4^{3/2}=8$.
5. Thus the rate increases by a factor of 8.

Concept tested: Effect of concentration on rate with fractional exponent

Published 10 October 2026 · How this solution was prepared and checked