For a first-order reaction , the concentration of A changes from to in . What is the exact instantan — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

The drop from 0.1 M to 0.025 M spans exactly two half-lives. Thus, $2 \times t_{1/2} = 40 \Rightarrow t_{1/2} = 20 \text{ min}$. The rate constant $k = 0.693 / 20 = 0.03465 \text{ min}^{-1}$. The rate at $[A]=0.01\text{ M}$ is $r = k[A] = 0.03465 \times 0.01 = 3.465 \times 10^{-4} \text{ M/min}$.