Question

a reaction was shown to follow second-order kinetics.
how much time is required for a to change from 0.500 m to 0.325 m? (k = 0.456 m⁻¹·s⁻¹)
answer:
______ s

Explanation:

Step1: Recall 2nd-order kinetics formula

The integrated rate law for a second-order reaction is:
$$\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt$$
where $[A]_0 = 0.500\ \text{M}$, $[A]_t = 0.325\ \text{M}$, $k = 0.456\ \text{M}^{-1}\text{s}^{-1}$

Step2: Calculate reciprocal concentration difference

First compute $\frac{1}{[A]_t} - \frac{1}{[A]_0}$:
$$\frac{1}{0.325} - \frac{1}{0.500} = 3.0769 - 2.0000 = 1.0769\ \text{M}^{-1}$$

Step3: Solve for time $t$

Rearrange the formula to $t = \frac{\frac{1}{[A]_t} - \frac{1}{[A]_0}}{k}$, then substitute values:
$$t = \frac{1.0769}{0.456} \approx 2.36\ \text{s}$$

Answer:

2.36 s