In a second-order reaction, the relationship between the concentration of reactants and time can be described by the equation:

\(\frac{1}{[A_t]} = k t + \frac{1}{[A_0]}\)

In this scenario, we are given the initial concentration \([A_0]\) as 0.7670 M after 300 seconds and the final concentration \([A_t]\) as \(7.3 \times 10^{-2}\) M after 750 seconds. To find the rate constant \(k\), we first need to determine the elapsed time:

Elapsed time = 750 seconds - 300 seconds = 450 seconds.

Now, substituting the known values into the equation:

\(\frac{1}{7.3 \times 10^{-2}} = k(450) + \frac{1}{0.7670}\)

Calculating the left side:

\(\frac{1}{7.3 \times 10^{-2}} \approx 13.6986\)

And the right side becomes:

\(13.6986 = k(450) + 1.3035\)

Next, we isolate \(k\) by subtracting \(1.3035\) from both sides:

\(12.3951 = k(450)\)

Now, divide both sides by 450 to solve for \(k\):

\(k = \frac{12.3951}{450} \approx 0.0272\) M^-1s^-1

Thus, the rate constant \(k\) for this second-order reaction is approximately \(2.72 \times 10^{-2}\) M^-1s^-1. The units for \(k\) in a second-order reaction are derived from the formula, where \(n\) is the order of the reaction. For second-order reactions, the units are M^-1s^-1, indicating that the rate constant depends on the concentration of the reactants and the time elapsed.