A first-order reaction has a half-life of 26.4 seconds. How long does it take for the concentration of the reactant in the reaction to fall to one-eighth of its initial value?

To determine how long it takes for the concentration of the reactant to fall to one-eighth of its initial value, we can use the half-life given in the question.

The half-life of a first-order reaction is the time it takes for the concentration of the reactant to decrease by half. In this case, the half-life is given as 26.4 seconds.

Since the half-life is the time it takes for the concentration to decrease by half, we can calculate the number of half-lives it takes for the concentration to fall to one-eighth (1/8) of its initial value.

To calculate the number of half-lives, we can use the formula:

Number of half-lives = log(Initial concentration / Final concentration) / log(2)

In this case, the initial concentration is 1, and the final concentration is 1/8 = 0.125.

Number of half-lives = log(1 / 0.125) / log(2) = log(8) / log(2) = 3 / 0.3010 ≈ 9.97

So, it takes approximately 9.97 half-lives for the concentration of the reactant to fall to one-eighth of its initial value.

To calculate the time it takes, we multiply the number of half-lives by the half-life:

Time = Number of half-lives × Half-life = 9.97 × 26.4 seconds ≈ 262.73 seconds

Therefore, it takes approximately 262.73 seconds for the concentration of the reactant to fall to one-eighth of its initial value.