91Ó°ÊÓ

In chemistry, half-life is an essential concept to grasp, especially when dealing with first-order reactions. It refers to the time required for the concentration of a reactant to decrease by half. For first-order reactions, the half-life is constant, meaning it remains the same no matter the initial concentration. This makes it particularly useful for predicting how a reaction progresses over time.

- The formula for half-life (\(t_{1/2}\)) in a first-order reaction is given by: \[ t_{1/2} = \frac{\ln(2)}{k} \] where \(k\) is the rate constant.

- For the problem at hand, we were given that the half-life is \(25 \text{ s}\). This indicates that every \(25 \text{ s}\), the concentration of \(\text{A}\) will reduce to half of its initial amount.

Knowing the half-life also makes it straightforward to predict how much time it will take for the concentration to reach other fractional values of its original concentration, such as one-fourth or one-eighth.

The rate constant (\(k\)) plays a pivotal role in characterizing how fast a reaction proceeds. In first-order reactions, \(k\) is derived from the half-life using the formula \(t_{1/2} = \frac{\ln(2)}{k}\). By rearranging this equation, you can solve for \(k\) as follows: \[ k = \frac{\ln(2)}{t_{1/2}} \] The rate constant provides insights into the speed of the reaction. A higher \(k\) value indicates a faster reaction, while a lower \(k\) suggests a slower one.

In the context of the given exercise:

• The half-life is known to be \(25 \text{ s}\).
• We can substitute this value into the formula to find \(k\).
• Thus, \(k = \frac{\ln(2)}{25} \approx 0.0277 \text{ s}^{-1}\).

Understanding \(k\) not only helps determine durations for half-lives but also aids in solving problems involving changes in concentration over time.

The integrated rate law for first-order reactions is a powerful tool for predicting the concentration of a reactant as a function of time. It can be expressed as: \[ \frac{[\text{A}]}{[\text{A}]_0} = e^{-kt} \] This formula correlates time \(t\), the initial concentration of the reactant \([\text{A}]_0\), its current concentration \([\text{A}]\), and the rate constant \(k\).

Here's how you can apply this law:

• For a concentration that is one-fourth of the original, it implies two half-lives have passed (i.e., \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)). Therefore, the time required is \(2 \times 25 \text{ s} = 50 \text{ s}\).
• For one-eighth, three half-lives are necessary (\(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)), resulting in \(3 \times 25 \text{ s} = 75 \text{ s}\).

Using the integrated rate law in conjunction with the knowledge of half-lives and the rate constant provides a strong framework for analyzing and predicting the behavior of first-order reactions over time.