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When examining a first-order reaction, one of the fundamental tasks is to determine the rate constant, known as \(k\). This rate constant is crucial because it helps us understand the speed at which the reaction occurs. To find \(k\), we utilize the first-order kinetics equation:\[\ln \left(\frac{[A]}{[A]_0}\right) = -kt\]where \([A]\) is the concentration at time \(t\), and \([A]_0\) is the initial concentration.
For the problem, the concentrations change from \(0.451\, M\) to \(0.235\, M\) over \(131\) seconds. Plugging these values into the equation gives us:- \(\ln \left(\frac{0.235}{0.451}\right) = -k \times 131\)- Solving for \(k\), we rearrange the equation to:\[k = -\frac{\ln(0.235/0.451)}{131}\]This calculation gives us a rate constant \(k\) of approximately \(0.005290\, \text{s}^{-1}\).
Knowing \(k\) allows us to predict future concentrations of the reactant at any given time, which is extremely valuable in various chemical applications.

First-order reactions are characterized by their predictable concentration changes. The concentration of the reactant decreases exponentially over time. In this context, the concentration change reflects the exponential nature of these reactions, making calculations straightforward when you have the rate constant.
To find how long it takes for a reactant to drop to a certain level, we use the same first-order equation:\[\ln \left(\frac{[A]}{[A]_0}\right) = -kt\]In our exercise, we previously found \(k = 0.005290\, \text{s}^{-1}\). Having determined \([A]_0 = 0.235\, M\) and \([A] = 0.100\, M\), we use the equation to find time \(t\) by plugging in these values:- \(\ln \left(\frac{0.100}{0.235}\right) = -0.005290 \times t\)- Solving for \(t\), we find:\[t = -\frac{\ln(0.100/0.235)}{0.005290}\]This results in \(t\) being approximately \(159.575\) seconds.
Therefore, it takes about \(160\) seconds to reach a new concentration of \(0.100\, M\) from \(0.235\, M\).

Exponential decay is a crucial concept in chemistry, especially in reactions involving the breakdown of reactants over time. In first-order reactions, this decay pattern describes how the concentration of a substance decreases at a rate proportional to its current concentration.
The formula \[\ln \left(\frac{[A]}{[A]_0}\right) = -kt\]governs this decay. The design of this formula illustrates that the relationship between concentration and time forms a straight line when plotted on a logarithmic scale. This linearity makes it easier to analyze and predict how long it will take for a substance's concentration to decrease to a particular level.
The implications of exponential decay extend beyond theoretical chemistry. It's influential in fields such as pharmacokinetics and environmental science. For example, knowing how long it takes for a drug to decrease to an ineffective concentration can help in designing proper dosage regimens. Similarly, understanding decay patterns can aid in predicting the persistence of pollutants in the environment.
In summary, mastering the concept of exponential decay in chemistry enables us to better predict and manage chemical processes, making it a fundamental tool in scientific research and practical applications.