Question

The first-order rate constant for the decomposition of a certain hormone in water at \(25^{\circ} \mathrm{C}\) is \(3.42 \times 10^{-4}\) day \(^{-1}\). (a) If a \(0.0200 \mathrm{M}\) solution of the hormone is stored at \(25^{\circ} \mathrm{C}\) for two months, what will its concentration be at the end of that period? (b) How long will it take for the concentration of the solution to drop from \(0.0200 M\) to \(0.00350 \mathrm{M} ?\) (c) What is the half-life of the hormone?

Short answer

Answer: The concentration of the hormone after two months is approximately 0.010 M. The time it takes for the concentration to drop from 0.0200 M to 0.00350 M is approximately 3427 days. The half-life of the hormone is approximately 2026 days.

Step-by-step solution

(Part a: Finding the concentration after two months)

You are given the initial concentration \([A]_0 = 0.0200M\) and the rate constant \(k=3.42\times10^{-4} \text{ day}^{-1}\). First, identify the time period (two months) and convert it to days: $$\text{2 months} = 2\times30 = 60 \text{ days}$$ Then, apply the first-order reaction formula, using the initial concentration, rate constant, and time: $$[A]_t = [A]_0 e^{-kt}$$ $$[A]_t = 0.0200 \times e^{-3.42 \times 10^{-4} \times 60}$$ Now, calculate the concentration after 60 days: $$[A]_t \approx 0.010 \; \text{M}$$ So, the concentration of the hormone after two months will be approximately \(0.010 \: \text{M}\).

(Part b: Finding the time for the concentration to drop)

We are given the initial concentration \([A]_0 = 0.0200 \: \text{M}\) and the final concentration \([A]_t = 0.00350 \: \text{M}\). Using the first-order reaction formula, we can find the time \(t\) required for this change: $$[A]_t = [A]_0 e^{-kt}$$ Next, rearrange the formula to solve for \(t\): $$t = \frac{\ln{\frac{[A]_t}{[A]_0}}}{-k}$$ Now, plug in the given values and calculate the time: $$t = \frac{\ln{\frac{0.00350}{0.0200}}}{-3.42 \times 10^{-4}} \approx 3427 \; \text{days}$$ Therefore, it will take approximately \(3427 \: \text{days}\) for the concentration of the hormone solution to drop from \(0.0200 \: \text{M}\) to \(0.00350 \: \text{M}\).

(Part c: Finding the half-life of the hormone)

To determine the half-life \(t_{1/2}\) for the hormone, use the first-order reaction half-life formula: $$t_{1/2} = \frac{\ln{2}}{k}$$ Now, plug in the given rate constant and calculate the half-life: $$t_{1/2} = \frac{\ln{2}}{3.42 \times 10^{-4}} \approx 2026 \; \text{days}$$ Thus, the half-life of the hormone is approximately \(2026 \: \text{days}\).

Key concepts

Rate Constant

In the realm of chemistry, the rate constant is a fundamental element that dictates the speed of a reaction. For first-order reactions, such as the decomposition of certain substances like the hormone in our example, the rate constant is especially significant. It is denoted by the symbol \(k\). This parameter has the unit of inverse time (e.g., \( ext{day}^{-1} \) or \( ext{second}^{-1} \)), reflecting that it describes how quickly the concentration of a reactant decreases over time.

For a first-order reaction, the relationship between concentration \([A]\) and time \(t\) is expressed through the equation \([A]_t = [A]_0 e^{-kt}\). Here, \([A]_0\) represents the initial concentration of the reactant, while \([A]_t\) is the concentration at time \(t\). The rate constant \(k\), therefore, serves as a multiplier that alters the exponent in this equation, impacting how fast the concentration diminishes.

In the exercise discussed, with a rate constant of \(3.42 \times 10^{-4} \text{ day}^{-1}\), we can quantify the hormone's decomposition rate at a specific temperature, ensuring precise calculations in predicting future concentrations.

Half-Life

The concept of half-life \(t_{1/2}\) is crucial when discussing first-order reactions. It refers to the time it takes for half of the reactant to be consumed or to decompose. For first-order reactions, this value is constant and does not depend on the concentration of the reactant, which makes it especially useful.

Mathematically, the half-life is calculated using the formula \(t_{1/2} = \frac{\ln{2}}{k}\), where \(\ln{2}\) represents the natural logarithm of 2, approximately equal to 0.693. This formula shows that the half-life is inversely proportional to the rate constant \(k\).

In the given problem, the half-life is calculated by substituting the rate constant: \(t_{1/2} = \frac{0.693}{3.42 \times 10^{-4}} \approx 2026\) days. This indicates that every 2026 days, the concentration of the hormone in the solution halves, allowing us to predict its stability over time and plan accordingly.

Concentration Change Over Time

Tracking how the concentration of a substance changes over time is a key part of understanding reaction kinetics. In first-order reactions, the concentration decreases exponentially with time, which means that the rate of decrease becomes slower as the concentration reduces. This is captured by the equation \( [A]_t = [A]_0 e^{-kt} \).

For practical tasks, like predicting the concentration of a reactant after a certain period, this equation is invaluable. Plugging in the initial concentration, the rate constant, and the elapsed time gives the remaining concentration. In our example, starting with an initial concentration of \(0.0200 \text{ M}\), we calculated the concentration after 60 days as approximately \(0.010 \text{ M}\).

Such calculations enable us to foresee how long a substance will remain active at a given concentration or when a critical lower limit might be reached. This is essential for industries like pharmaceuticals, where the potency and shelf-life of products are critical to effectiveness and safety.