Problem 41 - Chemistry The Molecular Science

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) A \(0.0200-\mathrm{M}\) solution of the hormone is stored at \(25^{\circ} \mathrm{C}\) for two months. Calculate its concentration at the end of that period. (b) Calculate how long it takes for the concentration of the solution to drop from \(0.0200 \mathrm{M}\) to \(0.00350 \mathrm{M}\). (c) Determine the half-life of the hormone.

Step-by-Step Solution

Answer

(a) Concentration is approximately 0.0160 M; (b) it takes roughly 195 days; (c) the half-life is approximately 2027 days.

Step 1: Understanding the First-Order Reaction

First-order reactions have a rate law of the form \( r = k[A] \), where \( k \) is the rate constant and \([A]\) is the concentration of the reactant. The relationship between the concentration and time for a first-order reaction is given by the equation \( [A] = [A]_0 e^{-kt} \).

Step 2: Calculate Concentration After Two Months

To find the concentration of the hormone after two months (approximately 60 days), use the equation for first-order kinetics.Given: Initial concentration \([A]_0 = 0.0200 \, \text{M}\), \( k = 3.42 \times 10^{-4} \, \text{day}^{-1} \), and \( t = 60 \, \text{days} \).Substitute these into the equation: \([A] = 0.0200 \, \text{M} \times e^{-3.42 \times 10^{-4} \, \text{day}^{-1} \times 60 \, \text{days}} \)After calculation, \([A] \approx 0.0160 \, \text{M}\).

Step 3: Calculate Time for Concentration to Drop to 0.00350 M

Use the integrated rate law equation again:\([A] = [A]_0 e^{-kt} \)Rearranging the formula to solve for \( t \):\(t = \frac{-\ln \left( \frac{[A]}{[A]_0} \right)}{k} \)Given \([A]_0 = 0.0200 \, \text{M}\), \([A] = 0.00350 \, \text{M}\), and \( k = 3.42 \times 10^{-4} \, \text{day}^{-1}\):\(t = \frac{-\ln \left( \frac{0.00350}{0.0200} \right)}{3.42 \times 10^{-4}} \)After calculation, \( t \approx 195 \, \text{days} \).

Step 4: Determine the Half-Life of the Hormone

For a first-order reaction, the half-life \( t_{1/2} \) is independent of the initial concentration and is given by:\(t_{1/2} = \frac{0.693}{k} \)Plugging in the value for \( k \):\(t_{1/2} = \frac{0.693}{3.42 \times 10^{-4} \, \text{day}^{-1}} \)After calculation, \( t_{1/2} \approx 2027 \, \text{days}\).

Key Concepts

Rate Constant

A rate constant is a pivotal concept in chemical kinetics, especially when dealing with first-order reactions. This constant, denoted as \( k \), signifies the speed at which a reaction proceeds. In simplest terms, it's like the throttle of a reaction, dictating how fast the reactants convert to products. For first-order reactions, the rate of reaction is directly proportional to the concentration of one reactant. The equation \( r = k[A] \) captures this idea, where \( r \) is the reaction rate and \([A]\) is the concentration. For our exercise, the rate constant \( k \) is given as \(3.42 \times 10^{-4} \text{ day}^{-1}\), a value indicating how quickly the hormone decomposes in water at \(25^{\circ} \mathrm{C}\). Understanding this value helps predict how the reactant concentrations will change over time.
Moreover, knowing the rate constant allows us to calculate critical aspects like how much of a reactant will remain after a certain period or how long it takes for the reactant concentration to drop to a specified level. As such, the value of \( k \) serves as the cornerstone for analysis and prediction in chemical kinetics.

Integrated Rate Law

The integrated rate law for a first-order reaction is a powerful tool that provides a relationship between the concentration of a reactant and the elapsed time. It translates chemical kinetics into an equation form, making it easier to calculate time-dependent concentration changes. The integrated rate law equation is: \[ [A] = [A]_0 e^{-kt} \]where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \( t \), and \( k \) is the rate constant.

• The equation expresses how the concentration declines exponentially over time.
• It allows us to calculate the concentration of a reactant remaining after any given time.
• It can also be rearranged to determine the time needed for the concentration to reach a certain level.

In our exercise, this equation was instrumental to determine the remaining hormone concentration after two months and to find out how long it takes for the concentration to decrease from \(0.0200 \mathrm{M}\) to \(0.00350 \mathrm{M}\).
By substituting known values into the integrated rate law, you can explore the behavior of the reaction through step-by-step calculations.

Half-Life Calculation

The concept of half-life is particularly intuitive in first-order reactions as it describes the time required for a reactant's concentration to decrease by half. This concept applies broadly, from radioactive decay to biological processes. For first-order reactions, the half-life formula is: \[ t_{1/2} = \frac{0.693}{k} \]This formula strikingly shows why half-life is unique in first-order reactions:

• It remains constant regardless of the starting concentration.
• It's solely dependent on the rate constant \( k \).
• This constancy makes it extremely useful for comparing different reactions or understanding how quickly a substance reduces to half its quantity.

In the provided exercise, substituting the rate constant \(3.42 \times 10^{-4} \text{ day}^{-1}\) into this formula resulted in a half-life of approximately 2027 days for the hormone.
This fact reveals a slowly degrading substance, highlighting the stabilizing nature and the long periods involved in the evolution of certain chemical and biological contexts. Recognizing the half-life allows comprehension of the persistence and time-resolved dynamics of a reaction.