Problem 99 - Chemistry The Molecular Science

Question

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is first-order with a rate constant of \(2.5 \times 10^{-4} \mathrm{~s}^{-1}\) (a) Calculate the half-life for decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (b) Calculate how long it takes for the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(\frac{1}{32}\) of its original value. (c) Calculate how long it takes for the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop from \(3.4 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) to \(2.3 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\).

Step-by-Step Solution

Answer

(a) 2772 s, (b) 13863 s, (c) 15020 s.

Step 1: Identify the formula for half-life of a first-order reaction

The half-life \( t_{1/2} \) for a first-order reaction is calculated using the formula \( t_{1/2} = \frac{0.693}{k} \) where \( k \) is the rate constant.

Step 2: Calculate the half-life

Substitute the given rate constant \( k = 2.5 \times 10^{-4} \mathrm{~s}^{-1} \) into the formula: \[t_{1/2} = \frac{0.693}{2.5 \times 10^{-4}} = 2772 \mathrm{~s}\]

Step 3: Recognize the formula for concentration decrease in a first-order reaction

For a first-order reaction, use the integrated rate law: \[\ln \left(\frac{[A]_t}{[A]_0}\right) = -kt\]where \([A]_t\) is the concentration at time \( t \) and \([A]_0\) is the initial concentration.

Step 4: Calculate time for concentration to drop to \(\frac{1}{32}\) of its initial value

Here, \(\frac{[A]_t}{[A]_0} = \frac{1}{32}\). Substitute and rearrange:\[\ln \left(\frac{1}{32}\right) = -2.5 \times 10^{-4} \times t -3.4657 = -2.5 \times 10^{-4} \times t t = \frac{3.4657}{2.5 \times 10^{-4}} \approx 13862.8 \mathrm{~s}\]

Step 5: Determine time for concentration change from \(3.4 \times 10^{-3} \mathrm{~mol/L}\) to \(2.3 \times 10^{-5} \mathrm{~mol/L}\)

Using the integrated rate law:\[\ln \left(\frac{2.3 \times 10^{-5}}{3.4 \times 10^{-3}}\right) = -2.5 \times 10^{-4} \times t \ln \left(\frac{2.3}{340}\right) = -2.5 \times 10^{-4} \times t -3.7549 = -2.5 \times 10^{-4} \times t t = \frac{3.7549}{2.5 \times 10^{-4}} \approx 15019.6 \mathrm{~s}\]

Key Concepts

Rate constant

The rate constant, often denoted as \( k \), is a crucial parameter in the study of chemical kinetics. Specifically for first-order reactions, the rate constant is a measure of how quickly a reaction proceeds. It has units of \( s^{-1} \) which signifies that it is a rate with respect to time. In the decomposition of \( \mathrm{N}_2\mathrm{O}_5 \), a rate constant of \( 2.5 \times 10^{-4} \mathrm{~s}^{-1} \) is given. This implies that the reaction proceeds at a certain speed which can be quantified and compared across different reactions or conditions. Understanding the rate constant helps you determine how factors like temperature or pressure can affect the reaction rate. A larger rate constant means a faster reaction, and vice versa. For first-order reactions, the rate constant stays independent of concentration, offering a simplified scenario for study.

Half-life

The half-life of a reaction is the time required for the concentration of a reactant to reduce to half its initial value. In a first-order reaction like the decomposition of \( \mathrm{N}_2\mathrm{O}_5 \), the half-life \( t_{1/2} \) can be calculated using the formula:\[ t_{1/2} = \frac{0.693}{k} \]This equation highlights that the half-life of a first-order reaction is inversely proportional to its rate constant. That means a higher rate constant will result in a shorter half-life. Using the given rate constant of \( 2.5 \times 10^{-4} \mathrm{~s}^{-1} \), the half-life for \( \mathrm{N}_2\mathrm{O}_5 \) is approximately 2772 seconds.One interesting aspect of first-order reactions is that the half-life remains constant throughout the reaction, regardless of the starting concentration.

Integrated rate law

The integrated rate law for a first-order reaction is a mathematical expression that relates the concentration of a reactant to time. It is given by the formula:\[ \ln \left(\frac{[A]_t}{[A]_0}\right) = -kt \]Here, \([A]_t\) is the concentration of the reactant at a specific time \( t \), and \([A]_0\) is the initial concentration. The integration of the rate law provides a direct relationship to calculate how concentrations change over time.This equation helps in figuring out the amount of time it will take for a reactant to reach a certain concentration or a specific fraction of its original amount. In practical terms, this is valuable for predicting how much time a reaction will need to proceed to any given point, such as when a reactant concentration is to fall to \( \frac{1}{32} \) of its initial concentration, as seen with \( \mathrm{N}_2\mathrm{O}_5 \).

Concentration change in reactions

In first-order reactions, changes in concentration over time can be accurately predicted using the integrated rate law. With this law, we can calculate the time required for a reactant to decompose to a particular concentration. Take, for example, the decomposition of \( \mathrm{N}_2\mathrm{O}_5 \) from \( 3.4 \times 10^{-3} \mathrm{~mol/L} \) to \( 2.3 \times 10^{-5} \mathrm{~mol/L} \).By substituting these values into the integrated rate law:\[ \ln \left(\frac{2.3 \times 10^{-5}}{3.4 \times 10^{-3}}\right) = -2.5 \times 10^{-4} \times t \]You can solve for \( t \) to find that approximately 15019.6 seconds are needed for this concentration change. This application showcases the strength of the integrated rate law in making concentration predictions over time and providing insights about how long a reaction takes, under specified conditions, to proceed to a certain extent.