Problem 25 - Chemistry and Chemical Reactivity

Question

The rate equation for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (giving \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) ) is Rate \(=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] .\) The value of \(k\) is \(6.7 \times 10^{-5} \mathrm{s}^{-1}\) for the reaction at a particular temperature. (a) Calculate the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\). (b) How long does it take for the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration to drop to one tenth of its original value?

Step-by-Step Solution

Answer

(a) The half-life is 10343 seconds. (b) It takes approximately 34395 seconds to drop to one tenth.

Step 1: Identify the Reaction Order

The rate equation given is \( ext{Rate} = k [ ext{N}_2 ext{O}_5] \), which identifies this as a first-order reaction. This type of reaction depends on the concentration of a single reactant raised to the first power.

Step 2: Apply the First-Order Half-Life Formula

For a first-order reaction, the half-life \( t_{1/2} \) is calculated using the formula: \[ t_{1/2} = \frac{0.693}{k} \] Substitute \( k = 6.7 \times 10^{-5} \text{s}^{-1} \) into the formula.

Step 3: Calculate the Half-Life

Substitute the given value of \( k \) into the formula: \[ t_{1/2} = \frac{0.693}{6.7 \times 10^{-5} \text{s}^{-1}} \] Calculate: \[ t_{1/2} \approx 10343 \text{s} \] Thus, the half-life of \( ext{N}_2 ext{O}_5 \) is approximately 10343 seconds.

Step 4: Use First-Order Reaction Formula for Concentration

The formula for a first-order reaction 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. We want the final concentration to be \( \frac{1}{10} [A]_0 \).

Step 5: Set Up the Equation for Concentration Drop

Substitute into the first-order formula: \[ \frac{1}{10}[A]_0 = [A]_0 e^{-kt} \] Cancel \([A]_0\) from both sides: \[ \frac{1}{10} = e^{-kt} \] Take the natural logarithm of both sides to solve for \( t \): \[ -kt = \ln\left(\frac{1}{10}\right) \] which simplifies to \[ t = -\frac{\ln\left(\frac{1}{10}\right)}{k} \]

Step 6: Calculate Time for Concentration to Drop

Now substitute the value of \( k \):\[ t = -\frac{\ln\left(\frac{1}{10}\right)}{6.7 \times 10^{-5} \text{s}^{-1}} \] Calculate: \[ t \approx \frac{2.302}{6.7 \times 10^{-5} \text{s}^{-1}} \] \[ t \approx 34395 \text{s} \] It takes approximately 34395 seconds for the concentration of \( \text{N}_2 ext{O}_5 \) to drop to one tenth of its original value.

Key Concepts

Rate Equation of First-Order Reactions

In chemistry, understanding the rate equation is crucial for analyzing how fast a reaction occurs. When looking at first-order reactions, like the decomposition of \( \text{N}_2\text{O}_5 \), the rate equation is expressed as: \( \text{Rate} = k[A] \). Here, \( [A] \) represents the concentration of the reactant, and \( k \) is the rate constant. This equation shows that for a first-order reaction, the rate is directly proportional to the concentration of one reactant. That means, as the concentration of the reactant decreases, the rate at which the reaction proceeds also decreases.

First-order reactions are quite common and are characterized by a linear relationship between the concentration of the reactant and time when plotted on a semilogarithmic graph. These reactions are different from zero-order or second-order reactions, where the relationship between the concentration of the reactant and the rate varies differently. The simplicity of the rate equation for first-order reactions is one of the reasons why they are often used as an example when learning about chemical kinetics.

Half-Life Calculation for First-Order Reactions

The concept of half-life is a significant aspect of first-order reactions. The half-life (\( t_{1/2} \)) is the time required for half of the reactant to be consumed in a reaction. Notably, for first-order reactions, the half-life is constant and does not depend on the initial concentration. This simplicity makes it easier to predict how long a reaction will take.

To calculate the half-life of a first-order reaction, you can use the formula: \[ t_{1/2} = \frac{0.693}{k} \] where \( k \) is the rate constant. For instance, if you have \( k = 6.7 \times 10^{-5} \ \text{s}^{-1} \), you plug that into the equation: \[ t_{1/2} = \frac{0.693}{6.7 \times 10^{-5}} \approx 10343 \ \text{s} \] This result tells you the half-life of \( \text{N}_2\text{O}_5 \) in this reaction scenario. Understanding the half-life helps predict how long the reaction will last and is particularly useful in fields like pharmacology or nuclear chemistry, where timing is essential.

Understanding the Rate Constant

The rate constant, represented as \( k \), plays a pivotal role in the kinetics of a reaction. Each reaction has its specific rate constant, influenced by factors like temperature and the presence of a catalyst. In the rate equation for a first-order reaction, \( \text{Rate} = k[A] \), \( k \) determines how quickly the reaction proceeds. The units of \( k \) in a first-order reaction are typically \( \text{s}^{-1} \), reflecting the time-based nature of these reactions.

It’s important to note that the rate constant is not constant across different conditions unless temperature and other influencing factors remain unchanged. Hence, if the temperature changes, you will likely need to recalculate \( k \) using experimental data. Moreover, the rate constant serves as a crucial piece of information when using the Arrhenius equation, which predicts how a reaction rate changes with temperature.

• The magnitude of \( k \) indicates the speed of the reaction: a larger \( k \) suggests a faster reaction, whereas a smaller \( k \) indicates a slower one.
• Calculating \( k \) accurately allows for better control and prediction of the reaction pathway.

In summary, the rate constant is essential for understanding the dynamics of first-order reactions and helps in designing experiments and predicting outcomes.