Problem 70 - Chemistry: Principles and Reactions

Question

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to \(\mathrm{NO}_{2}\) and \(\mathrm{NO}_{3}\) is a first-order gas-phase reaction. At \(25^{\circ} \mathrm{C}\), the reaction has a half-life of \(2.81\) s. At \(45^{\circ} \mathrm{C}\), the reaction has a half-life of \(0.313 \mathrm{~s}\). What is the activation energy of the reaction?

Step-by-Step Solution

Answer

Answer: The activation energy of the reaction is approximately 60,535 J/mol.

Step 1: Write down the given information

We are given the following information: 1. The decomposition of N2O5 is a first-order reaction. 2. The half-life at 25°C is 2.81 seconds. 3. The half-life at 45°C is 0.313 seconds. Let's denote the half-life at 25°C as t₁ and the half-life at 45°C as t₂. t₁ = 2.81 s t₂ = 0.313 s T₁ = 25°C + 273.15 = 298.15 K T₂ = 45°C + 273.15 = 318.15 K

Step 2: Find the rate constants (k)

The relationship between the half-life (t) and the rate constant (k) for a first-order reaction is given by: t = ln(2) / k We can use this equation to find the rate constants at both temperatures. k₁ = ln(2) / t₁ = ln(2) / 2.81 = 0.2468 s⁻¹ k₂ = ln(2) / t₂ = ln(2) / 0.313 = 2.2135 s⁻¹

Step 3: Write down the Arrhenius equation

The Arrhenius equation is given by: k = Ae^(-Ea / RT) where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin.

Step 4: Write down the ratio of the rate constants

Dividing the Arrhenius equations for both temperatures, we get: k₁ / k₂ = e^((Ea / R) * (1 / T₁ - 1 / T₂))

Step 5: Solve for the activation energy (Ea)

Rearrange the equation obtained in Step 4 to solve for Ea: Ea = R * ln(k₁ / k₂) / (1 / T₂ - 1 / T₁) Plug in the values and calculate the activation energy: Ea = 8.314 J/mol·K * ln(0.2468 s⁻¹ / 2.2135 s⁻¹) / (1 / 318.15 K - 1 / 298.15 K) Ea ≈ 60535 J/mol This is the answer: The activation energy of the reaction is approximately 60,535 J/mol.

Key Concepts

First-Order Reaction

A first-order reaction is one where the rate of reaction is directly proportional to the concentration of one reactant. This means only one mole of a reactant needs to break down over time. These reactions have a characteristic behavior where their half-life is constant regardless of the concentration of the reactant. The half-life, in this case, is the time it takes for half of the reactant to be consumed. This simple proportionality makes first-order reactions quite predictable and easier to analyze, especially when paired with mathematical models.

In the decomposition of \( \mathrm{N}_2\mathrm{O}_5 \), we find that the reaction is first-order, meaning the rate at which \( \mathrm{N}_2\mathrm{O}_5 \) breaks down is consistent and measured easily through its changing concentration over time.

Arrhenius Equation

The Arrhenius Equation is a fundamental relationship used to determine the effect of temperature on the rate constant of a chemical reaction. It is given by the formula \( k = Ae^{-Ea / RT} \), where:

• \( k \) is the rate constant, defining the speed of the reaction.
• \( A \) is the pre-exponential factor, which accounts for the frequency of collisions leading to a reaction.
• \( Ea \) is the activation energy required to initiate the reaction.
• \( R \) is the universal gas constant.
• \( T \) is the temperature in Kelvin.

The Arrhenius equation shows that even small changes in temperature can significantly impact the rate of a reaction. This relationship is crucial for understanding how reactions change under different conditions.

In calculating the activation energy for the decomposition of \( \mathrm{N}_2\mathrm{O}_5 \), the Arrhenius equation helps in connecting temperature and rate constant to reveal how energy barriers affect reaction speed.

Half-Life Calculation

Half-life is essential for understanding the kinetics of first-order reactions. For these reactions, the half-life (\( t_{1/2} \)) is calculated using the equation \( t = \frac{\ln(2)}{k} \). This formula shows that for first-order reactions, the half-life is independent of the initial concentration.

• At 25°C, the half-life was found to be 2.81 seconds.
• At 45°C, it was 0.313 seconds.

These values can be plugged directly into the formula to solve for the rate constant \( k \), reflecting how quickly a chemical species is being consumed. Understanding half-life simplifies tracking the decomposition over time and is a practical measure in predicting the life span of a reaction under set conditions.

Rate Constant Determination

Determining the rate constant \( k \) is crucial in analyzing reaction kinetics. For a first-order reaction, the relationship between the half-life and \( k \) is straightforward. With the given half-lives at different temperatures, we calculate:

• \( k_1 = \frac{\ln(2)}{t_1} = \frac{\ln(2)}{2.81} \approx 0.2468 \text{s}^{-1} \).
• \( k_2 = \frac{\ln(2)}{t_2} = \frac{\ln(2)}{0.313} \approx 2.2135 \text{s}^{-1} \).

These calculations provide insight into how quickly the reaction occurs under specific conditions. The rate constant offers a clearer understanding of the reaction’s dynamics and is indispensable for further analyses, such as calculating activation energy or predicting reaction timelines.