Problem 62 - Chemistry: The Central Science

Question

Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines. The decomposition of nitric oxide \((\mathrm{NO})\) to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is second order with a rate constant of \(0.0796 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(737^{\circ} \mathrm{C}\) and \(0.0815 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(947{ }^{\circ} \mathrm{C}\). Calculate the activation energy for the reaction.

Step-by-Step Solution

Answer

The activation energy for the decomposition reaction of nitric oxide (NO) to N₂ and O₂ is approximately -114.8 kJ/mol.

Step 1: Write down the Arrhenius equation

The Arrhenius equation relates the rate constant (k) to temperature (T) and activation energy (Ea): \(k = Ae^{-Ea / RT}\) Where A is the pre-exponential factor (or frequency factor), R is the gas constant (8.314 J/mol K), and T is the temperature in Kelvin. For this exercise, we have the rate constants and temperatures given, so we'll need to convert these temperatures to Kelvin and then find the activation energy (Ea).

Step 2: Convert the given temperatures to Kelvin

To convert Celsius to Kelvin, simply add 273.15 to the given temperatures: Temperature (T1): \(737^\circ \mathrm{C} = 737 + 273.15 = 1010.15\ \mathrm{K}\) Temperature (T2): \(947^\circ \mathrm{C} = 947 + 273.15 = 1220.15\ \mathrm{K}\)

Step 3: Use the Arrhenius equation for both temperatures

We can use the Arrhenius equation for both temperatures T1 and T2: \(k_1 = Ae^{-Ea / (R \cdot T_1)}\) \(k_2 = Ae^{-Ea / (R \cdot T_2)}\) Where \(k_1\) is the rate constant at T1 and \(k_2\) is the rate constant at T2.

Step 4: Divide the two Arrhenius equations

Divide the first equation by the second equation to eliminate the pre-exponential factor (A) and simplify: \(\frac{k_1}{k_2} = \frac{e^{-Ea / (R \cdot T_1)}}{e^{-Ea / (R \cdot T_2)}}\)

Step 5: Rearrange the equation for Ea

Next, we can simplify by taking the natural logarithm of both sides and rearranging the equation to solve for Ea: \(ln(\frac{k_1}{k_2}) = \frac{Ea}{R} (\frac{1}{T_2} - \frac{1}{T_1})\) Ea = R * ln(\(\frac{k_1}{k_2}\)) * \(\frac{T_1 \cdot T_2}{T_2 - T_1}\)

Step 6: Substitute the given rate constants and temperatures

Now, substitute the given rate constants and temperatures into the equation: Ea = 8.314 * ln(\(\frac{0.0796}{0.0815}\)) * \(\frac{1010.15 \cdot 1220.15}{1220.15 - 1010.15}\)

Step 7: Calculate the activation energy

Finally, calculate the activation energy: Ea ≈ 8.314 * -0.0247 * 5568.82 Ea ≈ -114.8 kJ/mol The activation energy for the decomposition reaction of nitric oxide (NO) is approximately -114.8 kJ/mol.

Key Concepts

Arrhenius Equation

The Arrhenius equation is a fundamental formula used in chemistry to describe how the rate constant of a reaction changes with temperature. It is expressed as:\[ k = Ae^{-Ea / RT} \]where:

• k is the rate constant.
• A represents the pre-exponential factor or frequency factor, which indicates the frequency of collisions that lead to a reaction.
• E_a is the activation energy, the minimum energy required for the reaction to occur.
• R is the ideal gas constant, approximately 8.314 J/mol K.
• T is the temperature in Kelvin.

This equation highlights the impact of temperature on the speed of chemical reactions. A higher temperature generally increases the rate constant, bringing about faster reactions.

Rate Constants

Rate constants are crucial to describing how quickly a reaction proceeds. They are specific to each reaction and depend on factors like temperature, pressure, and the presence of a catalyst.

The rate constant \(k\) is used together with the concentration of reactants in a rate law to predict the rate at which a reaction occurs. In the context of the Arrhenius equation, rate constants are linked to the activation energy and temperature.

For example, in the decomposition of nitric oxide, the rate constants at different temperatures help us understand how the reaction rate changes, assisting in calculations such as finding the activation energy.

Nitrogen Oxides Decomposition

The decomposition of nitrogen oxides, specifically nitric oxide ( O ), into nitrogen ( N ) and oxygen ( O ) gases is a crucial chemical process, often used as an example for studying second-order reactions.

This reaction's progress can significantly influence environmental conditions, since nitrogen oxides are pollutants found in vehicle emissions. Understanding the decomposition helps in devising methods to control these emissions, thus reducing pollution and protecting environmental health.

The rate law for this reaction is second order, meaning it depends on the concentration of nitric oxide squared, which is important for students to understand when performing kinetic analyses.

Temperature Conversion

Temperature conversion between Celsius and Kelvin is a straightforward but essential process in chemistry. To convert a temperature from Celsius to Kelvin, one simply adds 273.15 to the Celsius measurement. This is crucial in calculations involving the Arrhenius equation, where temperature must be in Kelvin.

For the problem of nitric oxide decomposition, converting the temperatures from Celsius to Kelvin allows accurate use of the Arrhenius equation, ensuring the right calculations for finding activation energy at different temperatures.

Learning how to switch effortlessly between these temperature scales is important for anyone studying chemistry, as many formulas require Kelvin.

Second Order Reaction

A second order reaction refers to a chemical reaction where the rate is proportional to the concentration of one reactant squared or the product of the concentrations of two reactants.

The rate law for a second order reaction is expressed as:\[ rate = k[A]^2 \]where:

• k is the rate constant.
• [A] represents the concentration of the reactant.

In the decomposition of nitric oxide, it is a second order in \( ext{NO} \), highlighting the importance of concentration changes in the rate of reactions. These reactions tend to slow down more noticeably as the reactant is consumed compared to first-order reactions.

Understanding the order of a reaction is fundamental for predicting how the reaction will proceed under different conditions and can be a powerful tool in studying complex chemical processes.