Question

Understanding the high-temperature formation and breakdown of the nitrogen oxides is essential for controlling the pollutants generated from power plants and cars. The first-order breakdown of dinitrogen monoxide to its elements has rate constants of \(0.76 / \mathrm{s}\) at \(727^{\circ} \mathrm{C}\) and \(0.87 / \mathrm{s}\) at \(757^{\circ} \mathrm{C}\). What is the activation energy of this reaction?

Short answer

The activation energy of the reaction is approximately 38.6 kJ/mol.

Step-by-step solution

Identify the Given Data

The problem provides two temperatures and their corresponding rate constants for the breakdown of dinitrogen monoxide (N\textsubscript{2}O) to its elements. Denote the rate constants as \( k_1 = 0.76 \, \text{/s} \) at \( T_1 = 727^{\text{ \text degree} }\text{C} \) and \( k_2 = 0.87 \, \text{/s} \) at \( T_2 = 757^{\text{\text degree}}\text{C} \). Convert these temperatures to Kelvin: \(T_1 = 727 + 273 = 1000 \, \text{K}\) and \(T_2 = 757 + 273 = 1030 \, \text{K}\).

Understand the Arrhenius Equation

The activation energy \(E_a\) can be found using the Arrhenius equation in its two-point form: \[ \frac{k_2}{k_1} = \frac{e^{ -\frac{E_a}{R T_2} }}{e^{ -\frac{E_a}{R T_1} }} \] Re-arrange it to: \[ \frac{k_2}{k_1} = e^{ \frac{E_a}{R} \big( \frac{1}{T_1} - \frac{1}{T_2}\big)} \]

Take the Natural Logarithm of Both Sides

Using the natural logarithm results in: \[ \text{ln}\bigg(\frac{k_2}{k_1}\bigg) = \frac{E_a}{R} \bigg(\frac{1}{T_1} - \frac{1}{T_2}\bigg) \] Known values: \( k_2 = 0.87 \, \text{/s}, \ k_1 = 0.76 \, \text{/s}, \ T_1 = 1000 \, \text{K}, \ T_2 = 1030 \, \text{K} \).

Calculate the Natural Logarithm Ratio

Plug in the rate constants: \[ \text{ln}\bigg(\frac{0.87}{0.76}\bigg) = \text{ln}(1.1447) \ = 0.1351 \]

Solve for Activation Energy

Plug in all values: \[ 0.1351 = \frac{E_a}{8.314} \bigg(\frac{1}{1000} - \frac{1}{1030}\bigg) \] Calculate the difference in temperatures: \[ \frac{1}{1000} = 0.001, \ \frac{1}{1030} \ ≈ 0.000970873786407769 \] Thus, the difference is approximately: \[ 0.001 - 0.000970873786407769 = 0.000029126213592231 \] or \( 2.9126 \times 10^{-5} \) Solving for \( E_a \): \[ 0.1351 = \frac{E_a}{8.314} \times 2.9126 \times 10^{-5} \] Re-arrange to solve for \( E_a \): \[ E_a = \frac{0.1351 \times 8.314}{2.9126 \times 10^{-5}} \] Thus, \[ E_a \ ≈ 38588.29 \ \text{J/mol} \ ≈ 38.6 \ \text{kJ/mol} \]

Key concepts

Arrhenius Equation

The Arrhenius equation helps understand how reaction rates change with temperature. It's written as: \[ k = A e^{ -\frac{E_a}{RT}} \] where:

• \text{\( k \): rate constant}
• \text{\( A \): frequency factor (pre-exponential factor)}
• \text{\( E_a \): activation energy}
• \text{\( R \): gas constant (8.314 J/(mol·K))}
• \text{\( T \): temperature in Kelvin}

We use this formula to find how varying temperatures affect reaction rates, and rearrange it to find unknowns like activation energy.

Nitrogen Oxide Breakdown

Nitrogen oxides (NOx) such as dinitrogen monoxide (\text{\( N_2O \)) are pollutants from combustion engines and industrial processes. Their breakdown is crucial in reducing environmental impact. For dinitrogen monoxide, it decomposes into nitrogen gas (\text{\( N_2 \)) and oxygen gas (\text{\( O_2 \)). Understanding its rate of breakdown helps in developing control strategies.

First-Order Reaction Kinetics

First-order reactions depend on the concentration of one reactant. For a first-order reaction, the rate law is: \[ \text{Rate} = k[A] \]Here, the rate is directly proportional to the concentration of substance \text{\( A \)). For example, the decomposition of dinitrogen monoxide has a rate constant indicating how fast the reaction occurs. This characteristic is exploited using the Arrhenius equation to determine the activation energy.

Temperature Conversion

Temperature needs to be in Kelvin for use in most chemical equations, including the Arrhenius equation. To convert Celsius to Kelvin, add 273: \[ T(K) = T(^{\text{\degree}}C) + 273 \]For example, converting 727° C to Kelvin:

• \( 727 + 273 = 1000 \text{ K} \)
• \( 757 + 273 = 1030 \text{ K} \)

Accurate conversion is essential for any temperature-dependent calculations.

Rate Constants

Rate constants (\text{\( k \)) are specific to a reaction and temperature. They quantify the speed at which reactants convert to products. In our exercise, the rate constants are 0.76 s⁻¹ at \text{1000 K} and 0.87 s⁻¹ at \text{1030 K}. These constants are integral to the Arrhenius equation and help calculate activation energy. They reflect how the reaction speed increases with temperature, indicating the process kinetics.