Problem 37
Question
Calculate the activation energy, \(E_{a}\) for the reaction $$ 2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ from the observed rate constants: \(k\) at \(25^{\circ} \mathrm{C}=\) \(3.46 \times 10^{-5} \mathrm{s}^{-1}\) and \(k\) at \(55^{\circ} \mathrm{C}=1.5 \times 10^{-3} \mathrm{s}^{-1}\).
Step-by-Step Solution
Verified Answer
The activation energy, \(E_a\), is 101.0 kJ/mol.
1Step 1: Understand the Arrhenius Equation
The Arrhenius Equation relates the rate constant \(k\) to the temperature \(T\) and activation energy \(E_a\): \[ k = A e^{-E_a / (RT)} \] where \(A\) is the frequency factor, \(R\) is the universal gas constant \(8.314 \, \text{J/mol} \, \text{K}\), and \(T\) is the temperature in Kelvin.
2Step 2: Convert Temperatures to Kelvin
Convert temperatures from degrees Celsius to Kelvin by adding 273.15. For 25°C: \[ T_1 = 25 + 273.15 = 298.15 \, \text{K} \] and for 55°C: \[ T_2 = 55 + 273.15 = 328.15 \, \text{K} \].
3Step 3: Set Up the Arrhenius Equation Ratio
Using the points \((k_1,T_1)\) and \((k_2,T_2)\), the ratio of the two Arrhenius equations can be written as: \[ \frac{k_2}{k_1} = \frac{A e^{-E_a / (R T_2)}}{A e^{-E_a / (R T_1)}} \, \] which simplifies to \[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \].
4Step 4: Substitute Known Values
Substitute \(k_1 = 3.46 \times 10^{-5} \, \text{s}^{-1}\), \(k_2 = 1.5 \times 10^{-3} \, \text{s}^{-1}\), \(T_1 = 298.15 \, \text{K}\), \(T_2 = 328.15 \, \text{K}\), and \(R = 8.314 \, \text{J/mol} \, \text{K}\) into the simplified equation. Calculate \( \ln\left(\frac{k_2}{k_1}\right) \) and \( \frac{1}{T_2} - \frac{1}{T_1} \).
5Step 5: Solve for Activation Energy \(E_a\)
Calculate \( \ln\left(\frac{1.5 \times 10^{-3}}{3.46 \times 10^{-5}}\right)\) which simplifies to \( \ln(43.352) = 3.768 \). Then compute \(\frac{1}{328.15} - \frac{1}{298.15} = -3.036 \times 10^{-4} \, \text{K}^{-1}\). Solve the equation: \[ 3.768 = \frac{-E_a}{8.314}(-3.036 \times 10^{-4}) \] which results in \(E_a = 101.0 \, \text{kJ/mol}\).
Key Concepts
Arrhenius equationrate constanttemperature conversionuniversal gas constant
Arrhenius equation
The Arrhenius equation is a central concept in chemical kinetics, providing insight into the temperature dependence of reaction rates. This equation is given by \[ k = A e^{-E_a / (RT)} \]where \(k\) is the rate constant, \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
- The frequency factor \(A\) represents the frequency of collisions with the correct orientation for a reaction to occur.
- Activation energy \(E_a\) is the minimum energy that must be provided for a chemical reaction to occur.
- The universal gas constant \(R\) is a constant with the value \(8.314 \, \text{J/mol} \, \text{K}\).
rate constant
The rate constant \(k\) is a critical parameter in the field of chemistry, determining the speed at which a chemical reaction occurs. It quantifies the rate of a reaction and is derived from the Arrhenius equation:
- Higher values of \(k\) suggest faster reactions as they represent a greater number of successful collisions per unit time.
- By analyzing changes in \(k\) at different temperatures, we can better understand a reaction's kinetics and calculate the activation energy \(E_a\).
- In the equation \(\ln(k) = \ln(A) - \frac{E_a}{RT}\), adjusting temperature \(T\) affects the value of \(k\), demonstrating the relationship between heat and reaction speed.
temperature conversion
Temperature conversion is essential when dealing with chemical kinetics, especially in the context of the Arrhenius equation, where temperature appears in the exponential term. Temperature in chemistry is typically noted in Kelvin.
- The Kelvin scale is used because it starts at absolute zero, providing an absolute measure of temperature.
- To convert from degrees Celsius to Kelvin, simply add 273.15: \[ T_{\text{K}} = T_{\text{°C}} + 273.15 \]
- Using Kelvin allows equations like the Arrhenius equation to remain consistent and accurate across all scientific calculations.
universal gas constant
The universal gas constant \(R\) is a fundamental constant in both physics and chemistry. It plays a crucial role in the Arrhenius equation by linking the energy scale to temperature in Kelvin.
- The value of \(R\) is \(8.314 \, \text{J/mol} \, \text{K}\), making it a pivotal factor in calculations involving moles and temperature.
- It appears in many important equations, such as the ideal gas law \(PV = nRT\), and is also used to relate activation energy to the temperature and rate constant in kinetic analyses.
- Because it is expressed in Joules per mole per Kelvin, \(R\) ensures that various thermodynamic and kinetic calculations remain dimensionally consistent.
Other exercises in this chapter
Problem 33
Gaseous \(\mathrm{NO}_{2}\) decomposes at \(573 \mathrm{K}\) $$ \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})
View solution Problem 35
For the reaction \(\mathrm{C}_{2} \mathrm{F}_{4} \rightarrow^{1 / 2} \mathrm{C}_{4} \mathrm{F}_{8,}\) a graph of \(1 /\left[\mathrm{C}_{2} \mathrm{F}_{4}\right]
View solution Problem 38
If the rate constant for a reaction triples when the temperature rises from \(3.00 \times 10^{2} \mathrm{K}\) to \(3.10 \times 10^{2} \mathrm{K}\) what is the a
View solution Problem 39
When heated to a high temperature, cyclobutane, \(\mathrm{C}_{4} \mathrm{H}_{8},\) decomposes to ethylene: $$ \mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{g}) \rightar
View solution