Problem 38
Question
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 activation energy of the reaction?
Step-by-Step Solution
Verified Answer
The activation energy is approximately 92.36 kJ/mol.
1Step 1: Identify what's given
The problem gives you: - The rate constant triples, meaning \( k_2 = 3k_1 \).- The initial temperature \( T_1 = 300 \ \text{K} \).- The final temperature \( T_2 = 310 \ \text{K} \).
2Step 2: Write the Arrhenius equation
The Arrhenius equation is \( k = A e^{-E_a/(RT)} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is activation energy, \( R \) is the gas constant \( 8.314 \ \text{J/mol K} \), and \( T \) is the temperature in Kelvin.
3Step 3: Use the Arrhenius equation for two temperatures
For two different temperatures, we have:\( \ln \frac{k_2}{k_1} = \frac{-E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \).Substitute \( \ln 3 \) for \( \ln (k_2/k_1) \) since \( k_2 = 3k_1 \).
4Step 4: Solve for activation energy \( E_a \)
Rearrange the equation to solve for \( E_a \):\( E_a = -R \cdot \ln 3 \left(\frac{1}{T_2} - \frac{1}{T_1}\right)^{-1} \).Calculate \( \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{310} - \frac{1}{300} = -0.00010774 \ \text{K}^{-1} \).Use \( R = 8.314 \ \text{J/mol K} \) and \( \ln 3 \approx 1.0986 \).
5Step 5: Perform calculations
Substitute back values:\( E_a \approx 8.314 \cdot 1.0986 \cdot \frac{1}{-0.00010774} \).Calculate inside the parentheses: \( \approx 92355 \ \text{J/mol} \).Convert to kJ/mol: \( \approx 92.36 \ \text{kJ/mol} \).
Key Concepts
Arrhenius EquationRate ConstantTemperature DependenceChemical Kinetics
Arrhenius Equation
The Arrhenius Equation is a fundamental formula in chemistry that helps us understand the temperature dependence of reaction rates. It is expressed as \( k = A e^{-E_a/(RT)} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) represents the activation energy, \( R \) is the universal gas constant \( (8.314 \ \text{J/mol K}) \), and \( T \) is the absolute temperature in Kelvin.
The Arrhenius equation shows that the rate constant \( k \) increases exponentially with an increase in temperature \( T \), or a decrease in activation energy \( E_a \). This means that higher temperatures can significantly speed up the rate of a chemical reaction.
This equation also implies that the rate constant is sensitive to changes in both \( T \) and \( E_a \), making it crucial for predicting how reaction rates will change with varying conditions.
The Arrhenius equation shows that the rate constant \( k \) increases exponentially with an increase in temperature \( T \), or a decrease in activation energy \( E_a \). This means that higher temperatures can significantly speed up the rate of a chemical reaction.
This equation also implies that the rate constant is sensitive to changes in both \( T \) and \( E_a \), making it crucial for predicting how reaction rates will change with varying conditions.
Rate Constant
The rate constant \( (k) \) is a pivotal element in chemical kinetics because it dictates the speed of a chemical reaction. In the Arrhenius Equation, \( k \) relates to the other variables as it describes how fast the reactants convert to products under specific conditions.
A higher rate constant typically means a faster reaction. The value of \( k \) depends on temperature and the presence of catalysts, which might lower the activation energy \( E_a \), increasing \( k \).
In many reactions, the rate constant can change dramatically with small variations in temperature, showcasing the interconnected nature of temperature and reaction rate. Understanding how to calculate \( k \) and how it varies allows chemists to control and optimize reactions effectively.
A higher rate constant typically means a faster reaction. The value of \( k \) depends on temperature and the presence of catalysts, which might lower the activation energy \( E_a \), increasing \( k \).
In many reactions, the rate constant can change dramatically with small variations in temperature, showcasing the interconnected nature of temperature and reaction rate. Understanding how to calculate \( k \) and how it varies allows chemists to control and optimize reactions effectively.
Temperature Dependence
Temperature dependence in chemical reactions refers to the principle that reaction rates typically increase with a rise in temperature. According to the Arrhenius Equation, even a small increase in temperature can lead to a significant increase in the rate constant \( k \), thereby speeding up the reaction.
To quantify this impact, chemists use the Arrhenius plot or graph, which is a plot of \( \ln(k) \) against \( 1/T \). This graph delivers a straight line, where the slope corresponds to \( -E_a/R \).
To quantify this impact, chemists use the Arrhenius plot or graph, which is a plot of \( \ln(k) \) against \( 1/T \). This graph delivers a straight line, where the slope corresponds to \( -E_a/R \).
- The gradient helps determine the activation energy \( E_a \).
- Such analyses are key in both industrial and laboratory settings to ensure reactions are conducted under optimal conditions.
Chemical Kinetics
Chemical Kinetics is the branch of chemistry that studies the rates and the steps involved in chemical reactions. It investigates how different factors such as concentration, temperature, and presence of a catalyst influence the speed of a chemical process.
Through kinetics, we can explore the mechanism—a step-by-step sequence of reactions that explain how the reactants turn into products. This helps chemists design better materials and processes.
Through kinetics, we can explore the mechanism—a step-by-step sequence of reactions that explain how the reactants turn into products. This helps chemists design better materials and processes.
- Key concepts in kinetics, like the Arrhenius Equation and rate constants, provide insights into reaction dynamics.
- They help determine the feasible conditions under which a reaction will proceed efficiently.
- Kinetics is applied across various fields, including pharmaceuticals, environmental science, and nanotechnology, to optimize reactions and develop new technologies.
Other exercises in this chapter
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