Chapter 14: Problem 108
A second-order reaction has a rate constant of \(8.7 \times 10^{-4} /(M \cdot \mathrm{s})\) at \(30^{\circ} \mathrm{C}\). At \(40^{\circ} \mathrm{C}\), the rate constant is \(1.8 \times 10^{-3} /(M \cdot \mathrm{s}) .\) What are the activation energy and frequency factor for this reaction? Predict the value of the rate constant at \(45^{\circ} \mathrm{C}\)
Short Answer
Step by step solution
Understand the Arrhenius Equation
Convert Temperatures to Kelvin
Apply the Arrhenius Equation for Two Temperatures
Calculate Activation Energy
Calculate Frequency Factor A
Predict Rate Constant at 45°C
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Arrhenius equation
- \( k \) is the rate constant, indicating the speed of the reaction.
- \( A \) is the frequency factor, reflecting how many collisions have the correct orientation to lead to a reaction.
- \( E_a \) is the activation energy, or the minimum energy required for the reaction to occur.
- \( R \) is the universal gas constant, valued at 8.314 J/(mol·K).
- \( T \) is the temperature in Kelvin.
rate constant calculation
temperature dependence of reaction rates
- **Increased molecular energy:** With higher temperatures, molecules move faster, leading to more frequent and energetic collisions.
- **Overcoming activation energy:** More molecules possess the minimum required energy, \( E_a \), to undergo a successful collision that results in product formation.
- **Exponential growth in reaction rate:** Even a small increase in temperature can significantly raise \( k \), thus accelerating the reaction.