Chapter 15: Problem 68
The gas-phase reaction $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ has an activation energy of \(103 \mathrm{kJ} / \mathrm{mol},\) and the rate constant is \(0.0900 \min ^{-1}\) at \(328.0 \mathrm{K}\). Find the rate constant at \(318.0 \mathrm{K}.\)
Short Answer
Step by step solution
Understand the Arrhenius Equation
Rearrange the Arrhenius Equation for Two Temperatures
Substitute Known Values
Calculate the Rate Constant at New Temperature
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Activation Energy
For example, in the provided gas-phase reaction, the activation energy is given as \( 103 \, \text{kJ/mol} \). This value indicates the amount of energy needed to initiate the conversion of \( \text{N}_2\text{O}_5 \) to \( \text{NO}_2 \) and \( \text{O}_2 \). Understanding and calculating activation energy can help predict the rate at which a reaction will occur.
Rate Constant
Temperature Dependence of Rates
The Arrhenius Equation provides a mathematical model this temperature dependence:\[ \ln \left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]This formula allows us to calculate a new rate constant \( k_2 \) at a different temperature \( T_2 \) using a known rate constant \( k_1 \) at temperature \( T_1 \). In the exercise provided, decreasing the temperature from \( 328.0 \, \text{K} \) to \( 318.0 \, \text{K} \) results in a new, slower rate constant of \( 0.00185 \, \text{min}^{-1} \). Understanding this relationship is key in controlling reaction speeds in industrial processes and laboratory settings.