Chapter 14: Problem 74
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 K. Find the rate constant at \(318.0 \mathrm{K}\).
Short Answer
Step by step solution
Understand the Arrhenius Equation
Use Arrhenius Equation Rearranged
List Known Variables
Calculate the Factor \( \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \)
Compute \( \ln \left( \frac{k_2}{k_1} \right) \)
Solve for \( k_2 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Activation Energy
- Think of it as the initial push needed to get a ball rolling down a hill.
- Chemical reactions rely on this energy to help break and form chemical bonds.
A higher activation energy implies a slower reaction at a given temperature, since more energy is needed to overcome the barrier, while a lower activation energy suggests a faster reaction.
Rate Constant
- It is influenced by factors like temperature and activation energy.
- The Arrhenius equation shows how \(k\) changes with temperature: \\[ k = A e^{-\frac{E_a}{RT}} \\]
In the exercise, the rate constant changes with temperature, showcasing its dependence on both temperature and activation energy.
Gas Constant
- The value of \(R\) is \(8.314\) J/mol·K.
- It is used to convert energy scales in the Arrhenius equation.
In calculations, \(R\) helps determine how factors like activation energy affect reaction rates.
Temperature Dependence of Reaction Rates
- The relationship is captured by the equation:\( k = A e^{-\frac{E_a}{RT}} \)
- An increase in temperature results in a higher rate constant \(k\), showing faster reactions.
In the example exercise, you can see how a small change in temperature (from 328 K to 318 K) affects the rate constant, highlighting the temperature sensitivity of reaction rates.