Chapter 13: Problem 67
A reaction of the form \(a \mathrm{~A} \longrightarrow\) Products is second order with a rate constant of \(0.164 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s}) .\) What is the half-life, in seconds, of the reaction if the initial concentration of \(\mathrm{A}\) is \(5.55 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?\)
Short Answer
Step by step solution
Understand the Concept of Half-life for a Second-order Reaction
Plug in the Values into the Half-life Formula
Calculate the Half-life
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Half-Life
- \( t_{1/2} = \frac{1}{k[A]_0} \)
Rate Constant
The measured value of \( k \) can offer insights into the environmental conditions and the energetic landscape of the reaction. For example, in the original exercise, \( k \) is provided as \( 0.164 \) L/(mol·s), which helps calculate the half-life when paired with the initial concentration.
- Higher rate constants indicate faster reactions.
- For second-order reactions, find \( k \) in the half-life formula: \( t_{1/2} = \frac{1}{k[A]_0} \).
Second-Order Kinetics
- \( ext{Rate} = k[A]^2 \) or \( ext{Rate} = k[A][B] \)
Unlike first-order reactions, second-order reactions involve two molecules interacting. Therefore, understanding these interactions is crucial for predicting how changes in concentration affect the system's behavior over time.
Chemical Kinetics
- Kinetics helps in identifying mechanisms and pathways of reactions.
- It guides us in modifying reaction conditions to optimize desired outcomes.
- Practical applications include the synthesis of pharmaceuticals, material design, and understanding biological processes.