Chapter 14: Problem 2
In the single-step, second-order reaction \(2 \mathrm{~A} \rightarrow\) products, how would a graph of [A] versus time compare to a plot of \(1 /[\mathrm{A}]\) versus time? Which of these would be the most similar to the same set of graphs for A during the single-step, second-order reaction \(\mathrm{A}+\mathrm{B} \rightarrow\) products? Explain.
Short Answer
Step by step solution
Consider the Reaction Type
Graph [A] vs. Time for 2A -> Products
Graph 1/[A] vs. Time for 2A -> Products
Compare to A + B -> Products
Analyze Similarities and Differences
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Reaction Rate
- A faster reaction has a higher reaction rate and occurs in a shorter time period.
- A slower reaction has a lower reaction rate and takes longer to occur.
For second-order reactions, which include reactions like \(2 \text{A} \rightarrow \text{products}\), the rate is influenced quadratically by the concentration of the reactant \(A\). The rate law for this second-order reaction would be expressed as \( \text{rate} = k[A]^2 \), explaining that as the concentration of \(A\) decreases, the rate of reaction slows down noticeably. This quadratic relationship means that small changes in \([A]\) can have a significant impact on the rate.
Rate Law
In the case of the second-order reaction \(2 \text{A} \rightarrow \text{products}\), the rate law is defined as \( \text{rate} = k[A]^2 \), where:
- \(k\) is the rate constant, unique to each reaction and depends on factors like temperature and pressure.
- \([A]\) is the concentration of reactant \(A\).
This law helps in predicting how the concentration of \(A\) affects the rate and is key to understanding the kinetics of the reaction. As \([A]\) decreases, the reaction rate decreases more rapidly due to the square relationship.
Concentration Changes in Reactions
For a second-order reaction like \(2 \text{A} \rightarrow \text{products}\), the concentration of \(A\) plays a pivotal role:
- The decrease in \([A]\) causes the reaction to slow down as there are fewer molecules available to react.
- The concentration decreases more sharply at the beginning compared to the end of the reaction.
Additionally, the plot of \(1/[A]\) versus time provides a linear graph, which reflects the second-order kinetics and can be used to determine the rate constant \(k\) visually from the slope of the line.
Kinetics of Multi-Reactant Systems
Consider the reaction \( \text{A} + \text{B} \rightarrow \text{products} \). If \([B]\) is in large excess or remains constant, the reaction can still follow second-order kinetics relative to \(A\), similar to \(2 \text{A} \rightarrow \text{products}\).
- When \([B]\) is constant, the overall reaction appears to depend only on \([A]\).
- This simplifies the kinetic analysis, allowing a linear plot of \(1/[A]\) versus time, just like in a single reactant second-order reaction.