Chapter 14: Problem 24
The rate constant for the second-order reaction $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is \(0.54 / M \cdot \mathrm{s}\) at \(300^{\circ} \mathrm{C}\). (a) How long (in seconds) would it take for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(0.62 M\) to \(0.28 M ?\) (b) Calculate the half-lives at these two concentrations.
Short Answer
Step by step solution
Understanding the Reaction Order
Using the Integrated Rate Law
Calculating Time for Concentration Change
Calculating Half-Life at \([NO_2] = 0.62 \, M\)
Calculating Half-Life at \([NO_2] = 0.28 \, M\)
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.
Rate Constant
Integrated Rate Law
- \([A]_t\) is the concentration at time \(t\)
- \([A]_0\) is the initial concentration
- \(k\) is the rate constant