The decomposition of ammonia on a platinum surface at 856 °C NH3 →1/2 N2 + 3/2 H2 is zero order in NH3 with a rate constant of 1.50×10-6 M s-1. If the initial concentration of NH3 is 1.18×10 M, the concentration of NH3 will be 3.45×103M after seconds have passed. Submit Answer Retry Entire Group 9 more group attempts remaining
The decomposition of ammonia on a platinum surface at 856 °C NH3 →1/2 N2 + 3/2 H2 is zero order in NH3 with a rate constant of 1.50×10-6 M s-1. If the initial concentration of NH3 is 1.18×10 M, the concentration of NH3 will be 3.45×103M after seconds have passed. Submit Answer Retry Entire Group 9 more group attempts remaining
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![The decomposition of ammonia on a platinum surface at 856 °C follows the chemical equation:
\[ \text{NH}_3 \rightarrow \frac{1}{2} \text{N}_2 + \frac{3}{2} \text{H}_2 \]
This reaction is zero order in \(\text{NH}_3\) with a rate constant of \(1.50 \times 10^{-6} \, \text{M} \, \text{s}^{-1}\).
If the initial concentration of \(\text{NH}_3\) is \(1.18 \times 10^{-2} \, \text{M}\), the concentration of \(\text{NH}_3\) will be \(3.45 \times 10^{-3} \, \text{M}\) after a certain number of seconds have passed.
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**Instructions:**
- Use the references to access important values if needed.
- Calculate the time it takes for the concentration to decrease from the initial value to the given concentration.
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Transcribed Image Text:The decomposition of ammonia on a platinum surface at 856 °C follows the chemical equation:
\[ \text{NH}_3 \rightarrow \frac{1}{2} \text{N}_2 + \frac{3}{2} \text{H}_2 \]
This reaction is zero order in \(\text{NH}_3\) with a rate constant of \(1.50 \times 10^{-6} \, \text{M} \, \text{s}^{-1}\).
If the initial concentration of \(\text{NH}_3\) is \(1.18 \times 10^{-2} \, \text{M}\), the concentration of \(\text{NH}_3\) will be \(3.45 \times 10^{-3} \, \text{M}\) after a certain number of seconds have passed.
---
**Instructions:**
- Use the references to access important values if needed.
- Calculate the time it takes for the concentration to decrease from the initial value to the given concentration.
**Interactive Components:**
- Submit Answer
- Retry Entire Group
- Remaining Attempts: 9
Options for navigation are present at the bottom:
- Email Instructor
- Save and Exit
This is displayed on a user interface provided by Cengage Learning, with options for technical support.
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