mmonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 50.0 L tank with 2.8 mol of mmonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 0.84 mol. alculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 gnificant digits. K = D x10 S ?
mmonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 50.0 L tank with 2.8 mol of mmonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 0.84 mol. alculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 gnificant digits. K = D x10 S ?
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...
Related questions
Question
![### Calculating the Equilibrium Constant for Ammonia Decomposition
**Problem Statement:**
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 50.0 L tank with 2.8 mol of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 0.84 mol.
Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.
**Reaction:**
Ammonia decomposes as follows:
\[ 2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2 \]
**Solution Steps:**
1. **Determine the change in moles of NH₃:**
Since 0.84 mol of N₂ is produced and the stoichiometric ratio between NH₃ and N₂ is 2:1, the moles of NH₃ that decompose:
\[ 2 \times 0.84 \text{ mol} = 1.68 \text{ mol} \]
2. **Calculate the moles of NH₃ at equilibrium:**
Initial moles of NH₃ – moles that decomposed:
\[ 2.8 \text{ mol} - 1.68 \text{ mol} = 1.12 \text{ mol} \]
3. **Calculate moles of H₂ produced:**
The stoichiometric ratio between NH₃ and H₂ is 2:3.
\[ 3 \times 0.84 \text{ mol} = 2.52 \text{ mol} \]
4. **Determine the concentrations at equilibrium:**
For the 50.0 L tank:
\[ [\text{NH}_3] = \frac{1.12 \text{ mol}}{50.0 \text{ L}} = 0.0224 \text{ M} \]
\[ [\text{N}_2] = \frac{0.84 \text{ mol}}{50.0 \text{ L}} = 0.0168 \text{ M} \]
\[ [\text{H}_2] = \frac{2.52 \text{ mol}}{50.0 \text{ L}} = 0.0504 \text{ M} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf399169-ef14-4107-9621-c5ec1888471d%2F673f6269-b3fe-4067-9997-180c7dfbe713%2F01p2f1q_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Equilibrium Constant for Ammonia Decomposition
**Problem Statement:**
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 50.0 L tank with 2.8 mol of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 0.84 mol.
Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.
**Reaction:**
Ammonia decomposes as follows:
\[ 2 \text{NH}_3 \rightarrow \text{N}_2 + 3 \text{H}_2 \]
**Solution Steps:**
1. **Determine the change in moles of NH₃:**
Since 0.84 mol of N₂ is produced and the stoichiometric ratio between NH₃ and N₂ is 2:1, the moles of NH₃ that decompose:
\[ 2 \times 0.84 \text{ mol} = 1.68 \text{ mol} \]
2. **Calculate the moles of NH₃ at equilibrium:**
Initial moles of NH₃ – moles that decomposed:
\[ 2.8 \text{ mol} - 1.68 \text{ mol} = 1.12 \text{ mol} \]
3. **Calculate moles of H₂ produced:**
The stoichiometric ratio between NH₃ and H₂ is 2:3.
\[ 3 \times 0.84 \text{ mol} = 2.52 \text{ mol} \]
4. **Determine the concentrations at equilibrium:**
For the 50.0 L tank:
\[ [\text{NH}_3] = \frac{1.12 \text{ mol}}{50.0 \text{ L}} = 0.0224 \text{ M} \]
\[ [\text{N}_2] = \frac{0.84 \text{ mol}}{50.0 \text{ L}} = 0.0168 \text{ M} \]
\[ [\text{H}_2] = \frac{2.52 \text{ mol}}{50.0 \text{ L}} = 0.0504 \text{ M} \]
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps

Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you

Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning

Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education

Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning

Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning

Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education

Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning

Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education

Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning

Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY