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
Hello,
how did you obtain the values of enthalpy change for the products and reactants?
Thanks!
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 1 images
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
Could you please show the work for the highlighted enthalpy values? I have the tabulated values for bond energies but I'd just like to see it step-by-step. Thank you.
Solution
by Bartleby ExpertFollow-up Question
I meant these values.
![**Explanation of Solution**
The given reaction is:
\[ 4\text{NH}_3(g) + 6\text{NO}(g) \rightarrow 5\text{N}_2(g) + 6\text{H}_2\text{O}(g) \]
The mathematical expression for the standard enthalpy change value at room temperature is:
\[ \Delta H^\circ_R = \sum nH^\circ(\text{products}) - \sum pH^\circ(\text{reactants}) \]
\[ \Delta H^\circ_R = 5 \times \Delta H^\circ_{f,\text{N}_2(g)} + 6 \times \Delta H^\circ_{f,\text{H}_2\text{O}(g)} - 6 \times \Delta H^\circ_{f,\text{NO}(g)} - 4 \times \Delta H^\circ_{f,\text{NH}_3(g)} \]
- The enthalpy change of \(\text{N}_2(g)\) = 0 kJ/mole
- The enthalpy change of \(\text{H}_2\text{O}(g)\) = -241.8 kJ/mole
- The enthalpy change of \(\text{NO}(g)\) = 91.3 kJ/mole
- The enthalpy change of \(\text{NH}_3(g)\) = -45.9 kJ/mole
Put the values,
\[ \Delta H^\circ_R = 5 \times (0 \text{kJ/mole}) + 6 \times (-241.8 \text{kJ/mole}) - 6 \times (91.3 \text{kJ/mole}) - 4 \times (-45.9 \text{kJ/mole}) \]
\[ \Delta H^\circ_R = -1816 \text{kJ/mole} \]
Now, value of \(\Delta n_g\) = (6 + 5) – (6 + 4) = 1
The mathematical expression of change in internal energy is:
\[ \Delta U^\circ_R = \Delta H^\circ_R - \Delta n_g RT \]
But the values...](https://content.bartleby.com/qna-images/question/32de9b67-12f8-4e34-b9e1-1600f61d5585/17f2994f-c3d1-450a-af11-6eb49d272806/9wr4opn_processed.jpeg)
Transcribed Image Text:**Explanation of Solution**
The given reaction is:
\[ 4\text{NH}_3(g) + 6\text{NO}(g) \rightarrow 5\text{N}_2(g) + 6\text{H}_2\text{O}(g) \]
The mathematical expression for the standard enthalpy change value at room temperature is:
\[ \Delta H^\circ_R = \sum nH^\circ(\text{products}) - \sum pH^\circ(\text{reactants}) \]
\[ \Delta H^\circ_R = 5 \times \Delta H^\circ_{f,\text{N}_2(g)} + 6 \times \Delta H^\circ_{f,\text{H}_2\text{O}(g)} - 6 \times \Delta H^\circ_{f,\text{NO}(g)} - 4 \times \Delta H^\circ_{f,\text{NH}_3(g)} \]
- The enthalpy change of \(\text{N}_2(g)\) = 0 kJ/mole
- The enthalpy change of \(\text{H}_2\text{O}(g)\) = -241.8 kJ/mole
- The enthalpy change of \(\text{NO}(g)\) = 91.3 kJ/mole
- The enthalpy change of \(\text{NH}_3(g)\) = -45.9 kJ/mole
Put the values,
\[ \Delta H^\circ_R = 5 \times (0 \text{kJ/mole}) + 6 \times (-241.8 \text{kJ/mole}) - 6 \times (91.3 \text{kJ/mole}) - 4 \times (-45.9 \text{kJ/mole}) \]
\[ \Delta H^\circ_R = -1816 \text{kJ/mole} \]
Now, value of \(\Delta n_g\) = (6 + 5) – (6 + 4) = 1
The mathematical expression of change in internal energy is:
\[ \Delta U^\circ_R = \Delta H^\circ_R - \Delta n_g RT \]
But the values...
Solution
by Bartleby ExpertKnowledge 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