Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Hello,
how did you obtain the values of enthalpy change for the products and reactants?
Thanks!
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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
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Follow-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 Expert
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 Expert
Follow-up Question
I meant these values.
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 Expert
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