2. Lead (II) sulfide is the principle source of lead metal. To obtain the metal, the sulfide is first heated in air to produce PbO. ΔΗ = .415.4 kJ PbS(+3/202) PbO () The oxide is then reduced to the metal with carbon. + SO2(g) PbO + C-Pb) + CO(g) AH = +108.5 kJ Calculate 4H for the reaction of one mole of PbS with oxygen and carbon, forming lead, sulfur dioxide and carbon monoxide. 3. Calculate AH for the formation of one mole of N₂O, from the elements at 25°C using the following data. 2H2+O2H₂O + H₂O - N₂O5 3.0), 1/2N2(g) + 3/202(g) +2H2(g) HNO3 AH = -174.1 kJ 2HNO₂ AH = -571.6 kJ AH = -73.7 kJ →
2. Lead (II) sulfide is the principle source of lead metal. To obtain the metal, the sulfide is first heated in air to produce PbO. ΔΗ = .415.4 kJ PbS(+3/202) PbO () The oxide is then reduced to the metal with carbon. + SO2(g) PbO + C-Pb) + CO(g) AH = +108.5 kJ Calculate 4H for the reaction of one mole of PbS with oxygen and carbon, forming lead, sulfur dioxide and carbon monoxide. 3. Calculate AH for the formation of one mole of N₂O, from the elements at 25°C using the following data. 2H2+O2H₂O + H₂O - N₂O5 3.0), 1/2N2(g) + 3/202(g) +2H2(g) HNO3 AH = -174.1 kJ 2HNO₂ AH = -571.6 kJ AH = -73.7 kJ →
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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![**Transcription and Explanation:**
**2. Lead (II) sulfide is the principal source of lead metal. To obtain the metal, the sulfide is first heated in air to produce PbO.**
The balanced chemical reaction is:
\[ \text{PbS}_{(s)} + \frac{3}{2}\text{O}_2 \rightarrow \text{PbO}_{(s)} + \text{SO}_2_{(g)} \]
\[
\Delta H = -415.4 \, \text{kJ}
\]
The oxide is then reduced to the metal with carbon.
The balanced chemical reaction is:
\[ \text{PbO}_{(s)} + \text{C}_{(s)} \rightarrow \text{Pb}_{(s)} + \text{CO}_{(g)} \]
\[
\Delta H = +108.5 \, \text{kJ}
\]
Calculate \(\Delta H\) for the reaction of one mole of PbS with oxygen and carbon, forming lead, sulfur dioxide, and carbon monoxide.
**3. Calculate \(\Delta H\) for the formation of one mole of \(\text{N}_2\text{O}_5\) from the elements at 25ºC using the following data.**
The reactions provided are:
\[ 2\text{H}_2_{(g)} + \text{O}_2_{(g)} \rightarrow 2\text{H}_2\text{O}_{(l)} \]
\[
\Delta H = -571.6 \, \text{kJ}
\]
\[ \text{N}_2\text{O}_5_{(g)} + \text{H}_2\text{O}_{(l)} \rightarrow 2\text{HNO}_3_{(l)} \]
\[
\Delta H = -73.7 \, \text{kJ}
\]
\[ \frac{1}{2}\text{N}_2_{(g)} + \frac{3}{2}\text{O}_2_{(g)} + \frac{1}{2}\text{H}_2_{(g)} \rightarrow \text{HNO}_3_{(l)} \]
\[
\Delta H = -174.1 \, \text{kJ}
\]
---
**Explanation of Calculations:**
- To solve](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07988e83-a3e9-4d52-ab01-33c4f0c30b8b%2F8c77e747-811a-4892-99e9-bb25e2381485%2Fxqc85if_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription and Explanation:**
**2. Lead (II) sulfide is the principal source of lead metal. To obtain the metal, the sulfide is first heated in air to produce PbO.**
The balanced chemical reaction is:
\[ \text{PbS}_{(s)} + \frac{3}{2}\text{O}_2 \rightarrow \text{PbO}_{(s)} + \text{SO}_2_{(g)} \]
\[
\Delta H = -415.4 \, \text{kJ}
\]
The oxide is then reduced to the metal with carbon.
The balanced chemical reaction is:
\[ \text{PbO}_{(s)} + \text{C}_{(s)} \rightarrow \text{Pb}_{(s)} + \text{CO}_{(g)} \]
\[
\Delta H = +108.5 \, \text{kJ}
\]
Calculate \(\Delta H\) for the reaction of one mole of PbS with oxygen and carbon, forming lead, sulfur dioxide, and carbon monoxide.
**3. Calculate \(\Delta H\) for the formation of one mole of \(\text{N}_2\text{O}_5\) from the elements at 25ºC using the following data.**
The reactions provided are:
\[ 2\text{H}_2_{(g)} + \text{O}_2_{(g)} \rightarrow 2\text{H}_2\text{O}_{(l)} \]
\[
\Delta H = -571.6 \, \text{kJ}
\]
\[ \text{N}_2\text{O}_5_{(g)} + \text{H}_2\text{O}_{(l)} \rightarrow 2\text{HNO}_3_{(l)} \]
\[
\Delta H = -73.7 \, \text{kJ}
\]
\[ \frac{1}{2}\text{N}_2_{(g)} + \frac{3}{2}\text{O}_2_{(g)} + \frac{1}{2}\text{H}_2_{(g)} \rightarrow \text{HNO}_3_{(l)} \]
\[
\Delta H = -174.1 \, \text{kJ}
\]
---
**Explanation of Calculations:**
- To solve
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