### Chemical Equilibrium and Reaction Yield **Question:** Which change would increase the production of water vapor (H₂O) in this reaction? \[ 2H_2S(g) + 3O_2(g) \leftrightarrow 2H_2O(g) + 2SO_2(g) \] **Options:** - O Adding SO₂ - O Removing H₂S - O Decreasing pressure - O Removing SO₂ **Answer Explanation:** In this chemical reaction, the formation of water vapor (H₂O) as a product is considered. To increase the production of water vapor, we need to shift the equilibrium position of the reaction towards the products side (right side of the reaction equation). By Le Chatelier’s principle, changes in concentration, pressure, and temperature can shift the equilibrium position in a predictable way: **Options Analysis:** 1. **Adding SO₂:** This would increase the concentration of a product, thereby shifting the equilibrium to the left, decreasing the production of H₂O. 2. **Removing H₂S:** This would decrease the concentration of a reactant, shifting the equilibrium to the left, and thus decreasing the production of H₂O. 3. **Decreasing Pressure:** The reaction has equal volumes of gaseous reactants (5 volumes) and gaseous products (4 volumes). Decreasing pressure favors the side of the reaction with more gas molecules, which is the reactants side in this case. Hence, it would decrease the production of H₂O. 4. **Removing SO₂:** This would decrease the concentration of a product, shifting the equilibrium to the right to produce more H₂O and SO₂. **Correct Option:** - **Removing SO₂** Removing one of the products will shift the equilibrium to the right, thus increasing the production of water vapor (H₂O). Students should consider the concepts of chemical equilibrium and Le Chatelier's principle when solving similar problems.
### Chemical Equilibrium and Reaction Yield **Question:** Which change would increase the production of water vapor (H₂O) in this reaction? \[ 2H_2S(g) + 3O_2(g) \leftrightarrow 2H_2O(g) + 2SO_2(g) \] **Options:** - O Adding SO₂ - O Removing H₂S - O Decreasing pressure - O Removing SO₂ **Answer Explanation:** In this chemical reaction, the formation of water vapor (H₂O) as a product is considered. To increase the production of water vapor, we need to shift the equilibrium position of the reaction towards the products side (right side of the reaction equation). By Le Chatelier’s principle, changes in concentration, pressure, and temperature can shift the equilibrium position in a predictable way: **Options Analysis:** 1. **Adding SO₂:** This would increase the concentration of a product, thereby shifting the equilibrium to the left, decreasing the production of H₂O. 2. **Removing H₂S:** This would decrease the concentration of a reactant, shifting the equilibrium to the left, and thus decreasing the production of H₂O. 3. **Decreasing Pressure:** The reaction has equal volumes of gaseous reactants (5 volumes) and gaseous products (4 volumes). Decreasing pressure favors the side of the reaction with more gas molecules, which is the reactants side in this case. Hence, it would decrease the production of H₂O. 4. **Removing SO₂:** This would decrease the concentration of a product, shifting the equilibrium to the right to produce more H₂O and SO₂. **Correct Option:** - **Removing SO₂** Removing one of the products will shift the equilibrium to the right, thus increasing the production of water vapor (H₂O). Students should consider the concepts of chemical equilibrium and Le Chatelier's principle when solving similar problems.
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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![### Chemical Equilibrium and Reaction Yield
**Question:** Which change would increase the production of water vapor (H₂O) in this reaction?
\[ 2H_2S(g) + 3O_2(g) \leftrightarrow 2H_2O(g) + 2SO_2(g) \]
**Options:**
- O Adding SO₂
- O Removing H₂S
- O Decreasing pressure
- O Removing SO₂
**Answer Explanation:**
In this chemical reaction, the formation of water vapor (H₂O) as a product is considered. To increase the production of water vapor, we need to shift the equilibrium position of the reaction towards the products side (right side of the reaction equation).
By Le Chatelier’s principle, changes in concentration, pressure, and temperature can shift the equilibrium position in a predictable way:
**Options Analysis:**
1. **Adding SO₂:** This would increase the concentration of a product, thereby shifting the equilibrium to the left, decreasing the production of H₂O.
2. **Removing H₂S:** This would decrease the concentration of a reactant, shifting the equilibrium to the left, and thus decreasing the production of H₂O.
3. **Decreasing Pressure:** The reaction has equal volumes of gaseous reactants (5 volumes) and gaseous products (4 volumes). Decreasing pressure favors the side of the reaction with more gas molecules, which is the reactants side in this case. Hence, it would decrease the production of H₂O.
4. **Removing SO₂:** This would decrease the concentration of a product, shifting the equilibrium to the right to produce more H₂O and SO₂.
**Correct Option:**
- **Removing SO₂**
Removing one of the products will shift the equilibrium to the right, thus increasing the production of water vapor (H₂O).
Students should consider the concepts of chemical equilibrium and Le Chatelier's principle when solving similar problems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1be645a2-1b37-480f-ac5f-87052d4711ee%2F28ab65c4-dbed-475e-a395-98bede8fd1bf%2Fielxqrf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemical Equilibrium and Reaction Yield
**Question:** Which change would increase the production of water vapor (H₂O) in this reaction?
\[ 2H_2S(g) + 3O_2(g) \leftrightarrow 2H_2O(g) + 2SO_2(g) \]
**Options:**
- O Adding SO₂
- O Removing H₂S
- O Decreasing pressure
- O Removing SO₂
**Answer Explanation:**
In this chemical reaction, the formation of water vapor (H₂O) as a product is considered. To increase the production of water vapor, we need to shift the equilibrium position of the reaction towards the products side (right side of the reaction equation).
By Le Chatelier’s principle, changes in concentration, pressure, and temperature can shift the equilibrium position in a predictable way:
**Options Analysis:**
1. **Adding SO₂:** This would increase the concentration of a product, thereby shifting the equilibrium to the left, decreasing the production of H₂O.
2. **Removing H₂S:** This would decrease the concentration of a reactant, shifting the equilibrium to the left, and thus decreasing the production of H₂O.
3. **Decreasing Pressure:** The reaction has equal volumes of gaseous reactants (5 volumes) and gaseous products (4 volumes). Decreasing pressure favors the side of the reaction with more gas molecules, which is the reactants side in this case. Hence, it would decrease the production of H₂O.
4. **Removing SO₂:** This would decrease the concentration of a product, shifting the equilibrium to the right to produce more H₂O and SO₂.
**Correct Option:**
- **Removing SO₂**
Removing one of the products will shift the equilibrium to the right, thus increasing the production of water vapor (H₂O).
Students should consider the concepts of chemical equilibrium and Le Chatelier's principle when solving similar problems.
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