2. Molecular nitrogen (N2) reacts with water vapor to form nitrogen monoxide (NO) and molecular hydrogen (H2). At some temperature a reaction system containing all four gases has been allowed come to equilibrium and the partial pressures of the gases are: PN2 = 0.12 atm, PH20 = 0.082, PNo = 0.55 atm, and PH2 = 0.76 atm. Another system contains the same reactants and products at the following pressures: PN2 = 0.15 atm, PH20 = 0.12, PNo = 0.87 atm, and PH2 = 0.76 atm. Is this second system at equilibrium? If not, how will the reaction shift to reach equilibrium? N2(g) + 2 H20(g) =2 NO(g) + 2 H2(g)
2. Molecular nitrogen (N2) reacts with water vapor to form nitrogen monoxide (NO) and molecular hydrogen (H2). At some temperature a reaction system containing all four gases has been allowed come to equilibrium and the partial pressures of the gases are: PN2 = 0.12 atm, PH20 = 0.082, PNo = 0.55 atm, and PH2 = 0.76 atm. Another system contains the same reactants and products at the following pressures: PN2 = 0.15 atm, PH20 = 0.12, PNo = 0.87 atm, and PH2 = 0.76 atm. Is this second system at equilibrium? If not, how will the reaction shift to reach equilibrium? N2(g) + 2 H20(g) =2 NO(g) + 2 H2(g)
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Equilibrium and Reaction Shift Analysis**
**Problem Statement:**
Molecular nitrogen (\(N_2\)) reacts with water vapor to form nitrogen monoxide (NO) and molecular hydrogen (\(H_2\)). At a certain temperature, a reaction system containing all four gases has been allowed to come to equilibrium. The partial pressures of the gases at equilibrium are as follows:
- \(P_{N_2} = 0.12 \, \text{atm}\)
- \(P_{H_2O} = 0.082 \, \text{atm}\)
- \(P_{NO} = 0.55 \, \text{atm}\)
- \(P_{H_2} = 0.76 \, \text{atm}\)
Another system contains the same reactants and products at the following pressures:
- \(P_{N_2} = 0.15 \, \text{atm}\)
- \(P_{H_2O} = 0.12 \, \text{atm}\)
- \(P_{NO} = 0.87 \, \text{atm}\)
- \(P_{H_2} = 0.76 \, \text{atm}\)
**Question:**
Is this second system at equilibrium? If not, how will the reaction shift to reach equilibrium?
**Chemical Reaction Equation:**
\[ N_2(g) + 2 \, H_2O(g) \rightleftharpoons 2 \, NO(g) + 2 \, H_2(g) \]
**Analysis:**
1. **Equilibrium Constant Expression (\(K_p\)):**
The equilibrium constant expression for this reaction in terms of partial pressures is given by:
\[
K_p = \frac{(P_{NO})^2 (P_{H_2})^2}{(P_{N_2}) (P_{H_2O})^2}
\]
2. **Calculate \(K_p\) for the Equilibrium System:**
Substitute the given equilibrium pressures into the expression to find \(K_p\).
3. **Calculate the Reaction Quotient (\(Q_p\)) for the Second System:**
Using the same expression, calculate \(Q_p\) using the pressures from the second system.
4. **Compare \(K_p\) and \(Q_p\):**
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4bda451-1bcc-473c-9a91-77116df85d00%2Fb5598c0f-8174-4cab-889f-32538798776b%2Fe7oluxv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium and Reaction Shift Analysis**
**Problem Statement:**
Molecular nitrogen (\(N_2\)) reacts with water vapor to form nitrogen monoxide (NO) and molecular hydrogen (\(H_2\)). At a certain temperature, a reaction system containing all four gases has been allowed to come to equilibrium. The partial pressures of the gases at equilibrium are as follows:
- \(P_{N_2} = 0.12 \, \text{atm}\)
- \(P_{H_2O} = 0.082 \, \text{atm}\)
- \(P_{NO} = 0.55 \, \text{atm}\)
- \(P_{H_2} = 0.76 \, \text{atm}\)
Another system contains the same reactants and products at the following pressures:
- \(P_{N_2} = 0.15 \, \text{atm}\)
- \(P_{H_2O} = 0.12 \, \text{atm}\)
- \(P_{NO} = 0.87 \, \text{atm}\)
- \(P_{H_2} = 0.76 \, \text{atm}\)
**Question:**
Is this second system at equilibrium? If not, how will the reaction shift to reach equilibrium?
**Chemical Reaction Equation:**
\[ N_2(g) + 2 \, H_2O(g) \rightleftharpoons 2 \, NO(g) + 2 \, H_2(g) \]
**Analysis:**
1. **Equilibrium Constant Expression (\(K_p\)):**
The equilibrium constant expression for this reaction in terms of partial pressures is given by:
\[
K_p = \frac{(P_{NO})^2 (P_{H_2})^2}{(P_{N_2}) (P_{H_2O})^2}
\]
2. **Calculate \(K_p\) for the Equilibrium System:**
Substitute the given equilibrium pressures into the expression to find \(K_p\).
3. **Calculate the Reaction Quotient (\(Q_p\)) for the Second System:**
Using the same expression, calculate \(Q_p\) using the pressures from the second system.
4. **Compare \(K_p\) and \(Q_p\):**
-
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