4. For the reaction below: 2 NO2 (g) + O2 (g) 2 NO3 (g) a. Write the law of mass action (the equilibrium expression for Keq) Keq [NO312 [0₂] [NO₂]² Keq b. If gas concentrations are as follows, 2.10 M NO2, 1.75 M O2, and 1.00 M NO3, calculate Keq = 0.130 M-¹ (NOTE: "M-¹" is another way to write "1/M") [1.00 M12 [1.75 M ] [2.10 M]² c. Based on the Keq value that you calculated in part b, are the reactants or products predominant (predominant means that there is a greater amount present)? REACTANTS: Keq is much less than 1
4. For the reaction below: 2 NO2 (g) + O2 (g) 2 NO3 (g) a. Write the law of mass action (the equilibrium expression for Keq) Keq [NO312 [0₂] [NO₂]² Keq b. If gas concentrations are as follows, 2.10 M NO2, 1.75 M O2, and 1.00 M NO3, calculate Keq = 0.130 M-¹ (NOTE: "M-¹" is another way to write "1/M") [1.00 M12 [1.75 M ] [2.10 M]² c. Based on the Keq value that you calculated in part b, are the reactants or products predominant (predominant means that there is a greater amount present)? REACTANTS: Keq is much less than 1
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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 Calculation for the Reaction
For the reaction:
\[ 2 \text{NO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2 \text{NO}_3 \, (g) \]
#### a. Law of Mass Action
Write the equilibrium expression, \( K_{eq} \), for the reaction:
\[ K_{eq} = \frac{[\text{NO}_3]^2}{[\text{O}_2][\text{NO}_2]^2} \]
#### b. Calculate \( K_{eq} \)
Given the concentrations:
- \([\text{NO}_2] = 2.10 \, \text{M}\)
- \([\text{O}_2] = 1.75 \, \text{M}\)
- \([\text{NO}_3] = 1.00 \, \text{M}\)
Substitute these values into the equation:
\[ K_{eq} = \frac{[1.00 \, \text{M}]^2}{[1.75 \, \text{M}][2.10 \, \text{M}]^2} = 0.130 \, \text{M}^{-1} \]
**Note:** "M\(^{-1}\)" means \( \frac{1}{\text{M}} \).
#### c. Predominance of Reactants or Products
Based on the \( K_{eq} \) value calculated (0.130 M\(^{-1}\)), determine the predominance of reactants or products. Since \( K_{eq} \) is much less than 1, the **reactants** are predominant in this equilibrium system.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdfd6ec15-f1ea-4f47-a8c3-1b2e52d65b25%2Fb2024a19-e4ab-4854-881b-f523d83e3629%2Fcj8cjcw_processed.png&w=3840&q=75)
Transcribed Image Text:### Equilibrium Calculation for the Reaction
For the reaction:
\[ 2 \text{NO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2 \text{NO}_3 \, (g) \]
#### a. Law of Mass Action
Write the equilibrium expression, \( K_{eq} \), for the reaction:
\[ K_{eq} = \frac{[\text{NO}_3]^2}{[\text{O}_2][\text{NO}_2]^2} \]
#### b. Calculate \( K_{eq} \)
Given the concentrations:
- \([\text{NO}_2] = 2.10 \, \text{M}\)
- \([\text{O}_2] = 1.75 \, \text{M}\)
- \([\text{NO}_3] = 1.00 \, \text{M}\)
Substitute these values into the equation:
\[ K_{eq} = \frac{[1.00 \, \text{M}]^2}{[1.75 \, \text{M}][2.10 \, \text{M}]^2} = 0.130 \, \text{M}^{-1} \]
**Note:** "M\(^{-1}\)" means \( \frac{1}{\text{M}} \).
#### c. Predominance of Reactants or Products
Based on the \( K_{eq} \) value calculated (0.130 M\(^{-1}\)), determine the predominance of reactants or products. Since \( K_{eq} \) is much less than 1, the **reactants** are predominant in this equilibrium system.
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