2HF (g) = H₂(g) + F₂ (g), Kc = 1.0 x 10-95 ibrium mixture contains mostly reactants. 2NO(g) + O₂(g) = 2NO₂(g), Kc = 6.0 x 10¹3 CO(g) + Cl₂(g) = COC12(g), Kc = 5.0 x 10 9 Equilibrium mixture contains mostly products.
2HF (g) = H₂(g) + F₂ (g), Kc = 1.0 x 10-95 ibrium mixture contains mostly reactants. 2NO(g) + O₂(g) = 2NO₂(g), Kc = 6.0 x 10¹3 CO(g) + Cl₂(g) = COC12(g), Kc = 5.0 x 10 9 Equilibrium mixture contains mostly products.
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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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![### Chemical Equilibrium and Equilibrium Constants
Understanding chemical equilibrium is essential for mastering the concepts of reaction rates and dynamics. Below are some key chemical equations with their respective equilibrium constants (Kc), which will help you determine whether the equilibrium mixture will contain mostly reactants or products.
#### Chemical Equations and Equilibrium Constants
1. **Equation 1**:
\[
2HF(g) \rightleftharpoons H_2(g) + F_2(g)
\]
\[
K_c = 1.0 \times 10^{-95}
\]
2. **Equation 2**:
\[
2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)
\]
\[
K_c = 6.0 \times 10^{13}
\]
3. **Equation 3**:
\[
CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)
\]
\[
K_c = 5.0 \times 10^{-9}
\]
Each of these reactions has a specific equilibrium constant, which indicates the ratio of the concentration of products to reactants at equilibrium.
#### Understanding the Equilibrium Constant (Kc)
- **Large Kc Value**:
- When Kc is significantly greater than 1 (for example, \( K_c = 6.0 \times 10^{13} \)), it suggests that the equilibrium position favors the formation of products. In other words, the equilibrium mixture contains mostly products.
- **Small Kc Value**:
- When Kc is significantly less than 1 (for example, \( K_c = 1.0 \times 10^{-95} \) or \( K_c = 5.0 \times 10^{-9} \)), it suggests that the equilibrium position favors the reactants. Hence, the equilibrium mixture contains mostly reactants.
#### Summary of Equilibrium Composition
- **HF(g) \rightleftharpoons H_2(g) + F_2(g)**:
- \( K_c = 1.0 \times 10^{-95} \)
- _Equilibrium mixture contains mostly reactants._
- **2NO(g) + O_2(g) \rightleftharpoons 2NO_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f1362fe-0b52-42c9-bff0-2cf928b098d5%2F4e851ec4-b99f-4169-b7e4-410f81e5dc80%2Fh51qql_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemical Equilibrium and Equilibrium Constants
Understanding chemical equilibrium is essential for mastering the concepts of reaction rates and dynamics. Below are some key chemical equations with their respective equilibrium constants (Kc), which will help you determine whether the equilibrium mixture will contain mostly reactants or products.
#### Chemical Equations and Equilibrium Constants
1. **Equation 1**:
\[
2HF(g) \rightleftharpoons H_2(g) + F_2(g)
\]
\[
K_c = 1.0 \times 10^{-95}
\]
2. **Equation 2**:
\[
2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)
\]
\[
K_c = 6.0 \times 10^{13}
\]
3. **Equation 3**:
\[
CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)
\]
\[
K_c = 5.0 \times 10^{-9}
\]
Each of these reactions has a specific equilibrium constant, which indicates the ratio of the concentration of products to reactants at equilibrium.
#### Understanding the Equilibrium Constant (Kc)
- **Large Kc Value**:
- When Kc is significantly greater than 1 (for example, \( K_c = 6.0 \times 10^{13} \)), it suggests that the equilibrium position favors the formation of products. In other words, the equilibrium mixture contains mostly products.
- **Small Kc Value**:
- When Kc is significantly less than 1 (for example, \( K_c = 1.0 \times 10^{-95} \) or \( K_c = 5.0 \times 10^{-9} \)), it suggests that the equilibrium position favors the reactants. Hence, the equilibrium mixture contains mostly reactants.
#### Summary of Equilibrium Composition
- **HF(g) \rightleftharpoons H_2(g) + F_2(g)**:
- \( K_c = 1.0 \times 10^{-95} \)
- _Equilibrium mixture contains mostly reactants._
- **2NO(g) + O_2(g) \rightleftharpoons 2NO_
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