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 Problem
#### For the reaction:
\[ \text{2HOCl(g) } \leftrightarrow \text{ H}_2\text{O(g) } + \text{ Cl}_2\text{O(g) } \]
Given that the value of the equilibrium constant (\(K\)) at 25°C is 11, calculate the value of the equilibrium constant in terms of partial pressure (\(K_p\)) for this reaction at 25°C.
\[K_p = \]
#### Explanation:
1. **Chemical Equation:** The reaction shows the decomposition of hypochlorous acid (\(HOCl\)) gas into water (\(H_2O\)) and dichlorine monoxide (\(Cl_2O\)) gases.
2. **Equilibrium Constant (K):** At 25°C, the equilibrium constant (\(K\)) is provided as 11. This value is dimensionless and represents the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants at equilibrium.
3. **Calculate \(K_p\):** To find \(K_p\), the equilibrium constant in terms of partial pressures, you'll need to use the relationship between \(K\) and \(K_p\):
\[ K_p = K(RT)^{\Delta n} \]
Where:
- \( R \) is the universal gas constant (0.0821 L·atm/mol·K)
- \( T \) is the temperature in Kelvin
- \( \Delta n \) is the change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants).
In this reaction:
\[ \Delta n = (1 + 1) - 2 = 0 \]
Therefore:
\[ K_p = K \times (RT)^{0} \]
\[ K_p = K \]
Since \( \Delta n = 0 \), the \( K_p \) will be the same as \( K \).
\[ K_p = 11 \]
Please fill in the value in the box provided accordingly.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5fe141b1-9817-4d6d-9b09-948fe13c457c%2F6ce3acd2-0cc3-4d2d-b8cc-374840a33860%2Fj5h6q1c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemical Equilibrium Problem
#### For the reaction:
\[ \text{2HOCl(g) } \leftrightarrow \text{ H}_2\text{O(g) } + \text{ Cl}_2\text{O(g) } \]
Given that the value of the equilibrium constant (\(K\)) at 25°C is 11, calculate the value of the equilibrium constant in terms of partial pressure (\(K_p\)) for this reaction at 25°C.
\[K_p = \]
#### Explanation:
1. **Chemical Equation:** The reaction shows the decomposition of hypochlorous acid (\(HOCl\)) gas into water (\(H_2O\)) and dichlorine monoxide (\(Cl_2O\)) gases.
2. **Equilibrium Constant (K):** At 25°C, the equilibrium constant (\(K\)) is provided as 11. This value is dimensionless and represents the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants at equilibrium.
3. **Calculate \(K_p\):** To find \(K_p\), the equilibrium constant in terms of partial pressures, you'll need to use the relationship between \(K\) and \(K_p\):
\[ K_p = K(RT)^{\Delta n} \]
Where:
- \( R \) is the universal gas constant (0.0821 L·atm/mol·K)
- \( T \) is the temperature in Kelvin
- \( \Delta n \) is the change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants).
In this reaction:
\[ \Delta n = (1 + 1) - 2 = 0 \]
Therefore:
\[ K_p = K \times (RT)^{0} \]
\[ K_p = K \]
Since \( \Delta n = 0 \), the \( K_p \) will be the same as \( K \).
\[ K_p = 11 \]
Please fill in the value in the box provided accordingly.
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