The elementary reaction 2 H,O(g) =2 H,(g) + 0,(g) proceeds at a certain temperature until the partial pressures of H,O, H,, and O, reach 0.0750 atm, 0.00550 atm, and 0.00500 atm, respectively. What is the value of the equilibrium constant at this temperature?
The elementary reaction 2 H,O(g) =2 H,(g) + 0,(g) proceeds at a certain temperature until the partial pressures of H,O, H,, and O, reach 0.0750 atm, 0.00550 atm, and 0.00500 atm, respectively. What is the value of the equilibrium constant at this temperature?
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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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![**Elementary Reaction and Equilibrium Constant Calculation**
The elementary reaction is given by:
\[ 2\text{H}_2\text{O(g)} \rightleftharpoons 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \]
This reaction proceeds at a certain temperature until the partial pressures reach the following values:
- Partial pressure of \(\text{H}_2\text{O}\) = 0.0750 atm
- Partial pressure of \(\text{H}_2\) = 0.00550 atm
- Partial pressure of \(\text{O}_2\) = 0.00500 atm
The equilibrium constant (\(K_p\)) at this temperature is calculated as:
\[ K_p = 0.004033 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe9fe32c-b492-41de-b627-5e5dec5bc081%2Fbf5f23eb-d17c-4a39-bbdc-5cd6bed51f8d%2Fpbvkwq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Elementary Reaction and Equilibrium Constant Calculation**
The elementary reaction is given by:
\[ 2\text{H}_2\text{O(g)} \rightleftharpoons 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \]
This reaction proceeds at a certain temperature until the partial pressures reach the following values:
- Partial pressure of \(\text{H}_2\text{O}\) = 0.0750 atm
- Partial pressure of \(\text{H}_2\) = 0.00550 atm
- Partial pressure of \(\text{O}_2\) = 0.00500 atm
The equilibrium constant (\(K_p\)) at this temperature is calculated as:
\[ K_p = 0.004033 \]
![**Chemical Equilibrium Problem**
Consider the reaction:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]
Given:
- Equilibrium constant (\( K_c \)) = 0.0420
- Concentrations at equilibrium:
- \([\text{PCl}_3]\) = 0.260 M
- \([\text{Cl}_2]\) = 0.140 M
**Problem:**
Determine the concentration of the reactant \([\text{PCl}_5]\) at equilibrium.
\[ [\text{PCl}_5] = \, \_\_\_\_\_\_\_ \, \text{M} \]
**Additional Tool:**
There is a tool provided for calculations, suggesting a scientific notation feature with \(x10^y\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe9fe32c-b492-41de-b627-5e5dec5bc081%2Fbf5f23eb-d17c-4a39-bbdc-5cd6bed51f8d%2F0f7k2q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Equilibrium Problem**
Consider the reaction:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]
Given:
- Equilibrium constant (\( K_c \)) = 0.0420
- Concentrations at equilibrium:
- \([\text{PCl}_3]\) = 0.260 M
- \([\text{Cl}_2]\) = 0.140 M
**Problem:**
Determine the concentration of the reactant \([\text{PCl}_5]\) at equilibrium.
\[ [\text{PCl}_5] = \, \_\_\_\_\_\_\_ \, \text{M} \]
**Additional Tool:**
There is a tool provided for calculations, suggesting a scientific notation feature with \(x10^y\).
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