A student ran the following reaction in the laboratory at 1080 K: 2SO3(9) ? 2SO2(g) + O₂(g) When she introduced SO3(g) at a pressure of 1.16 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of O₂(g) to be 0.343 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction. Kp =
A student ran the following reaction in the laboratory at 1080 K: 2SO3(9) ? 2SO2(g) + O₂(g) When she introduced SO3(g) at a pressure of 1.16 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of O₂(g) to be 0.343 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction. Kp =
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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 Experiment**
A student conducted an experiment in the laboratory at a temperature of 1080 K to observe the following chemical reaction:
\[ 2\text{SO}_3(g) \rightleftharpoons 2\text{SO}_2(g) + \text{O}_2(g) \]
**Procedure and Observations:**
- The student introduced sulfur trioxide (\(\text{SO}_3\)) gas at a pressure of 1.16 atm into a 1.00 L evacuated container.
- At equilibrium, the partial pressure of oxygen gas (\(\text{O}_2\)) was measured to be 0.343 atm.
**Task:**
Calculate the equilibrium constant, \(K_p\), for this reaction.
**Reaction Conditions:**
- Temperature: 1080 K
- Initial pressure of \(\text{SO}_3\): 1.16 atm
- Equilibrium pressure of \(\text{O}_2\): 0.343 atm
**Calculation of \(K_p\):**
\[ K_p = \] *(Enter the calculated equilibrium constant value here)*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8d8ce1c-d1ae-4fbf-bbe3-121c6915374e%2F8ab82d98-7659-4d68-8cea-03ffce10a33b%2Fuc45bcs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Equilibrium Experiment**
A student conducted an experiment in the laboratory at a temperature of 1080 K to observe the following chemical reaction:
\[ 2\text{SO}_3(g) \rightleftharpoons 2\text{SO}_2(g) + \text{O}_2(g) \]
**Procedure and Observations:**
- The student introduced sulfur trioxide (\(\text{SO}_3\)) gas at a pressure of 1.16 atm into a 1.00 L evacuated container.
- At equilibrium, the partial pressure of oxygen gas (\(\text{O}_2\)) was measured to be 0.343 atm.
**Task:**
Calculate the equilibrium constant, \(K_p\), for this reaction.
**Reaction Conditions:**
- Temperature: 1080 K
- Initial pressure of \(\text{SO}_3\): 1.16 atm
- Equilibrium pressure of \(\text{O}_2\): 0.343 atm
**Calculation of \(K_p\):**
\[ K_p = \] *(Enter the calculated equilibrium constant value here)*
Expert Solution
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Answer:-
This question is answered by using the simple concept of calculation of equilibrium constant using the equilibrium partial pressure and its expression using the equilibrium equation.
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