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An equilibrium is established for the reaction 2 CO(g) + MOO,(s) = 2 CO₂(g) + Mo(s). Use the expression for Kp from part a. A system which is not necessarily at equilibrium has a partial pressure of carbon monoxide of 1.67 atm and a partial pressure of carbon dioxide of 0.349 atm. What is the value of Q, the reaction quotient?

An equilibrium is established for the reaction 2 CO(g) + MOO,(s) = 2 CO₂(g) + Mo(s). Use the expression for Kp from part a. A system which is not necessarily at equilibrium has a partial pressure of carbon monoxide of 1.67 atm and a partial pressure of carbon dioxide of 0.349 atm. What is the value of Q, the reaction quotient?

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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
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Show work, thank you! It’s part c of question.
An equilibrium is established for the reaction 2 CO(g) + MoO,(s) = 2 CO₂(g) + Mo(s). Use the expression for Kp from part a. If at equilibrium the partial pressure of carbon monoxide is 5.21 atm and the partial pressure of the carbon dioxide is 0.733 atm, then what is the value of Kp? Your submission: 0.0198 Question 19.b of 38 Correct answer: Kp = Correct! PCO₂ Po 0.0198 Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. The equilibrium constant is based on the concentrations of the reactants and products at equilibrium. If the equilibrium partial pressures of CO and CO2 are 5.21 and 0.643 atm respectively: E (0.643)² (5.21)² Feedback: = 0.0152 Correct!
Transcribed Image Text:An equilibrium is established for the reaction 2 CO(g) + MoO,(s) = 2 CO₂(g) + Mo(s). Use the expression for Kp from part a. If at equilibrium the partial pressure of carbon monoxide is 5.21 atm and the partial pressure of the carbon dioxide is 0.733 atm, then what is the value of Kp? Your submission: 0.0198 Question 19.b of 38 Correct answer: Kp = Correct! PCO₂ Po 0.0198 Solution: The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. The equilibrium constant is based on the concentrations of the reactants and products at equilibrium. If the equilibrium partial pressures of CO and CO2 are 5.21 and 0.643 atm respectively: E (0.643)² (5.21)² Feedback: = 0.0152 Correct!
< Question 19.c of 38 An equilibrium is established for the reaction 2 CO(g) + MoO₂ (s) = 2 CO₂(g) + Mo(s). Use the expression for Kp from part a. A system which is not necessarily at equilibrium has a partial pressure of carbon monoxide of 1.67 atm and a partial pressure of carbon dioxide of 0.349 atm. What is the value of Q, the reaction quotient?
Transcribed Image Text:< Question 19.c of 38 An equilibrium is established for the reaction 2 CO(g) + MoO₂ (s) = 2 CO₂(g) + Mo(s). Use the expression for Kp from part a. A system which is not necessarily at equilibrium has a partial pressure of carbon monoxide of 1.67 atm and a partial pressure of carbon dioxide of 0.349 atm. What is the value of Q, the reaction quotient?
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