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?
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?
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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Part a answer is second image.

Transcribed Image Text: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. 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?

Transcribed Image Text:An equilibrium is established for the reaction
2 CO(g) + MoO₂ (s) = 2 CO₂(g) + Mo(s).
Which of the following is the correct expression for Kp?
Your submission:
D) see image
Correct!
Correct answer:
D) see image
Solution:
uilibrium constant is a ratio of the concentrations of products over reactants, including only the species in the gas or aqueous states of matter. For a pressure based
rium constant, Kp, we would use the partial pressures as our concentrations.
Kp =
Feedback:
Correct!
P² CO₂
Po
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