For the reaction CH4(g) + H2O(l) CO(g) + 3H2(g) if the system is at equilibrium and the volume of the container is then decreased, the reaction shifts A) toward products because it will shift toward the side with more total moles. B) toward products because it will shift toward the side with more gas phase moles. C) toward reactants because it will shift toward the side with fewer total moles. D) toward reactants because it will shift toward the side with fewer moles of gas. E) in neither direction because water is a liquid and therefore doesn’t show up in the equilibrium constant expression and therefore the reaction isn’t sensitive to pressure.
For the reaction CH4(g) + H2O(l) CO(g) + 3H2(g) if the system is at equilibrium and the volume of the container is then decreased, the reaction shifts A) toward products because it will shift toward the side with more total moles. B) toward products because it will shift toward the side with more gas phase moles. C) toward reactants because it will shift toward the side with fewer total moles. D) toward reactants because it will shift toward the side with fewer moles of gas. E) in neither direction because water is a liquid and therefore doesn’t show up in the equilibrium constant expression and therefore the reaction isn’t sensitive to pressure.
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 61QRT
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For the reaction CH4(g) + H2O(l) CO(g) + 3H2(g) if the system is at equilibrium and the volume of the container is then decreased, the reaction shifts
A) toward products because it will shift toward the side with more total moles. B) toward products because it will shift toward the side with more gas phase moles. C) toward reactants because it will shift toward the side with fewer total moles. D) toward reactants because it will shift toward the side with fewer moles of gas. E) in neither direction because water is a liquid and therefore doesn’t show up in the equilibrium constant expression and therefore the reaction isn’t sensitive to pressure.
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