For the system H2 (g) + CO2 (g) = H₂O(g) + CO (g) at equilibrium, the removal of some of the H₂O (g) would cause (according to LeChatelier's principle), at constant T: a) more H₂(g) to be formed. b) more CO₂ (g) to be formed. c) no change in the amounts of products or reactants, since T = constant. d) more CO (g) to be formed. e) the amount of CO (g) tp remain constant while the amount of H₂O (g) increases to the original equilibrium concentration.

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
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For the system H2 (g) + CO2 (g) = H₂O(g) + CO (g) at equilibrium, the
removal of some of the H₂O (g) would cause (according to LeChatelier's
principle), at constant T:
a) more H₂(g) to be formed.
b) more CO₂ (g) to be formed.
c) no change in the amounts of products or reactants, since T = constant.
d) more CO (g) to be formed.
e) the amount of CO (g) tp remain constant while the amount of H₂O (g) increases to the
original equilibrium concentration.
Transcribed Image Text:For the system H2 (g) + CO2 (g) = H₂O(g) + CO (g) at equilibrium, the removal of some of the H₂O (g) would cause (according to LeChatelier's principle), at constant T: a) more H₂(g) to be formed. b) more CO₂ (g) to be formed. c) no change in the amounts of products or reactants, since T = constant. d) more CO (g) to be formed. e) the amount of CO (g) tp remain constant while the amount of H₂O (g) increases to the original equilibrium concentration.
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