Question

The following reaction is at equilibrium at one atmosphere, in a closed container. NaOH(s)+CO_2

(g)↔NaHCO_3
(s) Which, if any, of the following actions will decrease the total amount of CO_2
gas present at equilibrium? decreasing the volume of the container removing half of the solid NaHCO_3
adding more solid NaOH adding N_2
gas to double the pressure None of the above

Answer 1

To decrease the amount of CO2 gas at equilibrium for the given reaction, decreasing the volume of the container is the action that will have the desired effect according to Le Chatelier's principle.

Step-by-step explanation:

To decrease the total amount of CO2 gas present at equilibrium for the reaction NaOH(s) + CO2(g) ↔ NaHCO3(s), we need to consider the effects of changes according to Le Chatelier's principle:

• Decreasing the volume of the container would cause an increase in pressure, favoring the formation of NaHCO3(s), a solid, and thus would decrease the amount of gaseous CO2.
• Removing half of the solid NaHCO3 would shift the equilibrium towards forming more NaHCO3(s) from NaOH(s) and CO2(g), thus potentially increasing the amount of CO2 gas until new equilibrium is reached.
• Adding more solid NaOH will not affect the amount of CO2 because the added solid does not directly participate in the gaseous equilibrium.
• Adding N2 gas to double the pressure will not affect the amount of CO2 directly, as N2 is inert and does not participate in the chemical equilibrium of the reaction.

The action that will decrease the total amount of CO2 gas present at equilibrium is decreasing the volume of the container.