I am fairly sure that a) is no change, and c) is shifts to the right...but I am not sure about b) and d). Thanks.
How does the equilibrium position shift as a result of each of the following disturbances?
Sodium bicarbonate undergoes thermal decomposition according to the reaction below.
2 NaHCO3(s) <~> Na2CO3(s) + CO2(g) + H2O(g)
How does the equilibrium position shift as a result of each of the following disturbances?
(a) 0.20 atm argon gas is added.
-shifts toward the left
shifts towards the right
-no change
(b) NaHCO3(s) is added.
-shifts toward the left
-shifts towards the right
-no change
(c) Mg(ClO4)2(s) is added as a drying agent to remove H2O.
-shifts toward the left
-shifts towards the right
-no change
(d) Dry ice is added at constant T.
-shifts toward the left
-shifts towards the right
-no change
(a) -no change
(b) -shifts towards the right
(c) -shifts towards the left
(d) -no change
Now, what did one mole of dry ice say to the other mole of dry ice?
"We carbon dioxide well together!"
(a) 0.20 atm argon gas is added.
The addition of argon gas, which is an inert gas, does not affect the concentrations of the reactants or products in the reaction. Therefore, there is no change in the equilibrium position. The correct answer is: no change.
(b) NaHCO3(s) is added.
By adding NaHCO3(s), you are increasing the concentration of this reactant. According to Le Chatelier's principle, if the concentration of a reactant is increased, the equilibrium will shift to the right in order to consume the additional reactant. Therefore, the equilibrium position shifts towards the right.
(c) Mg(ClO4)2(s) is added as a drying agent to remove H2O.
Adding anhydrous Mg(ClO4)2 as a drying agent removes water from the system. Since water is a product in the reaction, by removing it, you are essentially shifting the equilibrium towards the side with more reactants. Therefore, the equilibrium position shifts towards the left.
(d) Dry ice is added at constant T.
Dry ice is solid carbon dioxide (CO2). By adding it, you are increasing the concentration of CO2 in the system. Since CO2 is a product in the reaction, according to Le Chatelier's principle, increasing its concentration will shift the equilibrium towards the side with fewer moles of gas. In this case, it will shift towards the left, as NaHCO3(s) and CO2(g) combine to form Na2CO3(s). Therefore, the equilibrium position shifts towards the left.
To determine how the equilibrium position shifts as a result of each disturbance, you need to consider Le Chatelier's principle. Le Chatelier's principle states that when a system at equilibrium is subjected to a disturbance, it will adjust in a way that opposes the change caused by the disturbance.
Now let's analyze each of the disturbances and their effects on the equilibrium position:
(a) 0.20 atm argon gas is added.
Argon gas is an inert gas that does not participate in the chemical reaction. Adding an inert gas at constant pressure has no effect on the equilibrium position, so the answer is "no change".
(b) NaHCO3(s) is added.
Adding solid NaHCO3 to the system introduces more of the reactant. According to Le Chatelier's principle, the system will adjust in a way that consumes the added reactant. In this case, it means that the equilibrium position will shift towards the right to produce more Na2CO3(s), CO2(g), and H2O(g). Therefore, the answer is "shifts towards the right".
(c) Mg(ClO4)2(s) is added as a drying agent to remove H2O.
Mg(ClO4)2 is added as a drying agent to remove water (H2O) from the system. Since water is one of the products in the reaction, its removal will drive the equilibrium towards the product side. Therefore, the equilibrium position will shift towards the right. The answer is "shifts towards the right".
(d) Dry ice is added at constant T.
Dry ice is solid carbon dioxide (CO2). Adding more CO2 to the system increases the concentration of CO2. According to Le Chatelier's principle, the system will adjust in a way that opposes the increase in CO2 concentration. To reduce CO2 concentration, the equilibrium will shift towards the left, favoring the reactant side. Therefore, the answer is "shifts towards the left".
a and c are right.
b has no effect. You already have solid NaHCO3 present; adding a few more grams doesn't change anything. Remember solids do not show up in the keq expression. Or if you do want to include the solid, the (solid) = 1 by definition. Adding more solid and (solid) still is 1 by definition.
I suppose d, technically, may depend upon what the constant T is, but I'm inclined to think the dry ice will vaporize and that will increase (CO2) which makes it shift to the left. Or you can think pCO2 will be increased so pressure shifts it to the side with fewer moles which is to the left also. Nice that both gives the same answer.