Question

The answer is supposed to be (D), however not sure how to prove it? Would like to see and understand the math behind the answer.

2 BaO2(s) ⇄ 2 BaO(s) + O2(g)

ΔH° = 162 kJ/molrxn

A sealed rigid vessel contains BaO2(s) in equilibrium with BaO(s) and O2(g) as represented by the equation above.

Which of the following changes will increase the amount of BaO2(s) in the vessel?

(A)Removing a small amount of O2(g)

(B) Removing a small amount of BaO(s)

(C) Adding He gas to the vessel

(D) Lowering the temperature

Answer 1

Answer 2

The equilibrium constant for the reaction is given by the expression:

Kc = [BaO]^2[O2]/[BaO2]

At equilibrium, the concentrations of the reactants and products are constant. If we increase the amount of BaO2 in the vessel, the system will try to counteract this change by shifting the equilibrium to the left, towards the reactants. Similarly, if we decrease the amount of BaO2 in the vessel, the system will try to counteract this change by shifting the equilibrium to the right, towards the products.

To increase the amount of BaO2 in the vessel, we need to shift the equilibrium to the right, towards the products. According to Le Chatelier's principle, we can achieve this by decreasing the concentration of one or more of the products or by increasing the concentration of one or more of the reactants.

Now, let's consider each of the options:

(A) Removing a small amount of O2(g): This will increase the concentration of O2(g) in the vessel and shift the equilibrium to the left, towards the reactants. This will decrease the amount of BaO2(s) in the vessel, not increase it. Therefore, option A is incorrect.

(B) Removing a small amount of BaO(s): This will increase the concentration of BaO(s) in the vessel and shift the equilibrium to the right, towards the products. This will increase the amount of BaO2(s) in the vessel. Therefore, option B is correct.

(C) Adding He gas to the vessel: This will not affect the concentrations of the reactants or products and will not shift the equilibrium. Therefore, option C is incorrect.

(D) Lowering the temperature: According to Le Chatelier's principle, decreasing the temperature of an exothermic reaction will shift the equilibrium to the right, towards the products. As the given reaction is exothermic (ΔH° = -162 kJ/molrxn), lowering the temperature will increase the amount of BaO2(s) in the vessel. Therefore, option D is correct.

Therefore, the correct answer is (D) lowering the temperature.