After the equilibrium represented above is established, some pure O2 (g) is injected into the reaction vessel at constant temperature. After equilibrium is reestablished, which of the following has a lower value compared to its value at the original equilibrium? a. Keq for the reactionb. The total pressure in the reaction vesselc. The amount of SO3 (g) in the reaction vessel d. The amount of O2 (g) in the reaction vessele. The amount of SO2 (g) in the reaction vessel

Keq for the reaction (option a)

Explanation:

This is the equilbrium

2SO₂ (g) + O₂ (g) ⇄ 2SO₃ (g)

The expression for Kc is this:

Kc = [SO₃]² / [O₂]. [SO₂]²

If we add O₂, we are increasing reactant side, so when we reach the new equilibrium, Kc will be lower.

Kc ↓ = [SO₃]² / [O₂] ↑. [SO₂]²

- If we add O₂, total pressure will be increased.

- The amount of SO₃ will also increase. As we pertubate the reactant side, the reaction has to promote produc synthesis, to reach the new equilibrium.

Out of equilibrium we talk about Qc (quotient reaction) - The expression is the same but we apply Qc when we are not in eq.

Qc ↓ = [SO₃]² / [O₂] ↑. [SO₂]²

If Qc < Kc some reactants have to be transformed into products

- The amount of SO₂ is still the same.

- If we add O₂, partial pressure of it, will be increased, so the total pressure will also increase.