Video: EC17-18-S2-Q24

Consider the equilibrium system with the following equation. 2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g), ΔH = (−). State the effect of the following on the concentration of sulphur trioxide: i. removing oxygen from the reaction, ii. increasing the pressure.

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Video Transcript

Consider the equilibrium system with the following equation. 2SO₂ gas plus O₂ gas in equilibrium with 2SO₃ gas with negative change in enthalpy. State the effect of the following on the concentration of sulphur trioxide. Removing oxygen from the reaction, and increasing the pressure.

We’re being asked for the effect on the concentration of sulphur trioxide. This here is sulphur trioxide, SO₃. As we’re dealing with an equilibrium system, we’re going to be calling on Le Chatelier’s principle in order to work out the answers. Le Chatelier’s principle states that an equilibrium’s position will shift to oppose an external change. Let’s look at the first part of the question.

In this case, the change is the removal of oxygen. In this equilibrium, there are two reactions going on, the reaction of molecules of sulphur dioxide with oxygen forming sulphur trioxide and the decomposition of sulphur trioxide forming sulphur dioxide and oxygen. By continuously removing oxygen from the system, the forward reaction will slow down. Less oxygen to the system means fewer collisions with sulphur dioxide, means a lower production rate of sulphur trioxide. So relatively speaking, the removal of oxygen from the reaction will favour the reverse reaction. Meaning, more oxygen is going to be produced.

This means that a reaction that consumes sulphur trioxide is going to be favoured. Therefore, removing oxygen from the reaction mixture will cause the concentration of sulphur trioxide to decrease. What about increasing the pressure? All the components of this equilibrium are gases. And the pressure of a gas is proportional to the number of gas particles. In the forward reaction, the reaction of sulphur dioxide with oxygen, there are three equivalents of gas molecules whereas in the products, the sulphur trioxide, there are only two equivalents. Therefore, the forward reaction results in a decrease in the pressure.

Meanwhile, the reverse reaction, having only two equivalents in the reactants side and three equivalents of gas on the products side, serves to increase the pressure. Since the change in the system is increasing the pressure, Le Chatelier’s principle dictates that the equilibrium’s position will shift to favour the reaction that decreases the pressure. In this case, the forward reaction is favoured. This increases the concentration of sulphur trioxide.

Therefore, increasing the pressure in this system causes the concentration of sulphur trioxide to increase. So for the equilibrium between sulphur dioxide and oxygen and sulphur trioxide, the removal of oxygen from the reaction serves to decrease the concentration of sulphur trioxide. While increasing the pressure serves to increase it.

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