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How will reducing the partial pressure of \(\mathrm{O}_{2}\) affect the position of the equilibrium in the following reaction? $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$

Short Answer

Expert verified
Question: How will the equilibrium be affected if the partial pressure of O_2 is reduced in the following reaction: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ Answer: If the partial pressure of O_2 is reduced, the equilibrium will shift to the left, favoring the reactants side. This will cause the consumption of more SO_3 to generate more SO_2 and O_2, in an attempt to counteract the change and achieve a new equilibrium state.

Step by step solution

01

Understand Le Chatelier's Principle

Le Chatelier's principle states that if a change is applied to a system at equilibrium, the system will adjust to counteract the effect of the change and maintain equilibrium. To answer the given question, we need to consider how the reaction will adjust in response to a decrease in the partial pressure of O_2.
02

Identify the Reaction

The given reaction is: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$
03

Determine the Direction of Shift

We are given that the partial pressure of O_2 is decreased. According to Le Chatelier's principle, the reaction will shift in the direction that compensates for this change. In order to increase the partial pressure of O_2, the equilibrium will shift to the left, favoring the reactants side, which will consume the products and generate more reactants. This can be represented by: $$ 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \leftarrow 2 \mathrm{SO}_{3}(g) $$
04

Conclusion

By applying Le Chatelier's principle, we found that reducing the partial pressure of O_2 will cause the system to shift to the left, favoring the reactants side. The position of the equilibrium will change, consuming more SO_3 to generate more SO_2 and O_2, in an attempt to counteract the change and achieve a new equilibrium state.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
At the heart of many chemical processes is a state known as chemical equilibrium. This condition occurs when a reversible chemical reaction proceeds in both the forward and reverse directions at equal rates. In a visual sense, imagine a set of scales perfectly balanced; that's the essence of a reaction at equilibrium. The important point to remember is that even though the system appears static, with no observable changes, the reactants and products are continuously being converted back and forth.

For the reaction given in the exercise, 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), the equilibrium state is established when the rate at which sulfur dioxide (SO₂) and oxygen (O₂) combine to form sulfur trioxide (SO₃) is equal to the rate at which SO₃ decomposes back into SO₂ and O₂. It's important not to confuse equilibrium with the amounts of reactants and products being equal; it's about the rates of the forward and reverse reactions being equal.
Equilibrium Shift
With our system at equilibrium, what would happen if we altered the conditions? This is where Le Chatelier's principle comes into play. According to this principle, if an external change is applied to a reaction at equilibrium, the system will react in such a way as to counteract this change. These external changes can include modifications to concentration, pressure, volume, or temperature of the reacting system.

In our case, reducing the partial pressure of oxygen (Oâ‚‚) would cause such an equilibrium shift. Following Le Chatelier's principle, our reaction tries to oppose this reduction by producing more Oâ‚‚, hence shifting towards the reactants side (to the left). If, say, we were to increase the partial pressure of Oâ‚‚, the system would shift towards the products (to the right), trying to consume the excess Oâ‚‚. A simple tip to remember is that the reaction always shifts away from the side where you add something (like a reactant or product) and towards the side where something is removed.
Partial Pressure
The term partial pressure is crucial when discussing reactions involving gases. It refers to the pressure that a single gas component in a mixture of gases would exert if it occupied the entire volume on its own. The concept is tied intimately to concentration in reactions occurring in gaseous phases; as partial pressure increases, so does the concentration of that specific gas in the mixture.

Le Chatelier's principle interacts directly with changes in partial pressure to predict the direction of an equilibrium shift. So, when the exercise mentions reducing the partial pressure of oxygen (Oâ‚‚), it means that the concentration of Oâ‚‚ is being decreased. This reduction initiates a response from the system to restore balance, an intuitive dance governed by the laws of chemistry where the reaction seeks a new equilibrium by shifting as described earlier.

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Most popular questions from this chapter

The following reaction is carried out in a sealed, rigid vessel at constant temperature. $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}_{2}(g) $$ a. If the change in the partial pressure of \(\mathrm{O}_{2}\) is \(-x,\) what are the changes in the partial pressures of \(\mathrm{NO}\) and \(\mathrm{NO}_{2} ?\) b. As the reaction proceeds, what happens to the total pressure in the reaction vessel?

At equilibrium, is the sum of the concentrations of all the reactants always equal to the sum of the concentrations of the products? Explain why or why not.

Patients suffering from carbon monoxide poisoning are treated with pure oxygen to remove CO from the hemoglobin (Hb) in their blood. The two relevant equilibria are $$\begin{aligned} \mathrm{Hb}(a q)+4 \mathrm{CO}(g) & \rightleftharpoons \mathrm{Hb}(\mathrm{CO})_{4}(a q) \\ \mathrm{Hb}(a q)+4 \mathrm{O}_{2}(g) & \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q) \end{aligned}$$ The value of the equilibrium constant for CO binding to Hb is greater than that for \(\mathrm{O}_{2} .\) How, then, does this treatment work?

Write the \(K_{\mathrm{c}}\) expression for the oxidation of calcium sulfite to gypsum (calcium sulfate): $$ 2 \mathrm{CaSO}_{3}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CaSO}_{4}(s) $$

Why does adding an inert gas such as argon to an equilibrium mixture of \(\mathrm{CO}, \mathrm{O}_{2},\) and \(\mathrm{CO}_{2}\) in a sealed vessel increase the total pressure of the system but not shift the following equilibrium? $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g) $$

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