Consider the gas-phase reaction $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g) $$ Predict the shift in the equilibrium position when helium gas is added to the equilibrium mixture (a) at constant pressure and (b) at constant volume.

Short Answer

Expert verified
The addition of helium gas to the system does not shift the equilibrium position, be it at constant pressure or volume. This is because helium, being an inert gas, has no interactive effect on the reaction either by changing the pressure or volume of the system.

Step by step solution

01

Understanding Le Chatelier's Principle

In accordance with Le Chatelier's principle, the system will adapt to stressors applied to it. If a system at equilibrium is disturbed, the system will adjust itself to restore the equilibrium.
02

Evaluate the Addition of Helium Gas at Constant Pressure

At constant pressure, adding helium does not change the partial pressures or moles of the reactants or the products. Helium is an inert gas; it doesn't involve in the reaction. Therefore, it does not affect the equilibrium of the reaction. It does not provide any 'stress' that the equilibrium needs to counteract. So, according to Le Chatelier's principle, the equilibrium position will not change.
03

Evaluate the Addition of Helium Gas at Constant Volume

At constant volume, the addition of helium increases the total number of moles in the system but does not change the number of moles of reactants and products nor the volume in which they are contained. It also makes no difference if the additional substance is chemically inert, like helium. The increased total number of moles of gas would increase the total pressure, but the partial pressures of the reactants and the products (which depend only on their moles and the volume) remain the same, so, they have no effect on the position of the equilibrium per Le Chatelier's principle. Once again, the equilibrium position will not shift.

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Key Concepts

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

Chemical Equilibrium
In chemistry, chemical equilibrium is a state where the concentrations of reactants and products remain constant over time, as they are consumed and produced at equal rates. It's crucial to understand that equilibrium doesn't mean the reactants and products are present in equal amounts, but rather that their rates of formation are equal, leading to a stable ratio of concentrations.

Imagine a crowded dance floor (the reaction container) where dancers (molecules of reactants and products) are constantly switching partners (reacting). Equilibrium is reached not when the dancers stop moving, but when the number of partners changed (reactant-product transformations) per song (unit of time) reaches a point where it becomes constant. The dance goes on (the reaction continues), but no net change is observed.

In the context of the gas-phase reaction of carbon monoxide and oxygen to form carbon dioxide, \(2 \mathrm{CO}(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{CO}_2(g)\), the system achieves equilibrium when the rate at which CO and O2 combine to form CO2 is equal to the rate at which CO2 decomposes back into CO and O2.
Reaction Equilibrium Shift
The reaction equilibrium shift occurs when an external change is applied to a system that is at equilibrium, causing the system to adjust in response. Le Chatelier's Principle can help predict these shifts. For instance, changes in concentration, pressure, volume, or temperature can cause the equilibrium to shift to the left (favoring reactants) or to the right (favoring products).

In the provided gas-phase reaction example, if we increase the amount of CO, the system will shift towards the products to reduce this 'stress'. Conversely, removing CO2 would prompt the system to produce more CO2 to restore balance. Adding an inert gas such as helium, however, would not typically change the concentrations of reactants or products and thus not shift the equilibrium, as indicated in the solution of the exercise provided.
Partial Pressures
The term partial pressure refers to the pressure that each gas in a mixture would exert if it alone occupied the entire volume. In a reaction involving gases, the partial pressures play a significant role because they are directly related to the mole fraction and total pressure according to Dalton's Law of Partial Pressures. This law states that the total pressure in a mixture of gases is the sum of the partial pressures of the individual gases.

Let's use the analogy of balloons in a room where different colored balloons represent different gases. The pressure inside the room (total pressure) is a result of all the balloons pushing against the walls. The pressure each type of balloon exerts against the wall is its partial pressure, and it depends on how many such balloons there are and the size of the room.

In the mentioned exercise, adding helium to the reaction mixture at constant volume increases the number of 'balloons' without affecting how much the reactant and product 'balloons' push against the walls (their partial pressures). Consequently, this addition does not impact the equilibrium of the chemical reaction, affirming the learning that inert gases at constant volume do not shift reaction equilibria.

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Most popular questions from this chapter

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}\) : $$ \mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftharpoons \mathrm{NaCl}(l)+\mathrm{K}(g) $$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C}\), respectively.)

The equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$ 2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is \(4.17 \times 10^{-34}\) at \(25^{\circ} \mathrm{C}\). What is the equilibrium constant for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g) $$ at the same temperature?

Define equilibrium. Give two examples of a dynamic equilibrium.

The formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is an intermediate step in the manufacture of sulfuric acid, andit is also responsible for the acid rain phenomenon. The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is 0.13 at \(830^{\circ} \mathrm{C}\). In one experiment 2.00 moles of \(\mathrm{SO}_{2}\) and 2.00 moles of \(\mathrm{O}_{2}\) were initially present in a flask. What must the total pressure be at equilibrium to have an 80.0 percent yield of \(\mathrm{SO}_{3} ?\)

When a gas was heated under atmospheric conditions, its color was found to deepen. Heating above \(150^{\circ} \mathrm{C}\) caused the color to fade, and at \(550^{\circ} \mathrm{C}\) the color was barely detectable. However, at \(550^{\circ} \mathrm{C}\), the color was partially restored by increasing the pressure of the system. Which of these best fits this description? Justify your choice. (a) A mixture of hydrogen and bromine, (b) pure bromine, (c) a mixture of nitrogen dioxide and dinitrogen tetroxide. (Hint: Bromine has a reddish color and nitrogen dioxide is a brown gas. The other gases are colorless.)

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