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Chapter 14: Problem 61

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 at (a) constant pressure and (b) constant volume.

Short Answer

Expert verified
For both (a) at constant pressure and (b) at constant volume, the addition of helium gas does not shift the equilibrium position since helium is a non reactive gas and does not affect the balance between reactants and products in the reaction.

Step by step solution

01

Recall Le Chatelier鈥檚 Principle

According to Le Chatelier's Principle, when pressure is increased, the equilibrium will shift towards the side of the reaction with fewer gas molecules, and when pressure is decreased, the equilibrium will shift towards the side with more gas molecules. However, when a non-reactive gas such as helium is added to a system at equilibrium, it basically increases the volume of the container while the quantities (moles) of the reactants and products remain the same.
02

Applying Le Chatelier's Principle (part a)

When helium is added at constant pressure, the total pressure of the system is not changed but the partial pressure of the reactants and products decreases. According to Le Chatelier's principle, the system will shift in the direction where there are more moles of gas to compensate for this decrease in partial pressures. As the reaction has the same number of moles of gas on both sides (2 moles on either side), the addition of helium gas at constant pressure does not shift the equilibrium position as neither direction will reestablish the pressure balance.
03

Applying Le Chatelier's Principle (part b)

When helium gas is added to the equilibrium system at constant volume, the total pressure increases due to the addition of more gas particles, and so does the partial pressure of the reactants and products. However, helium does not participate in the reaction, it does not increase the pressure of only one side of the reaction and hence the equilibrium will not shift neither to the left nor to the right. Therefore, adding helium gas at constant volume also does not shift the equilibrium.

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Key Concepts

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

Equilibrium Position
Understanding the equilibrium position is crucial for predicting how a reaction will behave under different conditions. In a chemical reaction like the gas-phase one given, equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. This leads to constant concentrations of reactants and products over time. Le Chatelier鈥檚 Principle states that when a system at equilibrium is disturbed, it will adjust to minimize that disturbance and restore balance. In the context of this exercise, adding helium gas is a disturbance. However, because helium is inert and doesn't react with the involved gases, its addition does not change the equilibrium state in terms of concentration or the equilibriums of reacting gases. The principle is applied by considering changes in conditions that directly affect the substances in the reaction.
Gas-Phase Reaction
Gas-phase reactions involve substances in the gaseous state reacting with each other, often leading to a shift in the equilibrium position. For the reaction given:
  • The forward direction: 2 CO(g) + O鈧�(g) 鈫� 2 CO鈧�(g)
  • The backward direction: 2 CO鈧�(g) 鈫� 2 CO(g) + O鈧�(g)
In this reaction, both sides have the same total number of moles of gas. That is, 2 moles of CO and 1 mole of O鈧� results in 2 moles of CO鈧�. This balance in moles means changes in conditions must directly influence either side to cause a shift. For instance, suitable conditions like temperature change could favor one direction, but an inert gas like helium does not alter the gas-phase equilibrium either at constant volume or constant pressure.
Partial Pressure
Partial pressure is a significant concept in gas reactions. It refers to the pressure exerted by a single type of gas in a mixture. When helium is added while maintaining constant pressure, the partial pressures of reactant and product gases drop because the total volume effectively increases without an increase in the number of moles of reacting gases. Despite this, for the given reaction, the moles of gas are balanced on both sides. As a result, there鈥檚 no net effect on the equilibrium because the decrease in partial pressure affects all gases uniformly. This interaction demonstrates that inert gases don't impact the directional shift in equilibrium if they don't alter the partial pressures of reactants versus products distinctly.
Constant Pressure vs Constant Volume
Constant pressure and constant volume represent two different states under which reactions can be observed or evaluated.
  • Constant Pressure: When helium is added at constant pressure, even though it expands the total available volume for the gases, it doesn鈥檛 change their intrinsic reactions as the moles of reacting gases remain balanced.
  • Constant Volume: Adding helium in this condition raises the total pressure but not the reaction-specific pressures because helium does not react. Hence, it does not cause a shift in equilibrium.
Therefore, whether helium is added in a constant pressure or constant volume scenario, it remains that if the reaction involves an equal number of gas moles on either side, no shift is observed in the chemical equilibrium position due to simply maintaining pressure-versus-volume balance without altering reactant or product conditions.

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Most popular questions from this chapter

At a certain temperature the following reactions have the constants shown: $$\begin{array}{ll}\mathrm{S}(s)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g) & K_{\mathrm{c}}^{\prime}=4.2 \times 10^{52} \\ 2 \mathrm{~S}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) & K_{\mathrm{c}}^{\prime \prime}=9.8 \times 10^{128}\end{array}$$ Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the following reaction at that temperature: $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$

A 2.50 -mole quantity of \(\mathrm{NOCl}\) was initially in a \(1.50-\mathrm{L}\) reaction chamber at \(400^{\circ} \mathrm{C}\). After equilibrium was established, it was found that 28.0 percent of the NOCl had dissociated: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it speeds up both the forward and reverse rates to the same extent. To test this statement, consider a situation in which an equilibrium of the type $$2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g)$$ is established inside a cylinder fitted with a weightless piston. The piston is attached by a string to the cover of a box containing a catalyst. When the piston moves upward (expanding against atmospheric pressure), the cover is lifted and the catalyst is exposed to the gases. When the piston moves downward, the box is closed. Assume that the catalyst speeds up the forward reaction \((2 \mathrm{~A} \longrightarrow \mathrm{B})\) but does not affect the reverse process \((\mathrm{B} \longrightarrow 2 \mathrm{~A}) .\) Suppose the catalyst is suddenly exposed to the equilibrium system as shown here. Describe what would happen subsequently. How does this "thought" experiment convince you that no such catalyst can exist?

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. \()\)

At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.487 \mathrm{M} .\) The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) is \(4.63 \times 10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless and \(\mathrm{NO}_{2}(g)\) has a brown color. \(]\)
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