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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’s 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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Most popular questions from this chapter

Heating solid sodium bicarbonate in a closed vessel establishes the following equilibrium: $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)$$ What would happen to the equilibrium position if (a) some of the \(\mathrm{CO}_{2}\) were removed from the system; (b) some solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) were added to the system; (c) some of the solid \(\mathrm{NaHCO}_{3}\) were removed from the system? The temperature remains constant.

Based on rate constant considerations, explain why the equilibrium constant depends on temperature.

The dissociation of molecular iodine into iodine atoms is represented as $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ At \(1000 \mathrm{~K},\) the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(\mathrm{I}_{2}\) in a 2.30 -L flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?

For the reaction $$\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g)$$ at \(700^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.534 .\) Calculate the number of moles of \(\mathrm{H}_{2}\) that are present at equilibrium if a mixture of 0.300 mole of \(\mathrm{CO}\) and \(0.300 \mathrm{~mole}\) of \(\mathrm{H}_{2} \mathrm{O}\) is heated to \(700^{\circ} \mathrm{C}\) in a 10.0 -L container.

A sealed glass bulb contains a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases. Describe what happens to the following properties of the gases when the bulb is heated from \(20^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}:\) (a) color, (b) pressure, (c) average molar mass, (d) degree of dissociation (from \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) ), (e) density. Assume that volume remains constant. (Hint: \(\mathrm{NO}_{2}\) is a brown gas; \(\mathrm{N}_{2} \mathrm{O}_{4}\) is colorless.)
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