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Chapter 13: Problem 75

An important reaction in the commercial production of hydrogen is $$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ How will this system at equilibrium shift in each of the five following cases? a. Gaseous carbon dioxide is removed. b. Water vapor is added. c. In a rigid reaction container, the pressure is increased by adding helium gas. d. The temperature is increased (the reaction is exothermic). e. The pressure is increased by decreasing the volume of the reaction container.

Short Answer

Expert verified
Case (a): The system shifts to the right as more CO and H2O react to produce H2 and CO2. Case (b): The system shifts to the right, increasing H2 and CO2 concentration while decreasing CO concentration. Case (c): No shift in equilibrium as adding helium gas doesn't affect the number of moles of reactants or products. Case (d): The system shifts to the left, increasing CO and H2O concentration and decreasing H2 and CO2 concentration. Case (e): No significant shift in equilibrium, as the number of moles of gas on each side is the same.

Step by step solution

01

Case (a): Gaseous carbon dioxide is removed.

If we remove CO2 from the system, the reaction will shift towards the right side to make up for the loss and reestablish the equilibrium. This means more CO (g) and H2O (g) will react to produce H2 (g) and CO2 (g). In this case, the concentration of CO2 will increase, and the concentration of CO and H2O will decrease.
02

Case (b): Water vapor is added.

If we add more water vapor (H2O) to the system, the reaction will shift towards the right side to consume the excess reactant and reestablish the equilibrium. This means more CO (g) and H2O (g) will react to produce H2 (g) and CO2 (g). In this case, the concentration of H2 and CO2 will increase, and the concentration of CO will decrease.
03

Case (c): Pressure is increased by adding helium gas in a rigid container.

Adding helium gas does not directly involve the reaction, so it will not affect the number of moles of reactants or products. However, adding helium gas increases the total pressure in the container. Since all the substances involved in the reaction are gases, their concentrations will be affected by the change in pressure, but because the number of moles of gases is the same on both sides of the equation, the reaction will remain at equilibrium. As a result, there will be no shift in equilibrium in this case.
04

Case (d): Temperature is increased (the reaction is exothermic).

An exothermic reaction is one that releases heat. When we increase the temperature of this system, the reaction will shift to the left to counteract the change, thereby absorbing the extra heat and reestablishing the equilibrium. In this case, the concentration of H2 (g) and CO2 (g) will decrease, and the concentration of CO (g) and H2O (g) will increase.
05

Case (e): Pressure is increased by decreasing the volume of the reaction container.

When the pressure is increased by decreasing the volume of the container, the reaction will shift to the side with a lesser number of moles of gas to counteract the change in pressure, which in this case is the right side. Both sides of the balanced equation have the same number of moles of gas (one mole each), so the system will not shift significantly in either direction. As a result, the equilibrium will remain relatively unchanged.

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Most popular questions from this chapter

A gaseous material \(\mathrm{XY}(g)\) dissociates to some extent to produce \(\mathrm{X}(g)\) and \(\mathrm{Y}(g) :\) $$\mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g)+\mathrm{Y}(g)$$ A 2.00 -g sample of \(\mathrm{XY}\) (molar mass $=165 \mathrm{g} / \mathrm{mol} )$ is placed in a container with a movable piston at \(25^{\circ} \mathrm{C}\) . The pressure is held constant at 0.967 \(\mathrm{atm} .\) As \(\mathrm{XY}\) begins to dissociate, the piston moves until 35.0 mole percent of the original \(\mathrm{XY}\) has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of \(K\) for this reaction of $25^{\circ} \mathrm{C}$ .

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\) . After equilibrium is reached the total pressure is 1.5 atm and 16\(\%\) (by moles) of the original $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( has dissociated to \)\mathrm{NO}_{2}(g) .$ a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C} .\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g) .\) c. What percentage (by moles) of the original $\mathrm{N}_{2} \mathrm{O}_{4}(g)$ is dissociated at the new equilibrium position (total pressure \(=1.00 \mathrm{atm} ) ?\)

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is $$\mathrm{HCO}_{3}^{-}(a q) \leftrightharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) K=5.6 \times 10^{-11}$$ If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into 1.00 \(\mathrm{L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}^{2-2}\) ?

Explain why the development of a vapor pressure above a liquid in a closed container represents an equilibrium. What are the opposing processes? How do we recognize when the system has reached a state of equilibrium?

For the following reactions, predict whether the mole fraction of the reactants or products increases or remains the same when the volume of the reaction vessel is increased. a. $\operatorname{Br}_{2}(g)+\mathrm{H}_{2}(g) \leftrightharpoons 2 \mathrm{HBr}(g)$ b. $2 \mathrm{CH}_{4}(g) \leftrightharpoons \mathrm{C}_{2} \mathrm{H}_{2}(g)+3 \mathrm{H}_{2}(g)$ c. \(2 \mathrm{HI}(g) \leftrightharpoons \mathrm{I}_{2}(s)+\mathrm{H}_{2}(g)\)
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