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Chapter 13: Problem 65

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
a. The system will shift to the right, producing more H₂(g) and CO₂(g). b. The system will shift to the right, producing more H₂(g) and CO₂(g). c. The system remains at the same equilibrium point. d. The system will shift to the left, producing more CO(g) and H₂O(g). e. The system remains at the same equilibrium point.

Step by step solution

01

Case a: Gaseous carbon dioxide is removed.

When gaseous carbon dioxide is removed from the equilibrium system, according to Le Châtelier's principle, the system will shift to the side that produces gaseous carbon dioxide to counteract the disturbance and restore equilibrium. So, the system will shift to the right, producing more H₂(g) and CO₂(g).
02

Case b: Water vapor is added.

When water vapor is added to the equilibrium system, this disturbs the balance in favor of the reactants. According to Le Châtelier's principle, the system will shift to the side that reduces the amount of water vapor to restore equilibrium. Therefore, the system will shift to the right, producing more H₂(g) and CO₂(g).
03

Case c: Pressure is increased by adding helium gas.

Adding helium gas to a rigid reaction container increases the pressure, but it does not directly affect the number of moles of any component in the reaction. Hence, the change in pressure does not result in a shift in the equilibrium of the reaction. The system remains at the same equilibrium point.
04

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

Since the reaction is exothermic, increasing the temperature adds heat to the equilibrium system. According to Le Châtelier's principle, the system will try to minimize the disturbance by shifting to the side that consumes the added heat. In this case, heat is produced in the forward reaction (CO + H₂O → H₂ + CO₂), so the system will shift to the left, producing more CO(g) and H₂O(g).
05

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

Decreasing the volume of the reaction container increases the pressure in the system. According to Le Châtelier's principle, the system will shift to the side with fewer moles of gas to reduce the pressure and restore equilibrium. In this reaction, there are equal numbers of moles of reactants and products (1 mol of CO + 1 mol of H₂O ⇌ 1 mol of H₂ + 1 mol of CO₂), so the change in volume does not result in a shift in the equilibrium of the reaction, and the system remains at the same equilibrium point.

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Most popular questions from this chapter

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 \(1.0 \mathrm{~atm}\) (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 value of the equilibrium constant \(K\) depends on which of the following (there may be more than one answer)? a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.

For the reaction below, \(K_{\mathrm{p}}=1.16\) at \(800 .{ }^{\circ} \mathrm{C}\). $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ If a 20.0-g sample of \(\mathrm{CaCO}_{3}\) is put into a \(10.0\) - \(\mathrm{L}\) container and heated to \(800 .{ }^{\circ} \mathrm{C}\), what percentage by mass of the \(\mathrm{CaCO}_{3}\) will react to reach equilibrium?

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

Consider the following reactions. \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) and \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)\) List two property differences between these two reactions that relate to equilibrium.
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