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Ch.15 - Chemical Equilibrium
Chapter 15, Problem 157

The amount of carbon dioxide in a gaseous mixture of CO2 and CO can be determined by passing the gas into an aqueous solution that contains an excess of Ba(OH)2. The CO2 reacts, yielding a precipitate of BaCO3, but the CO does not react. This method was used to analyze the equilibrium composition of the gas obtained when 1.77 g of CO2 reacted with 2.0 g of graphite in a 1.000-L container at 1100 K. The analysis yielded 3.41 g of BaCO3. Use these data to calculate Kp at 1100 K for the reaction CO2(g) + C(s) ⇌ 2 CO(g).

Verified step by step guidance
1
Step 1: Write the balanced chemical equation for the reaction: CO2(g) + C(s) ⇌ 2 CO(g).
Step 2: Calculate the moles of CO2 initially present using its mass (1.77 g) and molar mass (44.01 g/mol).
Step 3: Determine the moles of BaCO3 formed using its mass (3.41 g) and molar mass (197.34 g/mol). This corresponds to the moles of CO2 that reacted.
Step 4: Calculate the moles of CO2 remaining in the equilibrium mixture by subtracting the moles of CO2 that reacted from the initial moles of CO2.
Step 5: Use the stoichiometry of the reaction to find the moles of CO produced at equilibrium, and then calculate the partial pressures of CO2 and CO to find Kp using the expression Kp = (P_CO)^2 / P_CO2.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. In this context, the reaction CO2(g) + C(s) ⇌ 2 CO(g) reaches equilibrium at a specific temperature, and the equilibrium constant (Kp) quantifies the ratio of the partial pressures of the products to the reactants at that state.
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Precipitation Reaction

A precipitation reaction occurs when two soluble reactants form an insoluble product, or precipitate. In this case, CO2 reacts with Ba(OH)2 to form BaCO3, which precipitates out of the solution. This reaction is crucial for determining the amount of CO2 in the gas mixture, as the mass of BaCO3 formed can be used to calculate the initial amount of CO2 present.
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Molar Mass and Stoichiometry

Molar mass is the mass of one mole of a substance, and stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. To calculate Kp, the moles of CO2 reacted can be determined from the mass of BaCO3 produced, allowing for the stoichiometric conversion to find the equilibrium concentrations of CO and CO2, which are essential for calculating the equilibrium constant.
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Related Practice
Textbook Question

Acetic acid tends to form dimers, (CH3CO2H2), because of hydrogen bonding: The equilibrium constant Kc for this reaction is 1.51⨉102 in benzene solution but only 3.7⨉10-2 in water solution. (b) Calculate the ratio of dimers to monomers for 0.100 M acetic acid in water.

Textbook Question
A 79.2 g chunk of dry ice (solid CO2) and 30.0 g of graphite (carbon) were placed in an empty 5.00-L container, and the mixture was heated to achieve equilibrium. The reaction isCO 1g2 + C s ∆ 2 CO g(b) What is the value of Kp at 1100 K if the gas density at 1100 K is 16.9 g/L?
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Textbook Question
A 14.58 g quantity of N2O4 was placed in a 1.000-L reaction vessel at 400 K. The N2O4 decomposed to an equilibrium mix- ture of N2O4 and NO2 that had a total pressure of 9.15 atm.(b) How much heat (in kilojoules) was absorbed when the N2O4 decomposed to give the equilibrium mixture? (Stan- dard heats of formation may be found in Appendix B.)
Textbook Question

Consider the sublimation of mothballs at 27 °C in a room having dimensions 8.0 ft ⨉ 10.0 ft ⨉ 8.0 ft. Assume that the mothballs are pure solid naphthalene (density 1.16 g/cm3) and that they are spheres with a diameter of 12.0 mm. The equilibrium constant Kc for the sublimation of naphthalene is 5.40⨉10-6 at 27 °C. C10H8(s) ⇌ C10H8(g) (a) When excess mothballs are present, how many gaseous naphthalene molecules are in the room at equilibrium?

Textbook Question

Consider the sublimation of mothballs at 27 °C in a room having dimensions 8.0 ft ⨉ 10.0 ft ⨉ 8.0 ft. Assume that the mothballs are pure solid naphthalene (density 1.16 g/cm3) and that they are spheres with a diameter of 12.0 mm. The equilibrium constant Kc for the sublimation of naphthalene is 5.40⨉10-6 at 27 °C. C10H8(s) ⇌ C10H8(g) (b) How many mothballs are required to saturate the room with gaseous naphthalene?