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Ch.19 - Free Energy & Thermodynamics
All textbooksTro 6th EditionCh.19 - Free Energy & ThermodynamicsProblem 94b
Chapter 19, Problem 94b

Consider this reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2( g) b. If BaCO3 is placed in an evacuated flask, what is the partial pressure of CO2 when the reaction reaches equilibrium?

Verified step by step guidance
1
Step 1: Identify the balanced chemical equation for the reaction. In this case, it is BaCO3(s) ⇌ BaO(s) + CO2(g).
Step 2: Recognize that the problem is asking for the partial pressure of CO2 at equilibrium. This means we need to use the equilibrium constant expression for this reaction, which is Kp = [CO2].
Step 3: Understand that the equilibrium constant Kp is related to the standard Gibbs free energy change (ΔG°) by the equation ΔG° = -RT ln(Kp), where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin.
Step 4: Look up the standard Gibbs free energy change (ΔG°) for the reaction. This value is typically found in a table in your textbook or online. Once you have this value, you can substitute it into the equation from step 3 to solve for Kp.
Step 5: Once you have the value of Kp, you can use it to find the partial pressure of CO2 at equilibrium. Since the reaction starts with only BaCO3 and an evacuated flask (meaning no initial pressure of CO2), the pressure of CO2 at equilibrium will be equal to the square root of Kp.

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Key Concepts

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

Equilibrium Constant (Kp)

The equilibrium constant for a reaction involving gases, Kp, relates the partial pressures of the products and reactants at equilibrium. For the reaction BaCO3(s) ⇌ BaO(s) + CO2(g), Kp can be expressed as Kp = P(CO2), since solids do not appear in the expression. Understanding Kp is essential for calculating the partial pressure of CO2 at equilibrium.
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Le Chatelier's Principle

Le Chatelier's Principle states that if a system at equilibrium is disturbed, the system will shift in a direction that counteracts the disturbance. In this case, if BaCO3 is placed in an evacuated flask, the system will respond by producing CO2 gas until a new equilibrium is established. This principle helps predict how changes in conditions affect the position of equilibrium.
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Partial Pressure

Partial pressure is the pressure exerted by a single component of a gas mixture. In this reaction, the partial pressure of CO2 is crucial for determining the equilibrium state. It can be calculated using the ideal gas law if the number of moles of CO2 and the volume of the flask are known, allowing for a quantitative understanding of the gas behavior in the reaction.
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Related Practice
Textbook Question

Consider this reaction occurring at 298 K: N2O(g) + NO2(g) ⇌ 3 NO(g) b. If a reaction mixture contains only N2O and NO2 at partial pressures of 1.0 atm each, the reaction will be spontaneous until some NO forms in the mixture. What maximum partial pressure of NO builds up before the reaction ceases to be spontaneous?

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Textbook Question

Consider this reaction occurring at 298 K: N2O(g) + NO2(g) ⇌ 3 NO(g) c. Can the reaction be made more spontaneous by an increase or decrease in temperature? If so, what temperature is required to make the reaction spontaneous under standard conditions?

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Textbook Question

Consider this reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2(g) a. Show that the reaction is not spontaneous under standard conditions by calculating ΔG°rxn.

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Textbook Question

Consider this reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2(g) c. Can the reaction be made more spontaneous by an increase or decrease in temperature? If so, at what temperature is the partial pressure of carbon dioxide 1.0 atm?

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Textbook Question

Living organisms use energy from the metabolism of food to create an energy-rich molecule called adenosine triphosphate (ATP). The ATP acts as an energy source for a variety of reactions that the living organism must carry out to survive. ATP provides energy through its hydrolysis, which can be symbolized as follows: ATP(aq) + H2O(l) → ADP(aq) + Pi(aq) ΔGrxn ° = -30.5 kJ where ADP represents adenosine diphosphate and Pi represents an inorganic phosphate group (such as HPO42-). a. Calculate the equilibrium constant, K, for the given reaction at 298 K.

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Textbook Question

Living organisms use energy from the metabolism of food to create an energy-rich molecule called adenosine triphosphate (ATP). The ATP acts as an energy source for a variety of reactions that the living organism must carry out to survive. ATP provides energy through its hydrolysis, which can be symbolized as follows: ATP(aq) + H2O(l) → ADP(aq) + Pi(aq) ΔG°rxn = -30.5 kJ where ADP represents adenosine diphosphate and Pi represents an inorganic phosphate group (such as HPO42-). b. The free energy obtained from the oxidation (reaction with oxygen) of glucose (C6H12O6) to form carbon dioxide and water can be used to re-form ATP by driving the given reaction in reverse. Calculate the standard free energy change for the oxidation of glucose and estimate the maximum number of moles of ATP that can be formed by the oxidation of one mole of glucose.

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