Skip to main content
Ch.18 - Free Energy and Thermodynamics
Chapter 18, Problem 90c

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?

Verified step by step guidance
1
insert step 1> Calculate the standard Gibbs free energy change (\( \Delta G^\circ \)) for the reaction at 298 K using the equation \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \), where \( \Delta H^\circ \) is the standard enthalpy change and \( \Delta S^\circ \) is the standard entropy change.
insert step 2> Determine the sign of \( \Delta G^\circ \) to assess the spontaneity of the reaction at 298 K. If \( \Delta G^\circ < 0 \), the reaction is spontaneous; if \( \Delta G^\circ > 0 \), it is non-spontaneous.
insert step 3> To find the temperature at which the partial pressure of CO2 is 1.0 atm, use the equation \( \Delta G = \Delta G^\circ + RT \ln Q \), where \( Q \) is the reaction quotient. Set \( Q = 1 \) for \( P_{CO2} = 1.0 \) atm.
insert step 4> Rearrange the equation to solve for the temperature \( T \) when \( \Delta G = 0 \) (equilibrium condition), which gives \( T = \frac{\Delta H^\circ}{\Delta S^\circ} \).
insert step 5> Analyze how changes in temperature affect the spontaneity of the reaction by considering the signs of \( \Delta H^\circ \) and \( \Delta S^\circ \). If \( \Delta H^\circ > 0 \) and \( \Delta S^\circ > 0 \), increasing temperature makes the reaction more spontaneous.

Verified Solution

Video duration:
8m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

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

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. In the context of this reaction, altering the temperature can affect the equilibrium position, favoring either the reactants or products depending on whether the reaction is exothermic or endothermic.
Recommended video:
Guided course
07:32
Le Chatelier's Principle

Gibbs Free Energy

Gibbs Free Energy (G) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. A reaction is spontaneous when the change in Gibbs Free Energy (ΔG) is negative. The relationship between ΔG, enthalpy (ΔH), and entropy (ΔS) is given by the equation ΔG = ΔH - TΔS, which is crucial for determining the spontaneity of the reaction at different temperatures.
Recommended video:
Guided course
01:51
Gibbs Free Energy of Reactions

Equilibrium Constant and Partial Pressure

The equilibrium constant (K) for a reaction at a given temperature is a measure of the ratio of the concentrations of products to reactants at equilibrium. For the reaction BaCO3(s) ⇌ BaO(s) + CO2(g), the partial pressure of CO2 is directly related to the equilibrium constant. At 1.0 atm of CO2, the temperature can be adjusted to find the specific conditions under which the reaction favors the formation of products, thus influencing the spontaneity of the reaction.
Recommended video:
Guided course
00:48
Partial Pressure Calculation
Related Practice
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?

499
views
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.

391
views
Textbook Question

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?

1692
views
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.

1492
views
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.

1285
views
Open Question
The standard free energy change for the hydrolysis of ATP was given in Problem 91. In a particular cell, the concentrations of ATP, ADP, and Pi are 0.0031 M, 0.0014 M, and 0.0048 M, respectively. Calculate the free energy change for the hydrolysis of ATP under these conditions, assuming a temperature of 298 K.
1
views