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Ch.18 - Free Energy and Thermodynamics
Chapter 18, Problem 65b

Consider the reaction: 2 NO(g) + O2(g) → 2 NO2(g) Estimate ΔG° for this reaction at each temperature and predict whether or not the reaction is spontaneous. (Assume that ΔH° and ΔS° do not change too much within the given temperature range.) b. 715 K

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Identify the given reaction: \(2 \text{NO}(g) + \text{O}_2(g) \rightarrow 2 \text{NO}_2(g)\).
Recall the Gibbs free energy equation: \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\).
Determine \(\Delta H^\circ\) and \(\Delta S^\circ\) for the reaction using standard enthalpy and entropy values from a data table.
Substitute the values of \(\Delta H^\circ\), \(\Delta S^\circ\), and the given temperature \(T = 715\, \text{K}\) into the Gibbs free energy equation.
Analyze the sign of \(\Delta G^\circ\) to predict spontaneity: if \(\Delta G^\circ < 0\), the reaction is spontaneous; if \(\Delta G^\circ > 0\), it is non-spontaneous.

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Key Concepts

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

Gibbs Free Energy (ΔG)

Gibbs Free Energy (ΔG) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. It is used to predict the spontaneity of a reaction: if ΔG is negative, the reaction is spontaneous; if positive, it is non-spontaneous. The relationship between ΔG, enthalpy (ΔH), and entropy (ΔS) is given by the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin.
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Enthalpy (ΔH) and Entropy (ΔS)

Enthalpy (ΔH) is a measure of the total heat content of a system, reflecting the energy required to break and form bonds during a reaction. Entropy (ΔS) quantifies the degree of disorder or randomness in a system. Together, these two properties influence the Gibbs Free Energy and thus the spontaneity of a reaction. A reaction with a negative ΔH and positive ΔS is more likely to be spontaneous.
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Temperature's Role in Spontaneity

Temperature plays a crucial role in determining the spontaneity of a reaction as it affects the balance between enthalpy and entropy. In the Gibbs Free Energy equation, increasing temperature (T) can enhance the impact of entropy (ΔS) on ΔG. Therefore, at higher temperatures, reactions with positive ΔS may become spontaneous even if ΔH is positive, as the TΔS term can outweigh ΔH.
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Related Practice
Textbook Question

Use standard free energies of formation to calculate ΔG° at 25 °C for each reaction in Problem 61. How do the values of ΔG° calculated this way compare to those calculated from ΔH° and ΔS°? Which of the two methods could be used to determine how ΔG° changes with temperature?

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Open Question
Using standard free energies of formation, calculate ΔG° at 25 °C for each reaction in Problem 62. How do the values of ΔG° calculated this way compare to those calculated from ΔH° and ΔS°? Which of the two methods can determine how ΔG° changes with temperature?
Open Question
Is the question formulated correctly? If not, please correct it. Here is the question: 'Consider the reaction: 2 NO( g) + O2( g) → 2 NO2( g). Estimate ΔG° for this reaction at each temperature and predict whether or not the reaction is spontaneous, assuming that ΔH° and ΔS° do not change significantly within the given temperature range. a. 298 K b. 855 K.'
Open Question
Consider the reaction: CaCO3(s) → CaO(s) + CO2(g). Estimate ΔG° for this reaction at each temperature and predict whether or not the reaction is spontaneous. (Assume that ΔH° and ΔS° do not change too much within the given temperature range.) a. 298 K b. 1055 K c. 1455 K.
Textbook Question

Determine ΔG° for the reaction: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) Use the following reactions with known ΔG°rxn values:

2 Fe(s) + 3/2 O2(g) → Fe2O3(s) ΔG°rxn = -742.2 kJ

CO(g) + 12 O2( g) → CO2(g) ΔG°rxn = -257.2 kJ

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Open Question
Calculate ΔG°rxn for the reaction: CaCO3(s) → CaO(s) + CO2(g). Use the following reactions and given ΔG°rxn values: Ca(s) + CO2(g) + 1/2 O2(g) → CaCO3(s) ΔG°rxn = -734.4 kJ, 2 Ca(s) + O2(g) → 2 CaO(s) ΔG°rxn = -1206.6 kJ.