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Using data from Appendix \(C\), calculate \(\Delta G^{\circ}\) for the following reactions. lndicate whether each reaction is spontaneous under standard conditions. (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g)\) (c) \(6 \mathrm{Cl}_{2}(g)+2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{FeCl}_{3}(s)+3 \mathrm{O}_{2}(g)\) (d) \(\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

Short Answer

Expert verified
(a) For the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\), we have calculated that \(\Delta G^\circ = -200.99\,\mathrm{kJ/mol}\). Since \(\Delta G^\circ\) is negative, the reaction is spontaneous under standard conditions.

Step by step solution

01

Write down the values of \(\Delta H^\circ\) and \(\Delta S^\circ\) from Appendix C for the given reaction

We have: \(\Delta H^\circ(\mathrm{SO}_{2}) = -296.8\,\mathrm{kJ/mol}\) \(\Delta H^\circ(\mathrm{O}_{2}) = 0\,\mathrm{kJ/mol}\) \(\Delta H^\circ(\mathrm{SO}_{3}) = -395.7\,\mathrm{kJ/mol}\) \(\Delta S^\circ(\mathrm{SO}_{2}) = 248.2\,\mathrm{J/mol\cdot K}\) \(\Delta S^\circ(\mathrm{O}_{2}) = 205.1\,\mathrm{J/mol\cdot K}\) \(\Delta S^\circ(\mathrm{SO}_{3}) = 256.1\,\mathrm{J/mol\cdot K}\)
02

Calculate \(\Delta H^\circ\) and \(\Delta S^\circ\) for the reaction

For the reaction, we have: \(\Delta H^\circ = 2\Delta H^\circ(\mathrm{SO}_{3}) - 2\Delta H^\circ(\mathrm{SO}_{2}) - \Delta H^\circ(\mathrm{O}_{2}) = 2(-395.7\,\mathrm{kJ/mol}) - 2(-296.8\,\mathrm{kJ/mol}) - 0 = -197.8\,\mathrm{kJ/mol}\) \(\Delta S^\circ = 2\Delta S^\circ(\mathrm{SO}_{3}) - 2\Delta S^\circ(\mathrm{SO}_{2}) - \Delta S^\circ(\mathrm{O}_{2}) = 2(256.1\,\mathrm{J/mol\cdot K}) - 2(248.2\,\mathrm{J/mol\cdot K}) - 205.1\,\mathrm{J/mol\cdot K} = 10.7\,\mathrm{J/mol\cdot K}\)
03

Calculate \(\Delta G^\circ\) using the formula and check for spontaneity

At standard conditions, \(T = 298\,\mathrm{K}\). Therefore, we can calculate \(\Delta G^\circ\) as follows: \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -197.8\,\mathrm{kJ/mol} - (298\,\mathrm{K})(10.7\,\mathrm{J/mol\cdot K}) = -197.8\,\mathrm{kJ/mol} - 3.19\,\mathrm{kJ/mol} = -200.99\,\mathrm{kJ/mol}\) Since \(\Delta G^\circ\) is negative, the reaction is spontaneous under standard conditions. (b-d) Similarly, we can calculate \(\Delta G^\circ\) for the other given reactions by following the same steps. I encourage you to perform these calculations yourself using the data from Appendix C and the steps outlined for reaction (a).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Gibbs Free Energy
Standard Gibbs Free Energy, denoted as \(\Delta G^\circ\), is a fundamental thermodynamic quantity that indicates the amount of free energy available to do work during a chemical reaction at standard conditions (usually 1 atm pressure and 298.15 K). It combines the system's enthalpy (\(\Delta H^\circ\)) and entropy (\(\Delta S^\circ\)) in the relationship, \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\), where \(T\) is the temperature in kelvins. A negative \(\Delta G^\circ\) value signals a spontaneous reaction under these conditions, meaning the reaction can occur without any additional energy input. Conversely, a positive \(\Delta G^\circ\) implies that the reaction is non-spontaneous and would require an external energy source to proceed. Calculating \(\Delta G^\circ\) helps chemists predict how reactions behave in a standard setting, guiding them in processes like the synthesis of chemicals or the design of energy-efficient systems.
Chemical Spontaneity
Chemical spontaneity refers to the natural tendency of a chemical reaction to proceed without needing an external force or energy once it has been initiated. This concept is intimately connected with the sign of the Gibbs Free Energy change (\(\Delta G\)). When \(\Delta G\) is negative, the reaction is considered spontaneous because it releases energy, meaning it can occur on its own under the given conditions. If \(\Delta G\) is positive, the reaction is non-spontaneous, indicating it requires outside energy to proceed. The spontaneity of a reaction doesn't necessarily relate to its rate; a reaction can be spontaneous and yet occur very slowly, depending on the activation energy or the presence of a catalyst.
Thermodynamic Properties
The thermodynamic properties of a system, such as enthalpy (\(H\)), entropy (\(S\)), and Gibbs Free Energy (\(G\)), describe the system's total energy, randomness, and available energy to do work, respectively. Enthalpy reflects the heat content of a system under constant pressure and is associated with the making and breaking of chemical bonds. Entropy is the measure of molecular randomness or disorder. As a natural tendency, systems evolve towards a state of higher entropy. The Gibbs Free Energy is derived from both enthalpy and entropy and serves as a criterion for spontaneity at a constant temperature and pressure. Understanding these properties and how they interrelate is essential to predict the direction of chemical reactions and the equilibrium position.
Entropy and Enthalpy
Entropy (\(S\)) and Enthalpy (\(H\)) are two critical concepts in understanding thermodynamics and Gibbs Free Energy calculations. Entropy describes the level of disorder or randomness in a system. During a chemical reaction, the entropy changes, indicated by \(\Delta S\), due to the rearrangement of atoms and molecules. A positive \(\Delta S\) value means there's an increase in disorder. On the other hand, Enthalpy represents the heat content of a system and changes with the absorption or release of heat during a reaction (\(\Delta H\)). For reactions under constant pressure, \(\Delta H\) indicates the difference in heat between the products and reactants. When heat is released to the surroundings, \(\Delta H\) is negative, and the reaction is exothermic; conversely, when the reaction absorbs heat, \(\Delta H\) is positive, and the reaction is endothermic. The interplay between entropy and enthalpy is crucial for determining the spontaneity and overall direction of a chemical reaction.

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Most popular questions from this chapter

(a) What is meant by calling a process irreversible? (b) After an irreversible process the system is restored to its original state. What can be said about the condition of the surroundings after the system is restored to its original state? (c) Under what conditions will the condensation of a liquid be an irreversible process?

Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta E\) zero for the process? Explain.

Consider the reaction \(6 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{P}_{4}(g) \longrightarrow 4 \mathrm{PH}_{3}(g)\). (a) Using data from Appendix C, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the reaction mixture consists of \(8.0 \mathrm{~atm}\) of \(\mathrm{H}_{2}, 0.050 \mathrm{~atm}\) of \(\mathrm{P}_{4}\), and \(0.22 \mathrm{~atm}\) of \(\mathrm{PH}_{3}\).

Calculate \(\Delta S^{\circ}\) values for the following reactions by using tabulated \(S^{\circ}\) values from Appendix \(C\). In each case explain the sign of \(\Delta S^{\circ}\). (a) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)\) (c) \(\mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{MgCl}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)\)

The \(K_{b}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{b}\). (b) By using the value of \(K_{b r}\) calculate \(\Delta G^{\circ}\) for the equilibrium in part (a). (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3}+\right]=\left[\mathrm{H}^{+}\right]=1.5 \times 10^{-8} \mathrm{M}\) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]=5.5 \times 10^{-4} \mathrm{M}\), and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.120 \mathrm{M} ?\)

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