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The following standard Gibbs energy changes are given for \(25^{\circ} \mathrm{C}\) (1) \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\) \(\Delta G^{\circ}=-33.0 \mathrm{kJ}\) (2) \(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(1)\) \(\Delta G^{\circ}=-1010.5 \mathrm{kJ}\) (3) \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})\) \(\Delta G^{\circ}=+173.1 \mathrm{kJ}\) (4) \(\mathrm{N}_{2}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=+102.6 \mathrm{kJ}\) (5) \(2 \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) \(\Delta G^{\circ}=+208.4 \mathrm{kJ}\) Combine the preceding equations, as necessary, to obtain \(\Delta G^{\circ}\) values for each of the following reactions. (a) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \quad \Delta G^{\circ}=?\) (b) \(2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(1) \quad \Delta G^{\circ}=?\) (c) \(2 \mathrm{NH}_{3}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(1)\) \(\Delta G^{\circ}=?\) Of reactions (a), (b), and (c), which would tend to go to completion at \(25^{\circ} \mathrm{C}\), and which would reach an equilibrium condition with significant amounts of all reactants and products present?

Short Answer

Expert verified
The ∆G° values are -105.8 kJ for reaction (a), -505.25 kJ for reaction (b), and -673.5 kJ for reaction (c). Therefore, all reactions would tend to go to completion at 25 ⁰C.

Step by step solution

01

Combine reactions to form the desired reaction (a)

The aim is to form the reaction \(N_{2}O(g) + 1.5 O_{2}(g) \rightarrow 2 NO_{2}(g)\). This can be achieved by reversing reaction (5) and adding it to reaction (4). The combined reaction is: \(2 N_{2}(g) + O_{2}(g) + N_{2}(g) + 2 O_{2}(g) \rightarrow 2 N_{2}O(g) + 2 NO_{2}(g)\). Simplifying gives the desired reaction: \(N_{2}O(g) + 1.5 O_{2}(g) \rightarrow 2 NO_{2}(g)\).
02

Calculate ∆G° for reaction (a)

The ∆G° for the combined reaction is the sum of the ∆G° values of the involved reactions, i.e., -∆G° of reaction (5) + ∆G° of reaction (4). Substitute the respective values: −(208.4 kJ) + 102.6 kJ = -105.8 kJ.
03

Repeat steps 1 and 2 for reactions (b) and (c)

For reaction (b), multiply reaction (2) by half: ∆G° is half of -1010.5 kJ, which equals -505.25 kJ. For reaction (c), subtract reaction (1) once from reaction (2) twice: calculating ∆G° gives -673.5 kJ.
04

Determine completion and equilibrium of reactions

From the final Gibbs energies of the reactions, if ∆G° < 0, the reaction will tend to go to completion. If ∆G° > 0, the reaction will reach equilibrium with significant quantities of both reactants and products. Since all ∆G° values calculated are < 0, all reactions (a), (b) and (c) will tend to go to completion at 25 ⁰C.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics focused on the study of energy transformations. It explores how energy changes from one form to another and how it influences matter. In chemical reactions, thermodynamics provides insights into whether a reaction can occur spontaneously or not. At the heart of thermodynamics are the three laws which describe energy conservation, entropy, and absolute zero. These laws help us understand how different substances will react under specific conditions.
  • First Law: Also known as the law of energy conservation, it states that energy cannot be created or destroyed, only transformed.
  • Second Law: Entropy of an isolated system always increases. This is a measure of the disorder in the system.
  • Third Law: As the temperature approaches absolute zero, the entropy of a system approaches a constant minimum.
Understanding these principles allows scientists to predict reaction outcomes based on energy changes. This is critical in developing efficient chemical processes and understanding natural phenomena.
Chemical Reactions
Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. Reactants transform into products through various pathways, and their feasibility depends on energy changes described by thermodynamics. Reactions can either release energy (exothermic) or absorb energy (endothermic).

  • Exothermic Reaction: Releases heat, resulting in a negative enthalpy change (\( \Delta H < 0 \)
  • Endothermic Reaction: Absorbs heat, resulting in a positive enthalpy change (\(\Delta H > 0\)).
The direction and extent to which a chemical reaction proceeds is dependent on factors like temperature, pressure, and energy changes. Catalysts also play a crucial role by lowering the activation energy, allowing reactions to proceed faster or under milder conditions.

Being able to understand the balance between enthalpy and entropy allows scientists to predict and manipulate reactions to obtain desired products.
Gibbs Energy Change
Gibbs Energy Change (\(\Delta G\)) is a measure of the maximum reversible work a thermodynamic system can perform, assuming constant temperature and pressure. It is a core concept for predicting the direction of chemical reactions. The Gibbs free energy equation is given by:\[\Delta G = \Delta H - T\Delta S\]Where \(\Delta H\) is the change in enthalpy, \(T\) is the temperature, and \(\Delta S\) is the change in entropy.

  • \(\Delta G < 0:\) The reaction is spontaneous and will proceed in the forward direction.
  • \(\Delta G > 0:\) The reaction is non-spontaneous, favoring the reverse direction.
  • \(\Delta G = 0:\) The system is at equilibrium, with no net change in the concentration of reactants and products.
In the provided exercise, calculating the Gibbs Energy Change for various reactions helps determine their spontaneity. Reactions with negative \(\Delta G\) values tend to go to completion, meaning they will proceed until the reactants are mostly converted into products.

This makes Gibbs Energy Change an essential tool for chemists in assessing reaction feasibility and optimizing processes for industrial and laboratory applications.

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Most popular questions from this chapter

If \(\Delta G^{\circ}=0\) for a reaction, it must also be true that (a) \(K=0 ;\) (b) \(K=1 ;\) (c) \(\Delta H^{\circ}=0 ;\) (d) \(\Delta S^{\circ}=0\) (e) the equilibrium activities of the reactants and products do not depend on the initial conditions.

Which substance in each of the following pairs would have the greater entropy? Explain. (a) at \(75^{\circ} \mathrm{C}\) and 1 atm: \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(1)\) or \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (b) at \(5^{\circ} \mathrm{C}\) and 1 atm: \(50.0 \mathrm{g} \mathrm{Fe}(\mathrm{s})\) or \(0.80 \mathrm{mol} \mathrm{Fe}(\mathrm{s})\) (c) 1 mol \(\mathrm{Br}_{2}\left(1,1 \text { atm }, 8^{\circ} \mathrm{C}\right)\) or \(1 \mathrm{mol} \mathrm{Br}_{2}(\mathrm{s}, 1 \mathrm{atm},\) \(\left.-8^{\circ} \mathrm{C}\right)\) (d) \(0.312 \mathrm{mol} \mathrm{SO}_{2}\left(\mathrm{g}, 0.110 \mathrm{atm}, 32.5^{\circ} \mathrm{C}\right)\) or \(0.284 \mathrm{mol}\) \(\mathrm{O}_{2}\left(\mathrm{g}, 15.0 \mathrm{atm}, 22.3^{\circ} \mathrm{C}\right)\)

Use data from Appendix D to determine values of \(\Delta G^{\circ}\) for the following reactions at \(25^{\circ} \mathrm{C}\) (a) \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})\) (b) \(2 \mathrm{SO}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\) (c) \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(2 \mathrm{Al}(\mathrm{s})+6 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})\)

The Gibbs energy change of a reaction can be used to assess (a) how much heat is absorbed from the surroundings; (b) how much work the system does on the surroundings; (c) the net direction in which the reaction occurs to reach equilibrium; (d) the proportion of the heat evolved in an exothermic reaction that can be converted to various forms of work.

The standard molar entropy of solid hydrazine at its melting point of \(1.53^{\circ} \mathrm{C}\) is \(67.15 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\). The enthalpy of fusion is \(12.66 \mathrm{kJmol}^{-1} .\) For \(\mathrm{N}_{2} \mathrm{H}_{4}(1)\) in the interval from \(1.53^{\circ} \mathrm{C}\) to \(298.15 \mathrm{K}\), the molar heat capacity at constant pressure is given by the expression \(C_{p}=97.78+0.0586(T-280) .\) Determine the standard molar entropy of \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})\) at \(298.15 \mathrm{K}\). [Hint: The heat absorbed to produce an infinitesimal change in the temperature of a substance is \(d q_{\mathrm{rev}}=C_{p} d T\).

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