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Determine \(\Delta \mathrm{G}^{\circ}\) for the reaction \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\ell)\) \(\Delta \mathrm{G}^{\circ} \mathrm{f}\) of \(\mathrm{NH}_{3}(\mathrm{~g})=-4.0 \mathrm{Kcal} / \mathrm{mole}\) \(\Delta \mathrm{G}^{\circ} \mathrm{f}\) of \(\mathrm{NO}(\mathrm{g})=20.7 \mathrm{Kcal} / \mathrm{mole}\) \(\Delta \mathrm{G}^{\circ} \mathrm{f}\) of \(\mathrm{H}_{2} \mathrm{O}(\ell)=-56.7 \mathrm{Kcal} / \mathrm{mole}\)

Short Answer

Expert verified
The standard Gibbs free energy change (ΔG°) for the reaction \(4 NH_3(g) + 5 O_2(g) \rightarrow 4 NO(g) + 6 H_2O(l)\) is approximately -241.4 Kcal/mole.

Step by step solution

01

Identify the reactants and products

In the given reaction: Reactants: 4 moles of NH₃(g) and 5 moles of O₂(g) Products: 4 moles of NO(g) and 6 moles of H₂O(l)
02

Write expression for ΔG°

According to the formula, we can write the expression for ΔG° as: ΔG° = [4 × ΔG°f(NO) + 6 × ΔG°f(H₂O)] - [4 × ΔG°f(NH₃) + 5 × ΔG°f(O₂)]
03

Substitute values

Now, we need to substitute the ΔG°f values for each species given in the problem: - ΔG°f(NH₃) = -4.0 Kcal/mole - ΔG°f(NO) = 20.7 Kcal/mole - ΔG°f(H₂O) = -56.7 Kcal/mole - Note that O₂ is an element in its standard state, so its ΔG°f value is 0. Substituting these values into the expression: ΔG° = [4 × 20.7 + 6 × (-56.7)] - [4 × (-4.0) + 5 × 0]
04

Calculate ΔG°

By solving the above equation, we get the value for ΔG°: ΔG° = [82.8 - 340.2] - [-16] ΔG° = (-257.4) + 16 ΔG° ≈ -241.4 Kcal/mole So, the standard Gibbs free energy change (ΔG°) for the given reaction is approximately -241.4 Kcal/mole.

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Key Concepts

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

Thermodynamics and Gibbs Free Energy
Thermodynamics is a branch of physics that studies how energy is transferred in the form of heat and work. A key concept within thermodynamics is Gibbs free energy, symbolized as \( \Delta G \), which is crucial for predicting the direction of a chemical reaction.

Gibbs free energy combines the system's enthalpy (\( \Delta H \)), its entropy (\( \Delta S \)), and the temperature (\( T \)) to calculate whether a process will occur spontaneously. The relationship is given by the equation \( \Delta G = \Delta H - T\Delta S \). If \( \Delta G \) is negative, the process or reaction is spontaneous, meaning it will proceed without outside intervention. In contrast, a positive \( \Delta G \) implies that the reaction is non-spontaneous, and additional energy is needed to make it proceed.

In the context of the exercise, the calculation of the standard Gibbs free energy change (\( \Delta G^\circ \)) for a chemical reaction gives us direct insight into the feasibility of the reaction under standard conditions (1 atm pressure and 298.15 K temperature). The negative value of \( \Delta G^\circ \) in the provided solution indicates that the reaction would spontaneously occur under standard conditions.
Understanding Chemical Reactions
Chemical reactions involve breaking chemical bonds in reactants and forming new bonds to create products. Each reaction is governed by the laws of thermodynamics, and tracking the energy changes during a reaction is central to understanding its progress and spontaneity.

In our exercise, we look at the reaction where ammonia (NH₃) reacts with oxygen (O₂) to form nitric oxide (NO) and water (H₂O). Knowing the standard Gibbs free energy of formation (\( \Delta G^\circ_f \)) for each reactant and product helps calculate the overall \( \Delta G^\circ \) for the reaction. This overall change represents the energy difference between the final and initial states of the system.

It is essential to recognize that the \( \Delta G^\circ_f \) for elements in their standard state, like O₂ gas in this problem, is defined as zero. This simplifies calculations and is a crucial detail in determining the overall Gibbs free energy change.
Enthalpy and Entropy
Enthalpy \( (\Delta H) \) and entropy \( (\Delta S) \) are fundamental concepts of thermodynamics related to the energy and disorder within a system, respectively.

Enthalpy is a measure of the total energy of a thermodynamic system, usually related to the heat absorbed or released during a reaction under constant pressure. A reaction that releases heat to the surroundings, known as exothermic, has a negative \( \Delta H \), while an endothermic reaction, which absorbs heat, has a positive \( \Delta H \).

Entropy, on the other hand, quantifies the amount of disorder or randomness. Systems tend to evolve toward a state of higher entropy, meaning that processes that increase the disorder of the system are generally favored. Entropy changes can be affected by changes in temperature, phase, and molecular complexity.

In the grand scheme of thermodynamics, while a reaction may require or release energy (enthalpy), an increase in disorder (entropy) might drive the reaction to be spontaneous; thus, both these quantities help in predicting the feasibility of a reaction when considering Gibbs free energy.

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Most popular questions from this chapter

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), can be synthesized in two ways. The first method involves reduction of oxygen by hydrogen, $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(\ell) $$ The second method involves oxidation of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{2}(\ell) $$ Find the free energy of formation, \(\Delta \mathrm{G}^{\circ}\), for both processes and predict which process is more efficient for the commercial preparation of hydrogen peroxide.

Calculate the equilibrium constant at \(25^{\circ} \mathrm{C}\) for the reaction: $$ \mathrm{S}+3 / 2 \mathrm{O}_{2} \rightleftarrows \mathrm{SO}_{3} $$ The heat formation of \(\mathrm{SO}_{3}\) at \(25^{\circ} \mathrm{C}\) is \(-94.45 \mathrm{Kcal} / \mathrm{mole}\) and the standard molar entropy changes for \(\mathrm{S}, \mathrm{O}_{2}\), and \(\mathrm{SO}_{3}\) at \(25^{\circ} \mathrm{C}\) are \(7.62,49.0\) and \(61.24 \mathrm{cal} / \mathrm{mole}^{\circ} \mathrm{K}\), respectively.

Sulfur exists as \(S_{2}\) vapor at temperatures between \(700^{\circ} \mathrm{C}\) and \(1500^{\circ} \mathrm{C}\). At \(1473 \mathrm{k}\) it combines with hydrogen according to the equation $$ \mathrm{H}_{2}(\mathrm{~g})+(1 / 2) \mathrm{S}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $$ At \(750^{\circ} \mathrm{C}\) the equilibrium constant is \(1.07 \times 10^{2}\) and at \(1200^{\circ} \mathrm{C}\) it is \(4.39\). Determine the heat of reaction in the temperature range \(750^{\circ} \mathrm{C}\) to \(1200^{\circ} \mathrm{C}\), and the change in free energy at each of these temperatures.

If the standard free energy of formation of HI from \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) at \(490^{\circ} \mathrm{C}\) is \(-12.1 \mathrm{~kJ} /\) mole of \(\mathrm{HI}\), what is the equilibrium constant for this reaction? Assume \(\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{deg}^{-1}\)

For sublimation of iodine crystals, $$ \mathrm{I}_{2}(\mathrm{~s}) \rightleftarrows \mathrm{I}_{2}(\mathrm{~g}) $$ at \(25^{\circ} \mathrm{C}\) and atmospheric pressure, it is found that the change in enthalpy, \(\Delta \mathrm{H}=9.41 \mathrm{Kcal} /\) mole and the change in entropy, \(\Delta \mathrm{S}=20.6 \mathrm{cal} /\) deg \(-\) mole. At what temperature will solid iodine be in equilibrium with gaseous iodine?

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