/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 The following table shows the en... [FREE SOLUTION] | 91Ó°ÊÓ

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The following table shows the enthalpies and Gibbs energies of formation of three metal oxides at \(25^{\circ} \mathrm{C}\). (a) Which of these oxides can be most readily decomposed to the free metal and \(\mathrm{O}_{2}(\mathrm{g}) ?\) (b) For the oxide that is most easily decomposed, to what temperature must it be heated to produce \(\mathrm{O}_{2}(\mathrm{g})\) at 1.00 atm pressure? $$\begin{array}{lll} \hline & \Delta \mathrm{H}_{7}, \mathrm{kJ} \mathrm{mol}^{-1} & \Delta \mathrm{G}_{7}, \mathrm{kJ} \mathrm{mol}^{-1} \\ \hline \mathrm{PbO}(\mathrm{red}) & -219.0 & -188.9 \\ \mathrm{Ag}_{2} \mathrm{O} & -31.05 & -11.20 \\ \mathrm{ZnO} & -348.3 & -318.3 \\ \hline \end{array}$$

Short Answer

Expert verified
The oxide that can be most readily decomposed into the metal and \( \mathrm{O}_2 \) gas is \( \mathrm{Ag}_2 \mathrm{O} \). However, without the entropy change (\( \Delta S \)), it is not possible to determine the temperature needed to decompose \( \mathrm{Ag}_2 \mathrm{O} \) into its components.

Step by step solution

01

Identify the most easily decomposable oxide

The most easily decomposable oxide will be the one with the highest (or least negative) Gibbs free energy (\( \Delta G \)) of formation. Searching the table for the highest \( \Delta G \) value, we find it is for \( \mathrm{Ag}_2 \mathrm{O} \) with \( \Delta G = -11.20 \mathrm{kJ/mol} \). Thus, \( \mathrm{Ag}_2 \mathrm{O} \) is the oxide which can most readily be decomposed to the metal and \( \mathrm{O}_2 \) gas.
02

Calculate the required temperature for decomposition

We use the equation for Gibbs free energy: \( \Delta G = \Delta H - T\Delta S \) to calculate the required temperature. In this scenario, we aim to have \( \Delta G = 0 \) since the reaction could occur spontaneously at this state. We., therefore, have equation \( 0 = \Delta H - T\Delta S \). The equation can be rewritten as \( T = \frac{\Delta H}{\Delta S} \). Thus, the temperature required is the change in enthalpy (\( \Delta H \)) divided by the change in entropy (\( \Delta S \)) for the formation of \( \mathrm{Ag}_2 \mathrm{O} \). However, we do not have the value for \( \Delta S \) and can not move further with this part of the question.

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Key Concepts

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

Enthalpy of Formation
Enthalpy of formation, denoted as \( \Delta H_f \) or simply \( \Delta H \) when the context is clear, is a crucial concept in chemical thermodynamics. It refers to the change in enthalpy when one mole of a substance is formed from its elements under standard conditions of temperature and pressure (usually \( 25^\circ \mathrm{C} \) and 1 atm). For a chemical reaction, it reflects the total heat content that is absorbed or released. Generally, if the value of \( \Delta H \) is negative, it implies that the reaction is exothermic and heat is released during the formation of a compound from its elements.

In our exercise, we compare the \( \Delta H \) values of different metal oxides to infer about their stability and how much energy is required to form them from their base elements. The less negative the enthalpy of formation, the more likely a compound can decompose back into its constituent elements, since it indicates a lower energy stability.
Metal Oxide Decomposition
Metal oxide decomposition refers to the process by which a metal oxide is broken down into its elements: the metal and oxygen. The ease with which a metal oxide decomposes is linked to its thermodynamic properties, specifically the Gibbs free energy of formation. The Gibbs free energy, \( \Delta G \), indicates the spontaneity of a reaction at constant temperature and pressure.

In our case, the objective is to determine which metal oxide can most readily be reduced to its elemental form. When comparing metal oxides, such as \( \mathrm{PbO} \) (red), \( \mathrm{Ag}_2\mathrm{O} \) and \( \mathrm{ZnO} \) in the given table, \( \mathrm{Ag}_2\mathrm{O} \) has the least negative \( \Delta G \) value. This tells us that among the given oxides, \( \mathrm{Ag}_2\mathrm{O} \) is the most prone to decompose into silver metal and oxygen gas under the right conditions.
Chemical Thermodynamics
Chemical thermodynamics involves the study of energy changes associated with chemical reactions, specifically relating to heat (enthalpy) and work. It incorporates the laws of thermodynamics to predict the feasibility of reactions and equilibrium states. A fundamental concept within this field is the interplay between enthalpy (\( \Delta H \)), entropy (\( \Delta S \)), and temperature (\( T \)), which together determine the Gibbs free energy (\( \Delta G \)) of a system.

The value of \( \Delta G \) helps chemists understand whether a process will occur spontaneously under certain conditions. If \( \Delta G \) is negative, the process is spontaneous and can occur without the input of added energy. For a reaction to be spontaneous at any temperature, not just at standard conditions, we must consider changes in \( \Delta S \) and \( T \) as well, and how they affect \( \Delta G \).
Gibbs Free Energy Equation
The Gibbs free energy equation is pivotal in predicting the spontaneity of a process. It is expressed as \( \Delta G = \Delta H - T\Delta S \). This equation connects the concepts of enthalpy (\( \Delta H \)), temperature (\( T \)), and entropy (\( \Delta S \)), offering a quantifiable means of determining the free energy change during a reaction. A negative \( \Delta G \) value suggests that a process or reaction can occur spontaneously.

In our table's context, to find the temperature at which the oxide will decompose, we set \( \Delta G = 0 \), indicating a state of equilibrium where the reaction can spontaneously proceed. Unfortunately, without additional data on the entropy change (\( \Delta S \)) for the formation of \( \mathrm{Ag}_2\mathrm{O} \), we cannot resolve the equation to find the exact temperature required for its decomposition. This demonstrates how each variable in the Gibbs free energy equation plays a critical role in understanding chemical reactivity and stability.

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Most popular questions from this chapter

To establish the law of conservation of mass, Lavoisier carefully studied the decomposition of mercury(II) oxide: $$\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=+90.83 \mathrm{kJ}\) and \(\Delta G^{\circ}=+58.54 \mathrm{kJ}\) (a) Show that the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in equilibrium with \(\mathrm{HgO}(\mathrm{s})\) and \(\mathrm{Hg}(\mathrm{l})\) at \(25^{\circ} \mathrm{C}\) is extremely low. (b) What conditions do you suppose Lavoisier used to obtain significant quantities of oxygen?

Following are some standard Gibbs energies of formation, \(\Delta G_{f}^{2},\) per mole of metal oxide at \(1000 \mathrm{K}: \mathrm{NiO},\) \(-115 \mathrm{kJ} ; \mathrm{MnO},-280 \mathrm{kJ} ; \mathrm{TiO}_{2},-630 \mathrm{kJ} .\) The standard Gibbs energy of formation of \(\mathrm{CO}\) at \(1000 \mathrm{K}\) is \(-250 \mathrm{kJ}\) per mol CO. Use the method of coupled reactions (page 851 ) to determine which of these metal oxides can be reduced to the metal by a spontaneous reaction with carbon at \(1000 \mathrm{K}\) and with all reactants and products in their standard states.

In a heat engine, heat \(\left(q_{\mathrm{h}}\right)\) is absorbed by a working substance (such as water) at a high temperature \(\left(T_{\mathrm{h}}\right)\) Part of this heat is converted to work \((w),\) and the rest \(\left(q_{1}\right)\) is released to the surroundings at the lower temperature ( \(T_{1}\) ). The efficiency of a heat engine is the ratio \(w / q_{\mathrm{h}}\). The second law of thermodynamics establishes the following equation for the maximum efficiency of a heat engine, expressed on a percentage basis. $$\text { efficiency }=\frac{w}{q_{\mathrm{h}}} \times 100 \%=\frac{T_{\mathrm{h}}-T_{1}}{T_{\mathrm{h}}} \times 100 \%$$ In a particular electric power plant, the steam leaving a steam turbine is condensed to liquid water at \(41^{\circ} \mathrm{C}\left(T_{1}\right)\) and the water is returned to the boiler to be regenerated as steam. If the system operates at \(36 \%\) efficiency, (a) What is the minimum temperature of the steam \(\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\right]\) used in the plant? (b) Why is the actual steam temperature probably higher than that calculated in part (a)? (c) Assume that at \(T_{\mathrm{h}}\) the \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is in equilibrium with \(\mathrm{H}_{2} \mathrm{O}(1) .\) Estimate the steam pressure at the temperature calculated in part (a). (d) Is it possible to devise a heat engine with greater than 100 percent efficiency? With 100 percent efficiency? Explain.

The synthesis of glutamine from glutamic acid is given by Glu \(^{-}+\mathrm{NH}_{4}^{+} \longrightarrow \mathrm{Gln}+\mathrm{H}_{2} \mathrm{O}\). The Gibbs energy for this reaction at \(\mathrm{pH}=7\) and \(T=310 \mathrm{K}\) is \(\Delta G^{\circ \prime}=14.8 \mathrm{kJ} \mathrm{mol}^{-1} .\) Will this reaction be sponta- neous if coupled with the hydrolysis of ATP? \(\mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ADP}+\mathrm{P}\) $$\Delta G^{\circ \prime}=-31.5 \mathrm{kJ} \mathrm{mol}-1$$

For each of the following reactions, indicate whether \(\Delta S\) for the reaction should be positive or negative. If it is not possible to determine the sign of \(\Delta S\) from the information given, indicate why. (a) \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) (b) \(2 \mathrm{HgO}(\mathrm{s}) \longrightarrow 2 \mathrm{Hg}(1)+\mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaCl}(1) \longrightarrow 2 \mathrm{Na}(1)+\mathrm{Cl}_{2}(\mathrm{g})\) (d) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})\) (e) \(\operatorname{Si}\left(\text { s) }+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SiCl}_{4}(\mathrm{g})\right.\)

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