The values of \(\Delta G\) are very important in metallurgy. The \(\Delta G\) values for the following reactions at \(1000 \mathrm{~K}\) are given as: \(\mathrm{S}_{2}(\mathrm{~s})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{2}(\mathrm{~g}) ; \Delta G=-544 \mathrm{~kJ}\) \(2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}) \rightarrow 2 \mathrm{ZnS}(\mathrm{s}) ; \Delta G=-293 \mathrm{~kJ}\) \(2 \mathrm{Zn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{ZnO}(\mathrm{s}) ; \Delta G=-480 \mathrm{~kJ}\) The \(\Delta G\) for the reaction: \(2 \mathrm{ZnS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{ZnO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{~g})\) will be (a) \(-357 \mathrm{~kJ}\) (b) \(-731 \mathrm{~kJ}\) (c) \(-773 \mathrm{~kJ}\) (d) \(-229 \mathrm{~kJ}\)

Short Answer

Expert verified
-502 \mathrm{~kJ}

Step by step solution

01

- Write the given reactions and their \(\Delta G\) values

The given reactions are: \[\mathrm{S}_{2}(\mathrm{~s})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{2}(\mathrm{~g}); \Delta G=-544 \mathrm{~kJ}\] \[2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}) \rightarrow 2 \mathrm{ZnS}(\mathrm{s}); \Delta G=-293 \mathrm{~kJ}\] \[2 \mathrm{Zn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{ZnO}(\mathrm{s}); \Delta G=-480 \mathrm{~kJ}\]
02

- Combine the half reactions to form the target reaction

We need to combine the given reactions in such a way that the resultant equation gives us the target reaction: \[2 \mathrm{ZnS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{ZnO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{~g})\] This can be done by reversing the second reaction and multiplying it by 2 followed by adding it to the first reaction multiplied by 2.
03

- Calculate the \(\Delta G\) for the target reaction

Since the Gibbs free energy change \(\Delta G\) is a state function, we can add or subtract the values for these reactions appropriately. For the reversed second reaction, the \(\Delta G\) value will have its sign changed.First reaction (doubled): \[-544 \mathrm{~kJ} \times 2 = -1088 \mathrm{~kJ}\]Second reaction (reversed and doubled): \[-(-293 \mathrm{~kJ}) \times 2 = 586 \mathrm{~kJ}\]The third reaction is not used because it is already in the correct form and does not need to be combined for our target reaction.The sum of the \(\Delta G\) values for the combined reactions gives us the \(\Delta G\) for the target reaction: \[-1088 \mathrm{~kJ} + 586 \mathrm{~kJ} = -502 \mathrm{~kJ}\]

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Key Concepts

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

Thermodynamics in Metallurgy
Thermodynamics plays a critical role in metallurgy, which is the branch of science and engineering concerning the properties of metals and their production and purification. In the context of metallurgical processes, thermodynamics helps us understand and quantify the energy changes that occur during metal extraction and refining.

At the heart of thermodynamics is the concept of Gibbs free energy, often denoted as \( \Delta G \). It's a measure of the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. This makes it an invaluable tool for predicting the direction of chemical reactions in metallurgy. If \( \Delta G \) is negative for a particular reaction, it signifies that the process can occur spontaneously under the given conditions.

For example, in the textbook exercise provided, we see that each individual reaction has a \( \Delta G \) value associated with it. These values help metallurgists determine which reactions will proceed, and are essential for designing processes that optimize the production of metals. In the steps provided, the calculation of \( \Delta G \) for the target reaction is effectively a thermodynamic analysis that informs us about the feasibility of that reaction occurring at 1000 K.

It is essential for students to understand that thermodynamics, and specifically the calculation of Gibbs free energy, is not just a theoretical exercise but a practical tool used in real-world metallurgical applications to ensure efficient and cost-effective production of metals.
Chemical Reactions in Metallurgy
When studying metallurgy, it's important to recognize that the production of metals often involves a series of intricate chemical reactions. Each reaction contributes to the overall process of extracting a metal from its ore and refining it to a usable form.

The exercise illustrates several chemical reactions that highlight critical aspects of metallurgy. For instance, the oxidation of sulfur \( \mathrm{S}_{2} \) to sulfur dioxide \( \mathrm{SO}_{2} \) and the formation of zinc sulfide \( \mathrm{ZnS} \) from zinc and sulfur. These reactions are simplified representations of the complex processes that occur in smelting operations.

Chemical reactions in metallurgy are chosen based on their energetics, often represented by the change in Gibbs free energy. By calculating the \( \Delta G \) for these reactions, metallurgists can determine their spontaneity. The combination of half-reactions to form a desired overall reaction, as seen in the textbook solution, is an application of stoichiometry and energetics to optimize the extraction and synthesis of the desired metal compounds.
Enthalpy and Entropy in Metallurgical Processes
Enthalpy \( \Delta H \) and entropy \( \Delta S \) are key components of Gibbs free energy, which is given by the equation \( \Delta G = \Delta H - T\Delta S \), where \( T \) represents temperature. In metallurgical reactions, enthalpy effectively measures the heat absorbed or released, while entropy measures the disorder or randomness of the system with respect to temperature and pressure.

In any metallurgical process, the change in enthalpy indicates whether the reaction is endothermic (absorbing heat) or exothermic (releasing heat). This is crucial because maintaining optimal temperatures is essential for efficient metal production. For the reactions given in the exercise, the negative values of \( \Delta G \) suggest that the reactions are exothermic; they release energy and can occur spontaneously at 1000 K.

Meanwhile, entropy changes during a reaction can provide insight into the degree of freedom or randomness of the particles involved. A positive \( \Delta S \) indicates an increase in disorder, which is common in reactions where gases are produced or the number of moles of gaseous products increases, as seen in the formation of \( \mathrm{SO}_{2} \).

To fully grasp these concepts, students should study how enthalpy and entropy changes affect the spontaneity of a reaction through the Gibbs free energy equation. Understanding the balance between enthalpy and entropy is fundamental for predicting reaction behavior and designing efficient metallurgical processes.

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Most popular questions from this chapter

The volume of a system becomes twice its original volume on the absorption of 300 cal of heat. The work done on the surrounding was found to be 200 cal. What is \(\Delta U\) for the system? (a) \(500 \mathrm{cal}\) (b) \(300 \mathrm{cal}\) (c) \(100 \mathrm{cal}\) (d) \(-500 \mathrm{cal}\)

Which of the following pair does show the extensive properties? (a) temperature and pressure (b) viscosity and surface tension (c) refractive index and specific heat (d) volume and heat capacity

An adiabatic cylinder fitted with an adiabatic piston at the right end of cylinder, is divided into two equal halves with a monoatomic gas on left side and diatomic gas on right side, using an impermeable movable adiabatic wall. If the piston is pushed slowly to compress the diatomic gas to \(\frac{3}{4}\) th of its original volume. The ratio of new volume of monoatomic gas to its initial volume would be (a) \(\left(\frac{4}{3}\right)^{\frac{25}{21}}\) (b) \(\left(\frac{7}{5}\right)^{\frac{3}{4}}\) (c) \(\left(\frac{3}{4}\right)^{\frac{21}{25}}\) (d) \(\frac{3}{4}\)

Two moles of an ideal monoatomic gas is heated from \(27^{\circ} \mathrm{C}\) to \(627^{\circ} \mathrm{C}\), reversibly and isochorically. The entropy of gas (a) increases by \(2 R \ln 3\) (b) increases by \(3 R \ln 3\) (c) decreases by \(2 R \ln 3\) (d) decreases by \(3 R \ln 3\)

A quantity of \(1.6 \mathrm{~g}\) helium gas is expanded adiabatically \(3.0\) times and then compressed isobarically to the initial volume. Assume ideal behaviour of gas and both the processes reversible. The entropy change of the gas in this process is \((\ln 3=1.1)\) (a) \(-1.1 \mathrm{cal} / \mathrm{K}\) (b) \(+1.1 \mathrm{cal} / \mathrm{K}\) (c) \(-2.2 \mathrm{cal} / \mathrm{K}\) (d) \(+2.2 \mathrm{cal} / \mathrm{K}\)

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