Problem 51 - A Complete Resource Book in Chemistry for JEE Main

Question

The factor of \(\Delta \mathrm{G}\) values is important in metallurgy. The \(\Delta \mathrm{G}\) values for the following reactions at \(800^{\circ} \mathrm{C}\) are given as \(\mathrm{S}_{2}(\mathrm{~s})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_{2}(\mathrm{~g}) ; \Delta \mathrm{G}=-544 \mathrm{~kJ}\) \(2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{ZnS}(\mathrm{s}) ; \Delta \mathrm{G}=-293 \mathrm{~kJ}\) \(2 \mathrm{Zn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{s}) ; \Delta \mathrm{G}=-480 \mathrm{~kJ}\) the \(\Delta \mathrm{G}\) for the reaction, \(2 \mathrm{ZnS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{~g})\) will be (a) \(-357 \mathrm{~kJ}\) (b) \(-731 \mathrm{~kJ}\) (c) \(-773 \mathrm{~kJ}\) (d) \(-229 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

The \( \Delta G \) for the reaction is -731 kJ, option (b).

Step 1: Understanding Given Reactions

We have three reactions with known \( \Delta G \) values given: \( \mathrm{S}_{2}(\mathrm{~s})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_{2}(\mathrm{~g}) \) with \( \Delta G = -544 \mathrm{~kJ} \), \( 2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{ZnS}(\mathrm{s}) \) with \( \Delta G = -293 \mathrm{~kJ} \), and \( 2 \mathrm{Zn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{s}) \) with \( \Delta G = -480 \mathrm{~kJ} \). We need to find the \( \Delta G \) for the reaction \( 2 \mathrm{ZnS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{~g}) \).

Step 2: Manipulating Given Reactions

We need to derive the desired reaction from the given ones. First, reverse the second reaction, changing its sign of \( \Delta G \): \( 2 \mathrm{ZnS}(\mathrm{s}) \longrightarrow 2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}), \Delta G = +293 \mathrm{~kJ} \).

Step 3: Creating Final Reaction

Add the reversed second reaction to the first reaction: 1. \( 2 \mathrm{ZnS}(\mathrm{s}) \longrightarrow 2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}), \Delta G = +293 \mathrm{~kJ} \) 2. \( \mathrm{S}_{2}(\mathrm{~s})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_{2}(\mathrm{~g}), \Delta G = -544 \mathrm{~kJ} \) Cancel the common terms on both sides to get: \( 2 \mathrm{ZnS}(\mathrm{s}) + 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{Zn}(\mathrm{s}) + 2 \mathrm{SO}_{2}(\mathrm{~g}) \); sum \( \Delta G = -251 \mathrm{~kJ} \).

Step 4: Adding the Third Reaction

Add the third given reaction to the result of Step 3: 3. \( 2 \mathrm{Zn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{s}), \Delta G = -480 \mathrm{~kJ} \) the addition of both reactions gives: \( 2 \mathrm{ZnS}(\mathrm{s}) + 3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{~s}) + 2 \mathrm{SO}_{2}(\mathrm{~g}) \), the final \( \Delta G = -251 \mathrm{~kJ} - 480 \mathrm{~kJ} = -731 \mathrm{~kJ} \).

Step 5: Selecting the Answer

Compare the calculated \( \Delta G = -731 \mathrm{~kJ} \) with the given options. The answer is (b) \(-731 \mathrm{~kJ} \).

Key Concepts

Thermodynamics in Chemistry

Thermodynamics in chemistry involves understanding how energy transformations appear in chemical reactions. This includes studying the principles that govern these transformations and the factors affecting them. At its essence, chemical thermodynamics helps us figure out whether a reaction occurs spontaneously.

Several key parameters play a vital role in thermodynamics:

• Enthalpy (H): It reflects the heat absorbed or released during a reaction.
• Entropy (S): It gives insight into the disorder or randomness in a system.
• Gibbs Free Energy (G): It predicts if a reaction is spontaneous at a constant temperature and pressure.

In the given exercise, Gibbs free energy (G) was pivotal in understanding how energy changes influence metallurgical processes, like extracting metals. This helps decide if a reaction will move forward and release energy or if it's non-spontaneous, requiring additional energy input.

Chemical Reactions and Gibbs Free Energy

Chemical reactions are driven by two major factors: the energy change associated with the reaction and the degree of randomness or disorder that results. Gibbs Free Energy (G) is a critical parameter used in determining the feasibility of a chemical reaction.

The Gibbs energy equation is:\[G = H - TS\]
where:

• H is the change in enthalpy.
• T is the temperature in Kelvin.
• S is the change in entropy.

A negative G indicates a spontaneous reaction, meaning the process can occur on its own under constant conditions of temperature and pressure.

In the context of the exercise, determining the G for the overall reaction helped in understanding whether zinc sulfide can convert to zinc oxide without external energy input. This information is crucial for designing efficient industrial processes.

Metallurgical Thermodynamics

Metallurgical thermodynamics is an area that applies the principles of thermodynamics to the field of metallurgy, focusing on the extraction and refinement of metals. Understanding thermodynamics is fundamental to predicting and optimizing metallurgical reactions, such as those required for smelting ores or creating metal alloys.

In metallurgy, lowering the Gibbs free energy (G) of a reaction pathway is often targeted to make a process more favorable and energy-efficient. This is important for cost savings and ecological considerations.

For example, the exercise demonstrates how calculating G for the reactions involved in transforming zinc sulfide to zinc oxide provides insights into the spontaneity and energy requirements of the process. This allows metallurgists to choose the most cost-effective and efficient methods for metal extraction and processing.

Thus, by examining reactions step by step and manipulating various parameters, metallurgists can innovate new, more effective techniques for metal recovery and use.