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

Question

The factor of \(\Delta \mathrm{G}\) values is important in metallurgy. The \(\Delta\) 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: Write the target reaction

We are tasked with finding 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: Identify given reactions and their \(\Delta G\) values

We have the following reactions and their \( \Delta G \) values: 1. \( \mathrm{S}_{2}(\mathrm{~s})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_{2}(\mathrm{~g}) \; \Delta G = -544 \mathrm{~kJ} \) 2. \( 2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{ZnS}(\mathrm{s}) \; \Delta G = -293 \mathrm{~kJ} \) 3. \( 2 \mathrm{Zn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{s}) \; \Delta G = -480 \mathrm{~kJ} \)

Step 3: Reverse and manipulate reactions

To obtain the desired reaction, we need to adjust the given equations: - Reverse reaction 2, which becomes: \( 2 \mathrm{ZnS}(\mathrm{s}) \longrightarrow 2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}) \; \Delta G = +293 \mathrm{~kJ} \) - Combine reactions using Hess's Law to form the target reaction.

Step 4: Combine reactions to form the target reaction

Combined reactions:1. \( 2 \mathrm{ZnS}(\mathrm{s}) \longrightarrow 2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}) \; \Delta G = +293 \mathrm{~kJ} \)2. \( \mathrm{S}_{2}(\mathrm{~s})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_{2}(\mathrm{~g}) \; \Delta G = -544 \mathrm{~kJ} \)3. \( 2 \mathrm{Zn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{s}) \; \Delta G = -480 \mathrm{~kJ} \)- Adding them up gives the target reaction: \( 2 \mathrm{ZnS}(\mathrm{s}) + 3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{ZnO}(\mathrm{s}) + 2 \mathrm{SO}_{2}(\mathrm{~g}) \) Total \( \Delta G = 293 \mathrm{~kJ} - 544 \mathrm{~kJ} - 480 \mathrm{~kJ} \).

Step 5: Calculate the \(\Delta G\) for the target reaction

Perform the calculation: \[ 293 \mathrm{~kJ} - 544 \mathrm{~kJ} - 480 \mathrm{~kJ} = -731 \mathrm{~kJ} \] Thus, the \( \Delta G \) for the reaction is \(-731 \mathrm{~kJ}\).

Key Concepts

Hess's Law

Hess's Law is a powerful tool in thermodynamics that allows us to determine the change in Gibbs Free Energy (\( \Delta G \)) for a chemical reaction, especially when dealing with complex reactions. The law states that if a reaction can be expressed as the sum of several sub-reactions, the total \( \Delta G \) for the main reaction is the sum of the \( \Delta G \) values for each sub-reaction. This principle is rooted in the fact that \( \Delta G \) is a state function, meaning it is path-independent and relies solely on the initial and final states.
To apply Hess’s Law, you often have to adjust, reverse, or multiply the equations before summing them up. For example, in our exercise, reversing the direction of a reaction changes the sign of \( \Delta G \), and when reactions are multiplied, their \( \Delta G \) value is multiplied by the same factor. This method provides a straight-forward and intuitive way to "build" the desired reaction from known reactions, allowing us to effortlessly calculate the \( \Delta G \) for reactions that may not be easy to observe directly.

• Formulate the reaction as a series of steps using available data.
• Adjust each step (e.g., reverse reactions) as necessary, modifying \( \Delta G \) values accordingly.
• Combine the adjusted reactions to find the total \( \Delta G \) for the overall reaction.

Metallurgy

Metallurgy is the science and technology of extracting metals from their ores and refining them for use. It involves various chemical processes, one of which heavily depends on the concept of Gibbs Free Energy (\( \Delta G \)). In metallurgical processes, reactions with negative \( \Delta G \) values are favorable under given conditions, resulting in spontaneous reactions where metals can be extracted or transformed efficiently.
Understanding \( \Delta G \) is crucial in assessing the feasibility of metal extraction operations, such as smelting, roasting, and electro-refining. For instance, a metallurgist would seek to optimize the conditions under which certain reactions have a negative \( \Delta G \), ensuring that the metal separation is energy-efficient and cost-effective.
In our given exercise, the calculation of \( \Delta G \) for the target reaction of converting zinc sulfide to zinc oxide using oxygen demonstrates how such a transformation is energetically favorable, which is a typical scenario in several metallurgical processes.

• Identify reactions with favorable \( \Delta G \) values for spontaneous metal extraction.
• Modify reaction conditions to achieve minimal energy expense while ensuring maximum extraction efficiency.
• Utilize Hess’s Law to calculate \( \Delta G \) for metallurgical processes when dealing directly with reactions is challenging.

Calculating ΔG

Calculating Gibbs Free Energy (\( \Delta G \)) for a reaction involves several steps to ensure accuracy and understanding of the underlying process. \( \Delta G \) provides insight into whether a reaction can occur spontaneously. For calculations, like in the exercise example, this process relies heavily on manipulating given reactions strategically to align with Hess's Law.
Consider the reaction calculations from the exercise: 1. **Identify all involved reactions and their \( \Delta G \).** Have a clear list of all relevant reactions and their associated \( \Delta G \) values given by experimental or theoretical data.2. **Reverse and adjust necessary reactions.** Reversing a reaction flips the sign of \( \Delta G \), while coefficients in reactions multiply its \( \Delta G \) correspondingly.3. **Sum the \( \Delta G \) values.** Add up all the \( \Delta G \) values from the steps to compute the overall \( \Delta G \) for the chemical process.
From our worked-out example, after organizing and combining given reactions, the calculated \( \Delta G \) was determined as \(-731 \text{ kJ}\), indicating the spontaneous nature of the reaction under the given conditions.

• Use known \( \Delta G \) values and the steps of Hess's Law to break down complex reactions into approachable components.
• Ensure the correct application of reaction manipulation when reversing or scaling equations.
• Summing these components gives precise \( \Delta G \) calculations for understanding the spontaneity of the reaction.