Problem 2 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For a reaction, \(4 \mathrm{M}(\mathrm{s})+\mathrm{nO}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{M}_{2} \mathrm{O}_{\mathrm{n}}(\mathrm{s})\), the free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which : (a) the slope change from negative to positive (b) the free energy change shows a change from negative to positive value (c) the slop changes from positive to negative (d) the slop changes from positive to zero

Step-by-Step Solution

Answer

Option (c) is correct: the slope changes from positive to negative.

Step 1: Understanding the Problem

We need to determine the temperature below which the oxide is stable based on changes in the free energy plotted against temperature. Stability of the compound is typically indicated by a negative free energy change (\( \Delta G < 0 \)).

Step 2: Interpreting Free Energy Changes

In thermodynamics, a reaction is spontaneous and the product is stable when the free energy change, \( \Delta G \), is negative. From the options given, we look for the one that indicates a transition from instability to stability.

Step 3: Analyzing the Options

(a) A change in slope from negative to positive means the rate of change of free energy becomes less negative or more positive. (b) A change in free energy from negative to positive means the reaction becomes non-spontaneous, which implies a loss of stability. (c) A slope change from positive to negative means the rate of change of free energy becomes more negative, potentially indicating increased stability. (d) A slope change from positive to zero could imply a minimum or maximum point.

Step 4: Selecting the Right Option

Since the oxide is stable when the free energy change is negative, we seek the option which indicates or leads to this condition. The change from positive to negative slope (option c) suggests that the free energy transition indicates growing stability, making it the correct indicator.

Key Concepts

Thermodynamics

Thermodynamics is a branch of physics that explores how energy changes in processes, particularly involving heat, work, and internal energy within systems. It's like understanding the "rules of energy play" in nature. The main laws of thermodynamics help explain phenomena that we see daily.

• **First Law**: Energy cannot be created or destroyed, but only transformed (conservation of energy).
• **Second Law**: Energy transfers or transformations increase the entropy (disorder) in the universe.
• **Third Law**: As a system approaches absolute zero, the entropy approaches a minimum.

In chemical reactions, thermodynamics allows us to predict whether reactions happen and whether they can do work. An essential concept in this context is Gibbs free energy (\( \Delta G \)), which combines enthalpy and entropy to show if a process can occur spontaneously. If \( \Delta G \) is negative, a reaction can spontaneously take place, potentially releasing energy and thus doing work.

Reaction Spontaneity

For a reaction to be spontaneous, the Gibbs free energy change must be negative. This is grounded in the principle that processes in nature favor lower energy states where the system becomes stable and balanced.
To determine spontaneity, one uses the formula:\[ \Delta G = \Delta H - T \Delta S \]where:

• \( \Delta G\) = Gibbs free energy change
• \( \Delta H\) = Enthalpy change, or heat content
• \( T\) = Temperature in Kelvin
• \( \Delta S\) = Entropy change, or disorder

A \( \Delta G < 0 \) suggests that the reaction is spontaneous at a given temperature. If \( \Delta G \) turns from positive to negative, it indicates a shift towards spontaneity and stability.

Temperature Stability

Temperature stability in a reaction is indicated by how \( \Delta G \) changes with temperature. The reaction that results in stable products operates where this free energy change is negative over a range of temperatures.
Imagine plotting \( \Delta G \) against temperature; where you see a slope from positive to negative, this can mean that as temperature rises, the reaction becomes more favorable energetically. A negative \( \Delta G \) is crucial because:

• It means the reaction releases free energy, beneficial for product formation and stability.
• Reaching a consistent low \( \Delta G \) value may signify the temperature threshold under which the reaction remains stable.

Finding this point is the goal in the exercise, highlighting which temperature ranges yield a stable oxide phase. Identifying the change from a positive to negative slope helps to pinpoint this region.