Question

For which of the following process entropy change ($\Delta S$) is negative ?
I) Sublimation of dry ice
II) Freezing of water
III) Crystallisation of the dissolved substance
IV) Burning of rocket fuel

• A. I, II only
• B. II, III only
• C. III, IV only
• D. I, IV only

Hint:

To easily determine the sign of $\Delta S$, look at the physical states involved in the transition:
- Phase changes from more ordered to less ordered states (solid $\rightarrow$ liquid $\rightarrow$ gas) have $\Delta S > 0$.
- Phase changes from less ordered to more ordered states (gas $\rightarrow$ liquid $\rightarrow$ solid) have $\Delta S < 0$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In thermodynamics, entropy (\(S\)) is a state function that quantifies the degree of disorder, randomness, or chaos within a system. The second law of thermodynamics suggests that for any spontaneous process in an isolated system, the total entropy tends to increase. However, within a specific system, entropy can decrease (\(\Delta S<0\)) if the process leads to a more ordered arrangement of particles. Generally, gases have the highest entropy because their molecules move freely and randomly. Liquids have moderate entropy, and solids have the lowest entropy because their atoms are locked in a rigid, repeating lattice structure. Therefore, any process that moves towards a more condensed or structured phase will typically have a negative change in entropy.
Step 2: Detailed Explanation:
Let's evaluate each given process one by one to determine the sign of \(\Delta S\):

I) Sublimation of dry ice: Sublimation is the direct conversion of a solid into a gas. Dry ice is solid carbon dioxide (\(CO_{2}\)). During sublimation (\(CO_{2}(s) \rightarrow CO_{2}(g)\)), the highly ordered solid molecules gain energy and escape into the gas phase where they move rapidly and randomly in all directions. This transition represents a massive increase in disorder. Thus, \(\Delta S>0\) (Positive).

II) Freezing of water: Freezing is the phase change from liquid to solid (\(H_{2}O(l) \rightarrow H_{2}O(s)\)). In the liquid state, water molecules can slide over each other and have significant translational kinetic energy. Upon freezing, they lose energy and settle into a fixed, highly structured crystalline lattice (ice). This transition from a random liquid state to a rigid solid state increases the order of the system. Thus, \(\Delta S<0\) (Negative).

III) Crystallisation of the dissolved substance: In a solution, solute particles (ions or molecules) are dissociated and randomly dispersed among the solvent molecules. They have a high degree of freedom. During crystallization, these particles come together to form a highly ordered solid crystal structure. The system moves from a state of high dispersion to a state of high organization. Thus, \(\Delta S<0\) (Negative).

IV) Burning of rocket fuel: Combustion is a chemical reaction that typically converts a small volume of liquid or solid propellant into a massive volume of hot, rapidly expanding gases (like \(CO_{2}\), \(H_{2}O\) vapor, \(N_{2}\)). The production of many gas molecules from fewer condensed molecules creates extreme molecular chaos. Thus, \(\Delta S>0\) (Positive).

Comparing our evaluations, processes II and III show a decrease in entropy.
Step 3: Final Answer:
The processes where the entropy change is negative are Freezing of water (II) and Crystallisation of the dissolved substance (III).