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 quickly determine the sign of $\Delta S$, look at the change in physical state. Phase changes towards "Solid" (Freezing, Condensation, Deposition) usually have $-\Delta S$, while changes towards "Gas" (Melting, Evaporation, Sublimation) have $+\Delta S$.

Answer 1

The Correct Option is B

Concept: Entropy ($S$) is a measure of the degree of randomness or disorder in a system. A negative entropy change ($\Delta S < 0$) indicates that the system is becoming more ordered.
• Gas > Liquid > Solid (Order increases $\rightarrow \Delta S$ is negative)
• Solid > Liquid > Gas (Disorder increases $\rightarrow \Delta S$ is positive)

Step 1: Analyze each process.

• I) Sublimation of dry ice: Solid CO$_2$ turns into gaseous CO$_2$. Disorder increases, so $\Delta S$ is positive.
• II) Freezing of water: Liquid water turns into solid ice. The molecules become fixed in a lattice, increasing order. Thus, $\Delta S$ is negative.
• III) Crystallisation of the dissolved substance: Solute particles move from a random state in a solution to a highly ordered crystalline solid. Thus, $\Delta S$ is negative.
• IV) Burning of rocket fuel: Chemical combustion typically releases a large volume of gases from solid or liquid propellants. Disorder increases significantly, so $\Delta S$ is positive.

Step 2: Conclusion.
Based on the analysis , processes II and III result in a decrease in randomness.