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

Step 1: Understanding the Concept:
Entropy (\(S\)) is a state function that measures the degree of randomness or disorder in a system.
A negative entropy change (\(\Delta S<0\)) occurs when a system becomes more ordered.
Order generally increases in the order: \(\text{Gas }>\text{ Liquid }>\text{ Solid}\).
Step 2: Key Formula or Approach:
Identify phase changes or chemical changes that reduce the number of microstates or freedom of movement.
Step 3: Detailed Explanation:
I) Sublimation of dry ice: Solid \(CO_2\) turns into gas. Randomness increases. \(\Delta S\) is positive.
II) Freezing of water: Liquid water becomes solid ice. Molecules move from random motion to fixed lattice positions. \(\Delta S\) is negative.
III) Crystallisation: Solute particles move from a random state in solution to a highly ordered crystalline solid. \(\Delta S\) is negative.
IV) Burning of rocket fuel: Chemical reaction produces high volumes of gas from solids/liquids. Disorder increases greatly. \(\Delta S\) is positive.
Thus, only processes II and III involve a decrease in entropy.
Step 4: Final Answer:
Processes II and III result in a negative entropy change.