Question

Identify the direction/process in which the entropy decreases

• A. \(H_2O(l) \longrightarrow H_2O(s)\)
• B. \(H_2O(l) \longrightarrow H_2O(g)\)
• C. \(H_2O(g) \longrightarrow 2H(g)\)
• D. \(\text{Crystal }(100\ K)\longrightarrow \text{Crystal }(200\ K)\)

Hint:

Entropy decreases when a system becomes more ordered, such as: \[ \text{Gas}\rightarrow\text{Liquid}\rightarrow\text{Solid} \]

Answer 1

The Correct Option is A

Step 1: Recall the entropy concept.
Entropy ($ S $) is the thermodynamic measure of disorder or randomness in a system. Processes that increase disorder (more arrangements, more freedom of motion) increase entropy ($ \Delta S > 0 $), while processes that bring order decrease entropy ($ \Delta S < 0 $).
Step 2: Analyze option (1) - Freezing of water.
$ H_2O(l) \to H_2O(s) $: Liquid water has molecules in disordered, freely moving state. Ice has molecules locked in a highly ordered crystal lattice. The transition from liquid to solid reduces freedom of motion and increases order. Therefore $ \Delta S < 0 $ - entropy decreases.
Step 3: Analyze option (2) - Vaporization.
$ H_2O(l) \to H_2O(g) $: Gas molecules have far more freedom (translational, rotational) than liquid molecules. This is a large increase in disorder. $ \Delta S > 0 $.
Step 4: Analyze option (3) - Bond breaking in gas phase.
$ H_2O(g) \to 2H(g) + O(g) $: One molecule becomes three separate atoms in gas phase. The number of particles triples, massively increasing disorder. $ \Delta S \gg 0 $.
Step 5: Analyze option (4) - Heating a crystal.
Crystal at 100 K to crystal at 200 K: Increasing temperature increases vibrational energy of lattice atoms, increasing microscopic disorder. $ \Delta S > 0 $.
Step 6: Final answer.
Only freezing ($ H_2O(l) \to H_2O(s) $) decreases entropy. \[ \boxed{H_2O(l) \to H_2O(s)} \]