During which of the following processes, does entropy decrease?
(A) Freezing of water to ice at 0°C
(B) Freezing of water to ice at -10°C
(C) N₂(g) + 3H₂(g) → 2NH₃(g)
(D) Adsorption of CO(g) on lead surface.
(E) Dissolution of NaCl in water
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Ask one question each time: does the system end up more ordered, or with fewer gas molecules?

1. What to judge. The question is about the entropy of the system. It falls when particles become more ordered or less free to move, for example liquid to solid, fewer gas moles, or gas stuck on a surface.

   Process	What happens to the particles	Δ S_(system)
   (A) Freezing at 0 °C	Liquid becomes an ordered solid	decreases
   (B) Freezing at -10 °C	Liquid becomes an ordered solid (the temperature does not change this)	decreases
   (C) N₂(g) + 3H₂(g) → 2NH₃(g)	4 mol of gas become 2 mol of gas	decreases
   (D) Adsorption of CO on lead	Free gas molecules get fixed on a surface	decreases
   (E) NaCl dissolving in water	Ordered crystal breaks into ions spread through the solution	increases

2. Freezing below zero. At -10 °C freezing is spontaneous, but only because the heat it releases raises the entropy of the surroundings. The water itself still loses entropy, so (B) counts.
3. So. Entropy decreases in (A), (B), (C) and (D), and only (E) is excluded. The answer is (c).

The idea to keep. Spontaneous does not mean the system's entropy goes up. Judge Δ S_(system) by order alone: solid formation, fewer gas moles and adsorption all lower it.

Answer from the NTA final answer key. Solution by PrepWiser.