Q.Match the following processes with entropy change:
Reaction

Concept: Entropy Change Prediction — Entropy ($S$) measures disorder. Phase changes, gas expansions, and reaction spontaneity all give predictable $\Delta S$ signs.

Reasoning:

1. Liquid vapourises — molecules go from ordered liquid to disordered gas. Disorder increases, so $\Delta S > 0$. Matches (b).
2. Non-spontaneous at all temperatures, $\Delta H > 0$ — for a reaction to be non-spontaneous everywhere, $\Delta G = \Delta H - T\Delta S > 0$ for all $T$. With $\Delta H > 0$, this requires $\Delta S < 0$ (so $-T\Delta S$ is positive, keeping $\Delta G > 0$). Matches (c). …

Entropy, denoted by $S$, is a measure of the randomness or disorder of a system. The change in entropy, $\Delta S$, tells us whether a process increases or decreases this disorder. A positive $\Delta S$ means the system becomes more disordered; a negative $\Delta S$ means it becomes more ordered. This is the core idea we will use to match each process.

Let’s examine each reaction and predict the sign of its entropy change.

1. A liquid vapourises

   When a liquid turns into a vapour, its molecules go from a relatively ordered, closely packed state to a highly disordered, widely dispersed gaseous state. The number of possible microscopic arrangements (microstates) increases dramatically. Therefore, the entropy of the system increases.

   $\Delta S > 0$, so this matches with (b).

2. Reaction is non-spontaneous at all temperatures and $\Delta H$ is positive

   The spontaneity of a reaction is governed by the Gibbs free energy change: $\Delta G = \Delta H - T\Delta S$. For a reaction to be non-spontaneous at all temperatures, $\Delta G$ must be positive for every value of $T$.

   • If $\Delta H > 0$ (given), then for $\Delta G$ to stay positive even as $T$ becomes very large, the term $-T\Delta S$ must be positive (i.e., $\Delta S$ must be negative). Why? If $\Delta S$ were positive, then at high temperatures, $-T\Delta S$ would become a large negative number, making $\Delta G$ negative and the reaction spontaneous.
   • The only way to prevent this is for $\Delta S$ to be negative, so that $-T\Delta S$ is always positive, adding to the already positive $\Delta H$. Hence, $\Delta S < 0$, matching with (c).
   Watch out

   A common mistake is to think that a non-spontaneous reaction always has a negative entropy change. This is not true. A reaction with $\Delta H > 0$ and $\Delta S > 0$ is spontaneous at high temperatures, not at all temperatures. The condition "non-spontaneous at all temperatures" is the key constraint that forces $\Delta S$ to be negative. …