Q.The entropy change can be calculated by using the expression ΔS = qrev/T. When water freezes in a glass beaker, choose the correct statement amongst the following :

Concept: Gibbs Free Energy — for a spontaneous process at constant $T$ and $P$, $\Delta G = \Delta H - T\Delta S < 0$. Freezing is exothermic ($\Delta H < 0$), so the surroundings gain heat.

Reasoning:

1. Water freezes spontaneously at $0^\circ\text{C}$ and 1 atm. The system (water) goes from liquid to solid — more ordered, so $\Delta S_\text{sys} < 0$.
2. The heat released by freezing ($q = -\Delta H_\text{fusion} > 0$) is absorbed by the surroundings, increasing their thermal motion. Hence $\Delta S_\text{surr} > 0$. …

When water freezes in a glass beaker, we are looking at a spontaneous process at a temperature below 0°C (assuming normal pressure). The key to understanding entropy changes here is to separate the system (the water turning to ice) from the surroundings (the beaker and the room).

The entropy change of the system is driven by the change in molecular order. Liquid water has molecules that slide past each other with considerable freedom; ice has a fixed, crystalline lattice. Freezing therefore decreases the number of accessible microstates — the system becomes more ordered. So ΔS_system is negative.

But the process is spontaneous. For a spontaneous process at constant temperature and pressure, the total entropy change of the universe must be positive:

If ΔS_system is negative, then ΔS_surroundings must be positive and large enough to outweigh it. How does freezing increase the entropy of the surroundings? Because freezing is exothermic — the water releases heat (its latent heat of fusion) into the beaker and the air. That heat increases the random thermal motion of molecules in the surroundings, raising their entropy.

Let’s walk through it step by step.

1. Identify the sign of ΔS_system Water → ice: from a disordered liquid to an ordered solid. Entropy of the system decreases.

2. Identify the sign of q for the system Freezing releases heat. The system loses energy to the surroundings.

3. Relate q to the entropy change of the surroundings For the surroundings, the heat gained is the negative of what the system lost (assuming no other heat exchanges):

At constant pressure (the beaker is open to the atmosphere), the entropy change of the surroundings is:

Since $q_{system} < 0$, this is positive.

4. Check spontaneity For freezing to be spontaneous below 0°C, the total entropy change must be positive: …