Question

Consider two processes: sublimation of I₂(s) and melting of I₂(s). (Note: the latter process can occur at the same temperature but somewhat higher pressure.) I₂(s) → I₂(g), I₂(s) → I₂(l) The process in which the magnitude of the entropy change will be greater is in the liquid.

A) True

B) False

Answer 1

The magnitude of the entropy change is greater in the sublimation of solid iodine to gaseous iodine compared to its melting to liquid iodine, thereby making the statement provided false. Sublimation results in molecules with significantly more freedom of movement, implying a larger entropy increase.

Step-by-step explanation:

The statement that the process of melting of I2(s) to I2(l) has a greater magnitude of entropy change than the sublimation of I2(s) to I2(g) is false. In fact, the sublimation process, which involves going from solid to gas, is accompanied by a greater increase in entropy because the molecules in the gaseous state have much more freedom to move and occupy a larger volume than in the liquid state.

In essence, during sublimation, the iodine molecules spread out more and can be found in many more microstates, which significantly increases the entropy (Sgas > Sliquid > Ssolid). The entropy change (ΔS) is positive for both melting and sublimation because both processes lead to an increase in entropy.

However, the transition from solid to gas (sublimation) involves the molecules completely overcoming intermolecular attractions, as opposed to only partially overcoming these attractions in the melting process. Consequently, sublimation results in a greater change in entropy than melting.