One of the skills of a scientist is to be able to verify mathematically what is understood conceptually.

For example, does the increase in entropy of the surroundings account for the fact that a liquid freezes when the temperature is lowered to below a certain temperature?

Calculate the change in entropy of the surroundings per mole of Hg when mercury freezes at 249°C; use ΔH_fus(Hg) = 2.292 kJ · mol^–1 at 249°C.

ANTICIPATE The normal freezing point of mercury is 238°C, so you should certainly expect it to freeze at a lower temperature. A contribution to this spontaneous freezing is the release of energy as heat to the surroundings, causing their entropy to increase (FIG. 4I.3).

PLAN To calculate the change in entropy of the surroundings when mercury freezes, write ΔH_freeze = 2ΔH_fus(freezing is the opposite of fusion) and convert the temperature to kelvins.

The change in entropy of the system, the mercury itself, is 29.782 J · mol^–1.

What should you assume? Assume that the temperature of the surroundings is constant.

Transcribed Image Text:

Entropy of the surroundings Heat E F (b) FIGURE 41.3 (a) In an exothermic process, heat escapes into the surroundings and increases their entropy. (b) In an endothermic process, the entropy of the surroundings decreases. The red arrows represent the transfer of heat between system and surroundings, and the green arrows indicate the entropy change of the surroundings.