Question

The value of `DeltaH_(vap)` of substance X is 45.7 kJ/mol and its normal boiling point is 72.5 degrees C. How do you calculate `DeltaS_(vap)`, `DeltaS_(surr)` and `DeltaG_(vap)` for the vaporization of one mole of this substance at 72.5°C and 1 atm?

Answer 1

The key here is to realize that phase transitions are phase equilibria, i.e. `DeltaG_"trs" = 0`.

Thus, immediately, `color(blue)(DeltaG_("vap","345.65 K")^@ = "0 kJ")` (where the `""^@` only implies standard pressure). This is the case if we simultaneously have `DeltaP = 0` and `DeltaT = 0`, and that is the case for phase transitions under "lab bench" conditions.

As a result, we also have the following relationship to consider:

`DeltaG = DeltaH - TDeltaS`

but...

`cancel(DeltaG_("vap","345.65 K")^@)^(0) = DeltaH_("vap","345.65 K")^@ - T_bDeltaS_("vap","345.65 K")^@`

Thus:

`color(blue)(DeltaS_("vap","345.65 K")^@) = (DeltaH_("vap","345.65 K")^@)/T_b`

`= (45.7 cancel"kJ""/"cancel"mol")/("345.65 K") xx "1000 J"/cancel"1 kJ" xx cancel"1 mol"`

`=` `color(blue)("132.2 J/K")`

In a situation where perfect conservation of energy holds (which we usually assume), we should consider whether the system is open to the air or not.

If the system is closed, insulated, and rigid, then `DeltaS_("surr","345.65 K")^@ = 0`, since nothing escapes into the surroundings, no molecules outside of the system are touched, and the surroundings' entropy is not altered.

If the system is open to the air, then `DeltaS_"vap" = -DeltaS_"surr"`, and

`color(blue)(DeltaS_("surr","345.65 K")^@ = -"132.2 J/K")`