# 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?

##### 1 Answer

The key here is to realize that phase transitions are phase equilibria, i.e.

Thus, immediately,

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

If the system is **open to the air**, then

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