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")#