How do you find the change in entropy of vaporization for water?
Vaporization is an equilibrium with constant pressure and temperature. When this is the case, we have that the Gibbs' free energy is zero, i.e.
#cancel(DeltaG_(vap))^(0) = DeltaH_(vap) - TDeltaS_(vap)#
This means that
We expect this to be a positive value, since we input energy at constant pressure (
Therefore, we have:
#color(blue)(DeltaS_(vap)) = ("40.7 kJ"/cancel"mol" xx cancel"1 mol" xx "1000 J"/cancel"1 kJ")/("373 K")#
#=# #color(blue)("109.12 J/K")#
As a reference, the actual value is around
So, our error is due to the number of decimal places we used.