# How do you find the change in entropy of vaporization for water?

##### 1 Answer

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.