Question

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

Answer 1

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 `(DeltaH_(vap))/T = DeltaS_(vap)`.

We expect this to be a positive value, since we input energy at constant pressure (`DeltaH_(vap) > 0`) to cause more disorder (`DeltaS_(vap) > 0`). Since we must use `T` in `"K"`, `T > 0` and everything is positive in this equation.

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 `"109.02 J/K"`. In fact, you can convince yourself that using the enthalpy of vaporization of `"40.68 kJ/mol"` from the reference as well as `"373.15 K"` instead of `"373 K"`, gives almost exactly `"109.02 J/K"`.

So, our error is due to the number of decimal places we used.