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

Apr 6, 2017

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.

${\cancel{\Delta {G}_{v a p}}}^{0} = \Delta {H}_{v a p} - T \Delta {S}_{v a p}$

This means that $\frac{\Delta {H}_{v a p}}{T} = \Delta {S}_{v a p}$.

We expect this to be a positive value, since we input energy at constant pressure ($\Delta {H}_{v a p} > 0$) to cause more disorder ($\Delta {S}_{v a p} > 0$). Since we must use $T$ in $\text{K}$, $T > 0$ and everything is positive in this equation.

Therefore, we have:

$\textcolor{b l u e}{\Delta {S}_{v a p}} = \left(\text{40.7 kJ"/cancel"mol" xx cancel"1 mol" xx "1000 J"/cancel"1 kJ")/("373 K}\right)$

$=$ $\textcolor{b l u e}{\text{109.12 J/K}}$

As a reference, the actual value is around $\text{109.02 J/K}$. In fact, you can convince yourself that using the enthalpy of vaporization of $\text{40.68 kJ/mol}$ from the reference as well as $\text{373.15 K}$ instead of $\text{373 K}$, gives almost exactly $\text{109.02 J/K}$.

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