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

##### 1 Answer
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.