# Gibbs free energy for the evaporation of water

### Problem

Find the value of $$\Delta G$$ for the evaporation of water at $$\pu{100 °C}$$ and $$\pu{1 atm}.$$

$$\pu{0.3 kJ mol^-1}$$

### Question

I have the following data:

$$\begin{array}{lcc} \hline \text{Compound} & \Delta H/\pu{kJ mol^-1} & \Delta S/\pu{J mol^-1 K^-1} \\ \hline \ce{H2O(l)} & –285.8 & 69.96 \\ \ce{H2O(g)} & –241.8 & 188.7 \\ \hline \end{array}$$

$$T = \pu{100 °C} + 273.15 = \pu{373.15 K}$$

I calculated $$\Delta G$$ of the product (gas) as

\begin{align} \Delta G &= \Delta H + T\Delta S \\ &= \pu{-285.8 kJ mol^-1} - (\pu{69.97 J mol^-1 K^-1})(\pu{373 K}) \\ &= \pu{-26395} \end{align}

Then I calculated $$\Delta G$$ of the reactant (liquid) as

\begin{align} \Delta G &= \Delta H + T\Delta S \\ &= \pu{-241.8 kJ mol^-1} - (\pu{188.7 J mol^-1 K^-1})(\pu{373 K}) \\ &= \pu{-70625} \end{align}

Finally,

\begin{align} \Delta_\mathrm{r}G &= \Delta G(\ce{H2O(g)}) - \Delta G(\ce{H2O(l)}) \\ &= -26395 + 70625 \\ &= -44230 \end{align}

• Well, to begin with, you are probably expected to know that $\Delta G$ is somehow related to $\Delta H$ and $\Delta S$. Rings a bell? – Ivan Neretin Mar 24 at 18:32
• I get -0.3 kJ/mol. Are you sure about the answer? – Buck Thorn Mar 24 at 18:33
• @IvanNeretin Yes, I'm aware of the formula ▲G = ▲H - T *▲S but I'm still getting it wrong ahaha – neavys Mar 24 at 18:37
• @BuckThorn Yes, I checked and it's positive. May I ask you how you got to that solution though? – neavys Mar 24 at 18:38
• @neavys It's Gibbs, not gibb's. And the main issue with your calculation is that you omitted units and allowed yourself to sum up numerical values for J and kJ, which you should've never done. I tried to brush up notations and math, but I cannot finish corrections since it would break initial intend. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. – andselisk Mar 24 at 21:29

The average heat capacity $$C_p$$ of liquid water and water vapor over the range between $$\pu{25^\circ C}$$and $$\pu{100^\circ C}$$ are, respectively $$4.18\ \pu{kJmol^{-1}K^{-1}}$$ and $$1.72\ \pu{kJmol^{-1}K^{-1}}$$. So the enthalpies of liquid water and water vapor at $$\pu{100^\circ C}$$ are:

$$H_l(\pu{100^\circ C})=-285.8+(0.018)(4.18)(75)=-280.2\ \pu{kJmol^{-1}}$$

$$H_g(\pu{100^\circ C})=-241.8+(0.018)(1.72)(75)=-239.5\ \pu{kJmol^{-1}}$$

So, at $$\pu{100^\circ C}$$, $$\Delta H=40.7\ \pu{kJmol^{-1}}$$

The entropies of liquid water and water vapor at $$\pu{100^\circ C}$$ are:

$$S_l(\pu{100^\circ C})=0.06996+(0.018)(4.18)\ln{(373.15/298.15)}=0.08684\ \pu{kJmol^{-1}K^{-1}}$$

$$S_g(\pu{100^\circ C})=0.1887+(0.018)(1.72)\ln{(373.15/298.15)}=0.1956\ \pu{kJmol^{-1}K^{-1}}$$

So, at $$\pu{100^\circ C}$$, $$\Delta S = 0.1088\ \pu{kJmol^{-1}}$$ and $$T\Delta S=40.6\ \pu{kJmol^{-1}}$$ So the change in Gibbs free energy between saturated liquid and saturated vapor at $$\pu{100^\circ C}$$ and 1 atm is $$\Delta G=40.7-40.6=0.1\ \pu{kJmol^{-1}}$$ which, to within roundoff error is zero (as expected).

The Spoiler answer is obviously incorrect.