# 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? Mar 24, 2021 at 18:32
• I get -0.3 kJ/mol. Are you sure about the answer? Mar 24, 2021 at 18:33
• @IvanNeretin Yes, I'm aware of the formula ▲G = ▲H - T *▲S but I'm still getting it wrong ahaha Mar 24, 2021 at 18:37
• @BuckThorn Yes, I checked and it's positive. May I ask you how you got to that solution though? Mar 24, 2021 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. Mar 24, 2021 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.