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


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


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

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


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

  • 1
    $\begingroup$ 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? $\endgroup$ – Ivan Neretin Mar 24 at 18:32
  • $\begingroup$ I get -0.3 kJ/mol. Are you sure about the answer? $\endgroup$ – Buck Thorn Mar 24 at 18:33
  • $\begingroup$ @IvanNeretin Yes, I'm aware of the formula ▲G = ▲H - T *▲S but I'm still getting it wrong ahaha $\endgroup$ – neavys Mar 24 at 18:37
  • $\begingroup$ @BuckThorn Yes, I checked and it's positive. May I ask you how you got to that solution though? $\endgroup$ – neavys Mar 24 at 18:38
  • 1
    $\begingroup$ @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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.