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Problem

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

Answer

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

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,

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

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    $\begingroup$ 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
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    $\begingroup$ @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
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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.

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