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 $\pu{1 atm}.$

Answer

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

Question
I have the following data:
$$
\begin{array}{lcc}
\hline
\text{Compound} &amp; \Delta H/\pu{kJ mol^-1} &amp; \Delta S/\pu{J mol^-1 K^-1} \\
\hline
\ce{H2O(l)} &amp; –285.8 &amp; 69.96 \\
\ce{H2O(g)} &amp; –241.8 &amp; 188.7 \\
\hline
\end{array}
$$
$$T = \pu{100 °C} + 273.15 = \pu{373.15 K}$$
I calculated $\Delta G$ of the product (gas) as
$$
\begin{align}
\Delta G &amp;= \Delta H + T\Delta S \\
  &amp;= \pu{-285.8 kJ mol^-1} - (\pu{69.97 J mol^-1 K^-1})(\pu{373 K}) \\
  &amp;= \pu{-26395}
\end{align}
$$
Then I calculated $\Delta G$ of the reactant (liquid) as
$$
\begin{align}
\Delta G &amp;= \Delta H + T\Delta S \\
  &amp;= \pu{-241.8 kJ mol^-1} - (\pu{188.7 J mol^-1 K^-1})(\pu{373 K}) \\
  &amp;= \pu{-70625}
\end{align}
$$
Finally,
$$
\begin{align}
\Delta_\mathrm{r}G &amp;= \Delta G(\ce{H2O(g)}) - \Delta G(\ce{H2O(l)}) \\
  &amp;= -26395 + 70625 \\
  &amp;= -44230
\end{align}
$$
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.