How to calculate vapor pressure of ethanol at given temperature?

The following is a part of the problem I am struggling with. I think what I need is a vapor pressure of ethanol at $\pu{25 °C},$ but I may be totally wrong.



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> A reactor is charged with $\pu{60 bar}$ of $\ce{C2H4(g)}$ and $\pu{40 bar}$ of $\ce{H2O(g)}$. The equilibrium mixture is cooled and the water and ethanol condense. The excess ethene is allowed to escape. What is the vapor pressure of the resulting liquid mixture at $\pu{25 °C}?$ What is the mole fraction of ethanol in the vapor in equilibrium with the condensed mixture at $\pu{25 °C}$?
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First, I calculated the equilibrium constant $K$ at $T = \pu{298.15 K}$ and $\Delta\_\mathrm{r}G = \pu{-8070 J mol^-1}$:


$$K= \exp\left(-\frac{\Delta\_\mathrm{r}G}{RT}\right) \approx 25.93.\tag{1}$$


Using this $K$ I calculated the partial pressure of each gas assuming they are not condensed:


$$
K = \frac{p(\ce{C2H5OH})}{p(\ce{C2H4})\cdot p(\ce{H2O})}.\tag{2}
$$
$$
\begin{align}
p(\ce{H2O}) &= \pu{2.43 bar},\\
p(\ce{C2H5OH}) &= \pu{37.57 bar}.
\end{align}
$$


I know the water and the ethanol are actually condensed (liquid) at this temperature with the corresponding molar fractions


$$
\begin{align}
x(\ce{H2O}) &= \frac{\pu{2.43 bar}}{\pu{2.43 bar} + \pu{37.57 bar}} \approx 0.061,\tag{3}\\
x(\ce{C2H5OH}) &= \frac{\pu{37.57 bar}}{\pu{2.43 bar} + \pu{37.57 bar}} \approx 0.939\tag{4}.
\end{align}
$$


Now all I need is the vapor pressure of the water and the ethanol at $\pu{25 °C}.$


I was able to find the vapor pressure of $\ce{H2O}$ at $\pu{25 °C}$ is $\pu{23.76 Torr}.$ But I don't know how to calculate the vapor pressure of ethanol at $\pu{25 °C}.$


I found a formula to calculate the vapor pressure of ethanol on [Wikipedia — Ethanol (data page)](https://en.wikipedia.org/wiki/Ethanol_(data_page)#Vapor_pressure_of_liquid):


$$p(\ce{C2H5OH}) = 10^{8.04494 - 1554.3/(222.65 + T)},\label{eqn:5}\tag{5}$$


where $T$ is in Celsius and $p$ is in torr. I plugged $T = 25$ to \eqref{eqn:5} and got $p = \pu{58.71 Torr}.$


$$
\begin{align}
p(\ce{mixture}) &= x(\ce{H2O}) \cdot p(\ce{H2O}) + x(\ce{C2H5OH}) \cdot p(\ce{C2H5OH}),\tag{6}\\
 &\approx 56.58 \pu{Torr}.
\end{align}
$$


But I don't think I am doing right. Is there a more appropriate way to find the vapor pressure of ethanol?


