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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.

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 $\pu{25 °C}$?

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):

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

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?

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    $\begingroup$ Why do you think ethanol vapour pressure is wrong and not other things? $\endgroup$
    – Poutnik
    Commented Feb 27, 2022 at 7:10
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    $\begingroup$ The Handbook of Chemistry and Physics gives the following values for the vapor pressure of ethanol. $1$ Torr at -$31.3$°C ; $10$ Torr at - $2.3$ °C ; $40$ Torr at $19$°C ; $100$ Torr at + $34.9$°C : $400$ Torr at$63.5$°C, and 760 Torr at $78.4$°C. So your calculation for $25$°C looks correct. $\endgroup$
    – Maurice
    Commented Feb 27, 2022 at 12:26
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    $\begingroup$ Please provide the rest of the problem statement. The part you are looking at seems like a VLE calculation. $\endgroup$
    – Chet Miller
    Commented Feb 27, 2022 at 12:31
  • $\begingroup$ Thank you very much @poutnik for updating my question and pointing out my wrong calculation. $\endgroup$
    – Jihyun
    Commented Feb 27, 2022 at 14:14
  • $\begingroup$ Thank you It's great to know the vapor pressure of ethanol is correct. I also added the rest of the question. $\endgroup$
    – Jihyun
    Commented Feb 27, 2022 at 14:15

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