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Liquid methanol is obtained with carbon monoxide gas and hydrogen gas in a reactor at $\pu{300 °C}$ and $\pu{250 atm}.$ $K_p = \pu{9.28E-3}.$ Find $K_c.$

With only these data, do I have to use the following equilibrium for the calculations

$$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)},$$

or do I have to suppose first methanol is obtained as a gas and then liquified:

$$\ce{CO(g) + 2 H2(g) <=> CH3OH(g)}?$$

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    $\begingroup$ To find out, it suffices to know the state of methanol under the reaction conditions: have you looked for it? $\endgroup$
    – Nicolas
    Commented Feb 19, 2021 at 11:06
  • $\begingroup$ Yes, I saw in a phase diagram of methanol that under these conditions it is a supercritic fluid (hybrid between gas an liquid...) $\endgroup$
    – fich
    Commented Feb 19, 2021 at 11:08
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    $\begingroup$ Carbon monoxide is CO. Further more, equations are not balanced, unless you managed conversion O -> 2 H $\endgroup$
    – Poutnik
    Commented Feb 19, 2021 at 11:10
  • $\begingroup$ Yes, sorry for the mistake $\endgroup$
    – fich
    Commented Feb 19, 2021 at 11:13
  • $\begingroup$ This question will be completely impossible to answer unless the text matches the chemical equations. are you talking about the reaction of carbon monoxide or dioxide? $\endgroup$
    – matt_black
    Commented Feb 19, 2021 at 11:14

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Note that under the given conditions, methanol cannot exist in the liquid state, however since you are given in the problem that methanol is obtained in liquid state, you will only consider the equilibrium (for problem solving purpose)

$$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)}$$

And since $K_p$ is related to $K_c$ by the relation

$$K_p = K_c(RT)^{\Delta n_\mathrm{g}},$$ where $\Delta n_\mathrm{g}$ represents the change in the number of moles of gaseous species, so here $\Delta n_\mathrm{g} = -3,$ Therefore $K_\mathrm{c}=K_\mathrm{p}(RT)^3=9.28\times 10^{-3}\times (300 \times 0.0821)^3 \approx \pu{138.65 atm^3 L^3 mol^-3}$.

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