# Equilibrium for the synthesis of methanol

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:

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

• To find out, it suffices to know the state of methanol under the reaction conditions: have you looked for it? – Nicolas Feb 19 at 11:06
• Yes, I saw in a phase diagram of methanol that under these conditions it is a supercritic fluid (hybrid between gas an liquid...) – fich Feb 19 at 11:08
• Carbon monoxide is CO. Further more, equations are not balanced, unless you managed conversion O -> 2 H – Poutnik Feb 19 at 11:10
• Yes, sorry for the mistake – fich Feb 19 at 11:13
• 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? – matt_black Feb 19 at 11:14

$$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)}$$
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 $$K_\mathrm{c}=K_\mathrm{p}(RT)^3=9.28\times 10^{-3}\times (300 \times 0.0821)^3 \approx \pu{138.65 atm^3 L^3 mol^-3}$$.