[Shadow]

Methanol is synthesized on a commercial scale by the reaction
CO(g) + 2H2(g)  CH3OH(g)
The equilibrium constant of this process at 500 K is Kp=6.10•10^–3 bar^–2. The equilibrium of this reaction can be shifted by varying the temperature, pressure, or the composition of the mixture.
A certain volume of syngas (a mixture of carbon monoxide and hydrogen gas used for the synthesis of methanol by the above reaction) was obtained by gasification of coal with superheated steam. Calculate the mole fractions of methanol in the equilibrium mixture in its synthesis at 500 K and a total pressure of 100 bar, assuming that the syngas contained no water.


Because both CO and hydrogen are made from the same reaction where their mole ratio is 1:1 they appear in equal quantities that I marked with x. But in the answer key, they assumed, just like that, that the number of moles of CO and hydrogen are 1 and 1 respectively. Mine and theirs final answer was the same, but I had to do a lot more calculation. How is their assumption valid if we know that pressure is a function of the quantity? For a given pressure of 100bar they say it is 1 mol of both reactant gases. I don't see why. Need help.

[Borek]

If I understand your question correctly, they simply assumed methanol synthesis doesn't consume enough CO/H₂ to change their pressures substantially. Taking into account rather low Kp value, that's not a bad assumption (although it should be checked after the calculations).

[Shadow]

No, the solution says that 0.436 moles of CO reacted which is not a small part. Here is what is written in the answer:
n(CO)=1-x
n(H2)=1-2x
n(CH3OH)=x
and then they did the standard calculation through Kp... , while I wrote that:
n(CO)=y-x
n(H2)=y-2x
n(CH3OH)=x
By solving the Kp equation for one unknown I was able to get the mole share of methanol which is the same as in the answer key. How could the method that uses the assumption of 1 mole give a correct answer? How can they assume that y=1 if the given pressure is 100bar (isn't the pressure dependent on the number of moles?)?

[kriggy]

Pressure is dependant on number of moles but also on temperature.

[Borek]

Still not clear to me.

However, why can't you assume 1 mole of reactants? You are not asked about result expressed in moles, but for a mole fraction - so the answer will be the same whether you use 1 mole, or 2 moles, or n moles (which would be the most general approach). In such a problem when the number of moles or volume is not given, you are free to choose any number that makes calculations easy.

[Shadow]

I thought that I can't assume 1 mole because I have a given pressure of 100bar. Isn't for 1mole of CO and 1 mole of hydrogen the pressure different?

[Borek]

But you can assume any volume - you can assume such a volume, that is occupied by exactly 1 mole. This is exactly equivalent of assuming 1 mole.

As I wrote - you can assume n moles, and do the calculations on symbols. Would that be OK with you?

[Shadow]

If there was a fixed volume the assumption would be invalid, but n could be calculated. I wrote that I did it with n moles.