[Shipwreck]

Hello again all.

Another question involving equilibrium constants:

A mixture of water and graphite is heated to 600 K. When the system comes to equilibrium it contains 0.18mol of H2, 0.18mol of CO, 0.82mol of H2O, and some graphite. Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.

Find the amount of CO in the flask when the system returns to equilibrium.

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I've tried to do this problem several different ways but without success. Here's how I went about my most recent attempt:

I wrote out the original equilibrium reaction

2H₂0(l) + 2C(s) <=> H₂(g) + CO(g) + H₂0 + C(s)

I then tried to figure out exactly how the adding of oxygen affects the overall reaction, and this is what I came up with...

2H₂0(l) + 2C(s) <=> H₂(g) + CO(g) + H₂0 + C(s)
2H₂(g) + O₂ => 2H₂O(l)
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H₂(g) + O₂(g) + 2C(s) <=> CO(g) + H₂0(l) + C(s)

So I think that E_eq = [CO] / [O₂] * [H₂]. I ignore the solids and the liquids.

I then tried to make an ICE table for my new equilibrium reaction, but if all the hydrogen gas is consumed in the reaction then the equilibrium concentration of hydrogen would be 0, which would give me an E_eq of infinity...

I know I probably haven't gone about this the right way. I'd really appreciate some guidance from you smart chem students out there! Thanks a loT!

[DrCMS]

Forget about how the system was originally made you have a mixture of:

0.18mol of H₂, 0.18mol of CO, 0.82mol of H₂O, and some graphite. If you react all the H₂ completely with the O₂ what will be left?

[Shipwreck]

Thanks for helping me DrCMS.

If I react all the hydrogen with the oxygen I will be left with 0.18 mol of CO, some graphite and some water (1.00 mol?). Are you saying the E_eq is then just the concentration of CO, since that is the only gas left?

I am assuming that the hydrogen gas will react with the oxygen to form water:

2H₂ + O₂ = 2H20

Thanks again for your hint. Would really appreciate a confirmation on this (there's no answer in the textbook).