Calculating equilibrium constant without volume?

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Xoster4
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#1
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Q.) Hydrogen can be produced by this reaction.

CO(g) + H20(g) ⇋ CO2(g) + H2(g)

In an experiment 4.20 mol of carbon monoxide were mixed with 2.00 mol of steam. When the raction reached equilibrium, 1.60 mol of hydrogen had been formed.

What is the value of the equilibrium constant, Kc, for this reaction?
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Pigster
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#2
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Have you accidentally muted it?

In this case:

Kc = n(CO2)/v x n(H2)/v all over n(CO)/v x n(H2O)/v

Since the v's cancel, you don't need to know it, just use the amounts in your ICE table, without converting to concs.
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Nathaniel911
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ATTENTION TO ANYONE NEEDING HELP POST 3 YEARS.

I know this person has probably finished his A-Levels but here is how I worked it out.

So first you know that Kc is total products over reactants. The reactants, in this case, is CO and H20 the products are H2 and CO2 gas.

We are told that 1.6 mol of H is made and since the products have a 1:1 ratio we know that 1.6 mol of CO2 is made.

Then we have a look at the reactants. CO must be 4.2mol-1.6mol and H2O must be 2mol-1.6mol too. That will give us values of 2.6 and 0.4 moles.

Then plugging it into the Kc equation you should ideally get 2.46 as the correct answer!
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Pigster
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#4
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(Original post by Nathaniel911)
ATTENTION TO ANYONE NEEDING HELP POST 3 YEARS.
But... your addition didn't help answer the OP's question.

They didn't want to know what the answer was, they wanted to know how they could do the calc. without knowing what the volume was.

The general advice is, if you ever find an old thread and don't know how to do something, just start a new thread of your own and let the dead lie.
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