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Should the concentration of liquid water be included in the $K_\mathrm{c}$ expression for the reaction $$\ce{2 NO2_{(g)} + 7 H2_{(g)} -> 2 NH3_{(g)} + 4 H2O_{(l)}}$$ if water is liquid and all the other compounds are gasses?

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  • $\begingroup$ I attempt to improve formatting. Yours was a bit "wierd". $\endgroup$ – M.A.R. Jan 23 '15 at 16:47
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In the equation that you stated initially, water would be omitted from the equilibrium constant expression.

Your second question (about esterification that you posted in a comment to one of the answers) is extremely interesting.

I have given it some thought: in the standard case of esterification water should be included because it is responsible for causing the reverse process ( hydrolysis of ester formed).

But in the case of zeolites, which by virtue of their structure, probably act as shape selective catalysts..ester hydrolysis by the water would be affected. (in some cases, I have read that selectivity towards ester formation is nearly 100%) In a case such as this, one can safely remove water from the equilibrium constant expression.

(I am not be a 100% correct, please excuse me, since I am just a student myself. Perhaps, someone else can provide a better/clearer explanation)

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No. If the water condenses it is effectively removed from the reaction equilibrium. Just as with solids that precipitate.

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  • $\begingroup$ What about the water produced during an esterification process? With zeolites it doesn't participate in Kc, but without it does, right? $\endgroup$ – RBW Jan 23 '15 at 16:44

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