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Suppose one has two chemical reaction equations:

$\ce{A_{(aq)} + B_{(aq)} <=>[H2O] C_{(g)} + D_{(g)}}$

and

$\ce{A_{(s)} + B_{(s)} <=> C_{(g)} + D_{(g)}}$

Suppose one puts 0.1 mol of A and 0.1 mol of B into 1 litre of water. The water is contained in a closed rectangular container that can hold up to 2 litres of material.

The produced gasses C and D are lighter than water and so float up out of the water into the top of the container where they can't react . So if some equilibrium is reached for A and B in the water where they do not react any more it has to be of the form $k_{\mathrm{eqc}} = \frac{1}{{c_{\ce{A}}} {c_{\ce{B}}}}$ unless A and B completely and totally react together into gasses or do not react at all to form gasses.

The produced gasses C and D float up out of the water and when they react in the air they produce heavier solids A and B which fall to the water at the bottom and then dissolve into ions. As there are no A and B products in the air any equilibrium for C and D has to be of the form $k_{\mathrm{eqp}} = {P_{\ce{C}}} {P_{\ce{D}}}$ unless C and D always completely and totally react together to form solids or do not react at all to form solids.

What happens here?

I think I'm making a mistake where I'm not factoring pressure in somehow.

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  • $\begingroup$ To begin with, solids have vapor pressure as well... $\endgroup$
    – Ivan Neretin
    Commented Mar 5, 2016 at 21:37
  • $\begingroup$ The gist is that thermodynamics still works. Since there are two phases in both reactions, both are really irreversible reactions and will go towards products and not have a reverse reaction. The "equilibrium" in the first case is really based on activities in solution not concentrations. The activity of the gas will be defined as 1 so there is a sort of "equilibrium." In the second reaction the solids would have an activity of 1, and there would be no reverse reaction in any practical sense. $\endgroup$
    – MaxW
    Commented Mar 5, 2016 at 21:56
  • $\begingroup$ So, @MaxW the reaction must be either $\ce{A_{(aq)} + B_{(aq)} -> C_{(g)} + D_{(g)}}$ or $\ce{ C_{(g)} + D_{(g)}} -> A_{(aq)} + B_{(aq)} $ or else I'd have a violation of thermodynamics? I see how at a macroscopic level this leads to a violation of thermodynamics but at a microscopic level why is that these reactions must be one-way? Also, for particular cases why would the reactions be one way or another? $\endgroup$
    – Ms. Molly Stewart-Gallus
    Commented Mar 5, 2016 at 22:20
  • $\begingroup$ Both reactions are irreversible because the reactants are in one phase and the products are in a different phase. // I asked essentially same question. chemistry.stackexchange.com/questions/47384/… $\endgroup$
    – MaxW
    Commented Mar 5, 2016 at 23:39

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