I understand why solids and liquids are not included in the $K_{eq}$ expressions. However, I'm wondering what the $K$ value for a reaction involving only solid or liquid reactants and products looks like. Theoretically, it'd be 1 no matter what since $1^n = 1$. Is this right? Also, do all such reactions go to completion?
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No, it definitly isn't. – Mithoron Feb 20 '15 at 11:19
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So then what is it? – user11629 Feb 20 '15 at 11:22
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Related: Why do liquids and solids have constant concentrations?, Why are solids and liquids not included in the equilibrium constant? What about in a reaction rate calculation? – Mithoron Feb 20 '15 at 11:41
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These do not address my question. I get that liquids and solids have non-variable concentrations and thus can be considered contributors to $K_{eq}$ via rearrangement, and thus they are not part of the changing expression. What is the K value for A + B --> C + D where A, B, C, and D are solids or liquids? – user11629 Feb 20 '15 at 11:57
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If they concentrations don't change, they aren't reacting and indeed you have K=1. – Mithoron Feb 20 '15 at 12:00
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Mithoron - this is not a duplicate. That question has nothing that addresses what I'm asking. – user11629 Feb 24 '15 at 04:24
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If something is reacting it's concentration will change, although it can be neutralised by regeneration, or external addition – Mithoron Feb 26 '15 at 16:57
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You could calculate an equilibrium constant at a given temperature for a reaction from the standard free energy change ($\Delta G^{\circ}$) for any reaction using:
$\Delta G^{\circ} = -RT \ln K$
$\Delta G^{\circ}$ can be calculated for reactions using tabulated standard free energies of formation ($\Delta G_{f}^{\circ}$) using the following equation:
$\Delta G^{\circ} = \sum{\Delta G_{f}^{\circ} \mbox{(Products)}} - \sum{\Delta G_{f}^{\circ} \mbox{(Reactants)}}$
If the reaction only contains solids and liquids, then K will still depend on the ratio between the product and reactant concentrations. However, these concentrations will not change as a reaction moves towards equilibrium, which implies that the reaction quotient ($Q$) remains constant
Tami M.
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