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This question already has an answer here:

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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marked as duplicate by M.A.R., Klaus-Dieter Warzecha, jerepierre, Philipp, bon Feb 24 '15 at 22:00

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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

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