1
$\begingroup$

I know it is convention to take liquid/solid concentrations as constants and lump it with the equilibrium constant. However, if we were trying to compare the equilibrium constants of two different reactions and if one of the reactions had a solid reactant, doesn't lumping the solid reactant concentration with the equilibrium constant make one reaction seem more product favoured than it actually is?

$\endgroup$
1
$\begingroup$

The equilibrium constant of a reaction containing solids or liquids doesn't contain hidden values for their concentrations.

The issue is that we use the concentration (or pressure) of a substance in the equilibrium expression because it is a shortcut. What we are actually putting into the expression is the activity of that substance, a more complicated idea. It very often reduces to the ratio of the concentration of a solution to the concentration at standard state. Since the concentration of a solution at standard state is 1M, there's no point in dividing through by 1 for each term.

For solids and liquids, the concentration at standard state is its molar density. ($C = \frac{mol}{L}$) This value doesn't depend on the quantity of substance available, so the ratio between the density at standard state and the density at any point in a reaction is 1. The values ARE in the equilibrium expression; they're just all 1's.

As to your concern about comparing equilibrium constants, we can only do that directly for reactions with the same number of reactants and products. Adding (or removing) any term to either side interferes with the comparison.

Compare:
$\ce{A(aq) +B(aq)->AB(aq)}$
$\ce{2A(aq) +B(aq)->A2B(aq)}$

We have two similar reactions but the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants. Similarly, with a constant solid/liquid concentration (activity = 1) the equilibrium constant effectively ignores that product or reactant.

It's just one of the limitations of the equilibrium constant.

Check out the wiki on Thermodynamic Activity for more information.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.