Water in Mechanism

[Leah Savage 2F]

If water is in one of the steps, should it be included in the equilibrium equation (K), or left out bc it is a liquid? If it is left out, does that mean it is also left out of the rate law?

[Anne 2L]

Solids and liquids are not included when determining the equilibrium constant K. I'm not sure if liquids are included in the rate law though.

[Angel Ni 2K]

If it is the solvent, then you can leave it out of the rate law. This is because solvents are present in such high concentrations that their concentrations don't change much during the reaction.

[Leah Savage 2F]

Would it tell us in the problem if water is the solvent? In one problem you had to assume water was the solvent, and then later they were like "now what if there was an organic solvent instead."

[Jennifer Ho 1K]

Water is left out of the equilibrium expression because it is a liquid, and liquids have constant concentrations. Generally this means it is included as part of the equilibrium constant because it's constant anyways. The same applies for the rate constant expression.

[GabrielGarciaDiscussion1i]

Water is left out of the equilibrium expression because it is a liquid, and liquids have constant concentrations. Generally this means it is included as part of the equilibrium constant because it's constant anyways. The same applies for the rate constant expression.

This is an extremely detailed and apparently competent answer. This is where I felt water stood in regards to reaction rate laws and rate constants, however, I was never 100% sure so if Chem_Mod could come through and help?
However, it stands to reason that most times water will work as a solvent for our problems and therefore can be left out. Not to mention it's pretty consistently constant concentration, I just never thought about including it as part of the equilibrium constant.