Solids and liquids in equilibrium constant (1 Viewer)

NexusRich

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I have been told by teachers and textbooks that when calculating the equilibrium constant, we do not take solids and liquids into account, but only the concentration of gases and aqueous molecules.
I've tried searching for an answer online, but most came up with vague answers like: the concentration is constant for solids and liquids, while others mentioned something about density that I couldnt understand.
Could someone please explain this, thanks so much !!!
 

Etho_x

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I have been told by teachers and textbooks that when calculating the equilibrium constant, we do not take solids and liquids into account, but only the concentration of gases and aqueous molecules.
I've tried searching for an answer online, but most came up with vague answers like: the concentration is constant for solids and liquids, while others mentioned something about density that I couldnt understand.
Could someone please explain this, thanks so much !!!
Yeah that’s correct with the concentration side of things. Basically solids and liquids don’t have actual concentration. I mean for example, take a piece of wood. That’s a solid, how does it have a concentration? It doesn’t really make sense to say 1 mol per litre of wood, especially when you wouldn’t measure wood in litres... and I would assume not moles either hahahah. With liquids this more so refers to pure liquids such as water which don’t really have a concentration. Using filtered water as an example, it is a clear substance which hasn’t been concentrated with anything else besides itself. In terms of expressing it as a concentration, it doesn’t really make sense to say 1 mol of water per litre of water. I hope that makes sense.
 

tito981

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Yeah that’s correct with the concentration side of things. Basically solids and liquids don’t have actual concentration. I mean for example, take a piece of wood. That’s a solid, how does it have a concentration? It doesn’t really make sense to say 1 mol per litre of wood, especially when you wouldn’t measure wood in litres... and I would assume not moles either hahahah. With liquids this more so refers to pure liquids such as water which don’t really have a concentration. Using filtered water as an example, it is a clear substance which hasn’t been concentrated with anything else besides itself. In terms of expressing it as a concentration, it doesn’t really make sense to say 1 mol of water per litre of water. I hope that makes sense.
well actually they do have a concentration, its just that having a constant concentration in the eq constant provides no additional information so they just disregard that info.
 

tito981

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quick proof:

c = n/v= n x 1/v = mass/MM X 1/v = m/v X 1/MM = 1/MM X P.
where P is density which is constant for solids and liquids and 1/MM is a constant as it is molar mass which is fixed. Thus concentration is constant.

sorry for poor formatting.
 

Eagle Mum

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Their changes in mass per volume as a response to changes to other conditions within the system are negligible, so for simplification are ignored.
 

CM_Tutor

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The proper answer to this question goes beyond school level, but the short version is this:

Equilibrium constants are not actually based on concentrations, they are based on the activity of the species. In most common circumstances, the concentration of species dissolved in a solvent or in gas state is a very good approximation for the activity, which is why the simplification to concentrations is used. Solids and pure liquids (solvents) have activities of 1 and so they are not so much disregarded as constant as they are irrelevant as their value is 1.

Note that the "disregard pure liquids" rule runs into major problems when pure liquids are mixed. For example, if I mix ethanol and acetic acid as pure substances and then add a trace of (say) HCl as catalyst, the equilibrium that is established with have ethyl acetate and water as the products, and all four species will occur in the equation for K. Water will not be the solvent, its concentration will change substantially if the system is disturbed, and you will have



Species where the state is "(sol)" (meaning dissolved in a solvent other than water) will also be included in expressions for K, while the solvent (with a "(l)" state) will be excluded. For example borane (BH3) is usually used as a complex with a solvent like THF:

BH3 (sol) + THF (l) <---> H3B(THF) (sol)

would have K = [H3B(THF)] / [BH3]

From an HSC perspective, @Eagle Mum's explanation is the closest to the actual reason... the the negligible changes in concentration approximate the actual fixed activities. :)
 

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