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Keq and rate
Started by chencxm
I wonder when is liquid water included in the equilibrium expression and rate. Is it when liquid water is a reactant/product or is it when water is not the solvent?
Thanks.
Pure solids and liquids are NEVER included in the equilibrium expression. If you see a reactant followed by (l) or (s)-don't include it. It's different for the rate law.
Pure solids and liquids are NEVER included in the equilibrium expression. If you see a reactant followed by (l) or (s)-don't include it. It's different for the rate law.
How is the rate law?
Pure solids and liquids are NEVER included in the equilibrium expression. If you see a reactant followed by (l) or (s)-don't include it. It's different for the rate law.
The last sentence is wrong. Chemicals are not included in a rate law or equilibrium expression if their concentration is essentially constant. This is true with pure liquids and solids, but doesn't hold in general for liquids. Concentration is essentially always constant for a solid, so they are not included.
Water may be included in a rate equation and equilibrium expression if it is not the solvent, and is either used or produced during the course of the reaction. If it is not used or produced in the reaction, then it is a catalyst, and may appear in the rate equation but not the equilibrium expression.
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The last sentence is wrong. Chemicals are not included in a rate law or equilibrium expression if their concentration is essentially constant. This is true with pure liquids and solids, but doesn't hold in general for liquids. Concentration is essentially always constant for a solid, so they are not included.
Water may be included in a rate equation and equilibrium expression if it is not the solvent, and is either used or produced during the course of the reaction. If it is not used or produced in the reaction, then it is a catalyst, and may appear in the rate equation but not the equilibrium expression.
Can you give an example where a liquid (not aqeuous solution) is included in the equilibrium expression?
Can you give an example where a liquid (not aqeuous solution) is included in the equilibrium expression?
Br2 + alkene ----> vicinal dibromide
rate ~ [Br2][alkene]
Br2 is a liquid.
Br2 + alkene ----> vicinal dibromide
rate ~ [Br2][alkene]
Br2 is a liquid.
That's the rate law. I meant the equilibrium expression.
That's the rate law. I meant the equilibrium expression.
Ok, same thing: Keq = [vic. dibromide] / [Br2][alkene]
How about another example, the oxidation of ethanol to acetic acid:
EtOH ---[O]----> AcOH
Keq = [AcOH] / [EtOH]
Both product and reactant are liquids, are you suggesting that you can't write an equilibrium expression for this?
Edit:
Think about why solids and pure liquids are usually omitted from the equilibrium expression. It's because their concentrations are effectively constant. In cases where this isn't true, they must be included.
Ok, same thing: Keq = [vic. dibromide] / [Br2][alkene]
How about another example, the oxidation of ethanol to acetic acid:
EtOH ---[O]----> AcOH
Keq = [AcOH] / [EtOH]
Both product and reactant are liquids, are you suggesting that you can't write an equilibrium expression for this?
Edit:
Think about why solids and pure liquids are usually omitted from the equilibrium expression. It's because their concentrations are effectively constant. In cases where this isn't true, they must be included.
I've always been taught that pure solids and liquids are omitted from the expression because their concentration is so great that it's assumed to be "1", which is why I find this weird. I've never come across a problem that included a liquid as you've done above in the equilibrium expression which is why I think it's a bit odd. I'll keep that in mind when I'm practicing.
I've always been taught that pure solids and liquids are omitted from the expression because their concentration is so great that it's assumed to be "1", which is why I find this weird. I've never come across a problem that included a liquid as you've done above in the equilibrium expression which is why I think it's a bit odd. I'll keep that in mind when I'm practicing.
Think for a second. Does it make sense to assume that a pure liquid's concentration is approximately 1? What is the concentration of liquid water? (~55M)
How about another example, a esterification:
AcOH + EtOH <--------> AcOEt + H2O (all liquids)
Think about the extreme case of the reverse reaction: you start with the ester under anhydrous conditions. What's the equilibrium concentration of acetic acid going to be? Since there's no water around to react, it must be zero. How starting with just acetic acid with no ethanol around, what is the equilibrium concentration of the ester going to be? Did you say zero again? Correct. You can do the same for all of the terms, and so the equilibrium expression must depend on all of the terms:
Keq = [AcOEt][H2O] / [AcOH][EtOH]
Now think about the reverse reaction where water is the solvent:
AcOEt + H2O <-----------> AcOH + EtOH
start: y 55M 0 0
end: y-x 55-x x x
Assuming dilute conditions (x << 55M), the concentration of H2O is constant.
Then, Keq = [AcOH][EtOH] / [AcOEt][H2O]
But since [H20] is essentially constant, it makes sense to just put that in with the equilibrium constant:
Keq' = [H20]Keq = [AcOH][EtOH] / [AcOEt]
Note that it isn't absolutely wrong to include water in the expression -- if you're consistent about it, you'll get the correct results. This is just a convention that is used because it leads to nice simplifications in many real systems. Since it's a convention, equilibrium constants given in references assume you're using the same conventions, so if you use literature values for equilibrium constants without this assumption, you'll get incorrect results.
Edit3: I still can't get the formatting to work for that table. Sorry.
Last edited:
You're thinking of activity, which is some sort of p-chem concept that is way beyond the MCAT.I've always been taught that pure solids and liquids are omitted from the expression because their concentration is so great that it's assumed to be "1", which is why I find this weird. I've never come across a problem that included a liquid as you've done above in the equilibrium expression which is why I think it's a bit odd. I'll keep that in mind when I'm practicing.
This whole discussion just confused me more. The consensus on when/if you can use a liquid in the equilibrium and/or rate law expression is what?
This whole discussion just confused me more. The consensus on when/if you can use a liquid in the equilibrium and/or rate law expression is what?
Never use solids, use liquids so long as they aren't the solvent.
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