Heterogeneous equilibrium Ignoring the solution or the solid state molecules when calculating the equilibrium constant.
- Let's say we wanted to figure out the equilibrium constant
- for the reaction boron trifluoride in the gaseous
- plus 3-- so for every mole of this, we're going to have 3
- moles of H2O in the liquid state-- and that's in
- It's going forward and backwards with 3 moles of
- hydrofluoric acid, so it's in the aqueous state.
- It's been dissolved in the water.
- If it wasn't dissolved, if it was in the solid state, you
- would call this hydrogen fluoride.
- Once it's in water, you call it hydrofluoric acid, and
- we'll talk more about naming in the future, hopefully.
- Plus 1 mole of boric acid, also in the aqueous state.
- It's dissolved in the water.
- H3BO3 in the aqueous state.
- So what would the expression for the equilibrium constant
- look like in this situation?
- So you might be tempted say, OK, that's easy enough, Sal.
- So the equilibrium constant, you just take
- the right-hand side.
- That's just the convention.
- There's symmetry here.
- I could've rewritten it either way, but let's just say you
- take the right-hand side and say, OK, this is dependent on
- the concentration of the hydrofluoric acid, the
- concentration of the HF, or the molarity of the HF, to the
- third power, times the concentration of the boric
- acid, H3BO3.
- And remember, this intuition of why you're taking this to
- the third power is what's the probability-- because in order
- for the reaction to go this way, you need to have 3
- molecules of hydrofluoric acid being very close to 1 molecule
- of the boric acid.
- So if you watched the last video I just made about the
- intuition behind the equilibrium constant, this is
- indicative of the probability of this reaction happening or
- the probability of finding all of these
- molecules in the same place.
- Of course, you can adjust it with a constant and that's
- essentially what that does.
- But that's on the product side, or the reactant,
- depending on what direction you're viewing this equation,
- divided by the molarity of the boron trifluoride times-- and
- I'll do this in a different color-- the molarity of the
- H2O to the third power.
- And that's, of course, the H2O liquid.
- So there you go.
- We'll just figure this out.
- And my rebuttal to you is I want you to figure out the
- molarity of the water.
- What is the concentration of the water?
- Remember, the concentration is moles per volume, but in this
- case, what's happening?
- I'm putting some boron trifluoride gas essentially
- into some water, and it's creating these aqueous acids.
- These other molecules are dissolved
- completely in the water.
- So what's the solvent here?
- The solvent is H2O.
- This might be how the reaction happens, but pretty much,
- there's water everywhere.
- The water is in surplus.
- So if you were to really figure out the concentration
- of water, it's everywhere.
- I mean, you could say everything but the boron
- trifluoride, but it's a very high number.
- And if you think about it from the probability point of view,
- if you say, OK, in order for this reaction to happen
- forward, I need to figure out the probability of finding a
- boron trifluoride atom or molecule-- actually,
- molecule-- in a certain volume, and it also needs 3
- moles of water in that certain volume.
- But you say, hey, there's water everywhere.
- This is the solvent.
- There's water everywhere, so I really just need to worry
- about the concentration of the boron trifluoride.
- So you could say the forward reaction rate, rate forward,
- is going to be dependent on some forward constant times
- just the concentration of the boron trifluoride.
- The water's everywhere, so you don't have to multiply it
- times the concentration of water, whatever that means,
- because the water's everywhere.
- So the denominator here, you do not put the solvent.
- So the correct answer for this one is you only put whatever
- is actually dissolved in the solution.
- Because frankly, the concentration doesn't actually
- makes sense for everything else, and if you think about
- it from the probability point of view, that also makes
- sense, because there's always water around.
- If you said, OK, what's the probability of finding water
- at any small volume of our fluid, it's going to be 1, so
- you could just multiply it by a 1 there, but that doesn't
- make a difference.
- Now, what about the following reaction?
- Any equilibrium where you have different states of matter is
- called a heterogeneous equilibrium.
- And so let me write another heterogeneous equilibrium.
- So let's say I have H2O in the gaseous state and that's
- essentially steam-- so it's not going to be the solvent
- this time-- plus carbon in the solid state.
- And let's say that that's an equilibrium with hydrogen in
- the gas state plus carbon dioxide in the gaseous state.
- This is a heterogeneous equilibrium because you have
- things in the gaseous and the solid state.
- And solid state, by definition, it can't be
- dissolved either into the gas or into the-- when we talk
- about solutions, we talked about colloids and suspensions
- and mixtures before, but we're talking about solutions.
- By definition, if this is in the solid
- state, it's not dissolved.
- If this was dissolved, we would write an aq here.
- It would be the aqueous state.
- So if you talk about the forward reaction, what's the
- forward reaction going to be dependent on?
- So the rate forward, well, the solid, there's a big block of
- carbon sitting there.
- There's a big cube of carbon there, and there's steam,
- there's water gas all around it.
- So if you pick any volume, especially if you pick some
- volume near the boundary of the carbon, you're always
- going to have carbon around.
- It's just what matters is the concentration
- of the water gas.
- That's what's going to drive the forward rate, so the
- forward rate is going to be dependent on some constant
- times the concentration of the water gas.
- And, of course, the backwards rate, so you need to get some
- H2, some molecules of-- let me draw it like that, because it
- has 2 hydrogen molecules plus a carbon dioxide, so maybe a
- carbon dioxide looks like that.
- So the reverse reaction, so rate, let's call that reverse,
- is going to be equal to some constant times the probability
- of finding both of these molecules in the same place.
- And, of course, the probability is related to or
- it's on a first-level approximation, depending on
- the concentration.
- So it's concentration of H2 times the concentration-- and
- to find both of them, you multiply the probability,
- because you need this and that-- times the
- concentration of CO.
- So when a reaction is in equilibrium, these two equal
- each other-- this is an r right here-- so this is going
- to be equal to the reverse rate of reaction H2 times
- carbon dioxide.
- Divide both sides by the K's, both sides by the H2O, and you
- get the forward coefficient or constant or whatever you want
- to call that, divided by the reverse constant-- I'm just
- dividing both sides by that-- is equal to this-- let me just
- copy and paste that-- is equal to that divided by this.
- You take that and you divide it by that.
- And so if we call this the equilibrium constant, because
- it's just two arbitrary constants, so we can just call
- this the equilibrium constant, you see that it actually makes
- a lot of sense to ignore the solid state in your
- equilibrium reaction.
- So the two takeaways here is when you're trying to
- calculate an equilibrium constant, you should ignore--
- especially when it's in a heterogeneous equilibrium--
- you should ignore the
- solution-- or not the solution.
- Ignore the solvent in that first example, where I did it
- with boron trifluoride with water.
- Water was the solvent, so I ignored it.
- Because water is everywhere, and you also
- ignore the solid state.
- Ignore the solid.
- Anyway, we'll probably use these in future things where
- we actually calculate the equilibrium constant.
- See you in the next video where we'll learn about Le
- Chatelier's principle.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
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This is great, I finally understand quadratic functions!
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At 2:33, Sal said "single bonds" but meant "covalent bonds."
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