Reactions in equilibrium | Chemical equilibrium | Chemistry | Khan Academy
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Reactions in equilibrium | Chemical equilibrium | Chemistry | Khan Academy

All of the reactions we’ve
looked at so far have been of the form lowercase a moles of
the molecule uppercase A, plus lowercase b moles of a
molecule uppercase B. They react to form the product
or the products. Let’s just say they have
a couple of products. I could have had as many
molecules here as I wanted. Let’s say c moles of the
molecule C plus d moles of the molecule capital D. And the idea here is that they
went in one direction. And if we did a little energy
diagram, just going off of the kinematics video we just did, if
that’s the reaction, or how the reaction progresses, you
could imagine that here you have it at a higher
energy state. You have lowercase a moles of
capital A molecule, plus lowercase b moles of capital B
molecule, and you have some activation energy. And then you get to a more
stable state or a lower energy level here, where it’s lowercase
c moles of C molecule plus lowercase d moles
of the D moles, and of course, you had some
activation energy. It only goes in this
direction. Once you get here, it’s
very hard to go back. So that if you came back and
looked at this– if you get enough of A and B– you’ll just
be sitting with molecules of C and D. It’ll only go in
this direction. But that’s not how it
happens in reality. In reality– well, it sometimes
happens like this in reality where the reaction can
only go in one direction. But in a lot of cases, the
reaction can actually go in both directions. So we could write, instead of
this one-way reaction, we could write a two-way
reaction like this. And not to confuse you too much,
these are the number of moles or the ratios of the
molecules I’m adding up, and they become relevant
in a second. So let’s say I have lowercase a
moles of this molecule plus lowercase b moles of this
molecule, and then they react to form lowercase c moles of
this molecule plus lowercase d moles of that molecule. Sometimes the reaction can
go in both directions. And to do that, to just show an
equilibrium reaction, you do these arrows that go
in both directions. That means that, hey, some of
this is going to start forming into some of this. But at the same time, some of
this might start forming into some of this. And at some point, I’m going to
be reaching an equilibrium. When the rate of reaction of
molecules going in that direction is equal to the number
of molecules going in the other direction, then I’m
going to reach some type of equilibrium. Now, why would this happen
as opposed to that? And I can think of
one situation. If we draw this energy
diagram again. Maybe both of these have similar
or not so different energy states. There could be other reasons,
but this is the one that comes to my mind. Maybe the energy states look
something like this. On this side, you have the A
plus B, and then you need some activation energy. And then maybe the C plus D,
maybe it’s a little bit of a lower potential, but it’s
not that much lower. So maybe they’re favored to go
in this direction, because this is a more stable state. So this is the A plus B, but
here you have the C plus D. But it’s not ridiculous
to go this way either. So most of it might go that
way, but some of it might go this way. If some of these molecules just
have the right amount of kinetic energy, they can
surmount this activation energy and then go backwards
to that side of it. And the study of this is called
equilibrium, where you’re looking at the
concentrations of the different molecules. And just to compare that to
kinetics, kinetics was how fast is this is going
to happen? Or what can I do to change the
activation, this hump here? Equilibrium is studying what
will be the concentrations of the different molecules that end
up, once the rate going in this direction is equal to the
rate going in that direction. And I want to be clear. Equilibrium is where the rate
going in the forward direction is equal to the rate going
in the reverse direction. It doesn’t mean that the
concentrations of the two things are equal. You might end up with 25% of
your eventual solution’s concentration to be A
and B and 75% here. All we know is that at some
point, you’ve reached an– equilibrium just means that
those concentrations won’t change anymore. And just to give you an example
what I mean here, I could have written– let’s
see, this is actually the Haber process. I could write nitrogen gas
plus 3 hydrogen gases. These are all in gas form, so
I can put a little g in parentheses. Actually, it’s an equilibrium
reaction, and it produces 2 moles of ammonia. It’s called the Haber process. We could talk about that
in another video. So in this case, we could say a
is just 1, this lowercase a. Capital A is the nitrogen
molecule. Lowercase b is 3. Uppercase B is the hydrogen
molecule. And then lowercase c is the
number of moles of ammonia and uppercase C is the ammonia
molecule itself. I just want you to realize this
is just an abstract way of describing a whole
set of equations. Now, what’s interesting in
equilibrium reactions is that you can define a
constant called the equilibrium constant. It’s defined as the constant
of equilibrium. Let me switch colors. I’m using this light
blue too much. The equilibrium constant is
defined as you take the products, or the right-hand
side– but if it goes in both directions, you can obviously
go in either direction. But let’s say that this is the
forward direction going from A plus B to C plus D. So you take the products, you
take the concentration of each of the products, and you
multiply them by each other, and you raise them to the mole
ratios that you’re taking. So in this case, it would be
the concentration of big C raised to the lowercase c power
and the concentration of big D raised to the
lowercase d power. And when I say concentration,
they usually– especially what you see in your intro chemistry
classes, the concentration is going to be
measured in molarity, which, just as a review, is
moles per liter. A couple of videos ago, when I
taught you what molarity was, I said, you know, moles per
liter— I don’t like it so much because the volume of your
fluid or your gas you’re dealing with is dependent
on temperature. So I didn’t like
using molarity. But in this case,
it’s kind of OK. Because this equilibrium
constant is also only true for a given temperature. We assume it for a given
temperature, and I’ll show you how we use it in a second. But it’s defined as the
concentrations of the products to the powers. And also, if I have time,
maybe I’ll do it in the next video. The intuition why you’re
raising it to the power divided by the concentrations
of the reactants, or the things on the left-hand side of
the equilibrium reaction. So capital A to the lowercase a
divided by capital B to the lowercase B. And what’s interesting about
this, and this is a bit of a simplification, because
this doesn’t apply to all reactions. But to most things that you’re
going to encounter in an intro chemistry class, this is true,
that once you establish this equilibrium constant for a
certain temperature– it’s only true for a certain
temperature– then you can change the concentrations and
then be able to predict what the resulting concentrations
are going to be. Let me give you an example. So let’s say that after you
did this equilibrium reaction– and actually, just
to make things hit home a little bit, let me take this
Haber process reaction and write it in the same form. So if I wanted to write the
equilibrium constant for the Haber reaction, or if I wanted
to calculate it, I would let this reaction go at
some temperature. So this is only true at– let’s
say we’re doing it at 25 Celsius, which is roughly
room temperature. So what I would do is I would
take the products. So the only product
is ammonia, NH3. I raise it to the power of the
number of moles that’s produced for every 1 mole
of nitrogen gas and 3 moles of hydrogen. So I raise it to
the power of 2. So that’s what that gets me. And I divide it by
the reactants. So 1 mole of nitrogen, so
I would just put the concentration of the nitrogen,
plus 3 moles of hydrogen– oh, no, no. I shouldn’t write
a plus there. It’s multiplied. So times the hydrogen, and I
raise it to the third power, because for every 1 mole of
nitrogen, I have 3 moles of hydrogen and then 2
moles of ammonia. And if I were to calculate this,
remember, when I put these in brackets, I’m
figuring out the concentration. So I would have to figure
out the moles per liter. Or sometimes they say, the
molarity of each of these things, and it’ll get
me some constant. If I change it, I can go and
calculate the rest, so let me just do an example right now. So let’s say I have 1 mole of
molecule A plus 2 moles of molecule B are in equilibrium
with 3 moles of molecule C. And let’s say that once we’re
in equilibrium, we go and we measure the concentrations,
and we figure out that the concentration of A is 1
molar, which is equal to 1 mole per liter. That’s the concentration of A. We figure out that once we’re
in equilibrium, the concentration of B is equal
to 3 molar, which means 3 moles per liter. And let’s say that once we’re
in that equilibrium, the concentration of C is equal to
point– well, I don’t want to do something too– let’s say
it’s equal to 1 molar as well. I should get rid of that point
there, because I don’t want to say 0.1 molar, so it’s
just 1 molar. So if we wanted to calculate the
equilibrium constant for this reaction, we just take C,
the concentration of C over here, so let’s see. The equilibrium constant is
equal to the concentration of C to the third power divided by
the concentration of A to the first power– because
there’s only 1 mole of A for every 3 of C and 2 of B– times
the concentration of– I’ll do it in that color–
B to the third power. So if we needed to calculate
this, concentration of C is 1 molar, and we’re raising it to
the third power, divided by concentration of A is 1 molar
times the concentration of B, which is 3 molars, to
the third power. So this is equal to 1/27. There’s a couple of things
we can think about. The fact that this is less than
1, what does that mean? Well, that means that our
concentration of our reactants is much larger than the
concentration of the products, where we view just the products
as whatever’s on the right-hand side of
the equation. So once this reaction goes to
equilibrium, we’re still left with a lot more of
this than this. And because we’re left with
a lot more of that, our equilibrium constant is less
than 1, which means that the reaction favors this
direction. It favors the backward
direction. Think about it. Because there’s more of this,
this must be happening more than the left-to-right
reaction. The left to right might be a
small direction like this, while more is happening there,
and that’s why we’re finding more reaction here, and that
causes the equilibrium constant to be less than 1. On the other hand, if the
equilibrium constant was greater than 1, that means that
this numerator is greater than this denominator. Which would imply that you have
more concentration– once you’re in equilibrium, you end
up with a lot more of the stuff on the right than you end
up with the stuff on the left, so then that means the
reaction would be going in the forward direction. The other interesting thing is
you can then figure out, well, what happens if I add another
mole of A to the reaction? So let’s say I throw some
A into the reaction. I add some concentration of A. So now my new A is equal to 2. Let’s say my new A
is equal to 2. Let’s say my new B, let’s say
that I want to– well, actually, we can figure out
the relation between the– actually, instead of going into
this situation where I change the concentration, let me
do that in the next video, because I just realized that I’m
running very low on time. But hopefully, you got a good
sense of what the equilibrium constant is all about
and how it’s measured or how it’s defined. And in the next video, we’re
going to talk a little bit about how else it
could be useful. In this video, you just said,
oh, if it’s less than 1, that means that the backward
reaction is favored. If it’s greater than 1, the
forward reaction is favored. In the next video, we’ll get a
little intuition, hopefully, on why it is defined this way as
opposed to, say, this way. My intuition said, hey, why
isn’t it three times the concentration of C divided by
one times the concentration of A plus three times the
concentration of B? This might have been more
intuitive to me, but this isn’t the case. This is what actually is
constant, regardless of how you change the concentrations
of the various reactants. So maybe we’ll talk a little bit
about why this is true and not necessarily this.

About James Carlton

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100 thoughts on “Reactions in equilibrium | Chemical equilibrium | Chemistry | Khan Academy

  1. The equilibrium constant used in the example should be 1/9 if I'm not mistaken
    [B] is 3 and the number of moles is 2 so the denominator should be 3^2 or 9

  2. also in the US…. this is university level chemistry that we take our junior year (11th grade) if we're advanced enough

  3. Yes I live in california and I am taking an AP course and i am a 10th grader in high school, others also end up learning this in 11th grade…

  4. Really? I am in Matric, and it is an essential and an important topic for me; it is in the syllabus! and I'm from Pakistan! Maybe you bunked the class o.O

  5. i have studied reversible reactions but not the topic this man is teaching in o didnt study in o level so be quiet.

  6. Who really cares when you learnt this. The way you behave I think you may still be in kindergarten.

  7. No he should be disappointed in you, and you should be too if you're watching chemistry videos if you're already a chemistry professor

  8. To all those people asking about H2 shouldn't it be 3/2 instead….of 3. The answer is no. You have 3 moles of hydrogen gas hence the (g). If it is H instead of H2 it is 6 which is correct. But an equilibrium is between aqueous and gas concentrations. It will always be products over the reactants. And the coefficient is your power.

  9. @***** There were 2 molecules of B, and the concentration of B was 3M. Hence, in the equilibrium expression, 3 should have been raised to the power of 2. This gives a value of 1/9 for Keq, so you and Sal were incorrect!

  10. Hi, just to point out a small honest mistake @ 12:10: You wrote 1A(M=1)+2B(M=3)=3C(M=1) so Keq should be [C=1]2/[A=1]1*[B=3]2
    So [3] is to the power of 2 not 3 because the moles of B=3.

  11. In the reaction 2O3(g) > 3O2(g) the rate of formation of O2 is 1.5×10^-3 mol/L*s. What is the rate of decomposition of O3?

    Need help with that chem question please. 

  12. Hey Khan, thanks for uploading the Chemistry playlist to watch this has been very helpful and I am sharing this link with my friends

  13. I still don't understand why in an equilibrium reaction aA + bB ↔ cC + dD, the concentration of reactants or products is their individual concentrations multiplied together and raised to their coefficients. Like why [aA + bB] = (([A]^a)([B]^b))??? I've watched the other khan video explaining it in terms of probability but honestly, that just confused me even more.
    So say the molarity of A is 1, and pretend there is no B. Then that means [aA] = [A]^a = [1]^a = 1… which is the same as the concentration of just one A! Which doesn't make sense if a>1 (so like 2A or 3A or 4A etc.) because wouldn't you expect the concentration to be greater??? HELP!

  14. Is it just me or is this guy saying something wrong at 12:42? Cause I learned that in this sort of reaction, everything is in a dynamic balance, and the reaction happens at the same rate both ways. This guy however says that the reaction happens more often from right to left, which makes no sense, because then you would run out of C, plus you can only do maths with this equasion when the reaction is in this dynamic balance state. PLEASE comment soon, got a test tomorrow

  15. @Frank Overbeeke he alludes to Le Chateliers principle. He is not saying that the reaction rates are different but the CONCENTRATION of the reactants and products are different. At equilibrium, products and reactants do not necessarily have the same conc. And he's saying that in this reaction, the reverse reaction is favored because the conc. of reactants are more.

  16. in india, we ought to know far more than this before earning a bachelor's degree. if anyone has a masters, he's given a hatsoff. 😛

  17. Why does increase in tempreture favor endothermic reaction(in reversable reactions) .I need an answer at a MOLECULAR level. I am familiar with le chatelier's principle,but i still need an EXPLANATION to it. Why doesnt increasing the tempreture favor exothermic ,it needs less activation energy (if we picturise the situation )compared to the endothermic reaction.

  18. When you say that the rates of reactions in both directions are equal in equilibrium, does that mean that the reactions occur in both directions simultaneously?

  19. 8:13 That reaction is actually only possible with 400 degrees celsius and 200 atm. The catalyst Fe could also hurry the equilibrium up a bit.

  20. I think B should be to the second power becaus there are 2 moles in the balanced equation: ([C]^3)/([A][B]^2)

  21. Hey, I have a question. At chemical equilibrium, the products are changing to reactants as fast as the reactants are changing to products. So, if you had 2A + 2B <=> 2C + 1D, wouldn't the chemicals have to be in the ratio of 2:2:2:1? Why isn't this the case? Thx!

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