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Chemistry: The Concept of Equilibrium

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  • Type: Video Tutorial
  • Length: 11:19
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 122 MB
  • Posted: 01/28/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry Review (25 lessons, $49.50)
Chemistry: Chemical Equilibrium (14 lessons, $20.79)
Chemistry: Principles of Chemical Equilibrium (3 lessons, $4.95)

Professor Harman explains the concept of equilibrium. In a dynamic equilibrium, though appearing static at the acroscopic level, the forward reaction equals the reverse reaction. To create this type of equilibrium, the system must be closed. This can occur in a chemical system, as well. In chemical equilibrium, the forward and reverse reactions are equal and the concentrations of products and reactants do not change. Professor Harman also explains how you can determine the direction of equilibrium mathematically. To do this, you can look at the partial pressures. As the pressures level out, the system reaches equilibrium. You can also look at the rates of the forward and reverse reactions. When the rates become equal, the system has reached equilibrium. Then, Professor Harman explains that if the equilibrium is disrupted, a new equilibrium will establish in which the overall ratios pf products and reactants are equal. These ratios help to determine the Equilibrium Constant, which reveals the direction of the overall balance of the chemical system.

Taught by Professor Harman, this lesson was selected from a broader, comprehensive course, Chemistry. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/chemistry. The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more."

Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

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Thinkwell
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Chemical Equilibrium
Principles of Chemical Equilibrium
The Concept of Equilibrium Page [1 of 2]
If I put a beaker of water here and I go home for the day, come back maybe a week later, what’s going to happen?
Well, intuition tells you or common experience tells you that that water is going to evaporate. It’s all going to be
released into the atmosphere. What if I do the same thing but instead I cover it? In fact, for the sake of argument, I
cover it with a perfect seal and then I go away for that same month. Well, you might be guessing—if you think about it
for a moment, maybe you think it’s going to evaporate, but of course, where can it go? It’s now in a closed system
and the water molecules that slowly are coming off of the surface of the water have no place to go but to build up in
this larger system. And eventually they reach a high enough vapor pressure that they’ll just return back to the liquid.
At that point we set up what we refer to as an equilibrium, or a dynamic equilibrium. It doesn’t mean that the water
molecules no longer leave the liquid and go into the gas phase. That process continues to happen. But what it does
mean is that the molecules that have gotten into the gas phase now are returning to the liquid phase that’s equal to
the rate that they’re leaving. We’ve set up, again, an equilibrium in the sense that we no longer have any observable
macroscopic changes: the volume of water, the gas pressure, the pressure of the vapor is exactly the same.
However, again on a molecular level, lots and lots is happening. Molecules are continually going into the gas phase
and coming back down to the liquid.
Now the same type of equilibrium is going to happen or can occur in a chemical system as long as we have, once
again, a closed system. So imagine for instance me mixing together molecules of hydrogen and of iodine, and
connecting them in a closed system and then raising the temperature to allow them to get over the activation barrier
so that these molecules can react with each other. So if I then wait for a long enough period of time—if I start out here
in my cartoon, the red circles indicate the bigger iodine molecules and the little blue circles here indicate the
hydrogen. If I again let that system go and go and go, eventually I reach an equilibrium state. Equilibrium in the
sense that my concentration of hydrogen and of iodine no longer changes anymore and my concentration of the
products, HI, no longer changes anymore. So my distribution might look like this now. Most of the molecules will
have converted over to hydrogen iodide, indicating the red and the blue connected together. But you’ll notice that I’ve
got a little bit left of I2 and H2 that seems to refuse to react with each other. So I have an equilibrium of concentration
of both I2 and H2. Now, it does not mean—just like when we talked about the water—that these guys won’t react with
each other. Indeed, they’re going to bump into each other eventually and react. However, what’s going on is that
while that’s happening, we also have two of the HI molecules reacting together in the back reaction, forming H2 and I2
again. So again, it’s a dynamic equilibrium in that, equilibrium in the sense that we reach a point that the
concentrations no longer change, but dynamic in the sense that on a molecular level the reaction continues to go
forward and backward. In other words, I should be pointing up here now. The reaction of H2 plus I2 continues to
make HI, and two molecules of HI continually come together to make H2 and I2.
Well, how do we quantify that? How can we describe mathematically—how can we describe in measurement terms—
exactly where that equilibrium lies? Does it lie more towards the products as I’m depicting here, or does it lie more
towards the reactants as I’m showing in this case? Clearly, the way I’m describing this equilibrium, it lies to the
product side in that I’m indicating that you wait enough time and this is where the reaction wants to go.
Well, we can measure the partial pressures of these materials. They’re all at a high temperature, at least that we’re
doing this reaction in. They’re all in the gas phase. So staying in the gas phase at this higher temperature, let’s follow
the partial pressures of each one of those components over time. For the H2 and the I2, let’s suppose that they start at
these arbitrary pressures. And again, over time we’ll find that those pressures decrease and eventually level out as
we approach equilibrium and maintain chemical equilibrium. On the other hand, the amount of HI starts out as zero.
We have no product initially. We just have H2 and I2. But the amount of HI rapidly increases and eventually, again,
levels out just like these level out. It levels out at its equilibrium pressure, just like these have now an equilibrium
pressure. So at this stage, I can wait 18 years and my concentrations or my pressures of these three components will
no longer change. But once again, on a molecular level, lots is happening. The reaction continues to go backwards
and forwards.
I can look at the forward rate of the reaction and measure that, and I can measure the back rate of the reaction.
Remember, that’s going to be defined as the disappearance of reactants or the appearance of reactants, depending
on how I want to define my rates. But if I look at what happens to the forward rate, the rate at which these two
combine together to make product, it starts out very fast but it drops down as the concentrations or the partial
pressures of these two components decreases. And eventually it slows down to this value. Meanwhile, the reverse
rate, which starts out at zero because there is no HI present initially so it can’t possibly react to form H2 and I2—but as
the amount of HI builds up, it’s capable of undergoing the back reaction, and so that rate increases, but eventually
Chemical Equilibrium
Principles of Chemical Equilibrium
The Concept of Equilibrium Page [2 of 2]
again levels out at it approaches equilibrium. At equilibrium, the forward rate becomes equal to the back rate, and at
that stage there is no longer any net change in the system because these two rates are exactly the same. So again,
on a molecular level it’s happening, but on a macroscopic level there are no more changes going on in this system.
Now, what happens if I come along and I say, “All right, I’m going to put in some more iodine”? I let the system reach
its equilibrium but then I come along and I inject more iodine. That’s going to cause the system to go out of
equilibrium, and as a result it’s going to respond to that. It’s going to undergo a reaction. And we’ll be able to predict
how it will respond to that in a little bit. But it will undergo a reaction, a net overall change—in this case, by the iodine
decreasing, by the H2 decreasing, and by the HI increasing a little bit—and then it levels out again. I have gone from
one equilibrium to another equilibrium, and in this case the values I have for the partial pressures at my new
equilibrium no longer are equal to the values I had at my original equilibrium.
So how is that possible? Again, I have an equilibrium here, and I threw in some more iodine and it reestablishes an
equilibrium, but at the new equilibrium the values of my pressures are completely different. Well, how did it know to
do that? What’s the relationship between these concentrations and these concentrations?
Here’s the interesting thing. If I look at the ratio of the products multiplied or taken to the exponent of their
coefficient—in other words, I’m going to square it because there are two products—and I divide by the concentrations
of reactants, the I2 and the H2–and by the way, I could do the exact same thing if I took the partial pressures of the
products squared divided by the partial pressure of the I2 and the partial pressure of the H2—if I take either of those
ratios at the temperature that I’m doing this experiment, which happens to be 425 degrees, I get a value of 56. So
what? What’s so special about 56? If I take the ratio of this guy squared divided by these two, I get 56. If I take this
number squared, divided by these two things, I get 56. So there’s the link. The equilibrium here and here are related
in that the ratio of products to reactants is exactly the same both here and here. But my individual pressures are
clearly very different.
So equilibrium is defined not by what the pressures are but what the ratio of pressures are of products over reactants.
For this particular system—and I have to emphasize that. For this particular system, it won’t make any difference
whether I take a ratio in terms of concentrations—in other words, moles per liter—or whether I take that ratio in terms
of pressures. I’ll get the same number in either case. That won’t always be the case but it is true in this system.
However, it is always the case that any chemical equilibrium, if I take my concentrations of products and divide it by
reactants, I will always get a constant number. As long as I don’t change my temperature, I’ll always get a constant.
That constant is the equilibrium constant. And what it tells us is, it tells us quantitatively where the equilibrium lies. If I
have an equilibrium that’s heavily favored by products, my numerator is going to be a lot bigger than my denominator
so my equilibrium constant is going to be a large number. That tells me right away where the equilibrium is. On the
other hand, if I have an equilibrium constant that’s a small number, it tells me right away that I have more reactants at
equilibrium than I do products, and so that tells me that the reactants side of the reaction would be favored. So it’s a
way numerically of describing where the equilibrium lies. Does it lie to the left side of the equation or to the right side
of the equation?
In this particular example, it lies to the right side. The right side is favored over the left. I simply have a lot more of
products than I do of reactants. But a lot of equilibria that won’t be true for. So again, this K, this equilibrium constant,
gives us information, telling us where the equilibrium lies—in other words, which side is favored—and by how much.
In general it depends very much on temperature. So the equilibrium constant will be different as we change
temperature. And we’ll have a lot more to say about that.
But what remains for us to do is now formally define the equilibrium constant for a variety of different reactions, and
we’ll do that next.

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