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About this Lesson
 Type: Video Tutorial
 Length: 12:36
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 135 MB
 Posted: 07/15/2009
This lesson is part of the following series:
Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Final Exam Test Prep and Review (49 lessons, $64.35)
Chemistry: Chemical Equilibrium (14 lessons, $20.79)
Chemistry: Using Equilibrium Constants (5 lessons, $8.91)
This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. 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, oxidationreduction 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 electronrich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.
Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies moleculebased magnetism.
Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.
About this Author
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We've been considering chemical reactions as equilibrium, and in particular the type of problem we've been looking at is trying to identify or determine the partial pressures of reactants or products, given the information of the equilibrium constant and starting concentrations or starting partial pressures. In the situation we just encountered, we looked at a reaction where the equilibrium constant told us that the reactants were heavily favored at equilibrium, yet in the starting conditions we initially had none of one of the products. That required us to run the reaction forward in order to reach equilibrium. Now the amount that it ran forward was very small because the equilibrium constant was so far to the left. In other words, that balance point heavily favored reactants. We started with no products; the reaction had to move a little bit towards the right to reach that equilibrium condition. Now let's look at a situation where we have the same reaction, the same equilibrium constant, but a different set of starting conditions such that our starting partial pressures are far from equilibrium. Remember, where equilibrium lies is to this side, favoring this heavily over products.
We're going to start out with a set of conditions now where we have no methane initially, so we're missing one of the reactants. We have 0.9 atmospheres of water, 3 atmospheres of carbon monoxide, and 4.2 atmospheres of hydrogen. So again let's assess the situation before we dive into the calculation. The equilibrium constant tells us that the left side of the reaction should be favored at equilibrium. We have both of the products in large amounts. We have none of one of the reactants. There's no question now what direction this reaction is going to have to go to get to equilibrium. If we calculated Q, Q would be infinity in this stage. It would have to move from right to left before we reach equilibrium, so we know that. But think about how far it's going to have to move. It's going to have to move a long way to the right before we reach equilibriumenough to the right, in fact, so that one of these guys is going to approach zero, because again the equilibrium lies so far to this side based on what that K is.
Let's try using an ICE diagram again, and I'm going to show you a potential problem with just walking into this problem blind without thinking about the equilibrium constant first. So setting up an ICE diagram the way we did in our last example, we would write down our set of initial concentrations or partial pressures. We then write down the chemical change that's going to occur to get to equilibrium. In this case, we identified that the reaction is going to go from the right to the left now, so we are going to gain methane. We're going to gain an equal amount of the water. We're going to be losing carbon monoxide, and we're going to be losing three times that much of the hydrogen. So that's just making sure that our stoichiometry is correct. And then finally, again we would write down the partial pressures at equilibrium; that in this case would just be X. The sum of these, 2.9 plus X. Three minus X, three minus the little amount that we have consumedor I shouldn't say "little amount"the amount of carbon monoxide that we've consumed. And then over here, again 4.2 atmospheres, what we started with, minus the amount that we consumed for this reaction to occur.
So at this stage we could go ahead and write down our concentrations, set up our equilibrium expression, and write down 4.2 atmospheres minus 3X, and 3.0 atmospheresthat's the hydrogen, so I have to cube that3.0 minus X is going to be our carbon monoxide. On the denominator we have 0.9 atmospheres plus X, and for methane we have Xand that, remember, is 1.8x10^7. Okay, now once again we have a real nasty looking equation, very similar to the one we started out with last time. But what saved us last time is not going to save us this time.
What saved us last time was, we said, well, okay, let's try the assumption that X is really small compared to these big numbers. Well, we could do that. We could say, let's suppose X is real small compared to 3, that it's small compared to 4.2, that X is real small compared to 0.9, and that leaves us just with one variable, this X. Looks easy enough. Reasonable to set up. But we know there's a problem. We know that in fact X isn't small. In fact, we know that because the equilibrium lies far to the left in this case, we know that one of these two things is almost going to be completely consumed, because that's what our equilibrium constant is telling us. So the problem is that we know that either 4.2 minus 3X is going to get very close to zeroin other words, X is going to be such that three times X is going to be really close to 4.2or we know that this X is going to approach very close to 3 such that this number becomes zero. So what we have to do next is identify which of these guys is going to be the ratelimiting material. In other words, I shouldn't say "ratelimiting." Which one of these is the limiting reagent, rather. One of these guys is going to be consumed before the other one, and whichever one is going to be consumed first, that final concentration will be fairly close to zero. So we're stuck. We've got a big, big problem here in that one of these two terms is going to be really close to zero and we need to know, how close to zero is it? So this is just not workable. We're going to have to go back to the drawing board and think about a different way to approach this problem, making assumptions perhaps, manipulating the situation a little bit perhaps, so that we end up with a more workable algebraic expression.
So let's go back to the drawing board here and say again, okay, those are our initial concentrations. Let's do something else. Let's acknowledge the fact that this set of starting conditions is indeed a long way from equilibrium. And let's go ahead and push it in the direction that we want it to go or that nature is going to want to take this. In particular, let's go all the way. Let's go ahead and combine these things completely until one of them completely runs out, and we'll just say that its concentration is zero or its partial pressure is zero. Just as a thought processwe wouldn't actually do this in the labbut if we just let this thing go all the way to the left until something was consumed completely, then we let the system come back and reach equilibrium, the equilibrium that we reach would in fact be exactly the same as if we started with these concentrations and let them go to equilibrium. We're banking on the fact that an equilibrium does not depend on where you start the reaction. As long as we have the same total molecules in there, then we're going to end up at the same place regardless of whether we started it on this side or on this side. So again, to simplify my problem, I'm going to try to take this reaction very close to the chemical equilibrium position.
Well, one of these guys is going to be consumed. My next step is to figure out which of those we're going to lose, so let's come back to this in a moment. Let me remind you that we have one part CO to three parts hydrogen that are going to in fact be consumed in this process. And so if I have 4.2 atmospheres of hydrogen initially, then I'm going to only need a third that much of the CO, so 4.2 divided by 3, that's only going to be 1.4 atmospheres of the CO that will be consumed by the time we completely consume all the hydrogen. So once I've completely removed all the hydrogen, I still will have 3 atmospheres minus 1.4 atmospheres, or 1.6 atmospheres left of the carbon monoxide.
Now, if I've lost you for a moment, hold on and we'll get back. Let's get rid of this because that didn't work. I'll get rid of that, too. Okay, so where are we? I looked at my initial conditions. Let's now look at my adjusted conditions. This S stands for stoichiometry. What we're going to do is run this reaction intentionally all the way to the left. Even though we don't do it in the lab, we can get away with that because, again, all we care about is equilibrium. So I'm going to run it all the way until we use up one of the two products, in this case hydrogen. I have how much carbon monoxide left? We did that calculation. We've got now 1.6 atmospheres left over by using up all the hydrogen. We've generated an additional 1.4 atmospheres of methane and an additional 1.4 atmospheres of water. So I add the 1.4 to the 0.9 and that gives me 2.3 atmospheres.
Now I have a problem that's going to be more familiar to you. I've got initial concentrations of these three guysor partial pressuresand no product. Well, we've seen that before. In fact, we've seen all of those numbers before. This is exactly the same problem that we worked previously. The numbers just happen to bein fact, I intentionally set up the problem so that it would work out this way, so we wouldn't use up the time recalculating this. But now we have a situation that we had last time where we have the two reactants and we have one of the products, and we're missing the other product. We have those concentrations. So just to remind you of what we did, we can now set up that equilibrium expression as we did before. And I'll just quickly remind you of what that is: 1.6 plus X, and multiply that by 3X cubed. On the bottom we had 1.4 minus X, and 2.3 minus X. We made the assumption that X is small, which it now is. The amount that the reaction runs in the opposite direction will be very small because we're already really close to equilibrium now. That's going to be 1.8x10^7, and we end up in this case evaluating, we get X = 2.4x10^3, or the partial pressure of hydrogen is going to be 7.1x10^3 atmospheres.
So what did we do? We had a real difficult problem that was unworkable algebraically without an advanced calculator at least or a computer. So what we did was we pushed the equilibrium intentionally all the way to the other side. We pushed the reaction all the way to the other side, where we knew we were close to equilibrium based on what the K was. And then after correcting for the different amounts, we had a new set of initial concentrations. And then starting with that set of initial concentrations, we evaluated the equilibrium expression to finally give us the partial pressure. In this case it was the same partial pressure as the previous problem. So if I lost you in this last step, you'll want to review the tutorial just before this where we go through in much more detail evaluating this expression.
Now, the last thing we need to do is to look at a type of a problem where we can't make any approximations and we're going to be forced to use the quadratic equation. So we'll give you an example of that next.
Chemical Equilibrium
Using Equilibrium Constants
Solving Problems Far from Equilibrium Page [1 of 2]
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