Le Chatelier's principle: Worked example
A worked example using Le Chatelier's principle to predict how concentrations will shift for different perturbations. Example includes changing reaction vessel volume, changing amount of solid product, adding inert gas, and adding a catalyst. Created by Yuki Jung.
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- At5:36there is a mention of when you add calcium carbonate (solid), you don't interrupt equilibrium. However thinking about that makes me wonder... For example, you add a scoop of calcium carbonate, which has a certain volume into this 'fixed' volume, isn't it so that it will change the pressure and/or volume slightly? And by doing so change the equilibrium a bit? Hope I'm clear. Thanks(42 votes)
- That's a great question Sander Sloof! The assumption made in the video, which I unfortunately failed to state, is that changing the amount of solid did not change the total volume of the container significantly. I apologize for any confusion! Your thinking is entirely correct-if enough solid was added to change the volume of the container, that would also change gas pressures and perturb the reaction from equilibrium. Thanks for commenting!(39 votes)
- At4:00, when Argon gas is added, the total pressure increases. Then shouldn't the reaction move backwards to balance the increase in pressure?
Like when Argon is added the volume available to CO2 decreases so the reaction move backwards
Or like if we add Ar wouldn't it be like the available volume to the other reagents is reduced ?(15 votes)
- I completely agree that the volume available to CO2 will decrease. But the total volume of the container still remains the same. That is why at4:47she says that we did not change the volume. So addition of Argon will not affect the direction of the reaction.
Coming to pressure, the total pressure did increase, but the partial pressure of CO2 did not. And we saw that Kp only depends on the partial pressure of CO2. So pressure too, will not affect the direction of the reaction.(5 votes)
- I couldn't understand that if concentration of CaCO3 is increased then how equilibrium is not broken? If CaCO3 is increased then concentration of Ca0 and CO2 will also increase then Equilibrium can break.Please make me understand.Thanks(11 votes)
- If you consider the following reaction:
CaO(s) + CO2(g) ----------> CaCO3(s)
Then what you are saying is correct. But we are adding CaCO3 as a finished product. We have already learnt in the previous videos that solids do not affect the equilibrium constant. So you don't have to worry about how CaCO3 is formed! Just notice in WHICH FORM the substance(any substance for that matter) is being added. Hope this helps!!(4 votes)
- At 4.56, she told adding Ar doesn't change PCO2. But, partial pressure is affected by total pressure, isn't it?(2 votes)
- Partial Pressure is the pressure a gas would exert by itself if it were the only gas in that volume. The partial pressure is not affected by total pressure because partial pressure stays the same for the same amount of gas.(4 votes)
- At3:23she mentions that adding more volume of co2, would favor forward reaction which doesn't make intuitive sense. According to the last video on Le chatelier's principle by Sal, he stated that when you have more of let's say products, then you would favor backward direction as to decrease the products and get them back at equilibrium. But here, it says that when you have more of products, then it would favor products. Doesnt that increase products even more and violate some conservation law? I hope I am clear(4 votes)
- I think you've confused with the volume of the container and volume of co2.
The speaker has said, increasing the VOLUME OF THE CONTAINER decreases the partial pressure of co2 since, volume of the container is at the denominator. As we increase the value of denominator, we decrease the whole value. So, as we increase the volume of the container, the partial pressure of co2 goes down. So, the equilibrium shifts to the right, which favors the products. So, the statement is actually correct.
Watch this video again.(1 vote)
- In the energy diagram, Why the size of activation energy barrier for forward reaction is greater than the backward reaction?(2 votes)
- The reactant molecules are at a higher energy level than the products, so they have further to climb up the energy hill.(4 votes)
- if adding argon increases prssure won't the reaction become fast?(3 votes)
- But changing the total pressure has NO effect on the partial pressure of CO2. thus the rate of forward or backward reaction is undisturbed.(1 vote)
- Imagine two scenarios: An infinitesimally small cube of calcium carbonate (CaCO3), and a large chunk of calcium carbonate. Obviously, the large chunk of calcium carbonate will have more surface area. Common sense would have me believe that the increase in surface area means an increase in the forward reaction: CaCO3(s) = CO2(g) + CaO(s).
Thus, by adding CaCO3(s), my intuition would be that this favors the forward direction. Le Chateleir's principle tells me this is not the case. "Because the solid is stationary and not dynamic", I understand that part, but still, we are ignoring surface area here. Help me conceptualize this idea.(1 vote)
- I assume that since you are invoking Le Chatelier's Principle that this is an equilibrium reaction.
Have you thought about what the increased surface area means for the reverse reaction?
Does that help you answer your question?(4 votes)
- At4:55, on adding argon how did the total volume not change?(2 votes)
- At6:15it is claimed that adding a catalyst does not introduce a shift. However, wouldn't there occur a difference in reaction rate and thus a new equilibrium if the "P" energy level is different from the "SM" energy level (as indicated in the figure) after a lowering of the threshold? My thinking is that the molecules on either side of the reaction has some statistical energy distribution (e.g. gaussian) which means that small differences in energy thresholds give an large (appr. exponential) impact on the number of molecules that manages to overcome the threshold. To me it would seem like a strange coincidence if any lowering of the energy level would lead to an equal reaction rate change regardless of the initial difference between the P and SM levels.(2 votes)
- [Voiceover] In this video we're going to go through an example reaction that uses Le Chatelier's principle. So what we're gonna do is, we're gonna apply Le Chatelier's principle to look at various changes to this reaction when we perturb our reaction from equilibrium. Just as a reminder, what do I mean when I say, This reaction is at equilibrium.? So what that means is we have a reversible reaction. We have the forward reaction which has some rate K forward. The reverse reaction has some rate K backward and a dynamic equilibrium. These rate are equal to each other, so all of the concentrations are going to stay constant, and then what we do is we decide to see what happens when we add some carbon dioxide gas. If we add carbon dioxide gas, the concentration of carbon dioxide gas will go up, or you can think about it as a partial pressure going up. And Le Chatelier's principle tells us, that if we had a reaction at equilibrium and then we perturbed it by adding more CO2, it will shift to try to reduce the effect of that change. It will favor the reverse reaction, so if we add CO2, what happens is, we favor our reactants. Another way we can see this is by looking at the equilibrium constant for this reaction. We can write our equilibrium constant K, where this is a capital K. It's kind of confusing, but I will try to make this look like a capital K. (laughs) We can write it in two ways, we can write it in terms of the molar concentration, and if we write Kc, the expression will be the product concentration, our CO2 gas concentration. And that's it, because when we write out Kc, we write out concentrations of gases and we write out concentrations of solutions, but we don't include solids. Kc is just the concentration of CO2 at equilibrium. I'm going to write an eq there just to show that's the equilibrium concentration. I said that you can also write it in terms of partial pressures, so there's our fancy capital K with a p subscript, which means that instead of concentrations, we're writing everything for gases in terms of partial pressures. We have the partial pressure of CO2 and again, that's it, because everything else is a solid, so we don't include those in our equilibrium expression. Writing these expressions out will be really helpful for our second condition. We're going to think about what happens, when you increase the volume of our container. We can rewrite the partial pressure, actually, in terms of the volume, so if you use the ideal gas law, the partial pressure of CO2 is equal to the moles of CO2 times RT divided by the volume. Similarly, we can rewrite molar concentration in terms of moles divided by volume. If we increase the volume of our container, increasing the volume, since it's in the denominator, will make our pressure go down, so our partial pressure of CO2 will go down and we won't be at equilibrium anymore and the same for the concentration, since our perturbation is decreasing the carbon dioxide concentrations, Le Chatelier's principle says, that our reaction will try to counteract that change. It will try to get back to equilibrium and try to get the CO2 concentration back up, so it will have to favor products. Our reaction will favor the products to try to get the moles of CO2 back up, so that we get back to the equilibrium concentration of CO2 and the equilibrium partial pressure. The third change we're gonna look at is what happens when you add argon gas. Argon gas is an inert gas. We don't expect it to react to anything. One thing that will happen when you add the argon gas is it will increase the overall pressure of your container, so it will increase the total pressure. But that actually doesn't tell us what it does to our equilibrium. Let's look back at our equilibrium expressions Kc and Kp. We can see that the partial pressure for Kp only depends on the moles of our CO2 and our volume. Since we didn't change the moles of CO2 and we also didn't change the volume. Even though we increased P-total, the partial pressure of CO2 stayed the same. That means we didn't perturb our reaction from equilibrium and since we didn't perturb it from equilibrium, there'll be no shift. We are still at equilibrium. The concentrations will still stay the same. What happens when we add more calcium carbonate? That's our starting material and it is a solid. Our equilibrium expressions are determined by our CO2 concentrations, so adding more calcium carbonate, which is a solid, isn't actually going to perturb our reaction from equilibrium. Our reaction is still going to be at equilibrium and we will get no shift in concentrations. We're actually gonna look at one more thing. We're going to think about what happens when you add a catalyst. Let's say we want to speed up this reaction. We can envision what's going on here when we add the catalyst by using an energy diagram. If we have an energy diagram. We have energy on the Y axis and we're looking at the difference in energy between our reactants, or starting materials, with our products and I just sort of made up these relative energies. The way that I have this drawn here, we can see that our product is lower energy than our starting material and our forward rate, Kf, which is is up here, is determined by the size of this activation barrier between our starting material and our transition state. Our backward rate, Kb, is determined by the size of this energy barrier, so the difference in energy between the product and our transition state. If we add a catalyst to our reaction, we can think about it as lowering the activation energy for our reaction and that means we have a lower energy barrier for our forward reaction, so our forward reaction is gonna get faster, but it's also gonna lower the rate of our backward reaction. K backward, or Kb, is also gonna speedup and since it speeds up both the forward and the backward reactions adding a catalyst also won't perturb your reaction from equilibrium. Adding a catalyst will result in no shift in concentrations. So the main things to remember from this problem, I think. the things I find most tricky anyway, are that adding an inert gas, it'll increase your total pressure, but it won't actually change any of your partial pressures. It won't shift your reaction from equilibrium and the same thing is true for solids and catalysts. All of those three things: inert gases, solids, and catalysts, will not shift your reaction from equilibrium.