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Chlorine and water react to form hydrogen chloride and oxygen, like this: 2 C12(g)+ 2HzC 0(g) -4 HCl(g)+02(g)Write the pressure equilibrium constant expression for ...

Question

Chlorine and water react to form hydrogen chloride and oxygen, like this: 2 C12(g)+ 2HzC 0(g) -4 HCl(g)+02(g)Write the pressure equilibrium constant expression for this reaction-

Chlorine and water react to form hydrogen chloride and oxygen, like this: 2 C12(g)+ 2HzC 0(g) -4 HCl(g)+02(g) Write the pressure equilibrium constant expression for this reaction-



Answers

Water from the combustion of hydrogen and pure oxygen is at $3800 \mathrm{K}$ and $50 \mathrm{kPa}$. Assuming we only have $\mathrm{H}_{2} \mathrm{O}, \mathrm{O}_{2}, \mathrm{OH},$ and $\mathrm{H}_{2}$ as gases with the two simple water dissociation reactions active, find the equilibrium composition.

In this problem we have. We want to consider the water gas reaction. In the example 15.4. You want to find the equilibrium constant at 500,000. 1214 100 Calvin, Uh, and what can we infer from the result? So, in this example, we had again this this, uh, this reaction here that we, um eso We have hr hydrogen molecules, carbon dioxide, um, and water and carbon monoxide. All kind of in our mixture here, going back and forth. And we don't have this in our tables, but we do have this one on. We do have this one on. We can see that this one is one half of this minus this again being the minus. Basically, flip these around so we can see we get our h 20 r c e o s. From There are CEOs from here, co two from here and our h 20 from there. All right, so we can look these up with the different temperatures and then calculate this is different temperatures. So at 500 Calver, we get that this value here is, um uh Well, we wind up with the pay anyway, is 0.0 77. Which means that we're going strong in that direction. Um, so we're gonna basically, uh, this we're going to get a lot of hydrogen now in a lot more hydrogen. Now, in 1000 Calvin, this goes up, and at one, we're basically half and half. Um uh, eso at at 1000. Calvin, Uh, we're still going this direction, but not as much. At 1200 Calvin, we're actually now moving over towards this direction. So we're getting mawr of each species. And at, um 1400 Calvin again, we're going Maurin this direction even stronger. So what we noticed is that as the temperature goes up, we're going We're getting less hydrogen. So if we wanted, if we're trying to make hydrogen here, we're tryingto convert. You know, some of the c e o into hydrogen that we can then use for something. Um uh, we want to do that at a low temperature. Um and I don't know if we went toe lower temperatures, I'm not sure if it would e mean at this temperature. It's pretty much going mostly all that way anyway, Onda after below this temperature, it's probably. You know, the percentages are probably the mole. Fractions are probably not going to change that much, because this thing is pretty low. At that point, eso we might lower temperatures will probably get more hydrogen, but not it would be kind of saturates. So at a certain temperature here, you can see how these, um just the way, the way, the way these things are, that as this gets smaller and smaller, it doesn't Basically, it's not a linear relationship. Anyway, that is, this thing goes so at this kind of value of K were almost all to that side. A Tenney, smaller values. Okay, we're even mawr almost all to that side, so there really isn't a whole lot of difference.

Okay. So in this reaction I'm gonna compare Q. Okay. And they're the same expression products Overreactions raised to their coefficient, but Q. Can be any conditions. Okay, They can be pressures and concentrations that don't necessarily have to be at equilibrium for K. You have to plug in conditions, pressures and concentrations at equilibrium. So by calculating Q. And comparing it to K. We can figure out whether or not something is at equilibrium, in which way it will shift. Okay, So to start with here, they've told us that we don't have any chlorine. Okay, So there's no chlorine present here. Mm Since chlorine is a product, the reaction has to shift to the right to make more product. Okay, so this reaction is definitely going to shift right? We didn't have to calculate much there because there was something missing. All right. And be though we're gonna go ahead and fill in for Q. Okay. So this is the partial pressure of chlorine squared forms of partial pressure of our water vapor squared. Mhm. Provided by the partial pressure of hcl to the fourth And the partial pressure of R. 02. So we'll go ahead and plug in our values. It's a .2 squared There's .35 squared Right to buy 3 to the 4th power And .15. So this is going to give us AQ. value of four. Mhm. Well, okay. Is 79. And cue is for so since Q is greater than K, it means I've got too much product. Right? Because these are products Overreactions. So I've got too much product to be at equilibrium. So my reaction is going to need to shift to the left in order to get back to equilibrium. Okay? So this will shift to the left to use up some of that extra product and return to equilibrium.

Okay. So in this reaction I'm gonna compare Q. Okay. And they're the same expression products Overreactions raised to their coefficient, but Q. Can be any conditions. Okay, They can be pressures and concentrations that don't necessarily have to be at equilibrium for K. You have to plug in conditions, pressures and concentrations at equilibrium. So by calculating Q. And comparing it to K. We can figure out whether or not something is at equilibrium, in which way it will shift. Okay, So to start with here, they've told us that we don't have any chlorine. Okay, So there's no chlorine present here. Mm Since chlorine is a product, the reaction has to shift to the right to make more product. Okay, so this reaction is definitely going to shift right? We didn't have to calculate much there because there was something missing. All right. And be though we're gonna go ahead and fill in for Q. Okay. So this is the partial pressure of chlorine squared forms of partial pressure of our water vapor squared. Mhm. Provided by the partial pressure of hcl to the fourth and the partial pressure of R. 02 So we'll go ahead and plug in our values. It's a 0.2 squared, there's 0.35 squared right to buy three to the fourth power and 0.15 So this is going to give us a Q. Value of four. Mhm. Well, okay. Is 79 And cue is for so since Q is greater than K, it means I've got too much product. Right? Because these are products Overreactions. So I've got too much product to be at equilibrium. So my reaction is going to need to shift to the left in order to get back to equilibrium. Okay? So this will shift to the left to use up some of that extra product and return to equilibrium.

In this problem. We have water from compression of hydrogen and pure oxygen is at 3800 Calvin and 50 k p a half an atmosphere and assume that we have water, oxygen and hydrogen and gasses. And we want to find the equilibrium composition. So we have Ah, water, um, dissociating into hydrogen and oxygen or oxygen. Hydrogen burning with oxygen to like water. So somewhere we're in here. Now we can, um, taken Look up. What? This is what L NRK is at this temperature on ditz minus 1.91 for water or for this association. Now again, we can printer by this so that we have We take, um two x from here, and we get to X over here and ex here. So our total number of moles after the perturbation is two plus x again because we have more molecules of stuff over here than we have over here. Um, the more ratios are, then thes and we can figure out the way can plug those into our formula for the, um for using K. And again, where are exponents? Appear is one. So we just have the ratio of the pressure um, that were here. Uh, we're at to that of atmospheric. Uh, and then this is, um this is Let's see here. This is theme. The mole ratio of 02 square. This is the mole ratio of now. This is the mole ratio of H two squared, and this is 02 squared, and this is one over. Um h H two squared so we can see some of these cancers out. Anyway, we witness equation, we get this quadratic equation for X, and we can find the route and the route that makes sense to keep everything positive. Um, again, we gotta be careful of making sure this day is positive. So X has to be less than one at least, and obviously greater than zero. So we figure out that in the root of this equation that makes sense than is X equals 0.54 And that means that we have mole fractions of water as 36.2% of oxygen as 21.3% and hydrogen at 42.5%. So again, and this this high temperature, we're not really, um we're actually have, um or, uh, kind of more on this side of the reaction than we are on this side. That's true. E mean. Anyway, we have more. We have more hydrogen than we have water. Um, so that's Ah, uh, kind of it. But we had a very high temperature. And then also this pressure change. This also shift things.


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