A reaction vessel at 27 °C contains a mixture of SO2 (P = 3.00 at...

Question

A reaction vessel at 27 °C contains a mixture of SO2 (P = 3.00 atm) and O2 (P = 1.00 atm). When a catalyst is added, this reaction takes place: 2 SO2( g) + O2( g) ⇌ 2 SO3( g). At
equilibrium, the total pressure is 3.75 atm. Find the value of Kc.

Accepted Answer

hi everyone for this problem, we're told a reaction chamber initially contains a mixture of nitrogen oxide gas and chlorine gas. At 1069 kelvin with pressures of 4. A. T. M and 2.8018 cm respectively. The following reaction occurs, calculate the value of K. C. If the total pressure at equilibrium is 5.45 A. T. M. So we want to calculate the value of K. C. So this is our equilibrium concentration, but they tell us what our pressures are. So the reaction that we can use to inter convert between K P and K C. Is this so we have K P. Is going to equal K. C times are T raised to delta N. Where K P represents our equilibrium pressures. Casey represents our equilibrium concentrations are is our gas constant, T. Is our temperature and delta N represents our change and moles of gaseous product. So let's write out what we know. We know that our r gas constant is equal to 0.08206. We know our temperature, we were given it it is 1069 Kelvin, Our Delta N. We're going to look at our change in moles of gaseous products. So our product, we have two moles and then our reactant. We have two moles here and then one mole here. So our change in moles of products to reactant is 2 -3. So we have negative one. Okay, and we need to rearrange this equation to solve for k C r equilibrium concentration. So let's go ahead and do that. So when we rearrange this equation we get K. C. Is equal to K. P. Over our tee, times our change in delta N. Okay, so let's move this over to the side and plug in what we know. So we don't we don't know K. P. Yet. That is something we're going to need to solve. So let's plug in what we do know. So we know we are R. Is 0.8 to 06. R. T. Is and our delta N. Is negative one. So that's the denominator and our numerator is K. P. But we don't know what K. P. Is. So that's what we're going to solve for it now. And once we do we'll be able to calculate the value of K. C. So R K. P. R equilibrium pressure. We're going to need to create an ice table in order to solve that. So let's go ahead and rewrite out our reaction. So we have our reaction was given. We're just going to copy that down okay? And we'll write out ice on the side because this is an ice table. So our iroh represents initial pressure. And so because we're dealing with pressure, we're calculating R. K. P. So I represents initial pressure and we were told what our initial pressures are. So we're going to write it down under what we have for our reaction. So for our nitrogen oxide we were told we have a initial pressure of 4.564 for a chlorine gas, we have an initial pressure of 2.8 and we have zero for our product. So that means our reaction is moving towards the right, we're moving towards the zero and so our change in concentration represents our C zero of our ice table. So because our reactant are decreasing, it's going to be minus. And we're going to look at the number of moles we have have for each of our reactant. So for nitrogen oxide we have two moles. So this is going to be minus two X. For chlorine gas we only have one mole. So this is going to be minus X. And because we're creating product, our product change is going to be a positive and our moles of our product is too. So this is going to be plus two X. For our equilibrium concentration, which is R E row. We're just going to combine our first two rows and write out what we have. So we have 4.564 minus two X. And then we have 2.801 minus X. And then for a product it's just going to be two X. Okay, so in order to find out what our equilibrium pressures are, we need to solve for X. And because we were told what our total pressure is, we can set everything in our equilibrium row equal to our total pressure to solve for X. So we're going to get our total pressure is equal to we're going to copy everything from our equilibrium row. So we have 4.564 minus two X plus 2.801 minus X plus two X. Is equal to our total pressure which we're told is 5.45 A. T. M. So now what we're doing is we're going to solve for X. So let's simplify this a little bit. Here are minus two X. And r plus two X. Are going to cancel. So we get 4. Plus 2.801 -1. Is going to equal 5.45. Okay, so let's simplify this more. And when we do we get 7.365 minus X is equal to 5.45. So now we can solve for X and we get X is equal to 1.915. So now that we know what our X. Is, we can figure out what our equilibrium pressures are because we have to just plug X in for our equilibrium row. So that means our equilibrium pressure for nitrogen oxide. Our equilibrium row of our ice table says 4.564 minus two X. So we know what X. Is. So let's go ahead and plug it in. So we have 4.564 minus two X. Which is 1.915 gives us an equilibrium pressure of 0.7348 E. M. For our nitrogen oxide. Our equilibrium pressure for our chlorine gas we wrote in our equilibrium row is 2.801 minus X. So let's plug in our X. Here. So we have four or we have 2. -1. gives us an equilibrium pressure of zero 886 A. T. M. And our equilibrium pressure of our product is going to be two times X. So when we plug in our X we get two times 1.915 equals and equilibrium pressure of 3.838cm. Okay, so now that we know what our equilibrium pressures are, we can solve for K. P. By using our equilibrium expression are so that expression is K. P. Is equal to our products over our reactant. Okay so we're going to plug in our pressures that we just calculated and we're going to raise it to the number of moles. Okay so we have what is our product? So it's going to be the pressure of our product which is N. O. C. L. And our reaction shows us that it has a two in front of it. So we're going to raise this to the second power and for our to react ints we have nitrogen oxide and this is going to be raised to the second power times our equilibrium pressure. So let me this should be pressure. Okay, so our pressure of our nitrogen oxide raised to the second power times are pressure of our chlorine gas. And we only have one mole of that so we can write it here just so we see it. But because it's one we don't need to put that in. So let's go ahead and plug in our equilibrium pressures. Okay, so for our product we set our pressure was 3.83. And remember we have to raise this to the second power For our first reactant nitrogen oxide. Our equilibrium pressure is 0.734. And remember we need to raise it to the second power times our equilibrium pressure of chlorine gas is 0.886. Okay. And that's raised to the first power. So we get a K. P. Of 30.73. Okay, so this value is going to be what we plug in to our equation that we wrote up above to solve for K. C. So let's go ahead and do that now. So we were missing that value. So now we plug in 30.73. So once we solve this now we're going to get our K. C. Value and that value is going to be 2.70 times 10 to the three. And this is our final answer. This is the K. C. value. If the total pressure at equilibrium is 5.45 a t. M. That is the end of this problem. I hope this was helpful.

A reaction vessel at 27 °C contains a mixture of SO2 (P = 3.00 atm) and O2 (P = 1.00 atm). When a catalyst is added, this reaction takes place: 2 SO2( g) + O2( g) ⇌ 2 SO3( g). At equilibrium, the total pressure is 3.75 atm. Find the value of Kc.

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