A mixture of CS21g2 and excess O21g2 is placed in a 10.0-L
reaction vessel at 100.0 °C and a pressure of 3.00 atm.
A spark causes the CS2 to ignite, burning it completely,
according to the equation
CS21g2 + 3 O21g2¡CO21g2 + 2 SO21g2
After reaction, the temperature returns to 100.0 °C, and
the mixture of product gases (CO2, SO2, and unreacted O2)
is found to have a pressure of 2.40 atm. What is the partial
pressure of each gas in the product mixture?

Question

A mixture of CS21g2 and excess O21g2 is placed in a 10.0-L 
reaction vessel at 100.0 °C and a pressure of 3.00 atm. 
A spark causes the CS2 to ignite, burning it completely, 
according to the equation 
CS21g2 + 3 O21g2¡CO21g2 + 2 SO21g2 
After reaction, the temperature returns to 100.0 °C, and 
the mixture of product gases (CO2, SO2, and unreacted O2) 
is found to have a pressure of 2.40 atm. What is the partial 
pressure of each gas in the product mixture?

Accepted Answer

Hi everyone. This problem reads a 25 liter reaction vessel containing a mixture of carbon dioxide and excess flooring gas is set up at degrees Celsius and 3.50 atmospheres of pressure according to the equation. Carbon dioxide plus flooring gas yields carbon touch of fluoride plus oxygen, die fluoride. A spark ignites carbon dioxide resulting in its complete combustion. The temperature returns to 125°C after the reaction and the pressure of the mixture of product gasses, carbon toucher, fluoride, oxygen, die fluoride and un reacted flooring gas is found to be 2. atmospheres. What are the partial pressures of the individual gasses in the product mixture? Okay, so the question that we want to answer is the partial pressure of the individual gasses in the product mixture. Okay, so we want to know what the partial pressures are for Carbon Toucher fluoride, oxygen, di fluoride and fluoride gas. Okay, so the first thing that we're going to want to do is we need to write down ideal gas law. Okay, so ideal gas law is PV equals N. R. T. And the first thing that we're going to want to do is we're going to want to calculate the total moles that we're starting with and the total moles that we end with. Okay, so we're going to calculate total moles, initial and final. Okay, so looking at our ideal gas law, we need to isolate our variable end because PV equals NRT represents pressure times volume equals number of moles times r gas constant times T temperature. So let's go ahead and divide both sides by R T. And we're going to get N is equal to P V over R T. So we're going to use this to calculate our initial moles and our final moles. Okay, so I'll go ahead and write down an initial Is equal to. So our initial pressure is 3.50 atmospheres. Our initial volume is 2. leaders and this is all over gas constant R which is 0.08206 leaders, atmosphere over moles, Kelvin Times temperature which our initial temperature is 398 Kelvin. So once we do this calculation, our initial total moles is 2. moles. Okay, so let's go ahead and do the same thing. But for our final moles so and final, our final pressure is Okay, one thing that we regard to do is convert that temperature to kelvin. Okay, so I wrote 398 kelvin and the reason that it's kelvin is because we need that unit in kelvin. So in order to convert degrees Celsius to kelvin, we're going to take 100 and 25 degrees Celsius plus Is 398 Kelvin. We add 273 2°C to convert it to Kelvin. Alright, so let's pick off where we left off. So we're going to convert or we're going to calculate our final total moles. So our final pressure Is 2.85 atmospheres. Our final volume is 2.50 L. And this is over gas constant, R 0.08206 leaders atmosphere over Mole Kelvin. And the final temperature is 398 Kelvin. So our final, our final Total moles is 2.1816 moles. Okay, so now that we know this, what we're going to do is create an ice table. We're going to create an ice table because this is going to help plus determine the amount of moles for each thing in our reaction. And the only way we'll be able to do that is by creating an ice table. So step two is an ice table. Okay, so let's go ahead and rewrite our reaction. So our reaction is we have carbon dioxide gas Plus four moles of flooring gas yields carbon tetra fluoride, Gas Plus two moles of oxygen. Di fluoride. Okay, so we're going to create an ice table. Alright, and in our ice table we're going to let y we're gonna use some variables here, we're going to let why equal are moles of carbon dioxide. So if we let y equals our moles of carbon dioxide, we're going to have 2. -. So this is our initial moles. Remember ice table for ice table I represent initial, so that means 2.6791, which is the value that we got when we calculated our initial moles that minus Y is going to equal our moles of flooring gas. So let's go ahead and plug in our values into our ice table. So for carbon dioxide for I we said we're going to let y equal or moles of carbon dioxide for our flooring gas, we said we're going to let it be represented by 2. minus Y. And we have zero product. Okay, so that means our reaction is moving in the forward direction. So see for our rice table represents change. So that means our reactant are being consumed and reforming product. Alright, so that means we're going to take minus X. Okay, And then minus four X. Because we have four moles of flooring gas and then on the product side we're going to have plus X and plus two X. We have two moles of oxygen die fluoride. The of our equilibrium row is just the combination of our eye and see rose. So when we bring everything down, we get y minus X. For carbon dioxide for flooring gas we get 2.6791 minus y minus four X. For carbon tetrachloride, it's X. And for oxygen die fluoride, it is two X. Okay, So now what we're gonna do is we're going to take the equilibrium row of our ice table and set it equal to the total moles final. Okay so let's go ahead and do that. We're going to take the equilibrium, we're going to take everything in our equilibrium row. So we're gonna take y minus X plus 2.6791 minus Y minus four X plus X plus two X. And we're going to set this equal to the total final. The total most final, which we calculated is 2. moles. Okay so now we're going to someplace fly. Okay so this y and minus Y cancel. Okay and this minus X and plus X cancel. So once we simplify we get 2.6791 minus or X plus two X. Is equal to 2. 816. Okay so we're going to simplify even more so we have minus four X plus two X. Okay so this minus four X plus two X. Becomes negative two X. Alright, so we're going to have negative two X. And we're going to take this 2.6791 and move it to the other side. We're going to subtract it from 2. because the 2.1816 is our total most final. So we're going to move that So we're going to have this is equal to 2. -2.6791. So we're going to take our total most final and subtract that 2.6791 from it. Okay, So remember we're trying to solve for X here. So let's go ahead and divide both sides by -2. Let's simplify first, before we divide both sides by negative two. So we have negative two. X is equal to negative 0. 75. Now we can go ahead and divide both sides by -2. Okay, so we get X is equal to 0.24875. So that was step three, solving for X. So let's go ahead and write that down. So we did our ice table and step three was solved for X. Okay, by solving for X. If you take a look at our ice table, we needed X in order to find out what the equilibrium amount for moles is for each thing. Okay, now that we know what X is, we can solve for the partial pressures of the individual gasses in the product mixture. So we know that X is 0.24875. Looking at our ice table, we see X represents the amount of moles for carp for carbon tetrachloride. Okay, so we can go ahead and say this equals the moles of carbon tetra fluoride. So now let's go ahead and calculate the pressures, the partial pressures for carbon touch of fluoride, oxygen, die fluoride and flooring gas. Okay, so this is going to be our last step. Alright, so let's go ahead and do that and we're going to add a page here. Alright, so we have remember ideal gas law is P. V equals N R T. Now we're going to solve for P because we want the partial pressures. Okay, so we're going to divide both sides by V. And we get pressure is equal to N. R. T over V. So let's start off with calculating the partial pressure for carbon tetrachloride. Okay, so we're going to write the partial pressure for carbon tetra fluoride is equal to N. Which is the number of moles we calculated is so if we go back to our ice table for carbon tetrachloride, the number of moles is represented by X. Okay, and we know that X is equal to 0. moles. Okay, our is our gas constant 0.8 to 06. Leaders atmosphere over mole Calvin temperature is 398 kelvin and this is all over volume, which is 25 liters. So once we do this calculation, we get the partial pressure of carbon tetrachloride is equal to 0.3 to five atmospheres. Okay, so that's one of our answers. So let's go ahead and highlight that. Alright, so now we're going to calculate the partial pressure of oxygen die fluoride. So remember P is equal to N. R. T over T. So our n number of moles. If we go back to our ice table? The number of moles for oxygen die fluoride is two times X. So we need to take two and multiply it by X. Which we said X is 20.24875. So two times X. Is zero point 4975 moles. Okay, Times R is 0.8206. Leaders atmosphere over mole kelvin temperature is 398 kelvin And this is all over volume, which is 25 liters. So our partial pressure of oxygen die fluoride is equal to 0.650 atmospheres. Okay, so we'll highlight that. And let's just make a note here that we got this because we said that the moles of oxygen die fluoride is equal to two X. That's what our ice table says. So we know that X. Is equal to So we're going to take two times zero point 875. Okay, so that's how we got that value for moles of oxygen. Di fluoride, lastly, we're going to calculate the partial pressure of flooring gas. Okay, so from our ice table we know that we need N. P. Is equal to NRT over V. So the moles of flooring gas based off our ice tables equilibrium road is represented by 2.6791 minus Y minus four X. Okay. So let's go ahead and plug in. So we have 2.6791 minus Y y is the moles of carbon dioxide? Remember we said why is the moles of carbon dioxide? And we have It's gonna be two 6791 minus Y. Alright, so we have y is equal to 0.24875. And this is going to be minus four times X and x is 0. 875. Okay, so we get the moles Is going to equal 1.43535. That's the moles of flooring gas. Okay. And now we have that so that we can plug it in. Okay, so let's go ahead and do that. So we get pressure is equal to n which is 1. moles times are 0.8206 liters atmosphere over mole kelvin times, temperature 398 kelvin and this is all over volume, Which is 25 L. So we'll go ahead and write down here that the partial pressure of flooring gas once we do this calculation is equal to 1. atmospheres. Okay, so that is it for this problem. We calculated the partial pressures for the individual gasses and the product mixture and that's it for this problem. I hope this was helpful

Verified Solution

Key Concepts

Ideal Gas Law

Dalton's Law of Partial Pressures

Stoichiometry of Chemical Reactions