A 0.831-g sample of SO3 is placed in a 1.00-L container
and heat...

Question

A 0.831-g sample of SO3 is placed in a 1.00-L container 
and heated to 1100 K. The SO3 decomposes to SO2 and O2: 
2 SO31g2 Δ 2 SO21g2 + O21g2 
At equilibrium, the total pressure in the container is 1.300 
atm. Find the values of Kp and Kc for this reaction at 1100 K.

Accepted Answer

Hi everyone for this problem. It reads in a two liter container 20.372 g sample of ammonium cyanide is heated to 870 kelvin, it decomposes into ammonia and hydrogen cyanide. What is the K. P. And K. C. Value it reaches if the total pressure of the container at kelvin is 8700.467 A. T. M. Okay, so what we want to do for this problem is find out the value of K. P. And K. C. These are both equilibrium constants. K C. Is defined by the molar concentrations of the gasses and K. P. Is defined by the partial pressures of the gasses. So let's go ahead and start off with K. P. Okay, so for us to find out the value of K. P, we need to write out our K. P. Expression and that expression is K. P. Is equal to the partial pressure of our products over the partial pressure of our reactant. So let's just take a look. So this is going to be the partial pressure of our products over the partial pressure of our reactant. Looking at our equation or reaction, we see that it's going to equal the partial pressure of ammonia times the partial pressure of hydrogen cyanide over the partial pressure of our ammonium cyanide. Okay, so in order for us to find out the K. P. We're going to need to know our partial pressures. Okay, and we're going to use the equation. We need to do an ice table because our ice table is going to give us the value that is going to help us solve for our K. P. Okay, so what an ice table. When we write out our rice table, we're going to write out our reaction. So let's take a minute to do that. I normally like to draw this line here. Okay, so our iro represents our initial, our initial pressure. Okay, so we don't know what our initial pressure is. So we need to find out that value. The equation that we can use to find out our initial pressure is our pressure is going to equal grams times are our constant times temperature over molar mass times volume. So all of these, all of this was given in the problem. So we just need to plug in so we can find out our initial concentrate our initial pressure. Okay, so we were told we have a 0. gram sample of our ammonium cyanide. Okay, so that is going to go here. Okay and then R is a constant that we should know. This is our universal gas constant, 0. Leaders Times atmospheric pressure over mole times kelvin. And this is going to be multiplied by our temperature in kelvin. So we need to make sure our units all match up and they do our temperatures in Calvin. And yes, so let's go ahead and move over to the denominator. We have our molar mass times volume. Our molar mass of ammonium cyanide is 44. g per mole. And our volume is two leaders. Okay, so we need to make sure our units cancel and because we're looking for a partial pressure are only unit remaining. Should be atmospheric pressure. Okay? So our grams cancel here. Our moles cancel our leader cancel, Calvin cancels. And we're left with just atmospheric pressure, which is what we want. Okay, so now we know our initial pressure Is equal to 0.3014. So we can go ahead and plug this into our ice table. Okay, so we have 0.3014. Its atmospheric pressure and we have zero for our product. So what this means is our reaction is going to want to shift to the right to establish equilibrium. Okay, Because we have no product. And so we're going to shift to the right. R. C. Represents our change. Okay? So we're going to look at the number of moles present. So we only have one mole of everything and we have X represent our change because our reaction is moving to the right, that means our concentration of reactant is going to decrease. So we put a minus here and an X. And then our reactant or our products is going to increase. So we put a positive and we only have one mole of everything. So that just remains X. E. Row of our ice table is our equilibrium room. So we're just going to combine the first two rows. So 0.3014 minus X. And this is just X. Okay, so our equilibrium row of our ice table is the row that's going to help us figure out our K. P. Value. But we don't know what the value of X is. So we need to solve for X in the problem. We're told that The total pressure of the container is .4678cm. So if we set up the equilibrium row of our ice table to equal our total pressure, we can solve for X. Okay, so let's go ahead and do that. So we know that our total pressure is equal to our equilibrium row. So we're going to take everything from our equilibrium rose. So 0.3014 minus X plus X plus X. And we're going to set this equal to our total pressure of 0.467 atmospheric pressure. Okay, so we can simplify here. This minus X and plus X will cancel. And so we'll have 0.3014 Plus X is equal to 0.467. Okay, so we can solve for X. And when we solve for X, we get X is equal to 0.1656. So now that we know what X is. We can take a look at our equilibrium row of our ice table and plug in X. To find out our pressures for our reaction and then plug in those values into our K. P. Expression that we wrote up above. Okay so let's go ahead and find out what is our equilibrium partial pressure for everything that is in our reaction. So we can start off with our ammonium cyanide. Okay, so our partial pressure of ammonium cyanide Based off the equilibrium of our ice table is 0. -1. We know what X. Is. So let's go ahead and plug that in. Okay, so we get the partial pressure of ammonium cyanide is 0.1358. So let's go ahead and do that for everything. So we have for ammonia. Our ice table says that it's just represented by X. So this is going to be the same for our hydrogen cyanide because they're both just equal to X. Okay, so that means the equilibrium pressure is 0. are ammonia and hydrogen cyanide. And this is both atmospheric pressure. Okay, so now that we know what those values are, let's go ahead and plug those values into our K. P. Expression. Okay, so we can make this a little smaller. So we have room. So we said our K. P. Is equal to the partial pressure of our products over the partial pressure of our reactant. So we have K. P. Is equal to 0.1656 times 0. over 0.1358. Okay, so once we do that calculation we get KP is equal to 0.202. Okay, so that is the first answer to our question. So our second part of the question asks us for K C. Okay, so remember we said K C represents our molar concentrations. It's defined by molar concentrations and there is an equation that relates K P and K C. Okay, so that equation is K P is equal to K C. Times r gas constant R times r temperature in kelvin raised to our change and moles. Okay, so R is a gas constant that we should have memorized and that is R is equal to let me do it in a different color. R is equal to zero .082 06 leaders times atmospheric pressure over mole times kelvin. And our change in N is our change in mold. So we're going to look at our final our product our moles of product minus our moles of reactant. So we have two moles of product and one moles of reactant. So our change in moles is one. Okay, and because we're interested in K. C. Here, we need to isolate K. C and rearrange this equation. And so when we when we rearrange it to solve for K C, we get K C. Is equal to K P over are times T times delta N. Okay, so we already know what R K. P value is. We know what R and T are and we know what our change in moses. So we have everything that we need to solve for K. C. So we get Casey is equal to zero .202 divided by 0. leaders times atmospheric pressure over mole times Calvin. And this is multiplied by our temperature, which we're told is 875 Calvin. And remember we have to raise this to our change in mold. So this is just one. So we get our K C. Is equal to 2.32 point Times 10 to the -3. Okay. Alright. So that is our value for K. C. So we're able to solve for both and we'll just right here. So it's easy to see KP is equal to 0. and KC is equal to 2. times 10 to the negative third. These are our final answers. Okay. And that is the end of this problem. I hope this was helpful

A 0.831-g sample of SO3 is placed in a 1.00-L container and heated to 1100 K. The SO3 decomposes to SO2 and O2: 2 SO31g2 Δ 2 SO21g2 + O21g2 At equilibrium, the total pressure in the container is 1.300 atm. Find the values of Kp and Kc for this reaction at 1100 K.

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