Skip to main content
Ch.15 - Chemical Equilibrium

Chapter 15, Problem 87

At 70 K, CCl4 decomposes to carbon and chlorine. The Kp for the decomposition is 0.76. Find the starting pressure of CCl4 at this temperature that will produce a total pressure of 1.0 atm at equilibrium.

Verified Solution
Video duration:
9m
This video solution was recommended by our tutors as helpful for the problem above.
1646
views
Was this helpful?

Video transcript

Hi everyone for this problem, we're told the decomposition of phosphorus Penta chloride produces phosphorous tri chloride, gas and chlorine gas, the equilibrium constant for this reaction at 523 Kelvin is 1.05. If the total pressure at equilibrium is 2.438 cm, calculate the starting pressure of phosphorus Penta chloride at this temperature. So we want to solve for the starting pressure of phosphorus Penta chloride. Okay, so we're going to need to set up an ice table in order to start this problem off. So let's first write out our reaction or we'll rewrite it out from what was given. Okay, and this is going to be our ice table. Set up. The reason I draw this line is to separate our products from our reactant. Okay, so we're going to assume that phosphorus we only have phosphorus Penta chloride present at the beginning. So we'll let why represent our initial concentration? And we are not going to have any products. So that means our reaction is proceeding towards the right, so that means our concentration of reactant is going to decrease in our concentration of products are going to increase. So for a second row for R. C it's going to represent our change in concentration or change in pressure in this case. So because we only have one mole of everything, this is going to be minus X. This is going to be plus X. And this is going to be plus X. R equilibrium row is going to be both of our our first two rows combined. So we have y minus X and then we have X and X. So in the problem they tell us that the total pressure at equilibrium is 2.43. So we're going to set our equilibrium wrote equal to this so that we can solve for why? Because we want to know what is our starting pressure of phosphorus pencil chloride. So we have our total pressure is going to be equal to y minus x. We're just writing out everything in our equilibrium row of our ice table. So we have y minus X plus X plus X. And we're going to set this equal to our total pressure which we're told is 2.43. So now we want to solve for Y here. So let's simplify this a little bit here are minus X and one of our plus X is going to cancel. And so we're going to get Y Plus X is equal to 2.43. And remember we're solving for Y here. So that means why is going to be equal to 2.43 minus x. Okay, so now what we're going to do here is we're going to solve for are partial pressure of phosphorus Penta chloride. So our pressure of phosphorus Penta chloride is going to be equal to let's take a look at our equilibrium bro here, it says y minus X. And we just solved for Y. Right here. So we have Y minus x. So are y is going to be 2.43 minus X. So that's the Y. And then we're going to minus x. Okay, so we just we just plugged in this value for y right here. So y is equal to 2.43 minus x. And then minus X. So if we simplify this this is going to be 2.43 minus two X. Okay so now that we know what that expression is, we can write out our equilibrium constant expression. So that is K C. Is going to equal our concentration of products over our concentration of reactant since and this is each race to the power of its geometric coefficient. So let's take a look at our table. So we're looking at our equilibrium row of our table here. So our products we have X and X. Okay so that means our products is going to equal X squared and our reactant is going to be y minus x. Okay so let me erase some things here so we can have a clearer picture. So we just copied everything from our equilibrium roll into our K C expression. So we have X squared over y minus X. And so now we can simplify this, we know what our K C value is because we were told in the problem that it is 1.5. Okay, so actually this should be K P not K C. Okay, so go ahead and make that correction there. So they tell us our KPR equilibrium constant is equal to 1.05 and we're going to copy what's on the right side is equal to x squared over Y minus X. And we know that our Concentration of our reactant Saar pressure of our reactant we said is 2. minus two X. So we're just plugging in what we know. So now what we're going to do is we're going to simplify this so that we are solving for X. So let's multiply both sides of our equation by 2.43 minus two X. And when we do that we get 1.5 times 2.43 minus two X is equal to x squared. So we can foil this out. And when we foil it out we get 2.5515 minus 2.1 X is equal to x squared. And we want to bring everything over to one side. So we'll get X squared plus 2.1 x minus 2.5515 is equal to zero. So now we can use the quadratic formula to solve for X. So I'll make a note here. So use the quadratic equation to solve for X. And when we use that quadratic equation we're going to get two values for X. We get X is equal to 0.8615 and X is equal to negative 2.9615. So we need to ditch determine which value is correct because pressure can't be negative. We know that the second X is not going to be correct. So our X value is going to be 0.8615. So now we can calculate why because we know what X is. So we said that our why was equal to 2.43 minus X. So now that we know what our X is, we can plug it in to solve for the starting pressure of phosphorus Penta chloride. So we get why is equal to 2. minus the X we just saw for which is 0.8615. So we get a final answer of why is equal to 1.57 A. T. M. So this is going to be our starting pressure of phosphorus pentacle ride. That's the end of this problem. I hope this was helpful
Related Practice
Textbook Question

Consider the endothermic reaction: C2H4(g) + I2(g) ⇌ C2H4I2(g) If you were trying to maximize the amount of C H I produced, 242 which tactic might you try? Assume that the reaction mixture reaches equilibrium. a. decreasing the reaction volume b. removing I2 from the reaction mixture c. raising the reaction temperature d. adding C2H4 to the reaction mixture

1517
views
Textbook Question

Consider the reaction: H2(g) + I2(g) ⇌ 2 HI(g) A reaction mixture at equilibrium at 175 K contains PH2 = 0.958 atm, PI2 = 0.877 atm, and PHI = 0.020 atm. A second reaction mixture, also at 175 K, contains PH2 = PI2 = 0.621 atm and PHI = 0.101 atm. Is the second reac- tion at equilibrium? If not, what will be the partial pressure of HI when the reaction reaches equilibrium at 175 K?

2876
views
Textbook 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.

3381
views
Textbook Question

The equilibrium constant for the reaction SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) is Kc = 3.0. Find the amount of NO2 that must be added to 2.4 mol of SO2 in order to form 1.2 mol of SO3 at equilibrium.

2170
views
Textbook Question

Carbon monoxide and chlorine gas react to form phosgene: CO(g) + Cl2(g) ⇌ COCl2(g) Kp = 3.10 at 700 K If a reaction mixture initially contains 215 torr of CO and 245 torr of Cl2, what is the mole fraction of COCl2 when equilibrium is reached?

3946
views