At 700 K, the equilibrium constant for the reaction
CCl41g2 Δ C1s...

Question

At 700 K, the equilibrium constant for the reaction
CCl41g2 Δ C1s2 + 2 Cl21g2
is Kp = 0.76. A flask is charged with 2.00 atm of CCl4,
which then reaches equilibrium at 700 K. 
(b) What are the partial pressures of CCl4 and Cl2 at equilibrium?

Accepted Answer

Hi everyone for this problem. It reads consider the following reaction at 100°C. We have di nitrogen tetroxide yields nitrogen dioxide at equilibrium. A reaction vessel initially contains 17.1 atmospheric pressure of di nitrogen tetroxide, calculate the partial pressures of di nitrogen tetroxide and nitrogen dioxide once the reaction reaches equilibrium. Okay, so our goal here is to calculate the partial pressure of di nitrogen tetroxide and nitrogen dioxide specifically when it reaches equilibrium. Okay. And so one important thing that was given here is the K. P. Value and K P represents our equilibrium constant and K. P. Is defined by the partial pressures of the gasses in our system. So our equilibrium constants have an expression that we write out. And so we know our K. P value is equal to 6.49. And what this translates to is our partial pressure of our products over the partial pressure of our reactant raised to their stoke eO metric coefficients. So let's take a look at our reaction. We have our partial pressure of our products, our product is this nitrogen dioxide. Okay, So we're going to put its the partial pressure of nitrogen dioxide raised to its Tokyo metric coefficient. So we have two moles of nitrogen dioxide. So we're going to raise this to the power of two and this is over our partial pressure of reactant. So we have di nitrogen tetroxide as our reactant and so it's going to equal the partial pressure of di nitrogen tetroxide raised to its Tokyo metric coefficient. We only have one mole of that. So that is just raised to the power of one. Okay, so this is what's going to allow us to find the partial pressures but we don't know what those partial pressures are. And the only way that we'll be able to do that is by creating an ice table. Okay, So let's go ahead and make our ice table where we write out our reaction, Okay? And we'll write out ice on the side. So the iro of our ice table represents our initial concentrations, our initial concentration. And in the problem we're told our reaction vessel initially contains 17.1 atmospheric pressure of di nitrogen tetroxide. So we'll go ahead and plug that in. So we have 17.1 atmospheric pressure and we have no product. So that means our reaction is going to shift to the right to create product. We're going to by shifting to the right, decrease our concentration of reactant and increase our concentration of products. So r C row of our ice table represents our change. And so we need to pay attention here to our stoke stoke geometry. So because our reactant is deke, it's going to be represented by minus and we only have one mole. So it's going to just be minus X. However, for our product we said we're creating product and so this is going to be represented by a plus sign and we have two moles. Okay, so this is going to be two X. Our equilibrium row is going to be the combination of the first two rows. So 17.1 minus X. And then this is just two X. Okay, so now that we have our ice table, our equilibrium row is what represents our partial pressures. Okay, so in order for us to calculate our equilibrium partial pressures, we need to figure out what is X. So that's going to be the next thing that we're going to do. So we're going to rewrite out R. K. P expression and plug in our equilibrium row into that expression. So we set our K P which is equal to 6. is equal to our partial pressure of our products over our reactant. So for our product we have nitrogen dioxide and the equilibrium row is represents two X. And we're going to raise this to the second power. Okay, over our partial pressure of our reactant which is represented by 17.1 minus X. Okay, so here we see, we have 6.49 is equal to two X squared over 17.1 minus X. We have what we need to solve for X. So let's go ahead and do that. But first we can determine whether or not X is negligible. So if X is negligible, that means we can ignore the X and our denominator and Just have 17.1 but we need to first see if it's negligible. It's negligible. If the value that we're about to calculate is greater than 500. So we're going to take our initial concentration so 17.1 divided by RK. P. Which is 6.49. Okay, so this is equal to or this is less than 500. So that means X is not going to be negligible. Okay, so X is not negligible. So that means we cannot ignore it. Alright, so let's go ahead and filter out here. So we have 6.49 is equal to four X squared over 17.1 minus X. So our goal here is to solve for X. So let's go ahead and multiply both sides of our equation by 17.1 minus x. So we get 6.49 times 17. minus x is equal to four X squared. We can filter out the left side. When we filter out the left side we get 100 and 10.979 minus 6.49 x is equal to four X squared. So we can go ahead and set this equal to zero by moving everything over to one side. And when we do that we get four x squared plus 6.49 x minus 110.979 is equal to zero. So with this we have we can use the quadratic formula to solve for x. Okay, so we'll use the quadratic formula. Okay and the quadratic formula is X is equal to negative B plus or minus the square root of B squared minus four A C. All over to a. So looking at our equation four X squared represents a 6.49 X represents B. And a negative 110.979 represents. See so we can just go ahead and plug everything in here. So we get X is equal to negative 649 negative 6.49 plus and minus 6.49 Squared -4 times four times negative 110.979. And this is all over two times a. So that's two times four. Okay, So if we simplify this a bit we get X is equal to negative 6. plus and -42.6355. All over eight. Okay, so we're going to get two values here since we have a plus and a minus. So the two values we're going to get is 4.5182. And X is equal to negative 6.1407. So our X can't be negative and so we can see here that our X. Is going to be 4.5182. Okay, because we can't have a negative concentration of something so that is our correct X. And remember we needed this value of X so that we can plug it in to our equilibrium row of our ice table to solve for those partial pressures. So now that we know the value of X. Let's go ahead and finish this off by solving for our partial pressures, which is what we were asked to solve for. So we know that our partial pressure of our of our di nitrogen tetroxide Based off the equilibrium or over ice table is 17.1 -1. So that is equal to 17. -4.5182. So our partial pressure for di nitrogen tetroxide Is equal to 12. atmospheric pressure. Okay. And we'll do the same thing for our nitrogen dioxide. So our partial pressure of nitrogen dioxide based off of our equilibrium rover ice table is equal to two X. So let's go ahead and plug in X here. So we get two times 4.5182, which gives us a partial pressure for nitrogen dioxide equal to 9. atmospheric pressure. Okay, so that is it for this problem. This is the answer to our problem. I hope this was helpful

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