For the reaction shown here, Kc = 255 at 1000 K. CO(g) + Cl2(g) ⇌...

Question

For the reaction shown here, Kc = 255 at 1000 K. CO(g) + Cl2(g) ⇌ COCl2(g)
If a reaction mixture initially contains a CO concentration of 0.1500 M and a Cl2 concentration of 0.175 M at 1000 K, what are the equilibrium concentrations of CO, Cl2, and COCl2 at 1000 K?

Accepted Answer

Hi everyone for this problem, we're told to consider the following reaction at degrees Celsius in a reaction vessel 2.50 moller S. 02 gas and 2.25 moller of cl two gas were mixed, calculate the concentration of all species once equilibrium is reached at 375 degrees Celsius. Okay, so our goal is to calculate the equilibrium concentrations. Okay, so we're going to need to set up an ice table in order to solve this problem. So let's go ahead and rewrite our reaction. So we have s. 0. 2 gas Plus cl two gas Yields S. 0 2, CL two gas. Okay, so I'm just going to draw a line here to separate our products from our reactant, and we're going to write out ice on the side. So R. I stands for initial concentration. So we need to put our initial concentrations in this row and we're told that we have an initial concentration of 2.50 molar for our S. 02 gas and we have 2. molar for a cl two gas. So that means we have zero product. And so right now our reaction is not at equilibrium. In order for it to establish equilibrium, it's going to shift in the four direction to make product. Okay, And so that means our concentration of reactant is going to decrease and our concentration of products is going to increase. So r c ro of our ice table represents that change. And so for our reactant, so we're going to have minus x. And for our product we're going to have plus X, we only have one mole of everything, so we don't have to add any coefficients here for equilibrium row, we're just going to add up our first two rows and bring it down. So we have 2.50 minus X and then 2.25 minus X. And then just X. So in order for us to calculate the equilibrium concentrations, we're going to need to figure out what is our X. Once we find out our value of X, we can plug it into our equilibrium row to solve for our equilibrium concentrations. So To solve for X, we're going to need to write out our equilibrium constant and its expression so that is KC is equal to the concentration of our products over the concentration of our reactant in the problem. They tell us what our KC value is, our equilibrium concentration. So that Casey is 25.9. Let's go ahead and put that here. So we have 25.9 is equal to concentration of products overreacting. So we're plugging in what's in our ice table for our equilibrium row here. So our product is S 02 cl two and that is represented by X and for our reactant we have two. So we're going to say it is 2.50 minus x, times 2.25 minus x. Ok. And so now our goal is to solve for X. So the first thing that we can do is check to see if the excess and our denominator are negligible. If they are negligible then we can go ahead and not consider it and this. So we're looking at these two bottom X is here. So the way that we figure out if it's negligible or not is by looking at Our initial concentration and dividing it by RK. C. Okay, so our smallest equilibrium concentration is 2.25. And when we divided by RKC of 25.9, we're going to get a value if the That value is greater than 500 than X is negligible, but if it's less than 500 then X is not negligible. So here we're going to get a value that's less than 500. So that means that X is not negligible. So we have to consider it. Okay, so let's go ahead and solve for X here. So we're going to simplify here and foil out our Denominator. So we're going to get 25.9 is equal to X over X squared minus four point 75 X plus 5.625. So we want to isolate X here. So we're going to multiply both sides of our equation by our denominator. Okay, so we're going to have 25.9 times eh X squared minus 4.75 X plus 5.625 is going to equal X. So now we want to set this equal to zero. So we're going to move that X on the right side over to the left side. So let's foil out first. What we have on the right side. When you boil it out we get 25.9 X squared minus 123.25 X plus 45.685 should be a + 45.68 equals X. So we're going to add this X on the right side over to the left side. And when we do we get 25.9 X squared minus 124. X plus 45.6875 equals zero. So now what we can do is we can use the quadratic equation to solve for X. And when we use the quadratic equation, we're going to get two values for x. So let's go ahead and do that. So the two values we're going to get is x equals 2. And x equals 2.066. So only one of these values can be correct. And what we need to do is look at our initial concentrations are value for X cannot be greater than the initial concentrations. So our 2.066 is less than both of our initial concentration. So this is going to be the correct value. The 2.7-2 is greater than our initial concentrations, so this is not going to be correct. Okay, so now that we know what the value of X is, we can go ahead and plug that X into the equilibrium row of our ice table to find out what is the equilibrium concentration for all of our species in our reaction. So let's go ahead and do that now. So for our concentration of S. O. To our equilibrium row of our table says it's going to be 2.50 minus X. So we just solve for X. So let's go ahead and plug that in. So we have 2.50 minus 2.66. And this is going to give us an equilibrium concentration of 0.434 moller for S. 02 for cl To our equilibrium rove our ice table tells us it's going to be 2.25 minus X. So let's plug in X here. So we're going to get 2.25 minus 2.66 is equal to 0.184 molar. And for our S 02 cl two, our equilibrium roll of our dice table says it's going to equal X. And so that is going to equal 2.7 molar. So this is going to be the equilibrium concentration for all of our species. Okay, so for S. 02, it's going to be 0.434 moller. For cl two, it is 0.184 moller, and for S 02 cl two, it is 2.7 moller. That's the end of this problem. I hope this was helpful.

For the reaction shown here, Kc = 255 at 1000 K. CO(g) + Cl2(g) ⇌ COCl2(g) If a reaction mixture initially contains a CO concentration of 0.1500 M and a Cl2 concentration of 0.175 M at 1000 K, what are the equilibrium concentrations of CO, Cl2, and COCl2 at 1000 K?

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