For the reaction shown here, K_c = 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?

Question

Accepted Answer

Welcome back, everyone. Let's look at our next problem for the reaction shown here. K subscript C equals 255 at 1000 Kelvin. And a reaction is co in the gas form plus CL two in the gas form reaches an equilibrium with COCL two in the gas form. If a reaction mixture initially contains ac O concentration of 0.2500 molar and AC L two concentration of 0.2750 molar at 1000 Kelvin. What are the equilibrium concentrations of Cocl two and COCL two at 1000 Kelvin. And then we have four different answer choices A through D with different concentration amounts. So we'll return back to them once we've worked through our math here. So we've got an equation and we have initial concentrations of some CO and some CO2. So our left side and nothing zero COCL two. So we only have one direction we can go if the reaction's going to proceed at all, which is to produce more product. So to more product, less reactants. So we know the change that's going to occur here. We have multiple unknown equilibrium concentrations. So we can't just solve for the desired concentration. We need to use the ice method. So that involves setting up our chart. So we first decide how many of our species go on to the chart. Do we have any pure liquids or p pure solids that we wouldn't consider? No, we don't. All of our products and reactants are in gas form. So they all grow across the top of our chart. So concentration of co concentration of CL two and concentration of Cocl two are our columns here and then down the left side to make our rows I ce so our ice chart. So our first row here I for the initial concentration for co we have 0.2500. I'm just going to put the number there just because I have a small box. It is units of molar. Of course, CL two, our initial concentration is 0.2750 molar. And of course, for COCL two, our initial concentration is zero. So when we think about the change, as we said, we will be losing reactants gaining product. So our change for co and CL two will be minus X fill that in and minus X and our change for coc two will be plus X. So then our E row equilibrium concentration four C will be 0.2500 minus X four C 20.2750 minus X. And for CC two, just X. So now with those, we have three expressions, there are three unknown concentrations. So we can go ahead and set up an equation based on our equilibrium constant. So we have KC, which equals 255 will equal the concentration of our product. So that would be X in the name rainer over the concentrations of our reactants. So in the denominator have in parentheses, 0.2500 minus X multiplied by 0.2750 minus X. So now we have to think is there a method of simplifying this or do we need to go to the old quadratic equation to solve it? Well, we don't have perfect squares in the numerator and denominator. So that's out. So let's just test, can we use a 500 approximation here? So 500 approximation, we need to look at dividing our initial concentration. In this case, we'll use the initial concentration of co over our equilibrium constant. So that will equal 0.2500. And we're going to divide that by 255. Well, I think that's pretty obvious to see it's much going to be much less than 500. It'll be about 0.001. So definitely less than 500 we can only use that approximation if this gives us a number greater than 500. So that's a big old no, on our 500 approximation. So we need to head into our quadratic formula or a quadratic equation. And we recall for that, we need our equation that we have up here to be equal to A X squared plus BX plus C equals zero. So we need it in that form. So let's scroll down and work on rearranging this equation. So first, we'll go ahead and multiply both sides by our denominator to, we don't have a fraction there. So we end up with 255 multiplied by 0.2500 minus X in parentheses, multiplied by 0.2750 minus X in parentheses equals X. So now in my next step, I'm going to expand those terms using boil and I will also subtract X from both sides so that I end up with that equal to zero on the right side. So I've got 255 and then multiplied by in parentheses, 0.06875 minus 0.2500 X minus 0.2750 X plus X squared, close parentheses minus X equals zero. So we'll go ahead and combine our terms our X terms in there. And then in the parentheses, we've got 0.06875 minus 0.525 X plus X squared, all of that times 255 minus X equals +01 thing to note in the previous step, be careful since you have two terms that are minus X that you make sure you end up with plus X squared. You're multiplying two negatives, easy to make a mistake that way. So now let's do that 255 multiplication there. And I've scrolled up a bit again. I have 17.53125 minus 133.875 X plus 255 X squared minus X equals zero. Let's combine our X terms. Please be careful. You've got two negative terms. So make sure you're taking that into account. And we've got 17.53125 minus 134.875 X plus 255 X squared equals zero. So we have the form we need just written in the order Cbxax squared. So I'm just going to put brackets around each of these terms and label them with A B and C to be used in my quadratic equation. So the 17.53125 has no X. It's RC term or C coefficient. Now notice this X coefficient has a negative sign. That's why I like to put these brackets here. So I don't lose that negative. So negative 134.875 is my B coefficient of X and then 255 is my A coefficient of X squared. So especially since these are written in backwards order rather than rewriting the whole thing I just label my A B and C. So let's remind ourselves of the format or of that formula, we're solving this equation which will be X equals negative B plus or minus the square root of B squared minus four ac divided by two A. And of course, remind ourselves that with that plus or minus, we will get two possible solutions. We have to go back to our original concentrations to determine which can actually be a possible solution. So we'll plug in our numbers here. So I've ended up with X equals 134.875. It's a positive since B with a negative number plus or minus the square root of. And then we've got negative 134.875 squared minus four multiplied by 255 multiplied by 17.53125 all that divided by two, multiplied by 255. So I'm going to simplify the expressions here. And for my two possible solutions, I have X equals 134.875 plus or minus 17.5895 divided by 510. So let's expand out each one or simplify each one to find our two possible X values. And when we do that, our first possible solution is X equals in parentheses 134.875 plus 17.5895 all that divided by 510 which comes out to 0.2990. Then my other possible solution X equals then in parentheses, 134.875 minus 17.5895 divide all that divided by 510 comes out to 0.2300. So now what we need to do to decide if these are possible solutions is to look back at our original concentration and see if this makes sense. So we know that our equilibrium concentration will be our initial concentration of our reactants minus X. So when we look back up at the E in our chart which still shows at the top of my uh screen here. So e and we can all recall that first column was a concentration of co a 0.2500 minus X and of the, the L two is 0.2750 minus X. However, we use our first solution where X equals 0.2990. This will give us a negative equilibrium concentration of our reactant will say that will generate negative uh reactants, which just doesn't happen in real life. So X the first X with the addition is not a possible solution. So therefore our solution here for X now note it's not the solution for the whole problem is our value of 0.2300 which is less than both of our initial concentrations. So that's a possible answer in the real world. So that's our X factor. But we still need to solve for our concentrations of our products and reactants. So we'll scroll up just a little bit and we're starting to run out of space here. So I'm just going to transfer down here. So we can see down below that we remind ourselves over in the corner, I'll draw a little square drawn from the E row of our chart that are equilibrium concentrations of our products and reactants, which is what we are looking for. We scroll up. Remember we're looking for equilibrium concentrations of COCL two and COCL two. So we determined from our ice chart that those concentrations were the concentration of CO 0.2500 minus XCL 2.2750 minus X and COCL two X. So we'll scroll up. Now we have that little box to refer to and using the X value that we solved for, we can calculate each of those concentrations. So we have our concentration of C and again, this is all at equilibrium. It's going to equal 0.2500 molar minus 0.2300 molar. We'll put the units back in now that we're coming to the end, which is going to equal 0.020 muller. And then for CL two, we have 0.2750 molar minus 0.2300 molar or X value. And we get 0.045 molar. And finally, the OCL two, the easiest one of all because it just equals X. So that would be 0.2300 molar. So here's our final answer. And our last step is just to go back to our multiple choice options and match this up with one of them, going to have to scroll up again. So we'll go back and forth a little bit. Let's start with COCL two equilibrium concentration because that is the easiest to remember there. 0.2300 molar. Scroll back up to our answer choices and looking at our COCL two concentrations. Choice A has 0.25 molar. So that's incorrect. Choice B has COCL two equilibrium concentration, 0.23 molar. So B is correct in that respect choice C so B is possibly correct. C says COCL two is 0.19 molar. So that's incorrect. And choice D is COCL two is 0.15 molar. So all of those have an incorrect value of X. Our only correct value of X is in choice B. So that is our answer. We can just double check our other answers. Co 0.020020020 molars which is correct. And cl two's concentration 0.045 molar also corresponds to what we calculate them. So after all that process. We have come to our final answer. We've determined our equilibrium concentrations. Everything here was at 1000 Kelvin. And we end up with a concentration of equilibrium concentration of co as 0.20 molar. An equilibrium concentration of cl 20.045 molar, an equilibrium concentration of COCL two as 0.23 molar. So answer choice B is our answer. See you in the next video.

For the reaction shown here, K_c = 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?

Verified Solution

Key Concepts

Equilibrium Constant (Kc)

ICE Table (Initial, Change, Equilibrium)

Le Chatelier's Principle

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?

Verified Solution

Key Concepts

Equilibrium Constant (Kc)

ICE Table (Initial, Change, Equilibrium)

Le Chatelier's Principle

For the reaction shown here, K_c = 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?