Consider the hypothetical reaction A(𝑔) + 2  B(𝑔) ⇌ 2 C(𝑔), for which 𝐾_𝑐 = 0.25 at a certain temperature. A 1.00-L reaction vessel is loaded with 1.00 mol of compound C, which is allowed to reach equilibrium. Let the variable x represent the number of mol/L of compound A present at equilibrium. (e) From the plot in part (d), estimate the equilibrium concentrations of A, B, and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

Video transcript

Hi everyone for this problem, we're told to consider the following hypothetical reaction. A one liter flask is charged with 1.50 mol of compound N. And allowed to reach equilibrium. The equation derived from the art ice chart is a cubic equation that cannot be solved in closed form, estimate the equilibrium concentrations of L. M. And N. By plotting the cubic equation. Note, make sure that the value of X will produce a positive concentration. Okay, So our goal for this problem is to estimate the equilibrium concentration of L. M. And N. And we're told to do so by plotting the cubic equation that we're going to derive. Okay. And so because we're dealing with a system that's at equilibrium and we're told the equation was derived from the ice chart, let's go ahead and create an ice chart for our reaction so that we can solve for our equal Librium concentrations. Okay, So we'll start off by rewriting out our reaction to create our ice table. Okay? And we'll go ahead and write out ice. And I normally like to draw a line here to separate our product from our reactant. Okay, so RI ro of our ice table represents initial concentration. So our initial concentration needs to be in moles over leader and we can solve based off of what was given in the problem. So we're told we have 1.5 mol of N. And we also have the volume of the flask. So that means we can solve for the initial concentration of N. And our initial concentration of N is going to equal 1.50 mole over one leader gives us a mill arat, E or a concentration of 1.50. So we can go ahead and put that into our ice chart for our initial concentration of N. Ok. And we have zero reactant. So what this means is our equilibrium system is going to want to shift to the left. Okay. And so what that means is our concentration of products will decrease and our concentration of reactant will increase. So for r C ro of our ice table, we're going to reflect that change by paying attention to how many moles we have. So for our product and we have two moles of that and our change is decreasing. So we have minus two X. For our m we just have one mole. So this is going to be plus X. And for R N for L we have two moles of it. So this is going to be Plus two X. Okay. And so our hero of our equilibrium concentration is a combination of our first two rows. So when we combine we have 1.50 minus X. And then we have X. And we have two X. Okay, So we were given a note in this problem and it said to make sure that the value of X will produce a positive concentration. So that means when we look at the equilibrium row for N we see that X is going to have to be less than 0.5 point 75. So that 1.50 minus two X will not be less than zero. Okay, so let's go ahead and write out our K. C expression because R K C expression is what's going to allow us to get that cubic equation. Okay, so R K C expression is our equilibrium constant. And what our equilibrium constant is defined by is by Molar concentration. So RKC. Which we're told the value is 0.85. We're going to set that equal to our concentration of products over our concentration of reactant since Okay, and we're paying attention to the equilibrium row of our ice table here. So what we see in this row is what we're going to plug in so that we can solve for For X. Okay, so let's go ahead and plug in. What's in our ice chart. Okay, so we're gonna have 0. is equal to our concentration of N. And we need to raise it to its Tokyo metric coefficient. So we have we have two moles of that. So this is going to be raised to the power of two over our concentration of reactant. So for our concentration of em we just have one mole of that. So we'll leave the exponent that way. And then for l we have two moles. So we'll raise this to the second power. Okay, so let's go ahead and plug in these values from our ice table. And so we have 1.50 - x squared over two X squared Times X. Okay, so we can go ahead and simplify this. So we get 0.85 is equal to 2. minus 3.0 X plus four X squared. And this is all over four X. Cute. Okay, so our goal is to filter out and solve for X here and we can do that by simplifying this this equation out. Okay, and when we simplify this equation out, we're going to get the cubic equation 3. X cubed minus four X squared Plus 6.00 x minus two point 225. Y. Okay, so that is going to be our cubic equation. And from that we're able to plot this cubic equation. And when we plot this cubic equation, this is what we're going to get. Okay, So that 2.225 Y is going to be equal to zero. Okay, so when we plot this out, what we can see from this graph Is that X is equal to 0.46. Okay, so we know that X is equal to 0.46 smaller. So now that we know the value of X based off what we just plotted. We can go ahead and find out the equilibrium concentrations for everything in our reaction. By plugging this value of X into the equilibrium row of our ice table. So let's go ahead and do that. The equilibrium row of our ice table tells us that the concentration of L. Is equal to two X. So if we plug in X we get our concentration of L Is equal to 0.92 molar. Okay. Our ice table tells us that our concentration of M. Is equal to X. So that means our concentration of M is equal to 0. molar. Okay. And our ice table tells us that our concentration of N is equal to 1.50 minus X. So we have 1.50 minus. No it's 1.50 two X. Excuse me. So our ice table should have two x reflected Not -1. Okay, so this is minus two X. Okay, so we have 1.50 minus two times 0.46. So that gives us the concentration of N equal to 0.58. So this is going to be our final answers are estimated equilibrium concentrations. Okay, so for L it is 0.92 moller four M. It's 40.46 Mohler and for N it's 0.58. That's the end of this problem. I hope this was helpful