For the equilibrium Br 2 (𝑔) + Cl 2 (𝑔) ⇌ 2 BrCl(𝑔) at 400 K, 𝐾 𝑐 = 7.0. If 0.25 mol of Br 2 and 0.55 mol of Cl 2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentrations of Br 2 , Cl 2 , and BrCl?

Hey everyone for this problem we're told that the K. c. for the following reaction is 37. If a one liter reaction flask is loaded with 10.550 moles of PBR three and 30.750 moles of B. R. Two. What would be the equilibrium concentration of PBR five when equilibrium is established. So here we're looking for equilibrium concentration. So that means we need to do an ice table. So let's go ahead and rewrite our reaction. So we have P. B. R. Three plus B. R. Two, Gives us p. b. r. five and we need to fill in our initial concentration our change and R. E. So let's go ahead and do that now. So we know that initial concentration is moles over leader. And here they tell us that our reaction vessel is one leader and they tell us how many moles. So let's go ahead and divide our moles by leaders to find the mill arat e. Of our initial concentration. And here it's simple because we're dividing by one leader. So for P. B. R. Three it's going to be 30.550 molar and four br two. It's going to be 20.750 molar. And that's because malaria T. Is equal to moles over leader. Okay, so let's go ahead and plug that in. So we have 0.550 and we have 0.750. And here it's a zero and we're moving towards the right because we are decreasing our concentration of reactant and increasing our concentration of products. So this reaction is moving this way. So that means our change is minus x minus X and plus X. So once we bring everything down we get 0.550 minus x 0.750 -1. And this is just X. Okay, so now that we have our ice table we can go ahead and figure out our K. P. So we know that our K. P. is equal to 37 and r K P. Is concentration of products over concentration of reactant. So using the hero of our ice table we get our K. P is equal to X over 0.550 minus x times 0.750 minus X. Now we can't square root both sides here. So we're going to have to do the quadratic equation which is going to be fair fairly long. But we'll go through it step by step. So we have to fall out our denominator. So when we foil out our denominator we get 37 is equal to X. Over 0.4125 -0.550 x -0.750 X plus X squared. So that is our denominator foiled out. And when we combine This right here this is going to equal negative 1.3 x. So let's multiply both sides of our equation by 37. So we can get rid of our fraction. So we'll do 37 times 0.4125 minus 1.3 X plus X squared is equal to x. And we can foil out our 37 by multiplying it by our 370.4125 or 1.3 and our X squared. So let's go ahead and do that. Once we multiply that out, we get 15 0.262 minus 48.1 X plus 37 X squared is equal to X. Now we need to set this equation equal to zero and the way that we do that is by moving everything to one side. So let's subtract X from both sides. And when we do that we get 37 x squared -49.1 x plus 15 point 262 is equal to zero. So this is our quadratic equation and each thing has a letter assigned to it. So this is our A. This is our B. And this is our C. So now we can use our quadratic equation to solve for R. X. And let's do that over here. So our quadratic equation tells us that X is equal to negative B plus and minus the square root of B squared minus four. All over to a. So this is our quadratic equation and our A B and C below is what we're going to plug into this to solve. So let's go ahead and plug in everything. So we get X is equal to Negative and RB is negative here, so negative 49.1 plus and minus the square root of B squared which is our negative 49.1. And make sure this is squared minus four times A. R A. Is 37 times C 15 point 262. And this is all over two times A. So this is over two times 37. So let's simplify a bit here. Once we simplify, we get X is equal to 49.1 plus and minus once. We simplify what's under our square root that equals 100 and 52.34. And this is all over two times 37 Which is 74. Let's simplify some more. We get X is equal to 49.1 plus and minus the square root of 100 and 52.34 is 12.33. And this is all over 74. Okay, so we just simplified this as much as we can and because we have a plus and minus, we're going to get two answers for X, but only one is going to be correct. So the first answer we're going to get is 49.1 plus 12.33 divided by 74. And when we do that calculation we get X is equal to zero point 830. And then our second one is going to be X is equal to 49.1 minus 12.33 divided by 74. And we get X is equal to 0.497. So now we need to determine which X is correct. Okay, so let's take a look at our ice table. We see that we have an initial concentration of 0.550 and 0.750. Now, For our X, if we subtract .830 that is going to not make sense, right? Because we're going to end up having something that's negative. So for PBR three, for example, if our exes 30.830 and we have 0.550 minus X, That is going to leave us with nothing. And so we need to choose the number that is less than or the X. That is less than the initial concentrations that we're starting with. And so that leaves us with a correct answer of X is equal to 0.497. Now, the question asked, what would be the equilibrium concentration of PBR five? And in our ice table we see that the concentration of P B r five is equal to X. And since we saw for X, we can say that the concentration of PBR five is equal to 0.497 and this is our final answer. This is our equilibrium concentration of PBR 51 equilibrium is established, I hope this was helpful, that's the end of this problem.

Ch.15 - Chemical Equilibrium

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Brown 15th Edition

Ch.15 - Chemical Equilibrium

Problem 57

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Chapter 15, Problem 57

For the equilibrium Br₂(𝑔) + Cl₂(𝑔) ⇌ 2 BrCl(𝑔) at 400 K, 𝐾_𝑐= 7.0. If 0.25 mol of Br₂ and 0.55 mol of Cl₂ are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentrations of Br₂, Cl₂, and BrCl?

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Identify the initial concentrations of Br_2 and Cl_2 by dividing the moles by the volume of the container: [Br_2]_0 = \frac{0.25 \text{ mol}}{3.0 \text{ L}}, [Cl_2]_0 = \frac{0.55 \text{ mol}}{3.0 \text{ L}}.

Set up an ICE (Initial, Change, Equilibrium) table to track the changes in concentrations as the system reaches equilibrium. Let x be the change in concentration of Br_2 and Cl_2 that reacts to form BrCl.

Write the expression for the equilibrium constant K_c: K_c = \frac{[BrCl]^2}{[Br_2][Cl_2]}.

Substitute the equilibrium concentrations from the ICE table into the K_c expression: K_c = \frac{(2x)^2}{([Br_2]_0 - x)([Cl_2]_0 - x)}.

Solve the equation for x, which represents the change in concentration, and use it to find the equilibrium concentrations: [Br_2] = [Br_2]_0 - x, [Cl_2] = [Cl_2]_0 - x, [BrCl] = 2x.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. In this state, the system is dynamic, meaning that reactions continue to occur, but there is no net change in concentration. Understanding this concept is crucial for analyzing how concentrations shift in response to changes in conditions, such as concentration or temperature.

Equilibrium Constant (Kc)

The equilibrium constant (Kc) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced equation. For the reaction Br2(g) + Cl2(g) ⇌ 2 BrCl(g), Kc = [BrCl]2 / ([Br2][Cl2]). This constant provides insight into the extent of the reaction and helps predict the direction in which the reaction will shift to reach equilibrium.

Concentration Calculations

Concentration calculations involve determining the molarity of substances in a solution, which is defined as the number of moles of solute divided by the volume of the solution in liters. In this problem, the initial concentrations of Br2 and Cl2 must be calculated based on the given moles and the volume of the container. These initial concentrations are essential for applying the equilibrium expression and finding the equilibrium concentrations of all species involved.