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When 3.00 moles of A and 3.00 moles of B are placed in acontainer and allowed to come to equilibrium, the resulting mixtureis found to contain 0.50 moles of D. What...

Question

When 3.00 moles of A and 3.00 moles of B are placed in acontainer and allowed to come to equilibrium, the resulting mixtureis found to contain 0.50 moles of D. What are the amounts of A andB at equilibrium? The reaction equation is: A (g) + 3B (g) ⇄ C (g)+ D (g)A. 2.50 mol of A, 2.50 mol of BB. 3.50 mol of A, 3.50 mol of BC. 2.50 mol of A, 1.50 mol of BD. 3.50 mol of A, 4.50 mol of BE. None of the above

When 3.00 moles of A and 3.00 moles of B are placed in a container and allowed to come to equilibrium, the resulting mixture is found to contain 0.50 moles of D. What are the amounts of A and B at equilibrium? The reaction equation is: A (g) + 3B (g) ⇄ C (g) + D (g) A. 2.50 mol of A, 2.50 mol of B B. 3.50 mol of A, 3.50 mol of B C. 2.50 mol of A, 1.50 mol of B D. 3.50 mol of A, 4.50 mol of B E. None of the above



Answers

Given: $\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \rightleftharpoons \mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{g})$ When one mole of $\mathrm{A}$ and one mole of $\mathrm{B}$ are mixed and allowed to reach equilibrium at room temperature, the mixture is found to contain $\frac{2}{3}$ mole of $\mathrm{C}$. (a) Calculate the equilibrium constant. (b) If two moles of A were mixed with two moles of $\mathrm{B}$ and allowed to reach equilibrium, how many moles of $\mathrm{C}$ would be present at equilibrium?

Or problem. Um 16.65 we have um the same equilibrium as 64 A plus B gives a plus D. And they asked you to calculate the equilibrium constant when one more of A and one more B is allowed to mix. To reach equilibrium the room temperature. And in the end you found that the equilibrium of this sea is to over three. And to better visualize this, I'm gonna put our information into a icebox. Um You add excellent products and subtract X to the reactant. And to find see Casey, we need to use the equilibrium concentration which is here. So uh this is a concentration of the types of concentration, see tons of concentration of a terms of concentration and G. And all of this is the equilibrium concentrations. And we know that the equilibrium concentration of C. Is to over three. And since you're adding X here, we can just calculate do simple maths and calculate that X is also to over three. So likewise, your d concentration would also be to over three. So when you calculate this uh you get to over three squares as the products and then um one minus 2/3 is just 1/3. So you get 1/3 squared here. And when you saw this fraction you get that your answer is for mm. Yes. Okay. So in part the they ask you if two modes of a is mixed with two modes of being allowed to reach equilibrium, how many moses he will be present at equilibrium. And also um I'm using the since the problem gives you most and um I'm gonna use most because um the Casey is just a ratio. So if you think like if you divide by any content of any type of volume you would still get like the same answers. It is just a ratio. So since they don't give you the volume, I'm just gonna let's say assume that it's one leader, it doesn't really matter what volume it is, as long as you stay consistent with your most unit. So yeah, so for the part, the onions set up another ice box and we have To most of a. and two most b. And for C. Um you just have 00 so you have plus X plus x. Okay sorry this would be minus X minus x, M plus X and plus x. Um Right okay so you plug this into our Casey formula which is X squared over two minus X square. And we found Casey in part is force of this way equal four. And to get with a square were you square? Were both sides? And you get X over two minus X equals two. And you simplify the proportion, you get X equals four minus two X. And you get the X equals 4/3, and you just plug this and back into the equilibrium expression of C. And you get that c concentration of our uh equilibrium, we have 4/3 mose. Uh See so that would be your answer for part B.

11. So the reaction given her as E guesses plus big as he gives reaction. See cases plus D guesses. So we'll have to calculate the equilibrium concentrations of this substance involved when the reaction is in the equilibrium moles and total volume. So as the total volume of the container is given, of what leader the moral concentration are equal to the most at equilibrium. So that's the equivalent concentrations for a will be 0.40. Given For me it will be zero and 40. M. Yeah, for C it will be It is one 60. Um and for B It's 1.16 good. So therefore the expression for the equilibrium constant of the reaction. As we know, the case goes to see B A B on B. So substituting the values of we'll get Kacey goes to 1.60 Into 1.60. Upon zero point 40 into zero and 40. Or we can just square them. So we'll get 16 after all in this. So that's the equilibrium constant. Casey For the reaction is 16 so no for the B part But 0.2 most of B. See each is added to the system, the B N. C. Increased by 0.60 m. And 1.8 Km. So first we'll have to calculate the act in question to know that which side is the which side the reaction church. So QE was too see the on A. B Subsequuting the values will get 1.80. 1.60 upon 0.40 In 20.60 Solving this will get 12. So since we can see that Q. Is smaller than Casey, the reaction proceeds to write site. No, the now to find the constant concentration tragic Librium assume X. Small of C exportable later upon of C. Or D. His phone means exports of A. Or B. Is consumed. So we'll have to substitute the values of equilibrium concentrations in the equilibrium expression. So we'll get 16 equals two When 180 plus x. We consider it as X. We'll have to make I see table also for this so make an icy table. He'll get the why I'm putting these eggs 1.60 plus six. X. Horn 0160 -6. So we'll get Simplifying. This will get 15 x minus uh cube minus 19 0.4 X plus 0.96 equals to zero. So the quadratic equation is solved using the quadratic formula. So we know the card direct formal that is minus B plus minus. And the route we know the formula. So we'll just replace the value. So this will get that's minus and the road for A. C. She is 19 point four square minus four into 15 into 0.96. So by solving this will get 1.24 Or the 2nd 1 will get 0.05. So the second solution of the quadratic equation is employed because the first solution gives a larger concentration than A. And B. So that's the nuclear free um consultations of A is Egg was 2 0.40 -6. Yeah. Which is equal to 0.40 m minus 0.05. Will get zero and 35 there to define a lot. Thanks.

Problem 17.64. We have the chemical um equilibrium A plus B equals C plus key. And for part A they want you to find the equilibrium constant for this reaction if you have yeah around 0.40 modes of A. M. B. And 1.60 modes of cnd. So the equilibrium equation is just the products Overreacting. So that would be the concentration of C. Times concentration of D. Over the concentration of A. Times the concentration of the. And this is given in the question. You have 0.4 modes of AMG and it's in a one leader solution. So that's why your concentration of a. And B would just be 0.40 and you square that and likewise you have 1.60 most square. Um And the answer you get for this Is that you get that the Casey is 16. And for part B. They ask you if 0.20 Mostel B and C are added to the system will be the new equilibrium concentration of A. And to do this you have to calculate the new concentrations of being seasons, you're adding um More most into that compound. So your new concentration of B would be 0.40 motor plus 0.20 motor which gives you 0.60 moller. And your new concentration of C is 1.60 motor plus 0.20 Mauler Equals 1.80 mother. I remember this is all in one leader solution. So that's why we can just use the number of what they give you since you just divide in my work. Um And if we should first find the Qc of this equation or of this new equilibrium. So we can see that if you're adding X. Or subtracting X from the reactions and products. So the Q. C. Equals C. Times the concentration the over the concentration of the times concentration of B. And in this case the concentration of see your new concentration is Um 1.80 times 1.60 which is the concentration of tea. And then 0.40 times uh 0.60. And this will give you 12. And because QC. Which is 12 is less than Casey which is 16, the forward reaction proceed. Uh So the equilibrium is still shifting to the right. So and this is why in this question you will be subtracting from the reactant and adding to the products. So we want to create an icebox next and um calculate the new concentration. Okay so to do that um you take the initial concentration of A. Which is 0.40 motor. An initial concentration of B which is 0.60 moller and see what you want. 0.0 and D. Which is 1.60 And you're subtracting X from the reactant and adding X to the product because the equilibrium shifts to the right. And you calculate you plug all this into the Casey formula which is um essentially either products overreacting. So you have You're Casey, you have is 16 which we found in part one. Um You just put it in here and then um you're you're playing your equilibrium expressions so you have 1.80 plus X. Which is the concentration of CIA equilibrium and then 1.60 plus X. Concentration of DEA equilibrium. And you have 0.40 minus X. And 0.60 minus X. Which equals six. Uh huh. Yeah I equals 16. And you just saw this quadratic equation or you saw this um fraction to get a quadratic equation and this will give you 15 X squared minus 19 from four X plus 0.96 equals zero. And you want to use the quadratic equation X equals negative B plus or minus squared B squared minus four A. C. Over to it. And that will give you 19.4 plus and minus square of 19 4th word -4 a. c. which is 15 times 0.96. All over to a which is two times 15. And when you simplify this you get that X. is either 0.050 or 1.24. And if you look at 1.24 um you would get a negative concentration for A. And B. When you put it into that equilibrium expression. So we're going to stick with that as our X. So the question asked you to find the equilibrium concentration of A. So we just do um concentration of a equilibrium equals zero, plenty of 40 minus X. Since that is what we get here, Um which is 0.40 -0.050, which equals 0.35 moller. So this will be your answer for part B.

L. M. You're given this chemical equation here And they say that one more of a. And one will be are placed into a 0.4 liter container. And after equilibrium has established their point to more of C. Is present in that container. So what you want to do is you want to calculate the concentrations and set up an icebox. So we know that the initial concentrations of AMG is 2.5 because you just do one developed by 0.4 and you find that the Equilibrium concentration of C is 0.5. Since the final, you have 0.20 more of c. So you do 0.2 divided by 0.4, which is the container size, which will give you 0.5. So based on this you can basically find that ex the change in concentration would just be 0.5 and you just want to plug that into all you have. So this will be d equilibrium concentration would just be one and then um for D. N. A. It would just be uh To so you just subtract 0.5 and to find the K. You just plug it into the formula products. Overreacting. So since there's a coefficient to in front of you have to do one square, which is just one times 0.5 Over the concentration of reactant, which is two squared, which is just four. So you just do .5 divided by four, which will give you 0.125. So this will be your K. value based on the information given.


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