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Consider the reaction:B = € + JDsolution was prepared by mixing 50.OO mL of [.00 x I0 'MA [00 mL 0f 2.00 10 'MB 10.0 mL OLMC and 75 mL of L50 < I...

Question

Consider the reaction:B = € + JDsolution was prepared by mixing 50.OO mL of [.00 x I0 'MA [00 mL 0f 2.00 10 'MB 10.0 mL OLMC and 75 mL of L50 < I6 MD. At equilibrium; the concentration of D was measured to be 6,0 x 10-M: Calculate the equilibrium concentrations of each speciesb) Calculate the equilibrium constant;

Consider the reaction: B = € + JD solution was prepared by mixing 50.OO mL of [.00 x I0 'MA [00 mL 0f 2.00 10 'MB 10.0 mL OLMC and 75 mL of L50 < I6 MD. At equilibrium; the concentration of D was measured to be 6,0 x 10-M: Calculate the equilibrium concentrations of each species b) Calculate the equilibrium constant;



Answers

Challenge In a generic reaction $A+B \rightleftharpoons C+D, 1.00$ mol of $A$ and 1.00 mol of $B$
are allowed to react in a $1-L$ flask until equilibrium is established. If the equilibrium concentration of $A$ is 0.450 $\mathrm{m} / \mathrm{L}$ , what is the equilibrium concentration of each of the other substances? What is $K_{\text { eq }} ?$

So we've got a balanced equation and an equilibrium constant. And we want to set up an icebox initial change in equilibrium. But first we're gonna need to find the initial concentration of our I. 04. So let's take a look. We are told it's point 9.05 more And we have .02 leaders. So that's .0181 malls of our sodium per I. O. Date. And then we're gonna take that Which is also .0181 moles priority. And we'll go ahead and divided by the volume to which it was diluted. And that will give us its initial concentration 2.07 24 moller. So we'll put this here. They were not worried about our liquid. And before we start we have none of this. So this would be minus X. And plus X. Because they're in a 1-1 ratio 0724 minus X. And X. So our Casey expression is our product. So the concentration of our product over the concentration of our per I. O. Date are reactant And Casey was .035. So that's going to be X Over .0724 -X. So we'll start by cross multiplying and then we'll go ahead and simplify And we'll get X. is going to be 24 Times 10 to the -3 More. And our X. Is our concentration That we're looking for the H. four. I owe six negative

Relevant. So for the reaction gases it gives big as his plastic cases. The expression for equilibrium constant for the reaction is Casey was to B. C. Honey. So the experience on the concentration term started in mind by the coefficients of the reaction. So we have a a equilibrium constant of the species in direction at 200 degrees Celsius. So we close to see it was 20.25 AM. The egg was 201 30 AM. Now substituting the values of consultation in the equilibrium expression will get Casey equals two 0.25 in 20.25 upon 0.13 we got 0.21. So Casey uh 200 degrees Celsius is 20 0.21. No, they be part. So if the volume is double, the concentration will be reduced in half. So the new initial concentrations of abc are equals two Sequels to 0.25 point two. That is 0.13 approx. Whereas A equals two given 30 30 upon tools. 0.15. So we'll have to calculate the reaction question is in the new initial concentrations the reaction question goes to being to see a body for this equation. So we'll get 0.11 after substituting the values. So because we can see that Q is smaller than Casey, the reaction proceeds to write to establish the equilibrium so no let X mol of B and C is formed. An X small upon later of A is consumed. So we'll have to concept I see table. So you guys will be going to construct an icy table and after concept in their table will have to substitute the values of consultation in the equilibrium expression and solve for X 0.21. It was too 0.13 plus X and 0.13 plus six upon 0.15 minus six. Well that X equals two. 0.0 to nine a.m. No use the value of extras. All the new equilibrium concentration so B equals two X. Big wants to see 20 0.13 plus 0.29 and equals to 0.16. So it was true, you know, you know 0.15 minus zero point 0 to 9. Um which is equal to 0.2 mm. So now for the c part. So in the C part we have been asked to half the volume. So if the volume is half then the concentration will redouble. So the new initial concentrations of A B and C will be close to you're going 50 M equals to 0.60 M. So now calculating the action questions using the new initial concentration will get B two B mm. Oh honey. So questions will get after replacing the values will get your 0.42. So because you can see Q is greater than Casey and the reaction proceeds to left to establish equilibrium. No moving further again level. Have to let X smaller moldovan later of B and C is consumed and X more literal X small upon leader of A is formed. So again we'll have to construct and I see table and after constructing it will have to substitute the values of the concentration in the equilibrium expression and solve for X 3.21 little 0.15 minus X. Your 0.50 minus six. This is 01 15 minus six. 10.60 ballistics following this will get X equals two. 0.11 X. So now use the value of X to solve the new concentration in the final concentration so big was to see Yeah, 0.50 will be minus from 0.11. And so we'll get 0.39 AM two equals to 0.60 lost 0.11 um which is equal to 0.71 M. And that's the answer. Finance. Thank you.

Probleble 17.66. We have this chemical equation here and for part A. We want to calculate K. C. When the concentration of A equals 0.30 motor concentration that be Equals a concentration of seat and i equals $0.25. And to find Casey you just do the concentration of products over the concentration reacting. So you do 3.25 sq. Because this is um concentration of B. Times concentration of C. Over the concentration of A. Which is just 0.30 And this will give you that the K. C. Is approximately 0.21 So this will be your answer for part A. For part B. Um They are asking if the volume of the container In which the system is an equilibrium is suddenly doubled at 200°C will be the new equilibrium concentration. So if you double the volume um then what happens is that the the concentration will be halved. So the concentration will be halved. And this is because if you remember the concentration is just mold over leaders which is a volume. So if you double this then he would have the total uh concentration here. So you want to calculate The concentration, the new concentration of a. Which is 1/2 0.30 which is zero point 15 moller. And then the concentration of B. And C Would be 1/2 0.25, which will be 0.125. So these will be your new concentration you'll be working with and Mhm. Um You want to set up an icebox. So this is your equation when you set up an ice box you have 0.15 for eight and 0.1254 BNC. And then you subtract X from the reactant, add X to the products because if you increase the volume then you also decreased the pressure. And if you decrease the pressure, the equilibrium will shift to the side with more modes of cast. So it will shift to the right so that's why you would add X to the um products and subtract X from the reactors. And when you saw for this You get that the K. C. Which we found here is 0.21 by I'm going to use the ungrounded forum. I have 0.20833 equals 0.125 plus X squared over 0.15 minus X. So you get this big quadratic equation you want to solve and what you get is um you cross multiply get rid of the proportion. You get 1/30 to minus 5/24 X equals X squared plus 1/4 X Plus 1/64. And you set this equal to zero and you put this into the quadratic equation which is just X. Equals negative B plus and minus square root of B squared minus four A. C over two. A. So you get X equals negative 11/24 plus and minus square root of 11/24 squared minus four A. Just one And see which is 1/64. And then all over to a. Which is one. So you get that x equals 0.03187 and also negative 0.4902. And you will not have a negative concentration. So we will just get rid of the second route here. So we have this as our X. We want to plug this into our equilibrium expressions. And and then we get that the concentration of B which also equals a concentration and see Is 0.125 plus X. Which we found a 0.031 nine. And this will equal around 0.12. Oh sorry This were equal around 0.16 moller. And then the concentration of a equals 0.15 -X. Which is um zero point 0319. And this will be around There are $2.12. So these will be your answers here. And for part C. They ask you um when the volume is suddenly halved will be the new equilibrium concentration. So and the last question we had the volume double. So this question when you have the volume the concentration will double. So what you get is the concentration of a Equals 0.25 times two. Which will be 0.5 more. And then the concentration of B. For the ego. Sorry that is the concentration of the and the concentration of C. The concentration of Would equal 0.30 times two which equals There are .6 more. So you want to use these concentration to set up in new icebox here. Like we did in the last question. So we have the equation here ice box. Um So the initial concentration of A. You're starting off with 0.6 moller and then for CMB are starting off with 0.5 Moeller. And then um in this case you are adding X. To the reactive than subtracting X to the product because you are decreasing the volume which will increase the pressure. So you shift to a side with more less moore's. So this and this question. Um The equilibrium will shift to the left so that's why you are adding to the reactors and subtracting to the products. So you plug this into your Casey formula Which is 0.5 um minus X squared over 0.6 plus X Equals Casey which is 0.21. Or you can use on rounded form. So when you basically simplify this into your quadratic equation you get the X square -1.21 x Plus 0.12 is zero. And you want to plug this into the quadratic formula to solve for X. And you get that X equals negative B. Which is 1.21 towards the minus school route for 1.21 square minus four. A. C little 40.12 All over to A. Which is a. Is just one. So you get two routes again 0.11 or 1.1. And you would discard 1.1 because if you do 0.5 minus 1.1, you will get like a negative answer and you cannot have a negative concentration. So you just get rid of that one. So You just plug in 0.11 into your equilibrium expression here and you get that um the concentration of A Equals 0.60 plus 0.11 which is around 0.71 motor. And then the concentration of B, which also equals the concentration of C Is 0.50 -0.11 which is around 0.39 older. So these will be your answers here. So let me try to say all this at one page.

In this problem, we have to go through a few steps to try to find an equilibrium. Concentration, were given a reaction and were provided with its equilibrium. Constant k. C. We want to do a nice table here, but we have to do a little bit of work before we can get to our initial conditions. We're told that we have a 0.905 Moeller solution of sodium io for the sodium will always go off on its own and be in in a plus I on so we don't have to worry about that. So this is in fact, the same as having an I 04 minus solution that is 0.905 Moller and concentration. We want to calculate our initial concentration, and we're told how much dilution happens. So we're going to use the equation. Him one V one equals m to be too solving this for him, too. In two equals in one V one over B two, that equals 0.905 Moeller times 25 mils over 500 mils. That gets us a concentration. The 0.453 molars so that is going to be our starting point for our ice table 0.453 And for this problem, we're in an angriest system. So there is a big excessive water, and we're just not going to worry about it Now. We're going to lose some of that concentration and gain it as h four. I owe six minus. Get these expressions for our equilibrium concentrations. Next, you're going to assume exit small, and we will check that later. K C equals 3.5 times 10 to the minus two, and that equals products over reactive. So here, X for H four Iove six minus over 0.453 minus X, which comes from their other equilibrium concentration. But we're assuming that extra small, so we're going to get rid of the minus X. If you do this, you find that X equals zero 0.16 Moeller. If you compare that number 20.453 you can see that it's a little smaller, but they're somewhat close to each other. So we're going to check this assumption that excess small by doing a little bit more math or once again going to start with the equilibrium constant of 3.5 times 10 to the minus two, once again solving for X. But this time we're going to plug in our old value of X to the denominator as an estimation of how much X would be lost from this system. And now, if you do this mass, so do this attraction on bottom first. Then multiply that over times 3.5 times 10 to the minus two. You will find that X equals 0.15 more so it's barely changed. If you wanted to verify this, you could do this process one more time. But we're a equilibrium here, so that is going to equal the equilibrium concentration with H four. I owe six minus.


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