A mixture Of 0.598 MH,O.0.3HMCL,O.and 0.690 M HCIO are enclosed in # vessel at 25 *HO(g) + C,0g) = 2HOCIg)O.0900 at 25"CCalculale the equilibrium concentratio...

The equilibrium concentrations in a gas mixture at a particular temperature are $0.13 \mathrm{M} \mathrm{H}_{2}, 0.70 \mathrm{M}_{2},$ and 2.1 $\mathrm{M} \mathrm{M}$ HI. What equilibrium concentrations are obtained at the same temperature when 0.20 $\mathrm{mol}$ of $\mathrm{HI}$ is injected into an empty 500.0 $\mathrm{mL}$ container?

So first let's find our pressures and then we can find K. P. So the partial pressure of hydrogen, we use the ideal gas law here. So we've got .112 moles .08- one. And our temperature is 731 Kelvin. Right, that was 458 Degrees C. Plus 2 73. And this whole thing is in five L. So our pressure of our hydrogen Is 1.34 ATM. And since we have the same number of moles of I to this is also the pressure of ri too. So let's find the pressure of R. H. I. So we've got 775 moles. So we're gonna have a larger pressure here. So that will be 9.29 ATM. That's our H. I. Yeah, I'm also gonna find the pressure in the flask after that we're adding. So the pressure of the H. I. That we're adding. It just makes everything easier if we're in pressures here. Okay, so then we can go ahead and find our K. P. Right? Based on our equation R. K. P. Expression is the partial pressure of the H. I squared Divided by the partial pressure of H two Times partial pressure of i. two. So this will give us 9.29 squared Divided by 1.34 squared. So our K. P. is 48.1. Mhm. So now we're gonna add that extra H. I. And find our new pressures. So we're gonna want to make an icebox. So let's start with our equation again. H. two plus I too Give Me two. H. I. Initial change equilibrium. So we're starting with 1.34 ATM of these two. And then for the H. I. We're going to have the 9-9 that we started with And then the 1.2 that we added. Okay so our initial pressure is bigger now. Therefore our reaction is going to shift to the left because we've got more product than we should at equilibrium. So we'll say plus X here and plus X And -2 x. So these will be 1.34 plus X. And this is going to be 1049 adding those two pressures together -2 x. Okay. But K. P. Hasn't changed right? R K. Is a constant and we haven't changed the temperature. So R. K. 48.1 Equals 1049 -2 x squared Divided by 1.34 plus X squared to simplify our math. Let's take the square root of both sides. So when we do that We're going to get six 0.94. It was 10.49 minus two X. Over 1.34. What sex? So let's start by cross multiplying. And then we'll simplify a little bit And then we'll go ahead and solve for X. And that will come out to the .134. So now we can find our partial pressures. So the partial pressure of H. two Will be the same as the partial pressure of i. two. And those will be 134 plus X. So that will give us 147 ATM. And that's the partial pressure of the H. two and the i. two. And then the pressure of the H. I. Will be 10 0.49 minus two X. Two times 134. And that will give us a partial pressure for the h. i. of 10 to 2 ATM.

So here we continue it on our work with the main group elements of the ice block and the P block. And so the equilibrium constant is given by the ratio of the concentration of the product and the reactant raids to their coefficient. So for the following reaction to s. 0. 3, I do too. What we have is to us. Yeah so to at the start to also to gives us to S. 03. And so the K. C. Is equal to s. 0. 3 squared divided by S. 02 squares not supplied by 02 concentration. And so we can draw a table that depicts the concentration of the irons. So initially the change um equilibrium. So with that so too it's one oxygen. It's five. So three of course is zero at the start, The changes -2 X -X two X. The equal agreements 1 -2, X five -X. and two x. So the equilibrium concentration of S. 03 is X. And 77.8% of S. 02 gets converted to S. 03 So therefore we can solve for X where X is equal to not 30.389 And then we can plug this value in two. Uh K C equation, we'll do that now. So what we have is K C. Is equal to two, multiplied by not 20.389 squared divided by one. Take away two X squared. Multiplied by well, we can plug in for the two acts now to lots of not point 389 multiplied by five take away 9.389. What we get is 2.7. And so the equilibrium constant for the reaction is 2.7.

Problem 89 in this problem first we need to calculate partial partial pressure of hydrogen hydrogen iodide. So according to I guess the question B. Bye and Equals two. Our tv So that equals two zero point 08-06 little radium divided by mall Calvin, multiplied by 7 31 Calvin. Do you buy? Did by five letter volume that is He calls to 11 point 99 Say what? ADM caramel. The airport Raise it up at two guests. It calls to press it up. I do this because both and both gases most given most are same. That is equals two 0.112 most multiply by present. But you need to move this 11.99. See one ATM caramel that equals two 1.344 approx. It goes to one three palladium. Okay. And parcel pressure up. I drew an iodide calls to 0.77 hide mall given multiply by pressure per unit move that is yet we can get out here. 11 point 997 A. T. M. Farmall That is equals to 9.298 approximately equals two 9.30 80. Um, Okay. Given that as to guests plus, I guess Holmes two hydrogen iodide gas. So keep people. This reaction calls to parcel press zero had risen iodide to the power to divided by Parcel place for up to two guests. Multiply by parcel press rob. I do. I guess that. Is it close to parcel president that I guess. Yeah, We calculated that is nine 298 to the power to divided where partial pressure of hydrogen and parcel petrova I guess is equal. So 1.3 for for to the power to that equals two 47.861 Approximately equals two 47.9 80 M. Sorry, 47.9. Okay. Given that parcel pace it up, hydrogen iodide parcel post pattern. I would date when 0.200 mi of hard dozen I guess is added. Okay. That is because two added mall. Your point multi multiply by this is razor per unit volume 11 point 997 ATM per unit mall. That is a cost to 2.3 99 port approximately equals two 2.40 idea. Okay, so we can write a question here again, hydrogen gas reactive it. I wouldn't guess and it will form two x. I guess. Okay. And in easily initially change and equilibrium. Okay. In Italy 1.34 18 hydrogen gas and one point report a team of violin gays and nine point 30 A. T. M. Plus. Good point. Poor doodle A. T. M. This partial pressure created by addition of 0.20 Mola. How do we are? Dead guest. So what does it send lettuce? We increase the product concentration. So reactant concentrations will be so reaction sibling backward direction. The airport tablets maximal of H two X X 80 M. O. As to guess. And exit demo. I do. This will form so change, you know, additional data guesses minus to work stadium. Okay, so at equilibrium Partial pressure of S. two guesses. 1.3 put Plus X 80 M. And one point three pour plus X 80. Um And Partial pressure of hydrogen ordered guess is 11 70 minus two. A idiomatic Libya. Therefore kippy For this reaction equals two 47 points 86 equals two parcel prisoner product that is 11 point seven year old minus two X. to the power to Divided by one point three port. Let's act to the power to so take the square root of both sides. Mm hmm. So big we can write after taking the square root six point nine 18 equals two 11.70 -2 x. Divided by one point three pour minus. Act. So after solving we can write as 8.9 18 X. Calls to it two point study 2.430. We can You can calculate here with simple mathematics. So X equals two zero point seven to Pour eight. That is approximately close to 0.2 they went to. Sorry. Okay, so at equilibrium Partial pressure up at two guests goes to partial pressure up. I guess That is equals two. We can calculate here. This That is equal to 1.34 plus x 80 m. Expedia. Okay. After calculating after putting the value, we get the value of partial pressure of oxygen. Partial pressure I do is one point 61 a t. M. Similarly parcel prison up hydrogen iodide partial pressure of hydrogen I read equals two. This parcel. This is parcel post partisan. I had already at equilibrium. Yeah, 70 minus to into G zero point 272 It's the this value of ACT. We can put your value packs That is equals two 11 point 6 18. This is our handling for this question, this and these.

Let's start by finding our initial concentrations and then we'll go ahead and set up an ice box to see what's happening. Okay so for the CO two we had one point five moles In 75 L. So that would be 2.00 more and H two. We had the same 1.5 Malls In 75 L. So both of these will start off as to moller. So our reaction is co two was H. Two gives us ceo and H 20 vapor. So these are all gases so they're all going to be important. So we'll set up our initial change equilibrium. We're gonna start with two moller of each of these and none of our product before the reaction happens. And then we're gonna react minus x minus X plus X and plus X. Because our ratio is all 1 to 1. So equilibrium we have two -X of these to react ints and we have X. Of our products. So our K. C. is given to us as .802. So that's equal. There are products over reactant. So that's going to be X squared Over two -X. All squared. So I think the simplest thing would be to go ahead and take the square root of both sides before to simplify the math. Who is going to give us .8 96 is X over two minus X. So we'll go ahead and cross multiply. Okay we'll go ahead and simplify and finally solve for X .945. So the concentration of our ceo Is the same as a concentration of our H. 20. And those are just X. So that's gonna be our 945 moller. And the concentration of our co. two Who's going to be the same as a concentration of our H2, And those are two -X. So those who both come out to be 1.05 five moller.