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Nz and Forate 3 the reaction: were equilibrium each Nz(g) M? concentration CzHz(g) of hydrogen Vi HCN; if the initialc 10+ concentrations at 300 3 2...

Question

Nz and Forate 3 the reaction: were equilibrium each Nz(g) M? concentration CzHz(g) of hydrogen Vi HCN; if the initialc 10+ concentrations at 300 3 2

Nz and Forate 3 the reaction: were equilibrium each Nz(g) M? concentration CzHz(g) of hydrogen Vi HCN; if the initialc 10+ concentrations at 300 3 2



Answers

The following equilibrium concentrations were measured at 800 $\mathrm{K}$ :
$\left[\mathrm{SO}_{2}\right]=3.0 \times 10^{-3} \mathrm{M} ;\left[\mathrm{O}_{2}\right]=3.5 \times 10^{-3} \mathrm{M} ;\left[\mathrm{SO}_{3}\right]=5.0 \times 10^{-2} \mathrm{M} .$ Calculate the equilibrium constant at 800 $\mathrm{K}$ for each of the reactions in Problem 13.1

Problem 53. 53 with a similar problem. Pity to another thing. Had to cover the pH at the end so way have hydro Gen Zia and I. Yeah, in what it associate become higher on young iron and sand. I No, I had was on it as it people cyanide. And this was that I see in his soul gin and echoed Miriam boy Oh, do five the zero, as you know, So lose us and lost as he had asked me ne Because Endres all the more the event. So final this left. And this a similar photo, the braver problem. We can set up the question right away. We have okay. A Usually when you see the constant less than 10 negative three, you can admit the sa denominator. However, I still show the the matter in the book. Here we have, boyo, Do fight. Write it down. Okay, well, and phone from this conclusion we can do for a boy. Oh, no, they tell you. Go. A square divided. Oh, fine. They show this equation. We heard people when you discovered, remember, cover the the expression inside a square. Okay. You cannot draw that. It and you let us number at this. You have tow cover all of them. And these people? 3.2 Thai, 10 and basic. It's so we have the concentration off, son. I I on equals concentration. Hide or new Mayan, this vehicle Three boy to tight end. And for the high road, Sandy. Because it we have hydrogen saw and I had boiled five lying us this number when I so this very small this number very small, right? So we had this little do five. And now we cover like Kiesche. So p is equal. Five boy 50 This passage is weak in a straw, right? Because the bs Ralph I So we're done.

So the equilibrium constant of a chemical reaction is the reaction quotient at equilibrium. So this is a state approached by a dynamic chemical system. After enough time has passed at which its composition has no measurable tendency towards further change and chemistry. Our concentration were fast to the amount of a substance that is in a defined space, where our concentration can be expressed in terms of must pay unit volume. So we're looking at the following reaction here of two eso To add 02 is an equilibrium with two eso three. So given the information in the following values, we will be determining a concentration value. So the equation that we're needing to use is R K equilibrium, constant equation where we have our product concentrations multiplied by the street. You metric coefficients over the starting material concentrations also multiplied by their strike you metric coefficients. However, this has been arranged. Where s 02 concentration is thesis object off our problem now where we have a streaking metric coefficient of two. So that means we square root are valued to generate no 0.52 as our missing folly

This problem gives us a reaction, and it gives us the equilibrium, um, constant. And then it also gives us the equilibrium concentrations of, um, and H three and of hydrogen gas. And we need to find the equilibrium concentration of nitrogen gas so we can use the equal equilibrium, constant equation, um, to solve for the equilibrium concentration of my string us. So Okay, thank you is equal to the products over the reactant. It's so in each three squared because you need to account for its coefficient times, nitrogen gas times, hydrogen gas, cute. And so we know these three values and then we can solve for nitrogen gets. So if we plug in our values, we get 6.59 times. Centenary of three is equal to 1.23 times 10 to native for squared over 2.75 times 10 to the of two huge times nitrogen gas concentration. And then we can continue to solve for this and we get nitrogen gas times 1.37 times 10 something of seventh, which is the product between this number, and this number is equal to 1.23 times 10 to 94 squared. All I did there was cross multiply. And then you get the nitrogen gas concentration, which is in moles per liter, or molar ity is equal to 0.110 polarities, polarity, and that's three sig figs to keep it constant with our, um, given values into this is our final answer.

So let's figure out what the equilibrium concentrations are for each of these substances by setting up an ice table. Now if we know what the initial concentrations are of the S. 03 and um the N. O. We can plug that into the initial line. But we plug in polarities. So we're told that we have 0.240 moles of each of these in a two liter container. So when we take molds and divide by leaders or 20.240 divided by two. Each of the polarities are .1-0. So make sure you put polarities into your ice table and we don't have any of our right side. So we're gonna have to make some right side. Which means that the right side is going to go up, the left side is gonna go down all of our coefficients in our chemical reaction or one. So let's look at the N. 02 1st I have to make some I don't know how much X. But I know for every N. 02 I make the same amount of S. 02 So that's gonna be X. I also know that for each N. 02 I make I have to use one N. O. Use the same amount and for the S. 03 of the same thing I use the same amount. So in equilibrium I started with .120. And it went down by a certain amount. Started with .120 went down by a certain amount started with none. Wind up by a certain amount started with none when up by a certain amount. Now I can write the law of mass action expression. So I can take my products and divide by my reactant since yeah I can now put those in X times X is X squared and 0.12 oh minus X times 0.12 oh minus X. 6.12 oh minus X Squared. And we know the value of the equilibrium constant is .5. All right. So why don't I put this to look like squares? Because now it's easy to just square root both sides. To make the math easy. So if I square root both sides, I get X Over .120 -1 Equals the Square Root of .5, which is .77. Now we'll go ahead and I'll cross multiply and then I'll collect life terms and solve for X. so .120 times .707 is point oh eight for eight And then .707 -1 is negative .707 x. And then when I cross multiply x times one his ex. So I'll collect like terms bring the X to the other side, Divide by 177. And I get X. Is equal to 004 nine seven. Okay so if I want to know what the concentrations are an equilibrium I have to go back to my ice table, N. 02 and S. 02 are represented by X. So in my polarity I have my concentrations of those two substances. So I know The concentration of n. 0. 2 And the concentration of S. 02. If I want the other ones I have to take .1-0 -1. Yeah. Yeah. Yeah. Mhm. Six. And that's point 0703 moles per liter. And that's the concentration of both My reactant. So is the concentration of S. 03. Mhm. And the concentration Havana. Yes.


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