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Determine the equilibrium constant for the following _ CUaq) 21(aq) reaction at 25 *C 2 Ct(aq) Vaq) E" +0.825 Vpolnt1.08 x 10 *9.25 x 1070.2170,825Given. the f...

Question

Determine the equilibrium constant for the following _ CUaq) 21(aq) reaction at 25 *C 2 Ct(aq) Vaq) E" +0.825 Vpolnt1.08 x 10 *9.25 x 1070.2170,825Given. the following half-reactions Ag (aq) Ag(s) E" +0.799 VAgCI(s) pout Ag(s) CI (aq) E" +0.222 V, calculate the molar solubility of AgCl1.34*10 'M4.46 * 10*M~0.577 M1.79 * 10 1 M

Determine the equilibrium constant for the following _ CUaq) 21(aq) reaction at 25 *C 2 Ct(aq) Vaq) E" +0.825 V polnt 1.08 x 10 * 9.25 x 107 0.217 0,825 Given. the following half-reactions Ag (aq) Ag(s) E" +0.799 VAgCI(s) pout Ag(s) CI (aq) E" +0.222 V, calculate the molar solubility of AgCl 1.34*10 'M 4.46 * 10*M ~0.577 M 1.79 * 10 1 M



Answers

Write equilibrium constant expressions for
the following reactions:
$$\begin{array}{l}{\text { a. } 2 \mathrm{NO}_{2}(g) \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{4}(g)} \\ {\text { b. } \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \operatorname{COCl}_{2}(g)} \\ {\text { c. } \mathrm{AgCl}(s) \rightleftarrows \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)}\end{array}$$
$$\begin{array}{c}{\mathrm{d} \cdot \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows} \\ {\mathrm \quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad {H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)}\end{array}$$

In this problem I can write the value of the north is equal to zero you know 591 by and log E two, Solving it further, I can write it as 0.22 is equal to 0.0 White 9 1 by two Long E que solving it further, I can write Love. G e q Is equal to two multiple tickets and 0.22 By 0.0 591, Which is equal to 7.44. On Further simplification, I gave the value. Okay, The issue is equal to 2.8 multiplication 10 to the Paul seven according to the option. Option is correct option. Age, correct answer for this problem. I hope you understand it.

Hello. So today we're going to be looking at different equilibrium A. And we're going to be writing that equilibrium, constant expression and calculating the equilibrium constant. So how about leave right? The equilibrium, Constant expression for this reaction right here. So it's products over react in. So we've got the science right here and has a to call efficient front. So Dsquared, then we've got our Nikolai on and over the reactant snick, Eliza solid. So we don't include it. And we have a two coefficient here, so we square. So how would we calculate this equilibrium? Constant? Well, we have an equation where sell potential is equal to gas constant times temperature over Z, which is the moles of electrons. Fair Day's number Natural log of K, which at 25 degrees Celsius simplifies to 0.0 257 older z natural log of K. So let's try to figure out Well, what is this cell potential? Because if we have the cell potential, we can figure the rest of this out. So how about we separate this equilibrium into you? Two separate half reactions. So nickel is being oxidized. And then the sign right here being reduced. So the production potential of this half reaction right here is negative. 0.2 55 And that's just for one electron being added. It's the same for two as well. So if we run this top half reaction and diverse so if a nickel was gaining two electrons, while the reduction potential for that reaction would be 0.2 57 So to figure out the cell potential, you would just have to dio 0.255 It's attracting negative 0.257 since it's what's being reduced, subtracting the opposite of what's being oxidized. So we would have 0.2 volts. So now that we have that we have 0.2 volts, 0.257 and we see here we have two electrons, so we have two moles of electrons moving on. Susie is too the natural log of K. And so if we do some math, you will see the natural log of K is equal to zero 0.1 five six. And so to find K, we just e to the 0.1 56 power and that is one point 17 So there we go. We have Okay, So now let's take a look at this equilibrium right here. So let's write our equilibrium expression. So products over react INTs. So we have man Guineas. I on water is a liquid, so we don't included. Then we have cloying gas men. Younis oxide is a solid, so we don't include it. You have the concentration of protons and has a foreign French. We put it to the fourth power. Then we have chlorine to the second power because as the two in front. Okay, so now how about we try to We're still using the same equation to figure out our K. So how about we find the scent cell potential for that? This reaction, Let's put it up into two half reactions. So we see that chlorine here is being oxidized. So to chlorine ions coming together to foreign chlorine gas and releasing two electrons. And then we have manganese oxide, you see, so we know that we can eliminate this. So we have to explain the rest of this equation. So man Guineas oxide plus for protons is becoming name Guinea's plus and water And if we take a look at this well, we will see the oxidation State of hydrogen is plus one. Oxygen is negative two and then Mannion eases plus four Hydrogen and oxygen in the products is the same. But we see that main Guineas change from plus four two plus two. So that means that we gain two electrons. So to find out our cell potential, the reduction potential of what's being reduced which in this case is this man Guineans up side. And the reduction potential for that reaction is positive. 1.23 Subtracting the reduction potential for the opposite of whatever is being oxidized in this case, chlorine. So for chlorine gaining two electrons, the reduction potential for that is 1.36 So we see that our cell potential is negative zero 0.13 So let's begin now. So 0.13 And so we have 0.257 over the moles of electrons, which in this case is too. And then the natural log of K. So it's stupid math and we see that Nigel log of K is equal to negative 10 0.1. So to figure out OK, we just need to do e to the negative 10.1 power, which is 4.0 for times 10 to the negative fifth. So there we go. You found the equilibrium constant for that expression. Now let's take a look here. So let's right the equilibrium expression for this equilibrium. And we see we have coin my owns. There's a two coefficient in front, so we just square it concentration of oxygen. And then the reactant switches the sign right here. And there's a to call efficient and front so square. So we need to figure out what the half reactions are so we could calculate the cell potential. So what exactly are the half reactions? Well, if we take a look at our table with although reactions in basic solution well noticed, we have chlorine oxide plus H +20 plus two electrons. He comes Coin Ryan, and to hydroxide it's and what will notice is? Well, this reaction would give us. It has the has one of the reactant, and it produces one of the products to, but only if we square, so we'll multiply everything by two. So let's see, we're going to have yes and we see now. Well, it kind of looks. It's starting to look like the equilibrium we have appear. The only thing is we don't have waters are reacted, and we have this hydroxide as a product which isn't in our original equilibrium. So then we look at the table and we will notice that there's a reaction which produces hydroxide. And if we write that reaction for reverse, it would be for hydroxide associating into oxygen, which hey, look is one of our products that we need. And it's also forming to water and releasing for electrons. So as we can see if we add Thies to half for actions together, let's see the hydroxide is cancel out. The water cancels out in the electrons, cancel out so that we get the equilibrium expression on top. So these are their half these air, the half reactions in our equilibrium so we can figure out the cell potential. So the reduction, the reduction potential for this reaction appear is positive 0.890 and the reduction potential for the opposite of distraction here. So water and, uh, water and oxygen combining with four electric mountains too create for hydroxide that the reduction potential for that reaction is zero point for 01 And so we see that our cell potential is 0.4 89 boats and so number. Before, we had our equation, where sell potential is equal to 0.25 seven over the most of electrons. Natural log of K. So now we know that self potential a 0.489 and we can see from our half reactions that we have four moles of electrons moving around, so zeroed before natural log of K. And so now you do some math and we see that the natural log of K is equal to 76 0.1. So K is equal to eat to the 76.1, which is 1.13 times 10 to the 33.

All right. Hello, everybody. Today we're going to be finding the equilibrium concentrations for some reaction. So let's get right into our first reaction is to seal minus plus B R. Two giving us seal to gas and to be armed. All right, pretty simple. Let's first make our half reaction. So starting with the chlorine of to seal minus a quiz produces cl to gas. We have, ah, negative to charge on the left side. So we're gonna add in two electrons on the right side. Easy does our second reaction with our bro me br to liquid produces two br minus a quiz. We have, ah, negative to charge on the right side. So we're gonna add in two electrons on the left. Easy and everything balances out two electrons on both sides. Um, you'll notice our topside electrons on the right. That means oxidation. And for bottom half reaction, electrons on the left means reduction. All right, now, let's look at our e. Not about so r e not for the reduction of, um, for the reduction of seal, too is equal to 1.36 volts. And are he not for the reduction off br to is equal to 1.8 volts, obviously are sealed to is being oxidized. So are you not for theocracy? Gatien of two c l minus is gonna be the negative off our reduction of seal to which is negative 1.36 polls adding them all together we get r e not total is equal to, um negative 1.36 folds plus 1.8 wolves, which gives us a sum total of sorry. I'm in a little bit of a break. Um, which gives us a sum total of negative zero point to eat. Awesome. All right, the finder equilibrium concentration. We just use our variant of nurse law, which says that l and Q Q being our equilibrium is equal to N e. Not over 0.257 Very handy with liquid. So we know and is too, because from up here we have two electrons being transferred so me of two times negative 0.28 over 0.257 which ultimately ends up giving us an equilibrium constant of negative 21.8. Uh, sorry. That's not actually the equilibrium constant because we still have our Ln que here, so q is gonna be equal to e to the negative 21.8, which will actually end up giving us 3.4 times 10 to the negative. All right. Um, and because our Q is less than one, that means that this reaction does not favor the products are you know, it favors the reactions. Awesome. All right, let's move on to our next one. Okay, so now we have f e two. Plus a curious plus a G plus a curious giving us f e three plus agrees. And Eiji solid. So strict. Us down we have are 2/2 reactions. F b two plus two feet three plus, um, we have a greater charge on our right side, so we need to add in one electron to fix that on the But on our second half reaction, we have 80 plus a quiz becoming a G solid and extra charge on our left side. So add in one electron to fix that. We have one electron in both half reactions, So we're going on that count. Okay, let's move on to our e nuts. So I e not original for F E three plus is going to be equal to 0.771 worlds. Um, coming back up here kind of forgot. Electrons on the right means oxidation. And for a second half reaction, electrons on left means reduction. All right, since this is since our first half reaction is oxidation, that means that 0.771 is actually going to be negative. 0.771 Meanwhile, there are e not for, um our silver is 0.799 horribles. That's the same because it's that stays the same, because of which means I e not total is going to be equal to 0.7994 volts minus 0.771 bolts, which is gonna be equal to 0.2 uh, eight volts. Notice we leave off this four cause of sick fix as All right. So now let's use our version of Nor Inslaw again. We have Ellen Q equals Andy. Not over a 0.257 Which is going to be our end is one, because one electron over here r e not is 0.2 a and over a 0.257 Which gives us e not our Sorry, None. Nina. A value of 1.0. Not in, actually. My bet. That should be 1.1 due to 666 Therefore, our Q is gonna be equal to e to the power of 1.1, which is 3.0 since our Q is greater than one. Therefore, this reaction does saver the products. Awesome. So that's your answer. Thank you very much. Have a good day.

And so far apart. A The given reaction is the result of the flowing reactions here. Edge See, too, is three or two, and it's positive plus C two h three or too negative. K is equal to one point it multiply 10 rest to the power minus eight. Yeah, and it's positive less or it's negative is to all k is equal to one upon. Okay, is it? We're doing a ban. Okay. W is equal to one. Multiply 10 dressed to dip our 14 and your age C two is three or two less Always negative c two h three or two they get you. Plus is to all care prime. Is it built to? And no and the dilithium constant off the net reaction is the product off the village Liam Constant off the steps. Hence the Bolivarian Constant is K is equal to 1.8 multiplied 10 rest to the power minus eight into one Multiply 10 dress to devour 14. Is it weird to 1.812 ply 10 raised to the power nine. So the value of K is is equal to 1.8 multiplied and rest with the power nine in solution for part B. The given reaction is the reverse off the organization off aesthetic is it? Hence the equilibrium constant should be the reciprocal of the original que hands scare is equal to one upon k A is equal to one of 11.8 multiply 10 rest to dip. Our minus five is equal to 5.6 month supply, then rest to devour for and you And so for part C for the reaction off strong. Is it in strong based? The new direction should be the information off water. It's positive plus or is negative is equal to is to all. This is reverse off the outer organization Constant kow here therefore K is equal toe one upon what organization? Constant kw here is equal to one upon one multiply 10 rest to dip. Our minus 14 is equal to one multiply 10 raised to the power 14


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