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Write the equilibrium constant expression; Kc for the following reaction: Please enter the compounds in the order given in the reaction: If either the numerator Or ...

Question

Write the equilibrium constant expression; Kc for the following reaction: Please enter the compounds in the order given in the reaction: If either the numerator Or denominator is 1 , please enter [NH,HS(s) = =NH3(g) HzS(g)KcSubmit AnswerRetry Entire Gtoupmore group attempts remaining

Write the equilibrium constant expression; Kc for the following reaction: Please enter the compounds in the order given in the reaction: If either the numerator Or denominator is 1 , please enter [ NH,HS(s) = =NH3(g) HzS(g) Kc Submit Answer Retry Entire Gtoup more group attempts remaining



Answers

Calculate the equilibrium constant $K_{c}$ and $\Delta G^{\circ}$ for the reaction between $\mathrm{I}_{2}(\mathrm{~s})$ and $\mathrm{Br}^{-}(\mathrm{aq})$.

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.

In this video we're going to write equilibrium constant expressions for the following reactions. Remember that if we have a solid or a liquid, which applies only the I two, we're going to admit it from the expression um we assume the concentration is constant. So for part a uh we'll do products Overreactions, these concentrations since we do have a quiz species, so we have concentration of cl two and then I minus. And that's going to be squared because of the coefficient. And then on the bottom we have C. L minus and it's squared for part B. The product is CH three, NH three plus And then we have on the bottom ch three and H. Two and then H plus last one for part C. On the top we have gold cyanide and on the bottom A. U. two plus gold ion and C N minus. And that's going to be raised to the fourth power due to its coefficient in the balanced equation. Mhm.

Okay, so we've got some balanced equations here, so we're gonna write some expressions for K. Okay, so Okay. Is always products Overreact mints raised to their coefficients. Okay. And we ignore solids and liquids. So the only thing we're going to pay attention to our gases and things that are Equus if it's a gas will use a partial pressure if it's Aquarius will do its concentration. So this first one looks like we've got a solid and a gas and gas and some acquis. Okay, So it looks like we're gonna go ahead and write the partial pressure of cl two right times the concentration of i minus and we'll square that. Mhm. Who will divide that by the concentration of C L minus squared, Which is supposed to be a quick it's not a gas. Okay, so in part b looks like we've got a couple, we've got three a career solutions, So go ahead and write K. We'll do our products Overreact ints of CH three. NH three plus, and we'll divide this by our reactant, which are what's it all mean, CH three, NH two and H plus. So then the last one, it looks like we've got some A creates things All Aquarius. So again, we're going to go ahead and write product. Okay, is this complex ion? It's an eye on. So we'll do concentration and we'll divide that by concentration of the gold to ion and the cyanide ion. They will raise that to the fourth because it had a coefficient of four.

So this question is asking us to write the equilibrium, constant expression for the following equilibrium. So to begin, we'll use the base Katya Que equation, which says that K a Q is equal to the concentrations of the products race to their coefficients over the concentrations of the reactant raised to their coefficients, plugging in my products and my reactant ce I get that the cake you is going to be equal toe end too. Times CEO too, which will be raised to the second because as a coefficient of two over an o squared times the concentration of C o squared and that is my cake, you expression.


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