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At a particular temperature; the following reaction occurs: 2A(g) 2B(s) + Clg) 2D(g) Kc 3.60 * 107 If the initial concentration of Dlg) is 0.250 M,and A, B and C ar...

Question

At a particular temperature; the following reaction occurs: 2A(g) 2B(s) + Clg) 2D(g) Kc 3.60 * 107 If the initial concentration of Dlg) is 0.250 M,and A, B and C are not present at the start of the reaction, calculate the equilibrium concentration of A_

At a particular temperature; the following reaction occurs: 2A(g) 2B(s) + Clg) 2D(g) Kc 3.60 * 107 If the initial concentration of Dlg) is 0.250 M,and A, B and C are not present at the start of the reaction, calculate the equilibrium concentration of A_



Answers

Consider the reaction:
Find the equilibrium concentrations of A, B, and C for each value
of Kc. Assume that the initial concentration of A in each case is
1.0 M and that the reaction mixture initially contains no products.
Make any appropriate simplifying assumptions.
a. Kc = 1.0
b. Kc = 0.010
c. Kc = 1.0 * 10-5

For this question, we have several different K values for the generic reaction A. Goes to be. And first well actually have one K value, we have different coefficient values. If both coefficients are one, then it's going to be the concentration of C. A. B divided by the concentration of a. Both raised to the one power. If we're starting out with one moller of a, then it's going to have to shift to the right will increase be by X and will decrease A by X. We solve for X&X 0.80 Mueller. So that means a is going to be 1 -18 or point to Mueller For the next one. They're both too. So we're gonna have to square both of them. We solve for X annex ends up being The concentration of be still in its .67 Mueller. And then a is going to be still one -X. So it'll be .33 Mueller For the last one. It's a little bit more challenging. We have two on B. I'm sorry, Yeah, two on B and a one on A. So as this shift to the right B is going to increase at twice the rate as a will decrease. So if we're defining the increase in B is X, which we're going to have to square because there will be a two here on little B, then the decrease in A is going to be one half X. So it will be one point oh minus one half X and the coefficient here is one. So we have a superscript one here, then we do a little bit of algebra and solve for X. And we get 1.24 Mueller for be for X. And then A is going to be one Mueller minus one half X or 10.38 Mueller.

This question says to consider this reaction or CEO in h two form seeds 30 H. It tells us an equilibrium mixture of this reaction at a certain temperature. Has the following concentrations ever hear of each of the components and ask us what the equilibrium constant is at this temperature? So is simply plugging in each of these values into the equilibrium, uh, constant equation. So the prop, the concentration of the products only one is sees through H. That's 10.1 85 and that value is not raised any exponents, since there's no term in front of this than the constellations of the other products. Actually, I don't need to do brackets anymore because these are concentrations is 0.105 and that does not have a term in front. So no exponents and now point 11 or but this does have a term in front. As to solar square. This if you work the Saudi calculator, you would find the answer is about 100 and 36

Given the reaction of three a plus b to become to see and also the usual even concentration all species over bkc. So we are going to provide logical deepened expression from our Kanko. Our you d be, um so we have four in equilibrium expression and you're going to So for that we should be able to find out the case equals to 1.5.

This question says that further reaction shown here, K C equals 0.513 that tells us of a reaction vessel initially contains. And to go for a concentration 0.5 Moeller, what are the equilibrium Concentrations of each of these components. So the two pieces of information we have for the initial concentrated into a four an equilibrium concert with respect to see. So if you want to find out the equilibrium concentrations um, uh, we need to soften equally when concentrations Um, so first thing we wanted d'oh is we want to at least get an equation for equivalent concentration. So we were right. The chains that happens here as negative x. Some concentration will be lost in 204 and some will be gained by the products. But the story geometric ratio between and No. Two and into a four is 2 to 1. So where is the reactant lose minus sechs products will gain two X. We can write these, um, equilibrium is as the initial with the change is the point. 05 minus X and zero plus two X, which is just two X. Now we have an equation to represent equilibrium concentrations, But we need to solve for X. Of course we do. Have Casey given us here, So let's plug in all of these patrol on arrow here. So we're gonna say that 0.5 13 equals construction N o to where say is to x, and that's gonna be squared because of this, um, term in front of n 02 Revived by the concentration of into a world said was point five minus sechs. And so, if you were to expand this out great is the quadratic formula. You would get four x squared plus 0.513 thanks, minus 0.2 five. That's six five equals zero. And if you were to solve this quadratic equation, which I won't do here, you'd get one possible exits positive. And one that's negative. The negative, of course. Can't be it because your concentration isn't gonna be negative. So it's gonna be the positive term. That positive term is X equals 0.3 eat five. Now that we have that, we can play it back into each of these equations. We can figure out that the concentration of N 204 equals 0.5 minus 0.3 eight five. And concentration, uh, two equals, um, two times 0.3 eight five. And these two values are 0.1 one is our 0.0 115 more and 0.7 seven Moeller. So it takes a lot of work to get here. But from just knowing the equilibrium, constant and one of the initial concentrations, we can show that at equilibrium the concentration of into a 4.115 moller and the concentration of 2.77


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