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X(g) + Y(g) = 22(9 Khen A O0 mol each ol Xlg} ad Y(@) &e placed Ir a 1.00 ( vessel and &lored react at constant len ptralure accordig lo Ine equatlon ebovc...

Question

X(g) + Y(g) = 22(9 Khen A O0 mol each ol Xlg} ad Y(@) &e placed Ir a 1.00 ( vessel and &lored react at constant len ptralure accordig lo Ine equatlon ebovc; 0.00 mol 0l Z(9) produced: Whal Is Ihe value the equlibrium constant K ?

X(g) + Y(g) = 22(9 Khen A O0 mol each ol Xlg} ad Y(@) &e placed Ir a 1.00 ( vessel and &lored react at constant len ptralure accordig lo Ine equatlon ebovc; 0.00 mol 0l Z(9) produced: Whal Is Ihe value the equlibrium constant K ?



Answers

The equilibrium constant $K_{P}$ for the following reaction is $4.31 \times 10^{-4}$ at $375^{\circ} \mathrm{C}$ : $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ In a certain experiment a student starts with 0.862 atm of $\mathrm{N}_{2}$ and 0.373 atm of $\mathrm{H}_{2}$ in a constant-volume vessel at $375^{\circ} \mathrm{C}$. Calculate the partial pressures of all species when equilibrium is reached.

Were given this reaction as well as a value for the equilibrium constant k p at a certain temperature. We're told that we initially have partial pressures of zero 0.862 atmospheres for into as well as 0.373 atmospheres for each to We can fill in this reaction people to determine the equilibrium partial pressures of all species so that we can use the given value of K P to solve. For those the values of those equilibrium, partial pressure is given their expressions. So based on the street geometry, the changes are minus X minus three x and plus two x So the equilibrium Personal pressure expressions, or a 0.862 minus x 0.373 minus three x in two x. So now we know that the equilibrium expression for KP based on the given reaction in terms of equilibrium, partial pressures is equal. Agreeing partial pressure of NH three squared, divided by the equilibrium, partial pressure of end to times, the equilibrium, partial pressure of H two cubed. And now these are the expressions for the equilibrium, partial pressures so we can plug those into our equilibrium expression for KP to for the numerator the equilibrium partial pressure of NH three as an expression of two x and we swear that term And then on the denominator, the equilibrium, partial pressure and to is zero 0.862 minus x and for H 20.373 minus three x and this is equal to KP and were given that value so we can condense this expression to give us in the numerator four x squared and the value of KP that were given is 4.31 times 10 to the power of negative four. We see that since this value of K P is very small, that the reactant will be heavily favored and so, therefore, to simplify the math that we have to do, we can assume that this change for both of the reactant is negligible so that the equilibrium concentrations do not change from their initial values for both of the reactions by more than 5%. So we can make that assumption to give us in the denominator zero 0.862 after we get rid of that minus X and 0.373 After we get rid of that minus three X and then we can easily solve for X by multiplying both of these constant terms in the dominator together and amount applying it over with the value of KP that were given. And then we divide over four from the numerator and take the square root to find X. And when we isolate and solve for X, we should find that it comes out to about 0.0 to one 95 and we can verify that the equilibrium partial pressure of each one of the reactor of the reactions did not change from their initial values by more than 5%. So for and to the change was from 0.862 atmospheres initially Teoh zero point 860 atmospheres divided by 0.862 You multiply it by 100%. We see that the change was 0.23% which is less than 5%. So that was a good assumption to make. And for H two, we initially had zero point 373 atmospheres, an equilibrium based on the value of X that we calculated after making that simplifying assumption the equilibrium Partial pressure is 0.3 66 Atmospheres only divide by that initial value in multiplied by 100%. We see that that change, responding to a 1.9% change, which is also less than 5%. So this was a good assumption to make again the way that we found. Well, both of these values were given 0.862 was the initial partial pressure for end to and 0.373 z initial value for for each two that was given and then to find 0.860 We looked at the equilibrium expression for the partial pressure of end to that 0.86 to minus that value of X that we calculated based on the assumption. So that's how we got to your 0.860 and same thing for H two. You did 0.373 minus three times that value of X, and that comes out to 0.366 atmospheres. So since that assumption was valid, we can say that the equilibrium partial pressure of N two is equal to 0.8 60 atmospheres in the equilibrium, partial pressure of each to is equal to zero point 366 atmospheres and the equilibrium partial pressure of NH three. It's equal to when we look at its equilibrium expression. It's two times X, so we multiply that value of X by two. And that comes out to about four 0.40 times 10 to the power of negative three atmospheres.

A little bit. So this is the given reaction and mass of H. I has given 75 points seven grounds. The Mueller wait for HIV is Mueller where he goes to mW vintage one of each plus one of eight. And for H we have the value 1.1 g per mole and for I we have 1 26.90 g so one when 26 point 90 mm so we'll get molar weight equals two 1 27.91 g thermal. So therefore the moods of H. I can be now find as follows. So N equals two number of musicals to 75 0.7. It is mass upon 1 27 0.91 g per mole that is smaller weight. So therefore N equals two your point 59 two months. So consultation of HIV will be find out using the formula M equals two months upon Leaders. Where M is the modularity. So polarity equals to move is we have to find out zero point 592 molds upon leaders. We have been given 1.50 liters. 1 50 leader. So 1.50 m. So no M equals two 0.395 Um So now to find the concentrations at the equilibrium, the reaction, somebody for this. I see, I see. Table will be like this to a T gases and to plus I too. So the initial concentration is given. This is initial confrontation. I read initial consultation zero point 395 So for 8 to 0 and for all to zero, no change due to reaction is minus two weeks. Yeah, plastics you're also plastics. Now the equilibrium uh concentration a light on the next page. So equilibrium concentration 0.395 minus two X. For H. Two. It's X. And for I do it also eggs. So no, we'll have to assume that X moles are consumed or processing the reaction, since the chemical activity of solids is always one. So the equilibrium constant for the reaction uh is Casey was two products upon nutrients productive as to and I too. And they react until I try to say sorry. So the exponents are determined by the coefficient in the reaction or substituting the value. We will in this situation we'll get Casey equals two X ex a bone 0.395 minus two X. Square. So we know that case equals to 0.830 And we'll have to find the value of X. So cases given zero point here today, zero he calls to X upon 0.395 minus two X. Square. So the value for X will be come out to be after. Uh So uh after solving all this value of X will become out I think 0.1 to seven I cannot show you the full multiplication because it is too big. So now we have we have find out the value of X. So far the equilibrium constant for each species in the reactions are for H. Two where it was X. So H two equals two 0.127 M. For A. Two, it was also X. Which is also it goes to 0.1 to seven a.m. And for H. I. Direction was the concentration was 0.395 minus two X 0.39 five minus data point 254 Following this will get at I equals two zero point 141 M. That's the answer. Thanks.

Problems 17.52. We have 75.7 g of H. I. Place in a 1.5 leader reaction. That's all. And it's allowed to its undergoing an equilibrium reaction written here and we want to find the concentration of each of the species at equilibrium. So first we need to convert the grams into more clarity. So you basically divided by the molar mass of H. I. Which is 107 127.904 g and won more of H. I. And you also divide by the volume, which is they say 1.50 leaders. And you get that Similarity of HIV is approximately 0.395. So you want to plug in this information into our icebox here, There are 2.395 and the initial concentrations of H two and I two. We're just going to assume to be zero. And you want to subtract two X from the reactor because there is a coefficient two in the front and add X. To their products. And you get that your equilibrium concentration will be these and you have to plug all this information into your K. C. Equation which is K. C. Which they give you a 0.830 equals arm concentration of H. I. Or concentration of products which is H. two times concentration of reactions which is to over the concentration of H. I. So this will be X squared times 0.395 minus two X. Squared. And don't forget to square the um reactions as well because of the coefficient to there. So you want to solve for X in this case. And thankfully we have an easy equation to solve because we can just square root both sides to get uh huh square with both sides and To get rid of the square. And then you this will just turn into 0.9110 X. over 0.395 -2 x. And when we saw for X we get that X equals 0.1274 So we want to plug in our X into the equilibrium expression we have written here to find the equilibrium concentrations. So you get that H. I. Concentration of H. Two equals a concentration of I. Two, which is just X. When we found that the excess 0.127 Moller. And then the concentration of H. I equals 0.3946 -2X. Which is third point 1274 approximately. And that will give you 0.40 And you want to just just to know you just want to round everything at the very end because um you might get rounding errors if you run early on in the question, so you have three sig figs, and these will be your answers right here.

Consider the reaction to S. 02 gas plus oxygen gas yields to s. Of three gas. And the K. P. Value at 950 kelvin is 9500.355. If you have a 2.75 later reaction vessel at 950 kelvin and initially contains 9500.1 moles of S. 02 and 20.1 moles of O. To calculate the total pressure in the reaction vessel when equilibrium is reached. So um I need to know because we're looking, we have K. P. I need to know the initial pressures of my S. 02 and 02. So we can use pivot to find that. Mhm. Um So pressure is what we're looking for. The volume is 2.75 liters. We have 0.100 moles are gas constant is 0.8 to one. And our temperature in kelvin is 950. If you saw for pressure, you get 2.84 80 M. And you know that it is the same for both 02 and for S. 02 because they have the same number of moles. Now we can go ahead and make an ice chart. We have the two S. 02 plus 022, S. 03 and I have 2.84 80 M for both of those and none of the S. 03. So I know the products will increase. The reactions will decrease and then it will uh increase and decrease strike geometrically like that. So at equilibrium, I know this is 2.84 minus two X. 2.84 minus X. And then just two X. If I write my K. P. Expression, I know that it is s the pressure of S. 03 squared over the pressure of S. 02 squared times the pressure of 02. I know my K. P value is 0.355. And when you have two X squared over 2.84 minus two X squared times 2.84 minus X. Now you need to solve for X. And you get 0.664. So if you put that back into our eyes chart, you get the values when you subtract two X. You get 1.51. When you subtract X, you get 2.18. And when you multiply by two you get 1.33. So all you need to do is add those three values together to get the total pressure at equilibrium. And you get 5.0 to 80 M total. So


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