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The equilibrium constant is .090 at 258C for the reaction:HzO(g) ClzO(g) = 2HOCI(g)For each of the following sets of initial conditions_ at 258C, in which direction...

Question

The equilibrium constant is .090 at 258C for the reaction:HzO(g) ClzO(g) = 2HOCI(g)For each of the following sets of initial conditions_ at 258C, in which direction will the reaction position shift (significantly)?RigntP(HzO) 429_ torr, P(ClzO) 88.50 torr, P(HOCI) 93.53 torr _RightP(HzO) 292.0 torr, P(ClzO) 35.60 torr, P(HOCI) 30.59 torr.A 5.0-L flask contains 2.455*10-2 mol of HOCI, 6.542x10-3mol of Clz0, and 6 550x10-1 mol of HzoRightP(Hzo) 490.0 torr, P(CIzO) 44.40 torr, P(HOCI) 30.97 torr.A

The equilibrium constant is .090 at 258C for the reaction: HzO(g) ClzO(g) = 2HOCI(g) For each of the following sets of initial conditions_ at 258C, in which direction will the reaction position shift (significantly)? Rignt P(HzO) 429_ torr, P(ClzO) 88.50 torr, P(HOCI) 93.53 torr _ Right P(HzO) 292.0 torr, P(ClzO) 35.60 torr, P(HOCI) 30.59 torr. A 5.0-L flask contains 2.455*10-2 mol of HOCI, 6.542x10-3mol of Clz0, and 6 550x10-1 mol of Hzo Right P(Hzo) 490.0 torr, P(CIzO) 44.40 torr, P(HOCI) 30.97 torr. A 2.0-L flask contains 4.700x10-2 mol of HOCI, 8.500x10-2mol of Cl20, and 1.708x10-1 mol of Hzo.



Answers

The equilibrium constant is 0.0900 at $25^{\circ} \mathrm{C}$ for the reaction $\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$. For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. $P_{\mathrm{H}, \mathrm{O}}=1.00 \mathrm{atm}, P_{\mathrm{C}, \mathrm{O}}=1.00 \mathrm{atm}, P_{\mathrm{HOC}}=1.00 \mathrm{atm}$ b. $P_{\mathrm{H}_{2} \mathrm{O}}=200 .$ torr, $P_{\mathrm{C}, \mathrm{O}}=49.8$ torr, $P_{\mathrm{HOC}}=21.0$ torr. c. $P_{\mathrm{H}_{2} \mathrm{O}}=296$ torr, $P_{\mathrm{C}_{2} \mathrm{O}}=15.0$ torr, $P_{\mathrm{HOC}}=20.0$ torr.

In this problem, we are looking at different conditions for the partial pressures of all of the gases. In this equilibrium reaction, we're going to figure out which of these equations o in the following conditions are at equilibrium and which wants him to shift in a certain direction to achieve equilibrium. So what we're going to be using an order to figure out how to solve this problem is que so Q is a lot like our K constant, except que is calculated at any condition. It does not have to be at equilibrium. So in this case, we're using partial pressure. So Q is going to be equal to the partial pressure of the product, which is h oh C L raised to the power of its coefficients that squared and then divided by the product of the partial pressure of its first reactant each to a water vapor times the partial pressure of its second reactant, which is C. L. 02 So again, Q is a lot like a, but it's not at equilibrium, necessarily, or at least not specified that we're calculating at equilibrium. Um, but it does look a lot like a so we're going to be comparing it to K to figure out whether we're at equal of them or not. And we're actually given in the problem K, which is actually KP here. But it shouldn't matter. The KP in the case, you should be the same, um, but the Kate. But that's only for this problem. Um, since there are the same number of moles on both sides of be equation, KP and Casey should be the same. Um, but in this case, K P is equal 2.900 So let's start plugging in. Our partial pressure is for Q. So in court A were given. That's our first set of conditions. We have one atmosphere of everything, so I have one a t m for water vapor, 14 c l 02 and then one for h o C l. So now let's plug in for a que que is going to be equal. Teoh, one 0.0 squared, invited by one times one. So it's going to be equal to point. Oh, sure, no one is going to be able to one which is greater than I should do. The right number of sick, big. So that's going to be three sig physics. One point is euros your own, and that is going to be greater than K, which is point 0900 just k. So if our Q is larger than our K, that means we have too much product. So how do we know that? Well, if we look at the equation for a que, we have the products partial pressure on top. So when we have an excess amount of product, it's going to have a higher partial pressure than it would have equilibrium. And that's going to make our quotient larger. Um, so in order to reverse that, we actually need to shift towards our reactant. So when we shift towards our reactant water vapour and Cielo to, um, when we do that, we lessen the amount of excess products and we increase the amount of reactant so or increasing the denominator and we're reducing our numerator. That's going to give us a smaller portion, eventually allowing us to reach K. So in order to reach equilibrium, meaning to shift towards the reactant since that's going to be shift left. So we're going to be doing the exact same thing for the next two parts of this question. So let's start with print. Be part B. In this case, we have our partial pressures and tour. You don't have to worry about the fact that we're in a different unit since que has no unit K and sister ratio. Um so is que We have 200 for water vapor for C l 02 you're gonna have 49.8 tour and then for h o C l. The partial pressure is going to be 21 went zero tour. So again we're going to plug in for Q. So he was going to be equal to the partial pressure who are products. So that's 21.0 square. It divided by be partial pressures. Um, they're the product of the partial pressures of the reactant since so it's going to be 200 multiplied by 49.8. So that is going to give us point zero who are 43 which is less than K, which is again call, even write it again 0.9 So since we have a cue that is less than K, that means that we're going to have an excess of reactant since. So the reason for that it's similar to the previous problem but the opposite when we have a large denominator in this case, when we have large partial pressures for the reactions, Um, that is going to give us a smaller portion. So in order to reverse that, we want to make the numerator bigger. So we want to shift towards the products that will decrease the denominator increased the numerator, allowing cute increase until it reaches K. The same value is K. So again, we're shifting towards the products. We are shifting to the right. Finally, we're gonna do see So we're gonna do this again the exact same way we did the others. This case we have 296 tour for water vapor for C l 02 will have 15 tour and then for h o C l. We will have 20. So it's going to give us que so we'll have 20 squared went top divide that by 2 96 times 15 that is going to give us 0.901 were actually end up having to round up, which is just a little bit bigger than K, so we'll have a little bit excess product just a little bit. So the equation will shift slightly to the left to decrease that extra product, get a sequel of Graham. That is how you answer these three parts of the question.

In this question we've been given to states that is the initial state and the final state. And have been asked to determine the standard and the change associated with that change from state from initial conditions up to final condition. So when we are given this type of information and the one equation that comes to mind is Lynn K two over K one is equal to the standard entropy change. Or the reaction divided by our this is negative, then we have one over T two, My nerves worn over to you. What? So all we want to do here is to calculate these standard and change. So the main task here is to be determining the unknowns here in this equation. So that we plug them in here for us to then make these standards and they'll be change the subject of the formula. So the first thing that we're going to do is to determine our K two and we know that kato K two. The K here is the equilibrium constant which relates the partial pressure of the products to that of the reactor. You've got extra being converted into X. Sure to mel's ex. So the K. P expression here is going to be partial pressure of X squared divided by the partial pressure of X two. So remember this is an equilibrium constant. So these are not just the partial pressures but the partial pressures at equilibrium. So the main task will be to determine these partial pressures at at equilibrium. This we're going to do for K1 and our K2 that were then playing to substitute into this formula for us to then determine the standard and hope for change of that reaction. So, first of all you have to have an I see your table when we have extreme And two moles of x. Remember we have the initial and we have a certain change and we use those two will add those to determine their pressure, it equilibrium. And if we are to look at this, we've been given to us. So we always have to make it happy to convert these into into atmospheres. So for example, P X two, this is going to be equal to remember this is the initial the initial pressure P X to not This is 7:55:00. And for us to convert to atmosphere, we divide by 76, Then this gives us 3.9 93 for two in must feed and P X P x p at equilibrium. This is going to be equal to 103, 103. You're divided by 760. So this is going to be 0.13 553. And this is also in in atmosphere. So if we have got an initial pressure here, this is going to be 0.99 342 atmosphere. And before this equation starts that is under initial conditions there won't be any products or this is going to be equal to zero. So if we are to look at this, Let's say this changes by a factor of X, we're going to say -1. This negative sign shows the partial pressure is decreasing with time because this is a reaction. So if we look at this we have extra that is forming two X. So if this changes, if X two changes by a factor of X, one more of X two is forming two moles of X. So if exchanges by a factor X, it means this is going to change by a factor of two X. Because we're looking at this documentary coefficients one institute, so this is going to be a positive. So at equilibrium we're going to have 0.99 three or four two minus X. And equilibrium, we're going to have the partial pressure of X is going to be 2X. But we know the pressure of of X at equilibrium. This is what will take your mind here. So what we're saying here is our two X is equal to 0.13553. So the value of X here is going to be equal to zero 0677 65. So the partial pressure off X two at equilibrium, this is going to be equal to we determine this B this expression right here. So this is going to be equal job 0.0.99345 -1. And this is equal to 0.99 345 0 0.0 67765. So the partial pressure of X to equilibrium, This is going to be equal to 0.1355. And this isn't atmosphere. So we now have we now have the partial operations at equilibrium. We can determine our K one Remember our K one. This is equal to the partial pressure of X. Raised to the power of two divided by the partial pressure of extra equilibrium. And this is going to be equal to zero 1355 three. And we have to square this divided by the partial pressure of extra at equilibrium. And this is going to be equal to 0.01 zero point This is 0.9 0.9 two. This is 0.92 56. Firefighter who make this subtraction here. This is going to be equal to 0.9 two. Funny 655. An atmosphere. So moving forward We now have our K one. Okay, one being equal to 0.0 1984. Yeah. Now that we have our K1. We then move on to determine our K two and we've been told that the initial the initial pressure of X two. Initial pressure of X- 27482. We divide this by 760 and that's going to give us 0.9 eight 4 to 1. And this is an atmosphere. And then pressure pressure of X. This is five 32 divided by 76. This is equal to 0.7 atmosphere. This is the partial pressure of X at equilibrium. This is the partial pressure of X two initial. So if we have our I see table were X to end X. The initial pressure here is going to be zero 98421 and zero. And these changes by a factor of negative X. And this will change by a factor of positive two X. So if we make this this is going to be equality. This is now at equilibrium. This is going to be a zero 98 4: one -X. And this is going to be two x. And the two X. Is the pressure at equilibrium which is zero point cinnamon atmosphere. Therefore X is equal to zero 35. So the partial operation of extreme at equilibrium Partial pressure of extra equilibrium is 0.9 84 2 1 -X. Which is open 984 2, -0.3. Right, and this gives us 0.634- one. So our K two is going to be equal to 0.7. This is the pressure equilibrium and we have to square this divided by pressure pressure of X. To its equilibrium Which is 0.634- one. So our K two here is going to be equal to 0.77261. Now that we have our K one and K two, we can then use this huge formula to determine our delta age standards. So if we make the delta age subject with the formula we're going to have and or we change the standard and are we change being equal to our multiplied by Lynn K two over K one. All of these defended pipe one over T 1 -1 over you too. And this is going to be equal to 8.314, Multiplied by Lynn K2, which we calculated above here, Which is 0.77 0.77 261 to fight it by Our K one which is 0.01 0.01 9844. and all of this is going to be divided by one over 29, -1/755. And at the end of the day our delta edge, the standard end will be changed standard. And are we change this is going to be equal to one faith 1.5, Multiplied by 10 to the powerful And this is going to be in joe's back. Yeah. So if we simplify this, we can say the delta edge standard, This is going to be 15 in killer Charles pan. Yeah. So if we are to look at this question, what we did here, the main formula that we're supposed to recall here this formula and were given to conditions of two different temperatures that is 2, 9, 8 and 75 fights. So under these conditions we are going to have different equilibrium constants Because this reaction in which X two is converted into two moles of X is dependent on temperature. So if we change the temperature, we are going to have different okay Values K one and K two. So it is this change of state that we want to determine the entropy that is associated with that change. So in some of what we have is we have an initial condition Where we're at 2 98 Kelvin and we have a final condition which are going to call condition to where we are now at 755 resource killed. So it is this change of state that we want to determine the entropy change the standard entropy change associated with this change of state. So to do that what we did here, we're supposed to look at this formula, that standard and will change with the temperatures. And unfortunately, this is also associated with different K values and we know that our K is the ratio of the partial pressure of the product to that of the reactant. So this is what we're doing. Well. First of all had to look at the initial condition where we're at 2 98 Kelvin to 98 Kelvin and It is its own K one associated with it within determined K two K, one of that, the initial conditions. And then, because we're given special operations those conditions, that is it 2 98 and 875. So after getting our K values, This is when we then use them to plug them into this formula, we have our is 8.314, and we finally get our standard and probably change about to change.

In order to solve for, uh the way the equilibrium shifts will use the equilibrium h 20 plus. He'll too. Oh, unequal room to a juicy L E equilibrium constants Expression is given, and we see that that's equal 2.900 Calculate Hugh, which is the quotient, or trial Kiki Value, which is based on the initial concentrations of each of the reactions from products involved in the equilibrium. For a we need to solve the initial values initial values are sold for by dividing each of them all amounts by the one leader container April CEO Initial eyes one Moeller seals who Oh uh, like Lawry a die chlorine oxide initial is 0.10 Moeller and the water vapor is 0.10 Moeller when we substitute these values into the que with three quotient which to find up above on substituting these values in and solving gives us a question Top 100. We take the question to compare this the actual equilibrium constant. When we go ahead and do that, we see that the quotient 100 is much greater than the equilibrium constant of 1000.900 Therefore, since Q is larger than K, the equilibrium shifts left for Part B. We have a different set of conditions initially, so we need to solve for the initial values, like so dividing each of them all amounts by two leaders. Because this is a two liter container. This time we take all of those values and we substitute those into the expression for the quotient, and that gives us a Q value of 0.9 zero. Compare this to the equilibrium constant. We see that this is exactly equal. Q is exactly equal to K, so in this case, if they're exactly the same, the system is at equilibrium. For part C, we again need to solve the initial values here. For the three. We take the three initial values you can see, given their dividing by 53 leader container. This time, substitute into the quotient expression, which gives us a question to a truck. He Q value of 110 Compare this to the equilibrium constant in 110 key was much greater than the 1100.9 Therefore, since Q is greater than K, the equilibrium shifts left


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