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For the reaction: 2 N 02(g) < NO(g) + 02(g), the partial pressure of 02(g) at equilibrium is 0.3500 atm. Ptotal 1.0866 atm. What is Kp for this reaction? A) 2.86...

Question

For the reaction: 2 N 02(g) < NO(g) + 02(g), the partial pressure of 02(g) at equilibrium is 0.3500 atm. Ptotal 1.0866 atm. What is Kp for this reaction? A) 2.86 B) 128 C) 0.350 D) 182 E) 66.9 Answer:

For the reaction: 2 N 02(g) < NO(g) + 02(g), the partial pressure of 02(g) at equilibrium is 0.3500 atm. Ptotal 1.0866 atm. What is Kp for this reaction? A) 2.86 B) 128 C) 0.350 D) 182 E) 66.9 Answer:



Answers

For the reaction $A(g) \rightleftharpoons 2 B(g)$, a reaction vessel initlally $\operatorname{con}$ tains only A at a pressure of $P_{2}-1.32$ atm. At equilibrium, $\mathrm{R}=0.25$ atm. Calculate the value of $\mathrm{K}_{\mathrm{p}}$. (Assume no changes in volume or temperature.) MissED THis? Read Section $16.6 ;$ Watch ME 16.5

In this question, we've been given a reaction between hydrogen and I obtained form too H. I. So if we are to look at this, we know that the equilibrium constants say K. P. We know this to be the ratio of the partial pressures of the product and the pressure pressure of the reactor much operation of three. And and we always have to remember to raise partial pressures to the power of theirs documentary. More efficient. So we have to square the partial pressure of H. So looking at this, if we couple it R K. P. We have K. P. Being equal to for the first set, we have 0.2 squares divided by 0.98 five times 0.877. So if we look at this hour K. P. It's going to be equal to 4.76 times 10 to the power negative four. Now, for the second part we can calculate our keeping and our QP is it is still he H I squared divided by p partial pressure of tyrosine, multiplied by the pressure pressure of I think. But we've been told that the pressure operation of 18 is equal to the pressure operation of hydrogen. So this is going to be equal to serve 101 squared divided by 0.6 to one square. So R Q. P. Is equal to 0.0 to six. Now, if we compare our QP. And our our K P can tell that R Q. P is greater than a P. So what this tells us is the reaction will not be equilibrium. Therefore it will be shifted towards the left. That is if we are to look at this reaction, if Q P is greater than K. P means it is not at equilibrium. So this reaction is going to be passed towards the left. That is we are going to form less off our products. That is a fact. So moving on, let's write our reaction. Say we've got each to end I too. And finally we have H if we are to just ride R I C. E table where we have the initial change and equilibrium. Remember these are always evaluated equilibrium. So the initial that we have this is 0.6 to one and we also have the 0.6 to one here. So initially before the reaction as the reaction progresses, the initial what we have here is zero point 101. So say hydrogen changes by effect of negative, it is decreasing hence the negative sign. If we look at our equation one more of hydrogen and one more off, I think they produced two moles of H I. So if hydrogen decreases by a factor of X, it means our team are also decreased by a factor of X. Mhm Our H I is going to increase by a factor of two. X. Because we are looking at this documentary coefficients, the moralistic geometric coefficients one more of these produces two moles of H R. So it equally from what we have is 0.20 point 6 to 1 minus X. This is the partial pressure of hydrogen and this is that off I 18. And at equilibrium we have zero point 101 plus two X. These are the partial pressures at equilibrium. Now if we put this into our remember we calculated our K. P. Here. So the K. P expression is still the same. This is the K. P expression and we calculated our K. P. Here. So if we substitute these equilibrium, we're going to have 4.76 times to enter the one negative four being equal to the partial pressure H. I squared which is zero point 101 plus two X. And we have to square this divided by zero point 6 to 1 minus X. And since these are the same, we can just square is So if we solve this, if we saw this, we get our X. Being equal to negative 0.0. Well now calculating the partial pressure of H. I. Partial pressure of H. I remember our 101 miners, this is plus two eggs. So the pressure pressure of H. I. This is going to be 0.10 one plus two, multiplied by the X. That we calculated there. And this is going to be a P. H. I. Is equal to 0.0 for one. And this isn't atmosphere

Look, we're told that the first reaction mixture is at equilibrium, so let's start with this. So the first reaction mixture, since it's at equilibrium that's calculates our equilibrium. Constant partial pressure of H I squared over the partial pressure of each two partial pressure of I, too, so we can plug these values in and calculator equilibrium constants. So this should be 0.20 squared 0.958 0.877 and calculating or equilibrium constant. We will find that it is equal to 4.76 times 10 to the minus four. So from the second reaction mixture were asked, Is this mixture at equilibrium? So what will have to do is we'll have to calculate the quotient from our initial values given. Let's go ahead and calculate the caution here. And this will be put 101 this week 0.6 to 1 in 0.6 to 1. This would be equal to 2.264 That's compared values quotient. 0 to 64 breath or equilibrium constant 4.76 times 10 to the negative four. Our quotient is larger then our equilibrium constant. So we're not at equilibrium. The reaction shifts left to reach equilibrium. And no knowing this. Let's determine the partial pressures, Uh, or the partial pressure of H I. Once equilibrium is reached. So we'll have to do set up a nice table here. Do you determine that our initial values we already have that were given to us? Um, 0.6 to 1 went 6 to 1 in 0.101 We know that it's going to shift towards the left because of our kosher that we just calculated so minus two x plus x plus x 20.6 to 1 plus x When. 61 plus x two x i equilibrium constant to find it again, and we know the value. It's 4.76 times 10 to the negative, for which we calculated previously two X squared over 0.6 to 1 plus x and 0.6 to 1 plus x solving. We get a quadratic equation here. Solving for X will yield a route here, uh, 0.433 and you can go ahead and verified that that is route that we get and then using that, the partial pressure of each eye at equilibrium back to our base table. We define that as two X. Therefore, this will be equal to one survey. Um, it's this wrong. I have the right answer here, but, uh, this is incorrect. Here. My, This is 0.101 minus two X 0.101 minus two x squared. That is rate. Uh, this is incorrect here. So everything else is right on this war. Go to put one's your one minus two X and therefore the partial pressure of H I at equilibrium 0.11 minus two X would be equal 2.0 144 atmospheres. So there is our answer to this question.

In this question, we're looking into the reaction where we're here and to reacting with or two at equilibrium with two miles off and off. So we'd like to calculate the ap and by definition the K. P. Is the ratio of the concentrations of products to that of the reactor. So if we are to look at this in terms of gases, this is going to be the pressure pressure of the product which is the partial pressure of an or divided by product of the partial pressure of the reactions. Which is into multiplied by partial pressure off. What so always helped her remember to raise the respective concentrations to there's documentary coefficients, the reaction we raise the partial pressure of nitrogen and oxygen to one. But here we've got two moles of N. O. So we raised this to the power of two. Are looking at the information that we've been given. We also we have already happened Is precious. So this is 0.05 squid divided by the pressure pressures of N. Two and 02. Which arm 0.155. Now our K. P. If we make this calculation, our K. P. Is going to be yeah 0.051

So we have the following reaction and the reaction. I equal a room at 175. Calvin is equal to or has a following, um, pressures of h two i to end it, John. Now we have a second reaction at 175 Calvin ago and were asked, is the second reaction at equilibrium. First calculates the KP constant or equilibrium constant for our actual You go with him so we have happy is equal to product supporter pressure of h I squared. I have cold French coefficient of two. An MP of I two p of h two. Well, what do we have for four wing pressure? The pressure of age eyes point girl to Girl Square, that pressure of eye to his point 877 and the pressure of I two. Excuse me? H do is 0.9 a. What? Well, that gives me 4.761 times 10 to the negative four for my equilibrium pressure. Now let's calculate our second reaction and see if it's at equilibrium. So we're actually just calculating the q P because it might not be at equilibrium, So we have the same formula. So that the pressure of each eye, which is 0.101 grid over the pressure of H two, which is point 6 to 1. And the same is for I just did it squared. So we get the whole point girl too. 65 Well, we see that the Q P value is actually greater than our K p value. So actually, our reaction is not any fool of them, and it's actually proceeding to the lock. So too, of h I. Wonder Reaction week is equal. Oh, Graham, we need to create a nice table. So we have. It's too I to and H Now we know our initial values. Their 0.6 to 1. What do you want? And 0.1 barrel one. Okay, so if you look back at our equation, we see that there's a two in front of age I. So let's multiply that by two in our product and are arbitrary change about in the consecutive X negative X symbols in our reaction site, you get points 6 to 1 minus 0.621 minus act 1.101 But to act okay, so our K P is equal to 0.1 there, one plus two acts squared, provided by 0.621 minus squared. And this holding is equal to our K p value that we found earlier, which is four point 761 times 10 to the fort. Now, when you help for on, it's the perfect squares involved. When did you take the square root of both sides to arrive at a linear equations? So if we take the square readable sites, we get 1.1 invited by 0.6 to 1. Mine, in fact, is equal to the square root of 4.761 time attention and a girl before now I'm gonna multiply our denominator in our lifetime side on both sides. So we're left with 14.1 plus two X is equal to the square, it times 0.6 to 1. Okay, now, if you were so Rex, you would get that X is equal to negative 0.0 for 3 to 5 30 and we get a negative because we know about our reaction is, um, shifting to the left. So we actually have to pick the actually when we saw for a quadratic equation that is negative. So once we have this ex ality we have, our pressure of H im is equal to what did we say it was equal to based on our table, that is 0.1 girl one plus two x plus two times or xlu, which gives me still 0.1458 p. M. And this is our partial pressure of eight.


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