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(a)At $500 mathrm{~K}$, the equilibrium constant for the reaction $mathrm{H}_{2(mathrm{~g})}+mathrm{I}_{2(g)} ightleftharpoons 2 mathrm{HI}_{(g)}$ is 24.8. If $frac...

Question

(a)At $500 mathrm{~K}$, the equilibrium constant for the reaction $mathrm{H}_{2(mathrm{~g})}+mathrm{I}_{2(g)} ightleftharpoons 2 mathrm{HI}_{(g)}$ is 24.8. If $frac{1}{2} mathrm{~mol} / mathrm{L}$ of $mathrm{HI}$ ispresent at equilibrium, what are the concentrations of $mathrm{H}_{2}$ and $mathrm{I}_{2}$, assuming that we started by taking HI and reached the equilibrium at $500 mathrm{~K} ?$(a) $0.068 mathrm{~mol} mathrm{~L}$(b) $1.020 mathrm{~mol} mathrm{~L}^{-1}$(c) $0.10 mathrm{~mol} mathrm{~

(a) At $500 mathrm{~K}$, the equilibrium constant for the reaction $mathrm{H}_{2(mathrm{~g})}+mathrm{I}_{2(g)} ightleftharpoons 2 mathrm{HI}_{(g)}$ is 24.8. If $frac{1}{2} mathrm{~mol} / mathrm{L}$ of $mathrm{HI}$ is present at equilibrium, what are the concentrations of $mathrm{H}_{2}$ and $mathrm{I}_{2}$, assuming that we started by taking HI and reached the equilibrium at $500 mathrm{~K} ?$ (a) $0.068 mathrm{~mol} mathrm{~L}$ (b) $1.020 mathrm{~mol} mathrm{~L}^{-1}$ (c) $0.10 mathrm{~mol} mathrm{~L}^{-1}$ (d) $1.20 mathrm{~mol} mathrm{~L}^{-1}$



Answers

At $350^{\circ} \mathrm{C}, K_{\mathrm{eq}}=1.67 \times 10^{-2}$ for the reversible reaction $2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}^{2}(\mathrm{g})+\mathrm{I}^{2}(\mathrm{g}) .$ What is the concentration of HI at equilibrium if $\left[\mathrm{H}^{2}\right]$ is $2.44 \times 10^{-3} M$ and $\left[\mathrm{I}^{2}\right]$ is $7.18 \times 10^{-5} \mathrm{M}$ ?

In this problem, we're taking a look at the equilibrium between two n o and into an 02 where you're given the equilibrium constant K C and the starting concentration of N O were asked to find the equilibrium concentrations of these species. One way to do this problem is to do a full ice table. You can set your initial concentrations knowing there's 0.175 Moeller of ino, and you start with zero of each into an 02 You then are going to lose some of your ino, which will write his two ex because it has a two coefficient here and gain into and go to the reaction ships to the right at equilibrium. You then have 0.175 minus two x, and you have X here and X here. At this point, you could go through and use the quadratic equation in combination with your equilibrium constant to solve for your final concentrations of into an 02 But there's also a faster way to find their final concentration. If we take a look at K C here, we will see that it is time to 0.4 times 10 to the third, which is a large number. It's greater than 1000. And so we're going to say We know this reaction falls most of the way on the side of the products. As a result, the two X that we lose here is going to be very close to the full original amount. So we're going to say two X is almost the same as their starting concentration of 0.175 roller. As a result, sex itself is approximately equal 20 point 087 and our final concentrations of into an 02 are just X as a reminder. This only works because we know our equilibrium constant is large. And so we assumed that we lose almost all of our N O to find the exact concentration of Nou would have to use the quadratic equation since we have a large and non negligible X here. But this is one way to think about solving the problem. If you need to quickly find out what the concentrations of into an 02 r

So here we've got an equilibrium expression that were given some starting concentrations for materials and were asked to calculate with the concentration of each species is gonna be once this reaction reaches equilibrium. So if we take a look at this closely, the values that were given a beginning are that the mill aridity of nitrogen oxide, his point to Mueller and that the concentration of oxygen gases 0.1 Mueller is our initial concentrations. But one thing that we should take into account with this problem is how large this casey value is. Casey values huge. It's the number of times 10 to the fifth. So what we can assume from that is instead of doing this as these values being are starting equilibrium, we should consider these, um, this reaction going totally to completion and actually consider it as if the reaction we're starting from the products and proceeding with reverse to their reactions. So to figure out our starting amount of nitrogen dioxide, we actually have to figure out which of these re agents are going to be limiting to ensure that we choose the right polarity value. So we're just gonna do some very quick straight geometry here. So point to O molds for leader uh, nitrogen oxide times a conversion factor of two moles of enter, too, for more. I know, for two moles of mm, Sorry. They're 1 to 1 ratio. So if this reaction, we're totally to go to completion, we would wind up with point to Mueller and over to. And if we look at this in terms of oxygen, what one No more oxygen for leader. Two moles of nitrogen oxide for one more a vote, too. And we also get point to a mall. So this is a case where the amounts of these that we have in our reaction vessel we're going to be equal so and this because they both produce the same amount instead of using This is our initial for a change to reach equilibrium, we can assume that we're starting with neither of these to start with 0.2 Moeller nitrogen dioxide. So now the steps are going to be very much the same to what we're used to do him. There's going to be some amount of change in the species. The easiest thing to do is to represent those as changes in X using the strike Yama trees in the equation. So here we're going to lose two moles of nitrogen dioxide for everyone. Mole of auction that we gain and every two moles of nitrogen oxide that we gain. This is our change, and then our final concentrations therefore going to be two x for the nitrogen oxide X for oxygen and point to minus two X for the nitrogen dioxide. So now the same way that we do in all of our problems, we're gonna take these values and take them as our concentrations and plug them into our equilibrium expression. So we have Our Casey is equal to 2.3 times 10 to the fifth, and we know that that's equal to the concentration of nitrogen dioxide squared on top, which here we're going to substitute out for 0.2 minus two x in that whole terms gonna be squared and we're going to divide that by the concentration of nitrogen oxide squared and the concentration of oxygen. No one important thing that we can do in this problem is because it's equally remiss so high we can assume that the shift backwards if we were to start with all nitrogen dioxide and shipped back towards towards the reactions, we can assume that this number here is gonna be very small compared to point to. So we're just going to assume that that number is zero, which is gonna leave us with 0.2 On top of this fraction, we simplify this bottom again, bottom a bit we're gonna have for X squared times X, which is for X cubed. Now, if we take these away through, we do some quick arithmetic and and, uh, rearrange a little bit, we will end up with this. Where excuse he put 1.74 times 10 to the negative seven. Finally, it would take this forward. We find that X is equal to 3.4 times 10 to negative third. So what that means is that the concentration that's going to be equal to the concentration of oxygen, the concentration of nitrogen oxide is gonna be two times that. So we're going to have 7.8 times 10 to the negative. Third is equal to the concentration of nitrogen oxide. And really, if you think about how small this value is in comparison to our starting concentration for Reggie and dioxide. Concentration of nitrogen dioxide really isn't going to change. Um, if you want to be very specific about it, it will drop it down to about 1.98 or so. 1492 Um, but again said, assuming that that change was negligible was very accurate. So these are the final concentrations for these species in this reaction, that equilibrium.

So it's start this question. We're told that a five litre reaction vessel is initially filled, so we've got each to initial be one more over five leaders. Give me a minority a 0.200 Moeller I to initial. It's also one Moeller five liter container went to users were Mueller and each eye you sure is 2.50 moles in the five liter container, which is 0.500 Moeller. Your equilibrium would be each too I to equilibrium to each. I guess equilibrium expression would be products of a reactant CE. And we're told that this is equal to 129. Since this value here is greater than one. We know that the products of favor so we can set up a nice table with our initial values. This was your 0.2000 point 2000.500 No, we're shifting towards the right minus x minus X plus two x Europe went to 00 minus sechs 0.2 zeros real minus X 3.500 plus two x. Let's leave the base there as we can come back and fill in those equilibrium values. Uh, plug this into our Casey expression. 0.500 plus two acts squared over 2.200 minus x 0.200 minus x. Let's go ahead. And so for X. Take a minute. Indeed, that sewing for X X will yield to roots 0.30 and through a 0.13 one to reject this route as it'll heal negative values. Let's take this street back up to our ice table and the equilibrium concentration of each to would be what? 200? Subtract 2000.13 minus X. This would give me 0.70 Moeller. This would be 0.70 Moeller be the same, and this would be a 0.5 plus two x, which would yield 0.76

To calculate KP for this reaction. First, we need to figure out what the equilibrium set up is gonna look like. So it's always products divided by reactant. And we have to raise each, um, concentration to the appropriate power based on the coefficients. So we have n o squared, sometimes be are too over an o b r squared. Now, based on the information we're given, we know the partial pressures of the different gases. So we have an O which is gonna be four sweet up to four squared times two for the protein over four squared for the N O V E r. Right. When we do this right, you could do this math. It's pretty simple to see that the four square it's work cancel out. So our actual answer here will be too


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