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Problem 82

The scenes below depict three gaseous mixtures in which $\mathrm{X}$

(orange) and $\mathrm{Y}_{2}$ (black) are reacting to form $\mathrm{XY}$ and $\mathrm{Y}$ . Assume that each gas has a partial pressure of 0.10 $\mathrm{atm} .$

(a) If $K=4.5$ , which mixture is at equilibrium? (b) Rank the mixtures from the most positive $\Delta G$ to the most negative $\Delta G .$

Answer

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## Discussion

## Video Transcript

were given information in this problem and a diagram to describe a gas phase reaction between two different gases. X and Y. We are given color codes for finding which one of these particles corresponds to x y to x y and why? And we're told inwards what the reaction is between between these so that we can write it out as X plus y to it was two x y plus y In part a were told that the equilibrium constant KP For this gas phase reaction is equal to 4.5 and we want to know which one of the molecular scenes were were given. 12 or three corresponds to this reaction. Being at equilibrium, we can write out the equilibrium expression KP using this equation that we have written out, we know that we need to have the partial pressure of each gas in the product side raised to its tricky metric coefficient, and we need to multiply that value for each gas in the products and then divide by those seem quantities for the gases on the reactant side. So all of the striking metric coefficients in this equation are one, so we just have to examine the partial pressure of each gas in the products, multiplying together and divide by the personal pressure of each gas in the reactant and multiply them together. We're told that the partial pressure for each one of the gases is equal to 0.10 atmospheres. And so, for each one of the molecular scenes, that is 12 and three in the diagram. We have to go through and count how many of yes, X Why to gas x y, I guess, and why, I guess that we have based on the color scheme that we are told in the problem. However, many of each one of those molecules we see in the diagram we can multiply by 0.10 to get the total partial pressure of the gas in that specific molecular scene. And when we do that, we can form the reaction quotient formulas for each one of those those expressions and determine after we perform the division, which one of them is equal to that equilibrium value of 4.5. So we start with KP. We know that from this expression we had, it would be the partial pressure of the products to X y gas times the partial pressure of why gas divided by the partial pressure of X times the partial pressure of why to And we know that at equilibrium this is equal to 4.5. So now if we look at the diagram in the 1st 1 and we count first the number of X wise that we see in the X wise in the problem or one orange Adam bonded to one black Adam and we multiply the total number that we see for or those by 0.1. You see, there are four of them total. So when we want to play it by 0.1, we have partial pressure of X Y In the first scene of 0.4 atmospheres. These air all pressure units, we can say or atmospheres. So I'm not going to write out the unit every time. But you can if you want to, and then we move on to the next term. The number of why gas particles. So how many of those single black Adam's do we see in that in that molecular scene? Then we multiply the total number that we see by 0.1, you see that there are two of them, so that comes out to partial pressure of why it 0.2. And we divide that by the number of X. So the number of these single orange Adams that we see in that scene and and that comes out to 0.1, since we would find that there is just one of those and then the partial pressure of why to so many of those two black Adams that are bonded together to form that die atomic why to? And there is just one of those. So thats 0.1 atmospheres, then we just find that ratio, and we see that it has a value of eight, and this is not equal to equilibrium value of 4.5. So we therefore know that the the molecular seen in the first given diagram is not an equilibrium. We use that same method to go through and sell for Q two and Q three. So again, when you count up the the number of each one of these molecules in two and three and multiply them by the partial pressure of 0.1 atmosphere, you should get work. You two 0.2, I'm 0.2 divided by 0.3. I am 0.2 and that should come out to about 0.67 Which again means that this is not at equilibrium, since it's not equal to 4.5 finally, for Q three should get 0.3. I am 0.3 divided by 0.2. I am 0.1 and it turns out that this does come out to 4.5 insistence that matches the equilibrium constant value of 4.5. We can conclude that the molecular seen the third diagram is the one at equilibrium. Out of the three now, her B we went to you have a ranking for the the highest changing Gibbs free energy down to the lowest. So this is the highest Delta G, and this is the lowest Delta G between the same three molecular scenes that we were shown for part A. And in order to determine that we we have this equation where we can compare the values of Q and K. You know, the K is always 4.5 and we calculated values of Q or 12 and three. And so based on this, we can see that if que is is greater than K, then that means that Delta G will be greater than zero. The reason for that is that if Q is greater than K, then this denominator value is greater than the numerator to make that that ratio come out to be less than one. When we take the Ln of a number that's less than one, it becomes negative. But with this negative sound sign out in front, that will be a positive value. Likewise, if Q is less than K, then Delta G is less than zero. And if Q is equal to K, then we are in equilibrium and Delta G equals zero. Now, if we go through starting with the first molecular scene, we see that key was eight and K is 4.5. So for the 1st 1 we say that Q is greater than K, and when Q is greater than K, Delta G is greater than zero. So all we can say qualitatively right now that we know Delta G is greater than zero or the first diagram going on to the next one. We see that Q 0.67 is less than K 4.5, so Q is less than K this time, and we know that that means that Delta G is negative. So Delta G is negative. And because of that, we can say that Delta G for the 1st 1 is greater than the 2nd 1 since the first ones positive positive in the second one's negative. Finally, for the 3rd 1 we know that this is an equilibrium, so Q equals K ensue. Delta G equals zero, so we can say that three is greater than two, since two is negative and 30 but three is less than one since one is positive and 30 So when we write that all out for our ranking, we have one has the highest Delta G since it's positive that has a greater Delta G than three, which is at equilibrium and has a delta G value of zero just greater than the lowest Delta G or the only negative. Delta G, which is present in diagram number two

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