Consider the gas-phase reaction of AB3 and A2 molecules:

(a) Write a balanced equation for the reaction. (b) What is the sign of the entropy change for the reaction?

Video transcript

well every once in this video, we're given this illustration here. This is the gas phase reaction between X two and gray and Y two and orange are being asked for part a to provide the balanced reaction equation ever be to predict the sign of the change of entropy. So delta S. For the reaction. So on my left here we can go ahead and count. So we already said that the gray is going to be X two. We have 12345 here. And then for my orange really said that is Y two. Let's go and count that as well. We have 12345678 and nine. Over to the right what this is, this is my XY three. And then this right here in gray is our X two. So for xy three we have 12345 and six. And then from X two S, we have one and two. So we see here that Y two is an excess. Let's go ahead and write out our chemical reaction. So maybe we'll just do this in red right next to our part. A question. So we can see here just from counting how many we have of each. We have five X two, which is in the gaseous state plus nine Y two. Again, that's in this gaseous state and this yields six moles of our X Y three as well as two moles of our X two, we go ahead and since we know it's an excess kind of subtract this out with this and this gives us three X two in its gaseous state directed with nine moles of Y two and its gaseous state to yield six moles of X. Y three again in its gaseous state. So the bass chemical reaction. If we simplify this by dividing each coefficient by three, he gets one mole X two, gas reacting with three moles of Y two gas which yields two moles of our xy three gaseous products. That's my answer for part A. Here. Then moving on to Part two B, which is to predict the sign of the chain of entropy for my reaction. So here what we're using us, the determinant factor is the phase. So the gas phase has the highest entropy followed by the liquid. And then of course our solid phase being the one with the lowest entropy. So again we're ranking from gas too liquid two solid. So before my reaction. So basically my starting material, so I just put S. M. We have 14 gas molecules And my reacting. So after I just put p. r. for product, we have a total of eight gas molecules. So we can see here if we're going from 14 to 8 of just gas molecules that we have decreased entropy because we have decreased kind of randomness. So this means that my delta s value will have a negative sign. So this right here is going to be my answer for part b. And this is my final answer for this entire problem.