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Ch.15 - Chemical Equilibrium

Chapter 15, Problem 48

The following picture represents the composition of the equi- librium mixture for the endothermic reaction A2 ∆ 2 A at 500 K. Draw a picture that represents the equilibrium mixture after each of the following changes. (b) Increasing the volume

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Welcome back everyone. The equilibrium mixture for the endo thermic process to X in equilibrium with X two at 600. Kelvin is shown in the following image, choose the image that illustrates the equilibrium mixture after increasing the volume. So what we want to reference to solve this problem is lush outliers principle. Recall that according to actually, if we disturb an equilibrium system, like our given reference to chemical reaction here, our reaction is going to readjust to maintain that equilibrium state. So we should recall that if we increase the pressure of the surroundings, that will correspond to a decrease in volume in our system. And we can understand that this will correlate to our equilibrium which favors the direction with less moles of gas. On the other hand, if we were to decrease the pressure of our surroundings, that will therefore result in an increase in volume within our system, which will correlate to equilibrium, favoring the direction where are moles of gas in our reaction are greater. So let's look at our reaction. We'll write it down to X in equilibrium with X two on the product side since we have just one mole of our product and two moles of our reactant, we can say that we have more or rather we'll use shorthand. So greater moles of gas on the reactant side. So we can actually write out a formal relationship where mold of our gaseous reactant is greater than our moles of gaseous product. In this case, X two. So according to the prompt, if we increase the volume, this will correspond to a decrease in pressure And therefore our reaction or will say equilibrium favors producing more of our X reactant. Since again, we have more moles of our gas X on the reactant side. And so therefore, equilibrium will shift left. So looking at our images, we can understand that the atoms that are two atoms bonded together or fuse together for one molecule will be our product X two. And then the single atoms isolated are our reactant X. So we want to count the image that has the more or rather larger prevailing of singular X atoms. Looking at our answer choices in choice one, we see that we have 12345678 moles of our gas X and two moles of X two R di atomic gas. In image two, we have 1234 moles of X. So we can already rule choice to out because again, we want equilibrium to favor the left direction, producing more of our moles of X. Looking at choice three, we have 123456 moles of X, which is still less than choice one. So we can rule out choice three. And looking at choice four, we have 12345678 moles of X And three moles of X two. So actually I made an error and in part one, it looks like there are exactly three moles of our X two product. And with that correction, then we would say that choice is one in four would technically both be correct answers. But notice that the image for box one is a lot larger. So the atoms are spaced out and that would definitely indicate a doubling of the original volume for our reaction. Whereas an image for the the molecules are a bit closer and our illustration is smaller as far as the arrangement in the box. And so we can actually go ahead and rule out choice for because as the prompt states the volume is being increased. So our only correct answer is going to be choice one, which is the only image that illustrates the equilibrium mixture after increasing the volume, which favors the production of higher moles of X our reactant. So I hope that this made sense and let us know if you have any questions.