Problem

How will an increase in temperature affect each o…

01:54

Problem 39 Hard Difficulty

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?
(a) $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \qquad \Delta H=92 \mathrm{kJ}$
(b) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad \Delta H=181 \mathrm{kJ}$ (c) $2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g) \quad \Delta H=-285 \mathrm{kJ}$
(c) $2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g) \quad \Delta H=-285 \mathrm{kJ}$
(d) $\mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CaCO}_{3}(s) \quad \Delta H=-176 \mathrm{kJ}$

Answer

a) decrease in volume of vessel will shift the equilibrium to left direction.
b) decrease in volume of vessel has no effect on equilibrium.
c) decrease in volume of vessel will shift the equilibrium to left direction.
d) decrease in volume of vessel will shift the equilibrium to right direction.

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Chemistry 102

Chemistry

Chapter 13

Fundamental Equilibrium Concepts

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Video Transcript

for this problem, we're given a list of equilibrium reactions. More has to consider how these equilibrium are going to be affected by an increase in temperature and buy a decrease in volume servers. Let's think about what these two, um, changes we're going to mean in general. So changes in temperature going to affect equilibrium based on the entropy for that reaction. So let's hypothetically consider for a second reaction that has a positive and they'll be. That means that the reactions endo thermic, and that means that heat is one of the reactant CE the least you can think about it this way, so because he doesn't react if we increased temperature, that's the same that's similar to adding heat. So we have heat is a reactant we increased temperature. That's going to drive the reaction in the forward direction by the same rule. If we remove heat and we think of heat is being a reaction that's going to cause the Q value to decrease below, the K is gonna shift the reaction to the left. Conversely, you think about the second part for this problem. We think about what a decrease in volume means decrease in volume is going to be proportional to an increasing concentration because we think about in instantaneous decrease in volume. Since the number of molecules and your reaction vessel haven't changed yet, same thing is condensing them tighter, so a decrease in volume is gonna cause an increase in concentration. So because of that, it's going to be dependent on the relative concentration of your products over your reactions to determine which direction the reaction is going to be pushed. And we can talk about that looking at our specific examples. So it was a report? A. We're talking about equilibrium between ammonia gas and nature din and hydrogen. So first we'll take a look at the puppy for this. So here the entropy is positive. That means is that heat is going to be reacting here because he doesn't reactant if we increase temperature. Maybe the same thing is adding any other reaction to any other type of equilibrium in the twin to cause the equilibrium to shift towards the right. So we're going to see more products being formed. Conversely, if we think about a decrease in volume again, that means we're looking at an increasing concentration, and if we think about what our, uh, k C. Is for this equation. We've got the concentration of nature dinner multiplied by the concentration of hydrogen to the third, divided by the concentration of ammonia. It's where so we think about this, this change in volume and we think about how that fortunately going to affect the concentrations of each of these species. But we effectively end up with is that the concentration for ammonia is gonna move up by one factor. The concentration of hydrogen is gonna move up by that same factor, but it's gonna be cubed. As far as this reaction expression is concerned, the concentration of ammonia is going to increase squared. So we think about this. In this abstract concept, we cross out some of these terms. What we see is that an increase decrease in volume in an overall increase in the concentration is ultimately going to lead to an excess in the concentrations of the product and what that means that an increase in a decrease in volume is going to cause the reaction equilibrium to shift to the left for this particular reaction. So if we move onto the next problem, look apart be again. Your, um your entropy is positive, which means that he is going to be one of your reactions. So an increase in temperature is going to cause the reaction to proceed forward. But not for this one. If we consider what an increase or decrease in volume is going to do, we're still going to have the decrease in volume causing an increase in concentration. But in this case, if we think of our equilibrium, the concentration of nature's and oxide just squared on top and the concentration of nature 10 times auction in on the bottom, everything about how this exist in terms of relative increase relative increase in concentration for the nitrogen oxides gonna be squared. You're gonna have that same rough relative increase in the others. What that means is that changing the volume in this case is not going to have a huge effect on the equilibrium. So a decrease in volume here is going to have no change on the equilibrium because ultimately, there isn't going to be a difference between the KC and the cue value here. So if we move on to part see again, so we look at our it'll be first on this case or any of these negative. So that means that he is going to be a product, which means that an increase in temperature for this reaction is actually going to drive, Uh, equilibrium in the reverse direction. It's the same thing is as adding a product we're going to see. A shift to the left is now. The products are too high. Concentration and Q is gonna be greater than K. Now, if we look at what effect the decrease in volume is going to have here, decrease in volume again is proportional to an increase in concentration. Here we consider it's ready to take. You see, it's not technically the Casey we look at our QC. We have the concentration of oxygen cubed, divided by the concentration of ozone squared began. If we do this volume decrease, it's gonna be the same thing. Is having the relative increased cubed on top? The relative increased squared on bottom, which is a net increase in concentration, meaning that in this case it's going to cause reaction to shift to the left. And finally, if we consider this last equilibrium expression that we have done here first again looking at the end will be here. The entropy is negative, which means that an increase in temperature is going to be forcing this reaction to move in the reverse reaction. Um, because the negative and the P implies that ah, he is being given off. And that is therefore a product so increasing temperature will ship this reaction to the left. And if we consider the changes in concentration here, we think about the equilibrium expression here and what our equilibrium quotient would be. We don't have any gas state, um, products in this reaction. We only have CEO, too, in the denominator of our expression. But that means is if we decrease volume, concentrations of increase, that means that QC is now going to be less than Casey because the only thing that's going to be going up is the concentration initially is going to be going up is the concentration of the carbon dioxide. So this is gonna dio is gonna shift our reaction forward as we try to get rid of some of this carbon dioxide

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