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Corisider the equilibrium: ZNO(g) 0,(g) ANO(S) How will the amount of chemicals at equilibrium be affected by: ZN,O(g) 0z(g)ANO(g)adding N,o removing 0z increasing...

Question

Corisider the equilibrium: ZNO(g) 0,(g) ANO(S) How will the amount of chemicals at equilibrium be affected by: ZN,O(g) 0z(g)ANO(g)adding N,o removing 0z increasing the volume of the container adding catalyst

Corisider the equilibrium: ZNO(g) 0,(g) ANO(S) How will the amount of chemicals at equilibrium be affected by: ZN,O(g) 0z(g) ANO(g) adding N,o removing 0z increasing the volume of the container adding catalyst



Answers

Explain the effect of a catalyst on an equilibrium system.

So in the context of chemical equilibrium, we have taken a look at Catalyst and how they are able to lower our activation energies on. Then in turn, we can allow our reactions to go forward with a lower activation energy. So in this scenario, we're considering whether a catalyst is able to affect the value, the numerical value off the equilibrium constant. So the answer here is no. A catalyst will not affect the value of our equilibrium constant where equilibrium constant is often denoted as que que so a catalyst will only lower the activation energy in a chemical reaction, and it will not have any effect on our equilibrium, constant value.

At 600 Calvin the K P is equal to 3.81 times 10 to the two and it 700 Calvin people have room. Constant is equal to 2.69 times 10 to the three. So in with an increase in temperature, we have an increase in the equilibrium constant r k p would be the partial pressure of the products over the partial pressure of the reactant CE. And if the equilibrium constant is increasing, that means the partial pressure of the products is increasing with translates, uh, as a shift towards the rates to reach equilibrium. Therefore, uh, the only way to have a shift right with a temperature increase is that the reaction must be endo thermic. So we've got P C 05 and beyond the reactant side PCL three and seal too. So with the temperature increase, we shift away from the increase and therefore the equilibrium constant goes up. That would be part a this question part B for this question. Um, how were the equilibrium amounts of reacting some products affected s So we used the equilibrium that we have there. We know that it is, um Endo thermic, and I'm going to remove our shift from earlier. Um, we're gonna increase the volume, so increase volume translates to a decrease in the pressure. And if we go back to her equilibrium, we've got one mole of gas. On the reactant side, we've got two moles of gas on the product side. This be my low pressure. Should be my high pressure. If we decrease the pressure, listen, ties. Principal wants to shift it toe high. Therefore, um, the, uh this would translate into a shift towards the rates, and the amounts of products would increase. So the products increase as a result of increasing the volume. The next stress on this equilibrium is in addition of an inert gas. In addition of an inner gas would not change any of the volume, so there would be no shift upon the addition of an inert gas and therefore thehe mount of reactions and products would stay the same. So no change and the amounts of reactant ce and products just three is the addition of a catalyst. In addition of a catalyst, would be no shift catalyst will speed up with four turnovers. Reactions and no shift would also be no change in the imbalance of reactant ce

Here we are asked to think about a reaction that occurs in the gas phase and his XO thermic for these five different conditions. How is the equilibrium constant? K impacted first, removing a reactor Well, we know from Lee Shot liaise principle that removing a reactant shifts the equilibrium of assistant itself. The equilibrium, constant itself K remains unchanged. That's because it's an equilibrium constant, and so it doesn't actually depend on the conditions you apply to it at a given time in terms of concentrations or, in this case, partial pressures of the gases in your system. Likewise, removing a product would invoke a leash Appy ace principle, but you still would see no change in K as it is an equilibrium constant. Similarly, if you decrease the volume of your system, we know that that would shift equilibrium towards the side that has fewer moles of gas. And so, while this might impact the exact composition of the system that you see the equilibrium, constant itself will not change. This is because equilibrium constance are only temperature dependent, not pressure, dependence. But they are temperature dependent. So now we have to think when we get to decreasing the temperature. We're told that the reaction is XO thermic. When we hear XO thermic, we can think of heat as a product if we add more heat than we're inhibiting this reaction from going forward, and so the reaction will actually shift back towards the starting materials. Therefore, the reaction constant K will decrease. Finally adding a catalyst and in a catalyst tends to speed things up. But ultimately there is no change in K since it's just a description of the ratios at equilibrium.

Hi guys. So problem 80 axe. How on increase in volume will affect this reaction based on the chart Ilias principle. So before we go into details off how to go about this problem, we need to I know what the chattel is principal is all about. So according to Lee, Chattel is principal When you have assist him at equilibrium, I'm on external stress is applied to the system. This system was shift in a direction that relieves they applied stress. We will explain that in a bit with this problem. So we are concerned about volume in this case, right? Remember from price law, we have pressure invest the relate active volume at constant temperature. What this means is is the same thing as saying what will happen? What will be the effect when you decrees pressure, right? Because they're mostly related. So when you increase the volume you have technically, Dick risen the pressure. So in this problem we have this used There's just an example to demo to illustrate we're trying to do, um we have this system in this room and then only increase the volume would have it like this. So this is the system. I'm all the gases contained in this system. So we have this assuming this from and then after increasing the volume, you have it in this form. So clearly you can see that we are decreasing the pressure. So what is really going on? Is the pressure exact exacted? Bye. The gases Continuing the system depends on the number off gases. Um, you have, which translates to the gas particles colliding with themselves on the walls, off the continue reversals. And the more you have the gas particles, he grew tired of pressure. Right? So now let's look at the equation that we have here. We have to most off as a two on one more off or two. Given the total off three moves off gas, it's that we have. And here we have two moors off gases in the product. So consider that the number off Patrick gas particles in the rack terms it's more than the number of God gas particles in the product. Um, by decreasing the pressure, we are actually force in this system to relieve that by moving to the direction that has the tendency off increasing the pressure so that the system will revert back to equilibrium. So we're decreasing the pressure right by increasing the volume. And when we do that, this system will revert. Move in the the reactant direction so that it relieves the decreasing pressure. So the system we're shift to the left. So at this point, it's important to mention that in a situation where you have have the same number off Moors on the reactant product sight, there's nothing this system can do. So there's no effect and change both volume or pressure when you have the same number off more on the reacting under product sites. That's one thing. The other thing is the reactant on products have to be, gosh ISS, so change in volume or pressure to have effect. So for this particular system, in order for this system to relieve the stress, this is them will shift to elect, which is towards new react in sight


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