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The gas phase decomposition of ozone is catalyzed by NzOs and the proposed mechanism is NzOs k1v NO3 + NOz NOz + 03 kz-- NO3 + 02 NO3 k3 -> NOz + 1/202 NOz + NOz...

Question

The gas phase decomposition of ozone is catalyzed by NzOs and the proposed mechanism is NzOs k1v NO3 + NOz NOz + 03 kz-- NO3 + 02 NO3 k3 -> NOz + 1/202 NOz + NOz k4-v N2O5 Find the rate law ofthe decomposition of 03 according to the given mechanism_

The gas phase decomposition of ozone is catalyzed by NzOs and the proposed mechanism is NzOs k1v NO3 + NOz NOz + 03 kz-- NO3 + 02 NO3 k3 -> NOz + 1/202 NOz + NOz k4-v N2O5 Find the rate law ofthe decomposition of 03 according to the given mechanism_



Answers

Ozone decomposes to oxygen gas. A proposed mechanism for this decomposition is What is the rate law derived from this mechanism?

For this question. We have a two step mechanism For ozone being converted into 0. 2 gas. The first step is fast and reversible. The second step is slow. So the second step is the rate determining step. And we can write the rate law for the overall reaction. From the rate law for this step. So rate is going to be equal to okay, multiplied by 03 concentration. Raised to the first power. We can do this for elementary steps in a mechanism where the order corresponds to the coefficient and then multiplied by oh raised to the first power. The problem with this is the overall reaction is simply 203 goes to three oh two. There is no oh anywhere. So we need to get rid of this in our rate long to do that. We go back to the first step because this is fast and reversible. We can assume that the rate in the forward direction equals the rate in the reverse direction. The rate in the forward direction is K. one multiplied by 03 to the first power. Then the rate in the reverse direction is K -1. As they are identified In the problem, multiplied by the 02 concentration, multiplied by the oh concentration. We can rearrange this and solve for the oh concentration as a ratio of the rate constants and a ratio of the concentrations. Then all we need to do is plug this in here into the true rate law and we get this and all of those caves can be reduced to a single K. Value. This then would be the rate law for the overall reaction.

We are given the to step mechanism for the decomposition of ozone in the atmosphere for step reversible. Second step for the intermediate of oh, were first asked to find an expression for the rate of decomposition of ozone. So the rate of information ozone to be team concentration of three for D. T. I think the negative. Okay, one concentration about three plus Okay, One prime concentration of 02 Oh, minus K two concentration of old Oh three Corpus equation one. And then we can also create an expression here. Rate of formation of oh, B T oh, over DT sequel The K one concentration of all three minus K one prime concentration of 02 Concentration of all minus K two. Concentration of all concentration of oh three. Applying the steady state approximation as oh is an intermediate. The over DT is equal to zero to substituting this into the above equation. We get k one 43 and this came one prime oh two Oh, minus k two Oh oh! Three equals zero K one or three equal to K one Prime Oh two Oh, plus Okay too Oh oh three came one or three equal to if we pull out. Oh, okay. One prime. Oh. Two plus K two oh three. Divide by, um, three. If we divide here and isolate for Oh, we would get K one concentration of all three over K one prime. Oh, two plus K two, demonstration of all three. And we're going to substitute consideration of all into equation one from up above. In doing so, give me its expression here minus K one three plus. Okay, One prime, two K one three over. K one prime. Oh. Two plus K two or three minus K two oh, three k one. Oh, three over. Okay, One prime or two plus K two. Oh, three. It should be equal to minus K one. Oh, three plus. Okay, One prime. Oh, two. He won. Oh, three minus K two. All three. Okay, one three came on prime. Go to plus K two. All three mm is equal to Okay. One minus cape. 103 times. K one prime. Oh, two plus K two oh, three over. K one, prime two plus K two oh three plus K one prime. Oh, two K one oh three minus K two oh three came one. Oh, three over. Okay, one prime. Oh, two plus K two oh three. This would be equal to minus K one. Hey, one prime or two or three minus K one K two oh, three squared minus K one K one prime. Oh, two oh three minus K one K two oh three squared over. K one prime too. Plus K two oh three This. Sorry, This would be a positive. Here. These would cancel and this would B minus two K one K two concentration of 03 squared over K one prime 02 plus K two oh three. So the rate of the reaction mhm, which is d X over D t, which is equal to minus d oh, three over. DT Z equal tube two K one K two concentration of all three squared over K one prime 02 plus K two oh three there would be are expression for the rate of decomposition of three. Now, the second part of this question, if Step two is slow, then K two oh three would be much less than K one prime 02 Therefore, the reeds is equal to DX over DT, which would be equal to two K one K two oh three squared over K one prime 02 And we can see here that the rate is second order with respect to ozone and forced order with respect two oxygen.

This question is a mechanism reaction. Specifically, they give us the mechanism for the decomposition of ozone. To go to it's proposed is a two step mechanism where ozone first decomposes likely due to, ah, high energy photon into 02 and a radical or free oxygen Adam. This is fast and reversible. Second process, then, is the radical oxygen that is very reactive will combine with ozone producing two oxygen molecules. This is slow, so this would be the rate determining step, the overall process that is going to be the some of these two reactions. And when we some these two reactions, you'll notice that oxygen is the intermediate. It was created and then later consumed, and we get to zero threes going to 30 twos. If the slow step is the rate determining step, then we can write the rate law from the molecular Garrity or the stoy geometry of this slow step. The rate is going to be equal to oxygen raised to the first power because it has a coefficient of one multiplied by the 03 concentration also raised to the first power because it has a coefficient of one. The problem with this rate law is it contains Thean term eDiets. You don't want the intermediate in our rate law. We only want reactant in our rate law or sometimes products, but definitely not an intermediate. If the first step is fast and reversible, then we can say that the rate in the forward direction because it's reversible, is equal to the rate in the reverse direction. The rate in the forward direction is going to be equal to some constant multiplied by the concentration of 03 and the rate in the reverse direction is going to be equal to some constant multiplied by the concentration of 02 multiplied by the concentration of just Oh, rearrangement of this expression allows us to determine what the concentration of oh is in as a function of the concentrations of the reactant 03 and the product so to this will then be helpful because we can now take this expression for, 02 and plug it into the rate determining step in order to get a better rate law for this mechanism. When we do that, we'll get rate is equal to a ratio of thes three K values which can all be combined into a single K value, which I'll do here in just a second multiplied by 303 Concentration squared this so three and this so three gives us both three squared and then we still have 02 in the denominator. And when we combine all the case together, then it will look something like this. You may choose to write it this way, or you may choose to put Theo to to the negative one power, so it would be an order of negative one. It then wants us to determine whether or not oxygen is the catalyst or the intermediate. We already defined it as three Intermediate. The last part of this question asks if instead of the reaction, if instead, the reaction occurred as a single step as shown here. If this were a single step, then would the rate law change? Will the rate lob the single step? If this were a single step, if the overall reaction were a single step, then we could write the rate law from its molecular arat e, and it would simply be equal to rate equals K ozone concentration squared. This is definitely different than what we determined down here because we do not have the 02 in our rate law. So yes, it is different.

This question is a mechanism reaction. Specifically, they give us the mechanism for the decomposition of ozone. To go to it's proposed is a two step mechanism where ozone first decomposes likely due to, ah, high energy photon into 02 and a radical or free oxygen Adam. This is fast and reversible. Second process, then, is the radical oxygen that is very reactive will combine with ozone producing two oxygen molecules. This is slow, so this would be the rate determining step, the overall process that is going to be the some of these two reactions. And when we some these two reactions, you'll notice that oxygen is the intermediate. It was created and then later consumed, and we get to zero threes going to 30 twos. If the slow step is the rate determining step, then we can write the rate law from the molecular Garrity or the stoy geometry of this slow step. The rate is going to be equal to oxygen raised to the first power because it has a coefficient of one multiplied by the 03 concentration also raised to the first power because it has a coefficient of one. The problem with this rate law is it contains Thean term eDiets. You don't want the intermediate in our rate law. We only want reactant in our rate law or sometimes products, but definitely not an intermediate. If the first step is fast and reversible, then we can say that the rate in the forward direction because it's reversible, is equal to the rate in the reverse direction. The rate in the forward direction is going to be equal to some constant multiplied by the concentration of 03 and the rate in the reverse direction is going to be equal to some constant multiplied by the concentration of 02 multiplied by the concentration of just Oh, rearrangement of this expression allows us to determine what the concentration of oh is in as a function of the concentrations of the reactant 03 and the product so to this will then be helpful because we can now take this expression for, 02 and plug it into the rate determining step in order to get a better rate law for this mechanism. When we do that, we'll get rate is equal to a ratio of thes three K values which can all be combined into a single K value, which I'll do here in just a second multiplied by 303 Concentration squared this so three and this so three gives us both three squared and then we still have 02 in the denominator. And when we combine all the case together, then it will look something like this. You may choose to write it this way, or you may choose to put Theo to to the negative one power, so it would be an order of negative one. It then wants us to determine whether or not oxygen is the catalyst or the intermediate. We already defined it as three Intermediate. The last part of this question asks if instead of the reaction, if instead, the reaction occurred as a single step as shown here. If this were a single step, then would the rate law change? Will the rate lob the single step? If this were a single step, if the overall reaction were a single step, then we could write the rate law from its molecular arat e, and it would simply be equal to rate equals K ozone concentration squared. This is definitely different than what we determined down here because we do not have the 02 in our rate law. So yes, it is different.


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