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Calctiotec 2 HzS 8 CH;OH () (g) occur AG" for the 2 Hz (g) 0,10 "spondleougye 2 2 Hzo (6) (5 reactions_ 3 (g) points 8 (ypea Also (6) 8 indicale whelher y...

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Calctiotec 2 HzS 8 CH;OH () (g) occur AG" for the 2 Hz (g) 0,10 "spondleougye 2 2 Hzo (6) (5 reactions_ 3 (g) points 8 (ypea Also (6) 8 indicale whelher you would (Ans_ (Ans (Ans expect lhe =

Calctiotec 2 HzS 8 CH;OH () (g) occur AG" for the 2 Hz (g) 0,10 "spondleougye 2 2 Hzo (6) (5 reactions_ 3 (g) points 8 (ypea Also (6) 8 indicale whelher you would (Ans_ (Ans (Ans expect lhe =



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For the reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) $$ the currently accepted mechanism is $$ \begin{aligned} \mathrm{N}_{2} \mathrm{O}_{5} & \rightleftharpoons \mathrm{NO}_{2}+\mathrm{NO}_{3} \\ \mathrm{NO}_{2}+\mathrm{NO}_{3} & \longrightarrow \mathrm{NO}_{2}+\mathrm{O}_{2}+\mathrm{NO} \quad \mathrm{slo} \end{aligned} $$ $$ \mathrm{NO}+\mathrm{NO}_{3} \longrightarrow 2 \mathrm{NO}_{2} $$ fas Determine the rate law for this reaction.

This problem, we're going to look at the slow step in each mechanism. Then we write the rate law based off for the slow step. So the first reaction or elementary step is reversible. So we can set the concentrations equal. So through substitution we can substitute the concentration of nitrate with the demonstration of these two, and then we get the rate law. Then we have the second reaction back. We based array off of the slow reaction. From the first reaction. We can equate these two, then we're going substitute and we get.

In this question were given to mechanisms that are proposed for a reaction and a observed rate law. We want to show that both of these mechanisms are consistent with the rate law. So, for the first mechanism we take the slower step and we go that we know therefore that rate equals K Times N 03 time Zeno, N No three is an intermediate. So we want to substitute that out. So looking at the first step of this reaction, we know that the ford reaction Represented by K. No Times 02 Should be equivalent to the reverse, which is K -1 Times and 03, Solving for an 03, We get that the concentration of n. should equal. Okay, over okay -1 time. Zeno times oh two substituting this back into orbit expression. We get that rate should equal K times K over K -1 times and oh times two times. And again these three case can all be combined into the overall rate constant. So then we get that rate equals K times and oh squared Times 02, which is consistent with our observers rate law For the 2nd Mechanism. Again, we look at the slower step so then we get that the rate should equal K times the reactant. And to go to and 02 And 202. In this case is a intermediate. So we look at for the first step to find that the forward reaction K of eno squared should be equivalent to the reverse reaction, Which the various reaction rate which is K -1 -1 times. And to go to. So for N 202 and you get that it should equal Okay over OK -1 times and squared, substitute that back into the right expression. And you get that rate should equal okay, Times K of OK -1 times N O squared times oh two. Again, these cases can be consolidated into the overall rate constant, so then you get that weight equals K time Zeno squared Times 02, which is consistent with the observed rate expression.

Giffen the you could even constant in concentration off reaction. Ah, decomposition Yashin off and the with you, we had to find out. Ah, you could even constant off information off a note here from Andal and future. Okay, so Ah, from the two reaction, we can see that the 1st 1 I'll be the first. The 2nd 1 the second we s and essentially is Ah, flip that They re worse. Reaction of the 1st 1 and also 1/2 their over everything. So, first of all, we, ah, from Russian want to rush into rehab fitted. And also we have to 1/2 of everything. So we a currently to the same calculation to the Casey. So we just need to take one over. You could even causing in concentration. And then we have everything in our coefficient. We're going to take square words off there, Casey. So we're one divided by one point in time. 10 to 6 and every scrap of I answers. And there we will be. Would you find it is roughly around 7 12 45 You were around sort of on five. Contender Power of two. And this is Theo. Option D, and this would be the answer

Mhm mm From the first part of the problem, it is clear that the reaction kinetics can be done. An initial stage of the reaction, A straight line is easily approximated in the initial stage hands, the initial rate can be calculated as the change in concentration of and or two divided by the change in time. From the first part of the problem, it is quite clear that the curve become almost flat at a very large instant of time. Therefore, as the time approaches infinity, the reaction rate we will be zero.


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