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Values of the rate constant for the decomposition of $\mathrm{N}_{2} \mathrm{O}_{5}$ gas at four different temperatures are as follows:
$$\begin{array}{cc}T(K) & K\left(S^{-1}\right) \\658 & 2.14 \times 10^{5} \\\hline 673 & 3.23 \times 10^{5} \\\hline 688 & 4.81 \times 10^{5} \\\hline 703 & 7.03 \times 10^{5} \\\hline\end{array}$$ a. Determine the activation energy of the decomposition reaction. b. Calculate the value of the rate constant at $300 \mathrm{K}$.

Okay, I'll show you one more time how to do it plotting. And then for the other ones that are left in this chapter, I'll show you how to do it a different way. So if you were to plot the data the natural log of K as the function of one over the Kelvin temperature, you get a graph. It looks like this with an equation. That is exactly this for this problem. Then we recognize that using the Iranians equation, the slope is equal to negative ea over our. So we'll take our slope value, which we determined from the equation of the line after plotting, said it equal to negative ea over our and solve for E A. Recognizing that are is 8.314 and we get e A of 9.79 times 10 of the fourth jewels. The Y intercept could be determined if we needed it by recognizing I'm sorry. The frequency factor could be determined from the y intercept if we needed it by setting the Y intercept equal to the natural log of the frequency factor. So we get 30.164 equal to the natural log of a A. Then if we take the anti natural audible sides, we get 1.26 times 10 to 13. Then to calculate the K value at any particular temperature we're going Teoh, use Theis equation of the line. Why is going to be equal to the natural log of K X Is gonna be one over the temperature. The Kelvin temperature, for which we need to calculate the K value is 300. So we'll get natural. Log of K called the negative 9.106 or K is equal to 1.11 times tend to negative four one over S.

In this question, we're gonna be taking a look at the distribution of scores on achievement test using a relative frequency table and a percentile graph or a relative frequency graph. So we're going to start with our data and we're going to construct our team a lot of frequencies and then calculate a relative frequencies to make our graph. So relative frequency is just a fancy way of talking about percents. Cumulative frequency means you're just gonna add these numbers up. So we're going to bring the five over, and then we're going to keep adding them as we got so five plus 17 it's 20 to 22 plus 22 is 44 out on another 48 we get 92 add on another 22 or upto 1 14 and at on six, and that gives us a grand total of 20. So that means we are represented representing 120 pieces of data. So that is our end in this case, to find a relative frequency, we're gonna take each one of these cumulative frequencies and divide by 1 20 So, for our first interval five divided by 1 20 is approximately 200.4 So just to show you where that's coming from, I'm rounding these because we're going to be graphing on this graph where everything is estimated, so we don't have to be that precise. Eso 22 divided by 1 20 is 18% or 180.18 44 divided by 20th. About 440.37 or 37% 92 divided by 1 20 is about 77%. 1 14 divided by 20 is 95% and 1 20 divided by 1 20 is one. Now we're gonna graph this over here and we're starting with 1 96 But there are no achievement test scores below 1 96.5 So that's gonna get a zero. There's nothing before. And then as we go to the end of each interval, we're going to continue to build up our graphics. We move forward. So at the end of the next interval work, four percents about here at the end of the next interval were at 18% so just below 20% about here at the end of the next interval were at 37%. So about here, the end of the next interval were at 77%. So about here, the end of the next interval were 95% almost to the tall. And at the end of the last interval, we are at 100% now. One of the things that's important to know is that each of these intervals represents even other kind of knot standard ways with numb label, a scale they do represent consistently spaced out intervals. So I want to try to connect these with a straight line as much as I possibly can free so and this is what we make our cumulative frequency graph. Now, the reason we connect these with a straight line is because we are assuming that all of the achievement tests in the interval in between are evenly distributed. That's our best way of making a guess here. That may not be actually the case, but that's what we're going to use to make our estimation. Okay, so now we have our frequency graph curved, and we can use our frequency curve draft on. We can use that to make our estimates. So first thing that we want to estimate is what is the approximate percentile for somebody who scored 2 20 So to 17 to 2 38 to 59. What? Um, what we need to kind of know is how big that interval is. So we did say that they were evenly spaced on. That means that in between each of those intervals, we're looking at about 21 points. So each one of those lines is representing seven. So to 20 plus seven is to 24. So when I'm looking for to 20 I'm gonna be looking at the score of 2 20 It's gonna be about here, so I'm gonna follow to 20 up to the curve, and then I'm gonna follow that line over 2% all trying to draw this line a straight is that possibly can, and it looks like it's between 5% and 10% a little bit closer to 5%. So let's estimate that at about the sixth percentile, question B is asking for the percentile associated with a score of 2 45 So to 45 is going to be about here, Right? So about 7 to 38.7 about 2 45 about there. So again, we're gonna follow from a test score up to the graph and then follow from the curve over to the percent axis. So that looks to be at approximately 25. So about the 25th percentile. Her question. See, we're looking for the percentile associated with score of 2 76 So to 76 is going to be somewhere over here. So a little bit below to 80. I'm so I'm gonna estimate about here. You should not quite so straight there. Tried to even it out some of the end on. Then we'll move that over a little bit below 70%. So I'm gonna estimate that to be about the 68th percentile. These are just estimates. So if you're not exactly spot on, um, that's okay, that's that's a reasonable, uh, thing to happen with this kind of graph. So now what is the percentile associated with to 80? So, I mean, that's just like here, right? So we can kind of see that there's kind of a big gap that's happening there. So we're all the way up here now. All right, so we should really be drawing this line all the way down. This isn't much better done with a ruler on paper, then digitally, unless you were doing something like on Power Point or something. But we're above 75%. So maybe, like the 76 there, 77th percentile, and then the final score we're going to try to estimate the percentile for is 300. So three hundreds gonna be a little bit below here, and we're gonna follow that up a bit wonky. But it's a little bit before this line hears about their and then we're gonna follow that over, and we know that we're below 95% but not a lot below 95%. So maybe 94 93%. Ah, percent. All right, now we've got all those lines on there that we're gonna have to try to ignore a little bit, but we're gonna work the other direction, and we're going to try to, um, estimate the test score now given a percent. So if we have the 15th percentile, what is the approximate test for that goes with that? So we're gonna work in the opposite direction. 15th percentile. Follow that line over to the graph, and then we're gonna come down here and estimate. So we remember we said that each one of these lines is about seven. So this would be like to 24. This would be like to 31 2 38 So somewhere between 2 31 2 38 little closer, 2 to 31. So maybe, like 232 G. Um, we're gonna estimate that 29th percent up, so you can see, like, all this estimation, all we're going to be able to do is kind of guess right, the 29th percentile. Follow that line right below the 30th percentile over Teoh about here, maybe, and I will follow that down. And so this is to 38 to 45. The next one would be, what to 52. So we're a little bit before 2 50 like maybe 2 49 The important thing to kind of remember is that when you have the percentile, you work. You started to have relative frequency and work to the data. When you have the data, you started the data and you work to the relative frequency. Okay, so, H, we are looking at the 43rd percentile so the 43rd percent tells somewhere between 40 and 45 and we're going to come over to the graph over here and then follow that line down. And if this is to 59 this would be to 66. And so we're a little bit I don't know in the low in the low to sixties. So let's say like to 64 I were looking for the 65th percentile. So 65th percentile is here. Follow that over to the curve and then from the curve down. So it's gonna be a little bit above this line right here. So that was 2 59 to 66 to 70 three. So a little bit above to 73 I don't know. Let's say to 74 and then finally, the last one the 80th percentile. So we'll follow from the 80th percentile to the curve, but here, then we'll follow that down and like to 87 a little bit below 2 87 So let's say to 85. So, as you can see is the percentile was increasing. So where the test scores and that makes sense because we're accumulating more and more test scores as we go. Um, and the same was true here. The lower the test score, the lower the percentile that it was associated with.

As we all know that dissociation constant, the association constant of water Is equal to one, multiplication 10 to the power -14. And dissociation constant of have you worked? There Is equal to 0.3, multiplication 10 to the power -14. Therefore, according to the option in this problem, option C. Age, correct them. Said of some seed, correct answer for this problem. I hope you understand the solution of this problem.

They were taking a look at the Iranians equation. So we have L n k is equal to l n a negative e a over are multiplied by one over T straight line equation is y equals MX plus C. And if Ellen K's plotted versus the inverse of the temperature, the slope is negative e a over our and the intercept will be l N A. So for the reaction, we have ch three and two ch three gives us two ch three radicals. Uh, and to So we plot Ellen K verses one over t we got a straight line. So the gradient is equal to negative e a over our, which is equal to negative 26.45 times 10 to the three Calvin. So therefore, if we put the value in for the our gas constant activation energy E A, it's equal to to 20 kg jewels from all yeah,


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