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Concentration Graphs Assignment 4Intthis tssienment imagine that €ach change is imposcd on the systcm in its equilbrium condition words, each othe reaction is...

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Concentration Graphs Assignment 4Intthis tssienment imagine that €ach change is imposcd on the systcm in its equilbrium condition words, each othe reaction is taking place in # closcd vessel t0 which you are makng . and fcconding thc resultng changcs and shifts on change (0r applying 4 suress) concentration graph You ae givcn thc initial cquilibnum concentrations; thcse are the starng points on your graph Iyou are adding Or rcmoving # species, the graph will show & immediate and the shift

Concentration Graphs Assignment 4 Intthis tssienment imagine that €ach change is imposcd on the systcm in its equilbrium condition words, each othe reaction is taking place in # closcd vessel t0 which you are makng . and fcconding thc resultng changcs and shifts on change (0r applying 4 suress) concentration graph You ae givcn thc initial cquilibnum concentrations; thcse are the starng points on your graph Iyou are adding Or rcmoving # species, the graph will show & immediate and the shift _ If you are changing thc stght Mp or doun change. tempcraturc or volume prexsure, the grph Will show the immediate shift In thls assignmcnt YoU are givcn thc Wuc applicd and you Will drew te concentrtion will change rclative t0 onc grph You must thow bowthc calculate cxact final conccntrations Hnorher , uaking mole ratios into accounL but You are Dot required t 4itcr hc shift Sce cramplcs of graphs in your Hebdcn Chemisty 12 Workbook[puges s0- 53) t0 help you


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Consider the following S$_N$1 reaction.
a. Draw a mechanism for this reaction using curved arrows.
b. Draw an energy diagram. Label the axes, starting material, product, $E_a$, and $\Delta H^\circ$. Assume that the starting material and product are equal in energy.
c. Draw the structure of any transition states.
d. What is the rate equation for this reaction?
e. What happens to the reaction rate in each of the following instances? [1] The leaving group is
changed from $I^-$ to $CI^-$; [2) The solvent is changed from $H_2O$ to DMF; [3] The alkyl halide is changed from $(CH_3)_2C(I)CH_2CH_3$ to $(CH_3)_2CHCH(I)CH_3$; [4] The concentration of $H_2O$ is increased by a factor of five; and [5] The concentrations of both the alkyl halide and $H_2O$ are increased by a factor of five.

Okay, since from this sort is using a graph given to answer some questions. Eso for a When did the maximum concentration occur? So the when is the X axis? That's gonna be around five hours. And the concentration in the blood stream looks to be around 20.28 That concentration is well, I just gives it a 0.28 Okay, So be they use it to compute the rate of change for the intervals from eight. To 10 and 22. 22. So for computing the rate, we're doing the change and see about about the change in time. Okay, so we need a couple of things. We need our concentration at eight. We need the concentration at 10. We needed the concentration at 20 and 22. So just keep in mind that as we play this into the calculator, um, all right, that form allows a TV H squared is just remained three squared plus h over H. Kids put 70 cancer cackling two different rates. So, as you put it in the calculators where we could go back to your entry has changed all of your ages from AIDS to whatever. Okay, so we're gonna have to. And eight. It's where it started by it's a turn. It's in my concentration. Uh, after eight hours is gonna be a 0.234 on the street itude through three decimal places. Now, just about 10 entries and change of those eights. Two tens. It's another concentrations down to 0.196 Okay, 20 right, 0.102 I'm doing some rounding here. So we're not getting exactly answers to what the book says. Just because we're just rounding it off because we're basically is getting the idea of what's happening here. I've been playing in 22 have 0.92 Okay, so the rates of change is are changing concentration divided by two because both of these over the course of two hours. So back to the calculator, everyone do. 0.234 We're gonna start trying 0.196 We're going to divide that about you. So the rate of change the loss of concentration is going to be negative. 0.19 concentrate is your 0.19 loss of concentration poor per hour, and then for the 20 to 22 you can see it only drop 0.1 That's gonna may 0.5 Okay, so we noticed that it's starting to level off, so the rate of change is decreasing. Oh, are getting closer to zero. I guess it's not decreasing cause it's, uh it was negative, but the rate of change is getting closer and closer to zero. Okay, Okay. So as, uh, exit purchase positive infinity are not access. It's a age. The concentration is going to approach zero. So the role of the H access place for this function it's our Assam tone so that it shows what the bloodstream concentration is getting closer to good. Thank you very much. I mean or guests in service time. So, like, as it lets us see, like the change over and but the concept of time so that we can actually see a change. So just going up and down and overlapping on itself

Hello there. Okay so for this exercise we are modelling the concentration of a drug in milligrams per liter of some drugs on the blood of a patient. So let's suppose that actually in this exercise this concentration is modeled as a function of two variables X. That determined the amount of uh in milligrams of the drug that were injected to the patient and t the time in hours after a science injection. So this function is the finest follows his tee times E. To the minus T. Five minus X. And X. is defined between zero and 4. And tease any positive value. So having this in mind we need to give plot The following function. So let's suppose that F. is given by 80 and we are going to plot for equals to one, 23 and four. So here we have the correspondent plots with different colours. And we need to give some analysis of this plot. So let's see let's remember that this function represent the concentration of the drug on the blood in milligrams per liter. Okay so milligrams per liter. So this the y axis of this plot represent uh the vertical access. This plot represents the amount of the concentration of the truck. So we can see the values of a represents the amount of drug that weird given to the patients. So this is the amount after drug. Okay, so it is clear that if we are rising the amount of product that we're injecting to the patient, the concentration is going to be vigor in blood. So that's why we have here. The pigs on each of the plots represents the maximum concentration and blood of course after some time. So this pigs increase as a increase. Okay, that means that we are reaching higher levels of concentration on the blood if we give more truck to the occasion. A second observation is that these pigs are displaced to the right. Okay. So we can observe some some change some shift to the right of the pigs of this of these curves. So that means let's remember that this. Mhm. Sorry. Here is not X. Had them a type of here. This is T. Okay. So the horizontal axis represents the amount in hours. Okay. So if we are shifting the peaks, that means that the maximum of concentration and blood take more time to reach a it takes it takes more time to reach the maximum of concentration on the patient. Okay. And a last observation is also the decrease. So we can see that other functions start to decrease after reaching the maximum that are the peaks of these curbs and then start to decrease. So it is also intuitive that if we give more concentrate more amount of drug to the patient, the concentration is going to uh diminish in longer time. So that's exactly where observing on the skirts, we see that for a equals to one oh in two hours we have almost not rock in the blood of the picture, right? The same happened for a prostitute after more or less 2.5 hour or maybe three hours. We have no drug in the patient For a three. This gets Prolonged to almost four hours or maybe five hours. That these this concentration start to To be near zero. So no more drugs and the blood of the patient. But when we give the maximum amount of drug that is equal to four, then the degrees of this of the concentration It prolongs too much. I mean it's almost a 10 that we reached zero. So here if we give too much drug to the patient That is a maximum. This case is four. Then the time that it takes to the body of the patient to eliminate the drug. It takes almost 10 hours it with respect to the other concentration, so the time eliminating the drug extent and also the concentrations rise, course of the concentrations increase the maximums. Yes the maximums of concentrations increase. Um The last observation is at the time to reach the maximum of concentration increase as well as a increase, thus increase as a increase. So these are the observations that we can give about this about the plots that we have seen here.

You are perfectly capable of using the T. I 83 plus program to do this question. The manual way to do it is by saying our two over our one, whether the two reaction rates is equal to the concentration of a whatever it is in this case, two over the concentration of a one to the X. Then you just solve for X here. Once you plug in all your values, so here for part A. When you do that, you get that X is equal the one or its first order for part B. You get second order for part C. First, order her apart. Do you get second order? And for part E, you get second order so these are most easily found by using that program. If you would rather use this methodology, it works perfectly fine. It's just not quite as fast

So we're given this function to model concentration over time and were given these constraints A B and K or positive. Constance and B is greater than K. So I just picked these random numbers that do work for these constraints. And I plugged him into the function and they got this function. So if you want to grab that, the first thing I'm gonna look at is the first derivative. So see of tea, See, prime of tea. So the first derivative is gonna be negative e to the power of negative T plus to e to the negative to t. Okay. And I'm gonna set this equal to zero to find my critical points. And I see that c is equal to Ellen of two. All right, so Ellen of two is approximately 20.693 So if we test values, this is Ellen, too. If we test values, um, into our first derivative for the first derivative test for less than Ellen to, we're going to see that it's increasing and anything greater than is gonna be decreasing. So it's a local Max. All right, So this could be useful for later and then see double prime toe. Look a ah kong cavity. See, double prime of tea. We're going to get e to the negative. T minus four e to the negative to t If we said that equal to zero, we're going to see, um, t equals to Ellen too, which is approximately, um, 1.39. And if we do the second derivative test of double prime and we test values around 1.39 we see that, um, it's gonna be con cave down. And then Khan gave up. So this is an inflection point. All right, so we're going to use this while we graph in just a moment, but I'm gonna look at ah, one more thing, because our graph actually does have an accent. Oh, it has. Ah. Horizontal ass in tow. So let's test, um, the limit, as are variable t approaches infinity over function, which is one times e to the negative one t minus e to the negative to t. All right. And we're gonna So let's just say that we do, um, the limit as t approaches infinity, I'm gonna rewrite this in a way that looks easier to understand. So this is the same thing as one over e to the t because it's, ah, negative exponents bring into the denominator. So one over E to the T minus one over each of the two tea. So this is the same thing, all right. And if you take a look at this, it's every substitute. Wherever we see tease, we're gonna substitute infinity. So as t approaches infinity, we see that one over. A very large number is close to zero. So this is gonna approach zero. So as T approaches infinity, our why values, which is concentration over time, is gonna approach zero. So we have a horizontal ask himto at zero. Okay, And now we can graph. And since we're looking at time, I'm just gonna show the positive part of this cause negative time doesn't really make sense to me. So this is concentration overtime for this access. And this is just time over here, and we know that we do have to say this is approximately one. We know we have a maximum of Ellen, too. And if you plug in Ellen to into the original function, you'll see that it's equal to 1/4. So we know that about Ellen two and 1/4 is our maximum and the sister of sketch. So I'm gonna say this is Ellen too. 1/4 at this point. And this is our maximum. Okay? And we see that, um, according to the second derivative test that it's gonna be conquered down. And then Khan gave up in our inflection point is 1.39. So that's what we're gonna seethe. Sign change at about 1.39. So let's say it's right there. Okay? So it's gonna be con cave down. Oh, and we also haven't Don't forget about the ass in tow. The horizontal asking to is right here. Okay, so we have, um, con cave down. It's gonna first of all, increase until it hits are maximum local, maximum, and then it's gonna starts a decrease after hits the max, but it's still gonna be con cave down until it hits our inflection point where it's gonna be Khan gave up an approach, but never touch are asking too. And this is our graph. All right, so this is our graph for concentration over time, and the question asked asks us, um, what's happening in this graph. So we see as timing is increasing. So as time goes on, um, concentration increases until we hit the. So as time goes on, concentration increases until we hit the maximum concentration, and then after that, the concentration continues to decrease forever.


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