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A chemical reaction begins when a certain mixture ofchemicals reaches 95°C. The reaction activity is measured in units(U) per 100 microliters (100 µL) of the ...

Question

A chemical reaction begins when a certain mixture ofchemicals reaches 95°C. The reaction activity is measured in units(U) per 100 microliters (100 µL) of the mixture. Measurementsduring the first 18 minutes after the mixture reaches 95°C arelisted in the following table.Chemical ReactionTime, t(minutes)Activity, r(U/100µL)00.1020.1040.2560.6081.00101.40121.55141.75161.90181.95(a) Find the function for the logistic model that givesthe activity of the chemical reaction in U/100𝜇L, where t i

A chemical reaction begins when a certain mixture of chemicals reaches 95°C. The reaction activity is measured in units (U) per 100 microliters (100 µL) of the mixture. Measurements during the first 18 minutes after the mixture reaches 95°C are listed in the following table. Chemical Reaction Time, t (minutes) Activity, r (U/100µL) 0 0.10 2 0.10 4 0.25 6 0.60 8 1.00 10 1.40 12 1.55 14 1.75 16 1.90 18 1.95 (a) Find the function for the logistic model that gives the activity of the chemical reaction in U/100𝜇L, where t is the time in minutes after the mixture reaches a temperature of 95°C, with data from 0 ≤ t ≤ 18. (Round all numerical values to three decimal places.) r(t) = 1.936 (1+29.06e−0.42t) U/100µL What is the limiting value for this logistic function? U/100µL (b) Use the model to estimate the average rate of change of the reaction activity between 3 minutes and 7 minutes. (Round your answer to three decimal places.) U/100µL per minute (c) Numerically estimate to the nearest thousandth by how much the reaction activity is increasing at 5 minutes. (Round your answer to three decimal places.) U/100µL per minute (d) Write sentences of interpretation for the answers to parts (b) and (c). (Round your answers to three decimal places.) Between 3 and 7 minutes after the mixture reached a temperature of 95°C, the reaction activity ---Select--- increased decreased by an average of U/100µL per minute. After 5 minutes, the reaction activity was ---Select--- increasing decreasing by U/100µL per minute.



Answers

In an enzyme-catalyzed reaction with stoichiometry $\mathrm{A} \rightarrow \mathrm{B}, \mathrm{A}$ is consumed at a rate given by an expression of the Michaelis-Menten form: $$r_{\mathrm{A}}[\operatorname{mol} /(\mathrm{L} \cdot \mathrm{s})]=\frac{k_{1} C_{\mathrm{A}}}{1+k_{2} C_{\mathrm{A}}}$$ where $C_{\mathrm{A}}(\operatorname{mol} / \mathrm{L})$ is the reactant concentration, and $k_{1}$ and $k_{2}$ depend only on temperature. (a) The reaction is carried out in an isothermal batch reactor with constant reaction mixture volume $V$ (liters), beginning with pure $A$ at a concentration $C_{\mathrm{A} 0}$. Derive an expression for $d C_{\mathrm{A}} / d t$, and provide an initial condition. Sketch a plot of $C_{\mathrm{A}}$ versus $t,$ labeling the value of $C_{\mathrm{A}}$ at $t=0$ and the asymptotic value as $t \rightarrow \infty$ (b) Solve the differential equation of Part (a) to obtain an expression for the time required to achieve a specified concentration $C_{\mathrm{A}}$ (c) Use the expression of Part (b) to devise a graphical method of determining $k_{1}$ and $k_{2}$ from data for In versus the pour plot should involve fitting a straight line and determining the two parameters $C_{\mathrm{A}}$ (int the parting of the partating and and the conting are a contation a conting from the slope and intercept of the line. (There are several possible solutions.) Then apply your method to determine $k_{1}$ and $k_{2}$ for the following data taken in a 2.00 -liter reactor, beginning with A at a concentration $C_{\mathrm{A} 0}=5.00 \mathrm{mol} / \mathrm{L}$ $$\begin{array}{|l|l|l|l|l|l|} \hline t(\mathrm{s}) & 60.0 & 120.0 & 180.0 & 240.0 & 480.0 \\ \hline C_{\mathrm{A}}(\mathrm{mol} / \mathrm{L}) & 4.484 & 4.005 & 3.561 & 3.154 & 1.866 \\ \hline \end{array}$$

Yeah. Solution Hair mhm. Ray. Tickle toe minus de by. Duty off. See? Equal to okay. See to the power one. Okay. Call to Okay. See you for first order. Mhm four. First order. No, for the next step. All right. See? A by a c A mhm. Hey called tau minus. Okay. Duty by integrating both side. Believe it. Both side we get Yeah, you know, see a note. But see you divide et off. Okay? Do you see right? Do I get by? See, um, sorry. Here it is. This here he called to minus Hello? Do you note Tohti. Okay, duty now has de yeah DX by X equal toe Ln X So here we get, you know. Mm. Helen? No. A Sorry at it. See you. Minus Helen. See? Not equal toe minus Katie. No. By solving this, we get Ellen. Yeah, mhm. See? You see a note equal to minus Katie. Mhm. So after that? Hey, Good. See you. Bye. See a note equal to hey, XP minus Katie on. This is the question first. Yeah, I will never even like now for those second step second. 4.0. Yeah. 0.279 equal to Yeah. See you. Nord on 21 0.3 0.0 toe 0.2 62 e call to see a Yeah. No. From here we find that delta de equal toe 17.3 putting in one, putting the value in one mhm putting and one new Christian one. So we get 0.0 262 to everybody. 0.2 79 Equal toe. I'm minus one sound point three k. Yeah. No, we did, Helen. Zero point 9 to 9. Sorry. It's 9. 30 nine Equal to minus 17 0.3. Okay. Oh, from here. Okay, Equal toe. Helen. Okay, 0.9. Titi nine. Do I did play minus 17.3, which will be equal to 3.63 into Tend to dip our minus three and Harris Sansa.

So continuing aren't words. Physical chemistry. The first thing we're looking to do here is sold for the relationship between CIA and see a no opti. So see a equal to one divided by one over. See a Not so that's the relationship. Negative, Katie. So in the second part, we are solving for the half life. So it's important to note that CIA not is equal toe any, nor over V, which is equal to p o of rt. By the ideal gas law on DSO, we can simplify this equation that we've just written on the screen. Eso for tea, half that is Artie over K p. Not so when 50% of eight conversion occurs, that is half the moles of a form half the moles of B and a quarter of the miles of sea. And so the total number of moles is 1.25 and a not the pressure we can calculate as follows. P one half is equal to 1.25 on a North Party over V P. One half is equal to 1.25 p. No so following this, we can plot the graph of one overpay not bus. It's t on half. We get straight line equation of why is equal to not point not not seven x at no point not not three. Good slope. We can calculate K Born 15 kelvin. So when I won five Calvin the rate constant is equal to 9.5 day three liters per mall to minus one seconds to minus one. What? Following this, we are able to make another plot. But this time we have al en t one half Pino over r t passes one over tea. It was straight line equation. Why is 29 937 x subtract 28.958 so we can use the winds up intercept calculate K not And so we consult the pre exponential factor as a result of this 3.77 times 10 to 12 liters per mall to minus one seconds minus one on the activation energy he act is 2.49 times 10 to the five joules per month. We can calculate the value of K at nine. 80 Calvin that is not 0.201 liters per mile per second and then we can use the more percentage to calculate the initial concentration of end to O, which is not point not one for 104 miles per liter. So the final concentration of n 20 when 90% has reacted is not 900.10 c a no. We substitute all the values and for the final equation to calculate t So we have the time required for a convulsion. And this equation 9.10 C a north is equal to 1/1 over C A no ad k teeth. So we solve for this equation to find t but T is equal to 4003 100 seconds or 71.8 minutes.

Chapter fourteen question for They're giving us a reaction and em as the products. They give us a graph off the concentration off M versus time. So this concentration off M is increasing until fifteenth minute and after they're just stops. So this means or hear. The reaction is completed and we no longer see the product formation. Is the reaction occurring at a constant rate from T zero to t fifteen. Since this is a linear line that suggests us this reactions occurring at a constant rate. So this is going to be correct. Is the reaction completed at fifteen? Thus, also, Cory, because it's tops right over here pulls the reaction. A supported here started its point twenty okay and point forty off. So that means if this reaction is going to want one after all this point twenties call seems what we're going to be left with point twenty off l. Because Trent tickle seems twenty off this guy and they're saying, say, after minutes thirty, we're adding an additional point twenty k. So what the plot would look like if we keep this plot moving from, Let's say go start of five Cols forty. Goes for two, five, fifty, fifty five. And finally sixty. The amount that we're consuming is going to be again twenty k. So this slope is going to look exactly like that for fifteen minutes until it reaches two point for the fire from thirty fifty minutes. And after here, Wolf, K and L are going to be called seem completely this time. So it's going to be and no reaction after that point. That's why we're going to see something like this right over here. So if you look at it, this part of the question until four to five, the slope is going to be the same, but then it will stop.

So, uh, we went to first draw a, uh, an energy profile for this reaction of hydrogen and flooring. And we know that there are a total of three steps, and so there should be an activation energy hump for each step. And we also know that the first step is thea rate determining step. Ah, or it has the highest activation energy, so that makes it three determining step. So the energy profile is going to look something like this with energy on the vertical and the reaction coordinate on the horizontal. So there will be reacting, set some level, and then a fairly high activation energy. And then, ah, we don't know anything about the activation energies for the other two. So we could ah, uh, come to a situation where the second step is very fast. And then the third step is not as fast. Uh, so something like this reactant and products. And then this is the transition state one for the first step in transition State two for the second transition state three intermediate one intermediate to and eso the big activation energy would be this first step. Then let's see, what else do we want to do here. Ah, reasonable mechanism. So we're looking at the reaction of ah h two and F two forming to H f. And so clearly, we've got a big change in the bonding here. The hydrogen atoms are bonded to each other, but they won't wind up that way so that H h Bond has to be broken. Ah, and likewise for the flooring Adams. They're bonded to each other and react. It's and bonded to the hydrogen send products, and so that FF bond has to be broken. So the bond energy data that they gave us here is of use because ah, the hydrogen. The hydrogen bond is very strong, but the flooring to flooring bond is very weak around 1/3 of the energy of the hydrogen bought. And so it's likely that the first thing that's gonna happen is the flooring molecule is going to separate into two flooring. Adams and then those may be fairly reactive toward the hydrogen eso that we get one HF there noticed the highest bond energy is the HF, so the tendency to form that bond is is great, is huge, is large. Ah, and that would leave us a free hydrogen. And then the flooring that's left over from step one would react with the hydrogen from step to make. The other H efforts of those would be our, uh, our steps. And this would be the slow step because, uh, the energy has got to be put in to break that bond. Part of the energy and breaking the age age bond is going to come from the flooring approaching one of those hydrogen atoms and forming that HF bond. And so that's why I wouldn't have as great an activation energy even though you're breaking, uh, those hydrogen hydrogen bonds. And then let's see, what else is there anything else we need to do or consider any other answer we need to give, um, of the reactant that was limiting in the experiments? Well, if you had a huge excess of flooring and you had small amounts of hydrogen ah, then even even at ah, that excess limiting relationship that we use mawr floor ain being involved in the slowest step, uh, would probably have and effect in increasing the reaction. But the same K was observed in ah, the same apparent great constant was observed in both runs and eso. Probably the hydrogen was the one that was in excess first and then in greater excess. And so it would have been the flooring. Ah, that would be limiting the one that we were actually watching, uh, reacting and decrease in concentration is three. Action went on.


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