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Part A"Acanan Inl-order reaction (A productb} has & ro*0 consta n d 6 IOxio $ reactont [A] Ja Djc= 6.2578 Ihe original poncentralin? Exproea nporcpnt 45 &...

Question

Part A"Acanan Inl-order reaction (A productb} has & ro*0 consta n d 6 IOxio $ reactont [A] Ja Djc= 6.2578 Ihe original poncentralin? Exproea nporcpnt 45 "C, How mani mnutes 00o2Loriho conccntilonLcatan Hint(e)8.56minYour an Nn= nqualto 1016 ol ttw T"o coo[" Udae[Vt @e Juko onu You MaY nbod @0 foviln# Convrti Dotwoon S0cAnJulmI Movloe No creditout Iry Apnndrop 6 26* ol tng oriqinnSubmltPmv oumannwingnPart 0,

Part A" Acanan Inl-order reaction (A productb} has & ro*0 consta n d 6 IOxio $ reactont [A] Ja Djc= 6.2578 Ihe original poncentralin? Exproea nporcp nt 45 "C, How mani mnutes 00o2 Loriho conccntilon Lcatan Hint(e) 8.56 min Your an Nn= nqualto 1016 ol ttw T"o coo[" Udae[Vt @e Juko onu You MaY nbod @0 foviln# Convrti Dotwoon S0cAnJulmI Movloe No creditout Iry Apnn drop 6 26* ol tng oriqinn Submlt Pmv oumannwingn Part 0 ,



Answers

Three different sets of data of $[\mathrm{A}]$ versus time are giv the following table for the reaction $A \longrightarrow$ prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ What is the approximate half-life of the first-order reaction?

All right. We have to complete the following nuclear reactions. There are four of them. Um, so essentially, we're just gonna balance or mass in our charge, and then we'll figure out what element we have to write as the new nucleus. Sound good. All right. So Boron 12 is going to carbon six. Sorry. Kharfen 12. Stupid me. Um, so are mass is the same here, Um, are mass is increasing by one. So we have to balance that by having a negative mass. Um, a negative charge. Sorry, s O. R. Mass is not increasing or decreasing our charges going up. I want So we have 12 and five on the top and bottom, respectively. So what particles? That that's going to be a beta particle. So you could write e r. You could also write B minus whatever you want to dio, um, act in IAM to 25 is going to France iam to 21. So, um, you're gonna notice some of these, um, some of these mass changes, right? That's a mass change of four. And a charge change of two. So what is that to you? That's gonna be a helium nucleus, which is Alfa. Um, that's what being out for the K. I can't write force that I am sorry. Rebounds are mass And charge. Yes, moving on. Nickel 63 is going to something and admitted the beta particle. Right, So same is a are massive staying the same are charges gonna go up by one. So now we have, um, 28 on the bottom on both sides. What is element 29? That's gonna be copper. Copper 29. Copper 63. I mean, sorry. Um, so that's the answer to see. And then, um that's beta decay. I know. We're not supposed We don't need to get in the case, but it's good practice. So then business to 12 is omitting an Alfa particle here, so that means that our mass is gonna decrease by four. Are charges gonna decrease by two? So I'm going to give us to await and 81. Um, what is Element 81? That is Sally. Um, t l all right. Mass and charge of balanced. Yes. OK. Looks like we're done here.

The best way to determine which of the following reactions based upon the data sets of concentration as a function of time is zero order. First order and second order is to go into excel and plot the concentration as a function of time. One over the concentration is a function of time and natural log of the concentration as a function of time for all three sets. This is going to give you a total of nine graphs. If you do this, you can then identify which one is first order. Well, which one is zero order? Because the concentration as a function of time graph will give you a straight line. You can also find out which one is first order because the natural log of the concentration as a function of time will give you a straight line and which one is second order because one over the concentration as a function of time will give you a straight line. So when you do this, as I have done here, you can see that mhm data set one, Yeah, gives a straight line According to an R squared value of 0.9996 I'm not showing all nine graphs here. I'm just showing the one that gives you a straight line. It's natural log as a function of time for the first one. Yeah, if you have all three graphs and you show the equation of the line and the R squared value, you'll notice that just one of them has r squared value closest to one, the one that has an R squared value closest to one. That's the one that's the best straight line for data set one. It's natural. Log is a function of time for data set to it's one over the concentration as a function of time. I'm sorry. It's just concentration as a function of time. And for data set three. It's one over the concentration as a function of time. So for the first Data set, its first order Second Data set it zero order. Third Data set its second order

For this next question. It asks us to calculate the rate for the second order reaction. In order to calculate the rate we need to revert back to the differential rate law differential rate law, sometimes just called the rate law is simply rate equals the rate constant multiplied by the concentration raised to the order. The order in this case because it's referring to the second order reaction would be to So if you haven't done question 27 yet, you need to do that by plotting concentration. Natural log of concentration, one over concentration as a function of time for all three experiments. Plot these, um, values as a function of time and determine which one gives you the best straight line based upon the R squared value. And you would identify that the second order reaction is the last one. Experiment three then, because it has an R squared value closest to one. If we had also plotted the natural log and went over, the concentration would actually plot all three for all of them. So with this being the second order reaction, this is the equation of the line. Then the K value simply corresponds to the slope for the second order reaction. So rate is going to be equal to K, which is the slope multiplied by the initial concentration, which is one up as shown here squared. So it's going to be 0.1 or more specifically, it would be point 00992 But we'll go with this as I assume the author was trying to make all of them 0.1 the K values for all of them one. So when we do that, we get our initial rate of point Oh one Mueller per second.


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