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02 Question (1 polnt) Cyclopentadlene (C; Hsh reacts with itselt toform dicyclopentad ene (CjaHus; 0 830 MsolutIcn o C H; meonltoter rcaclion proocco gaphot WcsHlwt...

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02 Question (1 polnt) Cyclopentadlene (C; Hsh reacts with itselt toform dicyclopentad ene (CjaHus; 0 830 MsolutIcn o C H; meonltoter rcaclion proocco gaphot WcsHlwtsu time vields straight IlneSaepage 660funclcnolilmenathIst attemptEs0 Pcriodic IabltSte HintHow many seconds wlll It take for the [CHc]tobe 0 AR The k for therosction [0.1414

02 Question (1 polnt) Cyclopentadlene (C; Hsh reacts with itselt toform dicyclopentad ene (CjaHus; 0 830 MsolutIcn o C H; meonltoter rcaclion proocco gaphot WcsHlwtsu time vields straight Ilne Saepage 660 funclcnolilmenath Ist attempt Es0 Pcriodic Iablt Ste Hint How many seconds wlll It take for the [CHc]tobe 0 AR The k for therosction [0.1414



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 initial rate of the second-order reaction?

For this next problem, you need to go back to problem 27. If you haven't done problem 27 you need to do problem. 27 1st in problem 27 you're going to plot the concentration as a function of time went over the concentration and natural log of concentration as a function of time for all three data sets. Once you've done that and you in Excel show the equation of the line, you can also ask Excel to show the r squared value. And I'm not showing all three graphs. You'd have three graphs here. Natural log of concentration, one over concentration and just concentration as a function of time. And you find that for data set, one natural log of concentration gives you the best straight line. You do all three here also, and find out that concentration versus time gives you the best straight line having the r squared value closest to one. That's how you determine if I had all three here. Which one is the best straight line is the one that has an R squared value closest to one, So its concentration is a function of time for the second one and one over. The concentration is a function of time for the third one. So natural law giving you the best straight lines. First order concentration is a function of time. Giving the best straight line is zero order and one over the concentration is second order. Then, with these graphs and the equations of the line, you can identify the the rate constant. The rate constant is the slope unless the slope is negative and then the rate constant is the negative of that negative slope or the positive value of the slope. Great constant is always positive, so the rate constant for the first one for first order is 0.1 Great constant for the second one's also 0.1 Great constant for the experiment three is also 30.1 or more like 0.992

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 question number 29 to determine the approximate half life of the first order reaction. It would be most helpful if we knew the rate constant for the first order reaction. In order to know the rate constant, we need to identify which of the three reactions is first order by plotting them. So if you haven't done step number or question 27 you need to do that. First you need to plot went over the concentration natural log of the concentration and concentration as a function of time for all three of them. You will then find out that we get a straight line of natural log of the concentration is a function of time for experiment one one over. The concentration is a function of time for experiment, too. I'm sorry. Concentration is a function of time for Experiment two and one over. Concentration is a function of time for Experiment three. The first order reaction is the one that gives us the best straight line with the natural log of the concentration of the function of time. So it's Experiment one after we have plotted it and identified that it gives us the best straight line. Then right from the equation of the line, we can get our K value. It's the negative of the slope. So our K value is 0.100 Then we can use the half life equation for a first order. Integrated rate law problems, which is half life, is equal to natural log of two divided by the rate constant or, in our case, natural log of two divided by 20.1, which is 69.3 seconds. And if we look at the data, we are about half that concentration somewhere before 75 so 69.3 makes sense.

In this question we've been given a reaction in which two malls off and or two are creating with um all of F. Two to produce two malls off and all to act. So this is the reaction that we're looking at and the rate law for this reaction is going to be in the form of great is the quantum Okay The concentration of one or 2 raised to the power of em multiplied by the concentration of F. Troop raised to the power of end. So the main goal here is to determine the order and the right expression. But since we want to determine our um our end and finally our K. For us to have a full rate expression of this reaction. So for us to determine the quarter of reaction with respect to an autumn, We can substitute the two initial concentrations of n. 0. 2 and the corresponding right that is let's say we have afraid being a courtroom. Okay prime multiplied by the concentration of an auto is to the pope. And so here our K. Prime is representing K multiplied by the concentration of F. Troop. So we're going to leave that for a moment. So we want to determine the rate the order of the reaction with respect to an auto. In other words, we want to determine our camp. So let's take for example a certain condition where we have our right one Corresponding to a concentration number one and we have our late true corresponding to the concentration of an autumn. It points to raised to the power of em. We have been given all this information. For example When the rate is equal 0.0-6, remember our key. Okay prime this is the same. So when the rate is equal 0.0-6 The concentration is going to be equal to a 7.1. And we raise this to the power of um And again when the rate is 0.051 the concentration is going to be equal to self control again or raise this to the power of him. So what we here at the end of the day is an expression of 7.5 B equal to 7.5. Place to the power of em. And if we introduce logarithms we're going to have our law 0.5 Being equal to M Milk. So we can conclude that our aim is equal to one achieves this reaction is first order first order with respect to with respect to and or two then walking on we now need to determine the reaction order with respect to F two. And for us to do that we're going to apply the similar approach to say our right Let's look at .2 being equal to K. Prime. The concentration of F two raised to the power And at point #2 and our rate say it point blank it's going to be equal to K. Multiplied by the concentration of afternoon. 8.3 Ways to the power of end Here our K. Is going to be equal to care the concentration of an auto raised to the power of em. Now that we all have this information, remember this is the same reaction. So these kinds of So we know that when the rate is 0.051 concentration is 0.1 and we raised this to the power of and And when the rate is 0.103 the concentration of F two is equal to 0.2. And we also raise this to the power of mm. What we now have is we also have a situation where we have 7.5 billion equal to 7.5 raised to the power And And again if we introduce logarithms, the log of 7.5 is equal to end log officer 15. Again our L is equal to one. That is it is first order with respect to F two. So now that we have this information we can then look at our overall expression and if we look at the overall right expression we have the rate being equal to this is the overwrite expression. Now that we have our and our end this rate is going to be returned in the form of great is equal to k concentration of an autumn, Multiplied by concentration of F two. And the overall right over order of this reaction is M plus M which is one plus one and this is true. So the overall order of this reaction is while trump. Now we have to move on to determine our cake K. Part of this expression. So to determine the value of K. Let's look for example, we know that our K. Is going to be equal to right Divided by the concentration of N. or two, multiplied by the concentration of F Troop. So at any point let's just take any point The same .1 When the rate is zero, our concentration of one or two is going to be served with one And that of F two is also going to the airport one. So our K. here is there is 2.6 and let's take another point. The rate is 0.051. The concentrations are going to be equal to 0.2 by 7.1 and we have K two of 2.5. And let's take another point. If our rate is 0.103 the concentrations are going to be 0.2 by 0.2 and this is Equal Truth 2.5. Okay. And another point they say careful, The rate is equal to 7.411 Concentrations are going to be a 0.4 x 0.4 and this is equal to two seven and if we look at these these units are pay more could be lead to pick Tessa meter or later the second Or 2.7. I am yes, they can. Now, if we look at this they are relatively the same. So what we can do is to just take the average of these K values to get an average off K being equal to if we make the average here, They say 2.6 plus 2.55 Plus right up to 2.7 and we divide this by four. Our key Is going to be 2.61 am this again. So this is our value. Okay, now that we have this, we can then write the full rate expression to say the rate of this reaction. The rate is equal to 2.7, multiplied by the concentration of N or two. And the concentration of actually, so this is the final solution And over order of this reaction is 2nd order. That is the order is true.


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