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Volume OT uie SUUUOIFinding the concentration of substance B in data set 1,2,and 3: To perform the "First Reaction" (for Data Set #1)- you measured out 10...

Question

Volume OT uie SUUUOIFinding the concentration of substance B in data set 1,2,and 3: To perform the "First Reaction" (for Data Set #1)- you measured out 10.0 mL of stock A and mixed it with 10.0 mL of Stock B Calculate the concentration of B in Data Set #1.b To perform the Second Reaction for Data Set #2, you chose different volumes of A and B to mix Calculate the concentration of B in Data Set #2.To perform the Third Reaction for Data Set #3,you chose different volumes of A and B to mi

volume OT uie SUUUOI Finding the concentration of substance B in data set 1,2,and 3: To perform the "First Reaction" (for Data Set #1)- you measured out 10.0 mL of stock A and mixed it with 10.0 mL of Stock B Calculate the concentration of B in Data Set #1. b To perform the Second Reaction for Data Set #2, you chose different volumes of A and B to mix Calculate the concentration of B in Data Set #2. To perform the Third Reaction for Data Set #3,you chose different volumes of A and B to mix Calculate the concentration of B in Data Set #3.



Answers

Consider the reaction $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q)$ The following data are obtained at a certain temperature: $$ \begin{array}{cccc} \hline \text { Expt. } & {\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}\right]} & {\left[\mathrm{OH}^{-}\right]} & \text {Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\ \hline 1 & 0.050 & 0.120 & 0.00158 \\ 2 & 0.050 & 0.154 & 0.00203 \\ 3 & 0.084 & 0.154 & 0.00340 \\ 4 & 0.084 & 0.200 & 0.00442 \\ \hline \end{array} $$ (a) Write the rate expression for the reaction. (b) Calculate $k$. (c) What is the rate of the reaction when $10.00 \mathrm{~mL}$ of $\mathrm{CH}_{3} \mathrm{COOCH}_{3}(d=0.932 \mathrm{~g} / \mathrm{mL})$ is added to $75.00 \mathrm{~mL}$ of $1.50 \mathrm{M} \mathrm{NaOH}$ ? (Assume that volumes are additive.)

So the first part of the question isn't to top were given plots of consciousness, a burst time, conservation of natural log of a verse time and one over the concentration of favors time. And so, you know, if you don't know what you're doing, then yeah, this is gonna be impossible. But you should know from the integrated rate laws that whichever one of these is linear will tell us what order is. And so it's very obvious this is the linear one. And that means that this is a second order reaction with respect to end. And that's that answers the first question. Um, now, um, to figure out the initial concentration raft into a little bit of work. Now, the first thing we have to do is we need to figure out what K is, and we can know from the integrator. A log in the slope of this graph if it's linear, is equal to K, and you can figure that out just graphically and looking at it that this is going to be 10 and the units will be leaders per mole per second. Now, from here, we can use the integrated Ray law and figure out some of the values that we're looking for. So the integrated rail off for a second or two reactions that one over the concentration of A In this point, I's gonna be Katie work a We know it t is the time that passes is equal to of a case. He plus one over the initial concentration. And so if we want to figure out the initial concentration, um, well, just pick any data point. So, for example, we can say, you know, we can use that first graph, which is concentration of a versus time at time T equals ah four, the concentration is 0.2 and we know it K is and so we can solve for the initial countries of A and that is 0.1 muller. Now that gives us the initial concentration and to figure out the concentration after time of nine seconds will do the same thing one over. Now the that's this over looking for the consolation of a K is still 10. The time is nine seconds plus one over then 0.1 Moeller. That's our initial concentration so we can solve for the concentration of a, um and at that time, we find that it is going to be equal to 0.1 more now, the last one you need to to figure out the half lives we need to know the half life formula, which comes from the integrated Ray Law. And that is that the half I was goes one over k times the initial concentration. And so you know, the initial concentration for the entire reaction is 0.1 Moeller. And if we know that this is 10 then very obviously we can see at the first half life. So it's a team won half subscript comma. One is one second than the second one. We're going to plug into the same Formula one over k times the initial concentration. But this initial concentration that we're looking at is no longer the 0.1. It's 0.5 Now. The reason is we've already gone through one half light. So one in that 10.1 now becomes a 0.5 But K is still 10. We have two seconds. You can see the pattern that it might be. You can check. You'll see that doubles every time. And so the first three half lives will be one second, two seconds, four seconds and then we're not ask for it becomes 878 seconds, 16 seconds and so on and so forth for a second order reaction. The half life increases with time over the course reaction.

To answer this question and determine the approximate concentration of a after 110 seconds for the zero first and second order reactions, we need to know what the rate constant is. So if you've done problem 27 you've plotted everything, you will get the rate constant for each of them from the slope. So Experiment one is first order its rate constants point to one experiment to zero order. Its rate constant is 00.1 an experiment three of second order and its rate constant is essentially 30.1 also so to determine the concentration at 110 seconds for the zero order reaction. Well, it goes to zero at 100 seconds. So if we're at 110 it's still nothing for B consulates. For first order, we use the first order integrated rate law, natural log of concentration at 110 seconds over, natural over a natural log of concentration at 110 seconds, divided by the concentration at time zero will be equal to negative K multiplied by t 110 seconds. Doing a little bit of algebra will find out that the natural log of concentration is negative 1.1 or 0.33 Moller. And that should make sense because here we see that at 100 seconds it's 1000.37 So it should be just a little bit less than that. At 100. And 10 0.33 is reasonable for the second order. Reaction will use the second order Integrated rate law one over. Concentration at time, zero minus one over concentration at time T equals positive. I'm sorry. Negative, Katie. I'm doing a little bit of algebra. Will see that one over. Concentration 110 seconds equals 2.9. Take the inverse of 2.9 and we get 0.48 Moller. And at 100 seconds for the second order we see it's 0.5. So it should be just a little bit less than that. 2.48 makes sense at 110 seconds.

For this question, we have the generic reaction of a single reactant goes to to be plus C. We then plot the concentration of a is a function of time, and we get this parabolic looking curve you the natural log of the concentration of a and it's still not quite a straight line, as shown in the graphs that air provided these air just reproductions. And then we plot one over the concentration of a as a function of time and we get a straight line because we get a straight line with one over the concentration as a function of time. Then this is going to be a second order reaction to determine the concentration at time. Zero. We simply extrapolate this back to time zero, and it falls pretty much right on 10. So that means that time 01 over the concentration is equal to 10, where the actual concentration is equal to 1/10 or 100.1. Moeller


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