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Consider the hypothetical reaction 2A -4C +DOver an interval of 3.0 $ the average rate of change of the concentralion of E was measured to be 0.0400 Mls: What is th...

Question

Consider the hypothetical reaction 2A -4C +DOver an interval of 3.0 $ the average rate of change of the concentralion of E was measured to be 0.0400 Mls: What is the final concentration of C at the end of this same interval if its concentration was initially 2.800 M?296 M264m244m3.16 M2840 M

Consider the hypothetical reaction 2A - 4C +D Over an interval of 3.0 $ the average rate of change of the concentralion of E was measured to be 0.0400 Mls: What is the final concentration of C at the end of this same interval if its concentration was initially 2.800 M? 296 M 264m 244m 3.16 M 2840 M



Answers

Consider the following hypothetical data collected in two studies of the reaction
$$
2 \mathrm{A}+2 \mathrm{B} \longrightarrow \mathrm{C}+2 \mathrm{D}
$$
In Experiment $1,[\mathrm{B}]_{0}=10.0 \mathrm{M}$
In Experiment $2,[\mathrm{B}]_{0}=20.0 \mathrm{M}$
$$
\text {Rate} =\frac{-\Delta[\mathrm{A}]}{\Delta t}
$$
a. Use the concentration versus time data to determine the rate law for the reaction.
b. Solve for the value of the rate constant (k) for the reaction. Include units.
c. Calculate the concentration of A in Experiment 1 at $t=30 .$ s.

So we have this, ah, data of concentration of a substance against time, and we need to make some plots and see what's linear in order to find out the order of the reaction with respect to substance. A. And so, uh, I typed in the time data into excel and the concentration data. Uh, and, uh, but I made a graph. It looked linear. And I thought, Huh, That I think it's the first time in this chapter that I've made a plot of the concentration against time. And it was linear. Ah, and so, Well, if it's ah if it zero order, then it's here. Order. And so I did the same thing with the second data set, and I got this graph and it clearly was not linear. And I guess that's one of the weaknesses of this technique you've gotta have. Ah, really good tight data. Set number one and you gotta have plenty of data number two. Otherwise, you can't distinguish a for example, a log plot being linear as composed to the original data being linear. But this clearly showed ah that it's not zero order. It was not linear for concentration against time withy Second data set. And so then I went on and tried the log plot. So here's Thea data set one taking the log of each value. If you were doing this by hand, it would be the same thing. You would need some good graph paper to make the graphs, and you would take the values, uh, like I did here. Ah, What I did is I told Excel to assign to each cell in this column Ah, value that's equal to the natural log of the corresponding value in this column. So natural log of one times 10 to the minus two is they get a 4.6517 and so on. And the natural log of five times 10 to the minus three is given here. And so you would just calculate each one of these with your calculator and write him in, and then you would go on, make a plot. And when I did that with data sent one, I got a nice line for the log of a versus tea for the first data set. And then when I did that same thing for the second day to set taking the natural log of each value in the second data set and made a plot of that. I also got a nice linear plot. And so that convinced me that the ah ah reaction is first order in a It's not zero order. Ah, I didn't bother to make the inverse bought once I get one that once I get one that looks linear. Well, just goes to show in this data set that you have to be careful about that because it can look a linear. One thing to notice here is the two endpoints were above the line and the two points in the midst were below the line, which suggests this kind of curvature. But, you know, you have to be able to see it might be a good idea to make all three plots. He would just make another column, copied the times over again, and, uh, make a column of of the universe. And, uh, and Joe hadn't plot that. See if it looks linear. Ah, but anyway, we have got it here. And so we've determined that the data is first order. The reaction is first order in substance A. In order to calculate the rate constant. Ah, we need to take account of the fact that in a log plot like this, the slope is equal to the negative of K, and that slope that you're getting is not the lower case K rate constant for the overall reaction. It's what we call K prime, which is related to the rate constant for the overall reaction has given here. And so we can calculate that by taking the observed K value with one substance divided by the concentration of the second substance. And so I did that for both plots for both slopes, and I got two different values. And that is an indicator that the assumption that you're making is not correct. Uh, not that it was wrong to do this a good idea to you. Go ahead and see what it looks like. But what you're doing is you're assuming that it's also first order and be. And if you do that and just divide the observed constant by the concentration of the other substance in that experiment, if you don't get K values that are equal to each other than that's an indication that you've tested the assumption that its first order, and that's not correct. So maybe it's second order. I didn't didn't think to see if it zero order first. I just tried. Well, clearly, it's not zero order because then you would divide by 10 to 0 in 22 0 which is one, and you'd get different values. So dividing by the square of the B concentration did give me within experimental error the same value of K. So that tells me that the rate law IHS a its first order in a and it his second order and be I could tell that again because considering be to be first order did not give me the same value for the rate constant from the two data sets, which it's the same reaction should it's constant is a constant should be the same. So here's the value of the constant. Not here, but about here. Uh, 0.1 point 73 times 10 to the minus four. Ah, and ah, we divide by polarity squared and the rate constant for the first orders per second, it was first order in a and so the units Ah, and then I can take the equation of the line and that. Why is the log of a We were plotting the natural log of a on the vertical and t for X on the horizontal. And so using that slope and intercept that air in the equation of the line are to just plug in the time and calculate the value of the log of a and so to get A and I take the exponential of the value that I had for the log of a Does that look reasonable? 5.95 times 10 to the minus three Moller. I mean, we've got a point right here at 30 and we're pretty close to um I'm sorry. Wrong plot. Right here at 30 we're pretty close to six times 10 to the minus three. We've got 5.95 Of course, it's not linear, but that gives us an idea of approximately where on this graph we would be Ah, a little bit below six and weird 5.95 So that that checks out. And so that is the concentration at 30 seconds

So Chemical kinetics is an area off our physical chemistry that is concerned with the understanding of rate chemical reactions as well as studying the chemical processes and transformations of reactant to products within a reaction vessel so fast we have l n concentration a equals negative k prime t at l n concentration A at time zero where we have a value for a value for T and a value for a zero. So then I have rearranged on plugged in my numbers undetermined what? K primers. And that is not 0.12 So moving on to the next part. Now we have the right que prime concentration A equals not 0.1 to 1 times 10 to the power minus two equals not point. Not not 12 miles per liters per second. So then the rate of K concentration a concentration be it's squared so K can be found using the following where K equals not point not 13 So next weaken Continue on to the second part. So we have our equation of concentration of a is the negative. Okay, prime t add L. A and concentration at Time zero, where we have k we have concentration of a concentration off a at time zero. We re arrange for tea, which is time, and it is time to be 5.8 seconds. Next, we've got concentration of a cause. Negative K prime t add Ln concentration A at time zero where we have our cake. Prime value R T value on a North valley. We re arrange for concentration of a on that is determined to be no point, not not to one. So this is just the last part. Here we have the concentration off. See, that is concentration of T C. Subtract two Delta concentration of a where the concentration of a is one point not times 10 to the power minus two subtract nor point nor nor to one which generates a value off no point, no, No. Eight. So that's concentration of C is concentration of c. Subtract two Delta A, which is to

In this problem for this hypothetical reaction, the rate law is given, and the initial consideration of the reactant are also given The reaction started and after eight seconds, the concentration off a becomes 3.8 in to 10 to the power minus three Mueller. First, we have to find out the rate constant for this reaction. No, since in this reaction we see that the concentration of we is much, much greater than the concentration off initially. So the concentration of be almost remains constant. And the reaction is a studio first reaction. The read can be written as we're key. Dish is que in tow. Concentration off beer is to power to No, we'll use the given equation to find out that read constant substituting all they will use. We can calculate the red constant and this comes out Toby 0.12 per second. Next, we have to find out there half life off the reaction since half life is related to it constant by this equation. So half life comes out, Toby 5.8 2nd. Next, we have to find out the concentration off E. After 13 seconds on solving, we get the consultation off a as 2.1 into 10 to the power minus three molars. Next, we have to find out the concentration off. See, after 13 seconds, no sense amount off. A reacted he's given us. So amount off, See? Reacted will be amount off. See remaining to Mueller as expected. So the concentration off he remains constant during the course of reaction.

In this problem we have to calculate the value of rate constant for the given reaction. two moles of B give see and to most of the and also we have to get great half life for this reaction. Suppose initial concentration of B is indicated by be not, which is equal to zero point 50 Moeller. Final concentration or B at any instant. Tea is denoted by a beauty Which is equal to zero. Going to zero Molar According to question. It takes 4.4 into 10 days. Three power two seconds for an initial concentration of B to go from 0.50 Mueller to 0.20 Mueller. Therefore, elapsed time the is equal to 4.4 into 10 days to the power two seconds. Right? Given that it is a zero order reaction, so four zero order The accent Yeah rate constant. Okay, denoted by K. Is equal to initial concentration of compound minus final concentration of compound at time. T divided by time taken for last time. Now substitute the value of being on tv TNT in this expression. Mhm $0.5 billion minus 0.20 Mueller, divide by 4.4 in to 10 days. To the power to second. We get K equals to 6.8 in to 10 days. To the power minus four moller per second. This is answer for rate constant. Okay. Yeah now we have to calculate half life for the reaction. Yeah so half life. Uh huh. Of the re accent is in order by T half. And for Gerada reaction it is equal to initial concentration divided by to into the great constant. Now substitute the value of Bennett and carrying this expression $0.05 billion divide by two into 6.8 into changes three power -4 Moeller per second. With the help of what we can get. This value equal to 364.64 seconds, Which is approximately equal to 365 seconds. This rate constant care is equal to 6.8 in two tenders report minus four Mueller course second and T half is equal to 3 65 seconds. Yeah. Yeah. This is our final answer for this problem. Thank you.


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