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QuESTion 16For an investigation into kinetics, You measured the times _ took t0 reach the endpoint versus with the concentration of iodate Ion in solution? If the c...

Question

QuESTion 16For an investigation into kinetics, You measured the times _ took t0 reach the endpoint versus with the concentration of iodate Ion in solution? If the collapsed rate constant isTale K'[IO31PHow would you find the order olreactonp?Graph logltelative rate) (X) loelconceteatlon| W)and Wnd the spo. Gmph relative rate W) concentration ( Jand fndthe -lope Graphrelative Fale () concentratien (I and the slopr Wateys Hecot ntratlon (), and find the slope Graph Lfrelatve rate] (V} vs. top

QuESTion 16 For an investigation into kinetics, You measured the times _ took t0 reach the endpoint versus with the concentration of iodate Ion in solution? If the collapsed rate constant is Tale K'[IO31P How would you find the order olreactonp? Graph logltelative rate) (X) loelconceteatlon| W)and Wnd the spo. Gmph relative rate W) concentration ( Jand fndthe -lope Graphrelative Fale () concentratien (I and the slopr Wateys Hecot ntratlon (), and find the slope Graph Lfrelatve rate] (V} vs. toplconcentcation] (xl, and frd the ~lope Graph lorlrelative



Answers

Using the graph from Problem $13.53,$ determine the time required for the $\mathrm{SO}_{2} \mathrm{Cl}_{2}$ concentration to drop from $0.100 \mathrm{~mol} \mathrm{~L}^{-1}$ to $0.050 \mathrm{~mol} \mathrm{~L}^{-1}$. How long does it take for the concentration to drop from $0.050 \mathrm{~mol} \mathrm{~L}^{-1}$ to $0.025 \mathrm{~mol} \mathrm{~L}^{-1}$ ? What is the order of this reaction?

Problem. Party doubling and no concentration. Okay, doubling no concentration while holding or to constant increase the three by a factor poor. Okay. in experiment one an experiment to Now doubling, doubling or concentration of co two while holding concentration of eno constant. Double the rate double buried in experiment two. And experiment three. So the reaction is second order in a new and first ordering or two. Therefore red equals two K. Into. And not to give power to multiply by or to the power Radicals to multiply by concentration of an L. to the power to multi private concentration of co two. This answer. Poor party. Now in part B. First we deal part C. Okay from experiment first kevin equals two. 1.41. Offly use this formula. So rate given question that is one point poor. one into 10. to the power -2 M / 2nd divided by concentration of a new. That is zero 0126 to the power to into Judo Point Judo one 25 I am after calculating we get 710 five. That is approx madly equals two. Seven point of 11 into 10. To the power three M. to the Power -2. And as to Devour -2. No K two equals two. Similarly given rate is zero point 11 three M per second, divided by zero point 02 pipe to. It is a concentration of N. O. So well 0.025 to the power to multiply their concentration of or two. That is 0.02 pipe zero. After calculating we get 7118. That is approximately close to 7.12 into 10. To the power three M. to the Power -2. As to the power managed to K three. Similarly use this formula that is 5.6. put value of Read, it's 5.6. put into 10 to the power minus two. I am worse. Again, divided by concentration of a new. That is 0.0 252 to the power to multiply by zero point 125 After calculating we get 71 05. That is approximately equals two seven point 1 1 into 10. to the power three a.m. to the power managed to. And as to the power managed to the airport Kiev risk equals two. We add All three value. Okay, there it is. 7105. Bless 7118 plus seven. divided by three. That is equals two 7109. The approximately equals two 7.11 Into 10. To the power three M. to the power -2. And as to the power managed this answer part, part C and answer for part B. Is that is and sell for part B. That is what is the unit of rate constant. So you need to operate constant this you need up this in interpret constant M. to the Power -2. As to the Power -2. This answer part part of me now in part of the red equals two. We apply this formula simply put the value value. Okay we already calculated that is seven point 109 into 10 to the power three M. to the Power -2. And as to the power. Mhm. and to develop -1 surely surely hear this problem. Something Little bit mistake there's a part of one not to please correct it minus two man sorry again minus one minus one minus one minus one minus one. Okay So red equal to 7.1 09. In return to the power 3 -1. Multiply by concentration of and no there it is 0.0750 to the power to multiply by concentration of or two. That is zero point 0100 mm. This also I'm after calculating We get 0.399 that is approximately close to zero point Pour €0 for a second. This answer for part of the Now in the party, the data given in terms of the disappearance of a new Use. A question 14.42 Related to the disappearance of a no to the disciplines of auto. Therefore we can write us minus delta concentration of and no divided by two delta T. It calls to minus Changing the concentration of or two divided by delta T. For the concentration given in the part of the that is delta and no divided by party. There is a concentration of and no divided by Elta equals two zero point four double zero and per second. Okay, now simply put the value, so there is a concentration of or two divided by Delta T equals two. We put this value in this equation, so Jiro Point ford. Jiro, Jiro am per second into two, That is equals two zero point 200 I am per second is the answer for party.

So here's a rough sketch of the concentrations versus times. Um For more precise one, you can look at the at the textbook. Um And looking at the more precise one, it's apparent that the ending concentration, like the way that B be slopes down A seems to slope up twice as fast. So That means that you have, like the ratio is going to be one B. Produces to A. Or one B. U. Requires to a. Now it's a matter of this to a form B. Or just be form to a. Well we see that the concentration of B goes down over time, meaning it's used up concentration of a goes up over time, meaning that it's formed. So the one that's used up is the reactant and then the one that's formed as the product. So it's going to be be makes to A. And that corresponds to a choice for Yeah. Then for part B. We want to know the the rate in terms of appearance or disappearance. So that means for the reaction, the rate is going to be the negative change in the concentration of the over time, and then for the product you have a positive sign since the the amount is going to increase. So that's the change in the concentration of a over time. And we have to also divide by two because two units are quantities of A are produced for every quantity unit of the reaction, so we divide it out um to express the rate in terms of the reaction itself.

This question is a review of several concepts that were introduced in this chapter. The first is in reference to relative rates. If we know that the observed rate for the disappearance of Eno is 9.3 times 10 to the negative five Mueller per second than anything with the same coefficient would have the same rate. So the rate of appearance of No two would be the same at 9.3 times 10 to the negative five Mueller per second. However, in looking at the coefficient on oxygen, it's one half the other coefficients, so its rate of disappearance will be one half the rate of the other two or 4.7 times 10 to the negative five molar per second. Next, it wants you to calculate the rate constant. The rate is always equal to one over the coefficient, multiplied by the rate of appearance or disappearance. So we can use the rate of Eno multiplied by one half to get the rate of the reaction set that equal decay, multiplied by the concentration of N. O squared, multiplied by the concentration of 02 Rearrangement gives us .831 over Mueller squared seconds. Because it's third order overall, we need to have polarity squared in the denominator, and the last question is in reference to what happens if we increase the concentration of N. O by a factor of 1.8? Well, because the rate is a function of the square of the concentration, Then we would square the increase in the concentration to get the increase in the rate, which equates to 3.24 times quicker.

So the first thing we're doing here is determining the order off reaction. So we plug our known values into the second order rate equation because it looks as though it could be a second order reaction. So the equations up on the screen just for your reference, in case you're not aware of it, swept in the first two examples where I plug my numbers into that equation. So I have not 0.892 are not 0.870 So the values of Cape are similar here by using our second order rate equation. And so the reaction is second order. Next, we're determining value K, which is our rate constant. So what we need to do is continue with the final two examples calculations that I haven't completed above but uses the exact same formula as the two that I have completed. We then add them all together, divide by four to generate an average value for K of not 40.872 So next we're trying to determine Time T when the concentration of C four h six is not point, not not for 23 So the equation we need is upon the screen. I plug in my numbers where I plug in my desired concentration. That is, at time t that we don't know. And then we also plug in the concentration at times several. We also plug inthe e k value that we determined that was above, as you can see, and then we determine it to be two times 10 to the power. Two minutes. So just for the final example now we're finding a time again when the concentration off C four h six is at no point. Not not 50 So we're using the exact same equation. We plug in our desired concentration. We plug in our initial concentration. We plug in our rate constant. They're the only three figures that we need here on. Then we determine time to be 1.6 times 10 to the power, two minutes


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