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Experiment 16 Determination of Rate Law Prelaboratory Questionscontuimiu 0.00502 M SO;" (UMIS Icsl [uuc blue at 55.332 $. What is + thc rate = the renction? 05...

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Experiment 16 Determination of Rate Law Prelaboratory Questionscontuimiu 0.00502 M SO;" (UMIS Icsl [uuc blue at 55.332 $. What is + thc rate = the renction? 0504 50 3 ML 55 ra ( [so 5 5 3 2 07 .10 ~ Jm &experiment between tWO gases During 10 * Mof the product foned _following datarecorded = when ]0543 B2E+F[A Jinjt molesIL[BJinit molewLTm required form L0xIO* M of FsecundsRate Wrt molcyiasBNxIQ Luriu=A AoMuoSSOrder of _Order of B3J-7 3.33 39 Rate constant 3 ? 3'|0 = 5 K = 33.3 ~ 3

Experiment 16 Determination of Rate Law Prelaboratory Questions contuimiu 0.00502 M SO;" (UMIS Icsl [uuc blue at 55.332 $. What is + thc rate = the renction? 0504 50 3 ML 55 ra ( [so 5 5 3 2 07 .10 ~ Jm & experiment between tWO gases During 10 * Mof the product foned _ following data recorded = when ]0 54 3 B 2E+F [A Jinjt molesIL [BJinit molewL Tm required form L0xIO* M of Fsecunds Rate Wrt molcyias BNxIQ Luriu= A Ao Muo SS Order of _ Order of B 3J-7 3.33 39 Rate constant 3 ? 3'|0 = 5 K = 33.3 ~ 3 '2Jjs 9? | Rate law expression: 10 - 5 (to " ? )3 3 3 Y("> 7V 7 k 3.33,L0- 5 33 3 10 f = 3 3 9 [ 9]' [ 6] Oi7



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The following data were collected for the destruction of $\mathrm{O}_{3}$ by $\mathrm{H}\left(\mathrm{O}_{3}+\mathrm{H} \longrightarrow \mathrm{O}_{2}+\mathrm{OH}\right)$ at very low concentrations: $$ \begin{array}{llll} \hline \text { Trial } & {\left[\mathrm{O}_{3}\right](\boldsymbol{M})} & {[\mathrm{H}](\boldsymbol{M})} & \text { Initial Rate }(\boldsymbol{M} / \mathrm{s}) \\ \hline 1 & 5.17 \times 10^{-33} & 3.22 \times 10^{-26} & 1.88 \times 10^{-14} \\ 2 & 2.59 \times 10^{-33} & 3.25 \times 10^{-26} & 9.44 \times 10^{-15} \\ 3 & 5.19 \times 10^{-33} & 6.46 \times 10^{-26} & 3.77 \times 10^{-14} \\ \hline \end{array} $$ (a) Write the rate law for the reaction. (b) Calculate the rate constant.

Okay, Problem. 87. 1st off all, you have to run again. We have to rise on the table and a table. Okay? And now you see the equation. 03 plus hydrogen brought you oxygen and Oesch. So our purpose e to find a and B and right with the red color right here. So we know the race for this equation is K the concentration of ozone and the Power A and B. Okay, hydrogen concentration of hydrogen and Bobby. And now we just substitute the data we have. And you have tow realize how can we do it? Just to the right number to solve the problem and let me do the first. I still rate 3/3 1. I forgot greater here. Okay. Okay. So Okay, Cancel, Cancel. And for this, you will have 37 Great one here and one a about a one game. For this. We will have this about Do right? Eighth, be a b. Okay. And now we got about two p ico too, right? So that's about equal to on. And that's why be must be equal one. Unless we do we fight a I smoked beast. Let me by AI to re to over a one general. Rather, you just subject you and find a good answer. So you done. E must gay here. I felt that. And three boys. 25 Okay, we must be here, huh? And the great for this we have five boy. 17 Okay, perhaps. And it's like three boy to to. Okay, so for this, we almost the same. We cancel. I can sell this to and for the right side. We have about one half right. One half. Hey. Hi. And this not be equal a 9.944 And for this you have a bow. Fine Thai boy. Right? This might be about or fine. So one hour, half a equal 10.5. It must be equal One. Right? Okay. Night for this. But April 1 and then you get right, you know, super equal case or threes? One we don't need to write. Okay. And be so get the right constant. And now we have to call it a rate constant K Mhm. Yeah. It can be calculated from the formula there. So gay, equal raised, divided by oh, threes. Hi, hydrogen. Right. We put anything up from the table A bit for night, boy for four. So maybe I told this one from two. Hmm. Okay, I'm gonna get okay. So what? They were about one boy. 12 Okay. Yeah, we got more right here. I from 11 So one more. One more. Here. So good. Minus one. Here. The right constant. Thank you. We can put any that upon table, so you have to see which one is right for you. So you saw the problem?

So to determine the order with respect to its reaction, we're going to look at the exponents. So for part A we see no exponent, which means the order and it is one. And then for be the order is three according to the expo on. The overall order is going to be the some of the orders for all the reacted. So one plus three is equal to four. For part B. We see no exponents, which means it's a one so one and one, and then the overall order is one plus one or two. Um For part C we have first order and A. We don't see be written, which means the rate is independent of the so zero order and be the overall order is one plus zero or one. And for part D. Its third order in a first order and be and three plus one is four for the overall order.

So to determine the order with respect to its reaction, we're going to look at the exponents. So for part A we see no exponent, which means the order and it is one. And then for be the order is three according to the expo on. The overall order is going to be the some of the orders for all the reacted. So one plus three is equal to four. For part B. We see no exponents, which means it's a one so one and one, and then the overall order is one plus one or two. Um For part C we have first order and A. We don't see be written, which means the rate is independent of the so zero order and be the overall order is one plus zero or one. And for part D. Its third order in a first order and be and three plus one is four for the overall order.

To write the rate law. We simply need to know the order of the reaction with respect to each reactant and that was given to us. It's first order with respect toe on first order. With respect to an 02 we sum up the total order, the overall order being too. And then what we do is we have the rate constant be equal to one over time where the substance of the exponents on the polarity and the denominator is one minus the total order. So the total order is to the exponents. Then on em would be one minus two, which is one we go to the next one. So rate is equal to K second order with with respect to Eno First order with respect to cl to some those up that gives us three. So that means the rate is the rate constant is going to be Have units of one over polarity squared two on one gives us 33 minus one is to one over. Polarity squared over seconds. Then for the next one we've got rate is equal to K concentration of C h C. L three. First order on the concentration of CL to race to the one half. So we some those up, we get 1.5, subtract off one, we get one half. So it's one over polarity to the one half seconds. The last one ozone is to the second order oxygen atoms is to the negative one order that gives us a total of one appear. So one minus one is zero. So it's one over m 20 which makes the M go away and the units on care just one over seconds.


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