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Tabulated below are initial rate data for the reaction2 Fe(CN) 32 1-12 Fe(CN)1zInitial [I-1] [Fe(CN) '1Run [Fe(CN)s"3][V]Rate (Mls)0.01 0.01 0.02 0.02 0.0...

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Tabulated below are initial rate data for the reaction2 Fe(CN) 32 1-12 Fe(CN)1zInitial [I-1] [Fe(CN) '1Run [Fe(CN)s"3][V]Rate (Mls)0.01 0.01 0.02 0.02 0.020.01 0.02 0.02 0.02 0.020,01 0.01 0.01 0.02 0.020.01 0.01 0.01 0.01 0.02Ix 10-$ 2x 10-$ 8 x 10-$ 8x 10-5 8x 10-$3The value of k is:107 M-s 8 "1 10* M-3 $-! 10 M-2 s-! 50 M-2 $-1 none of thesePick one

Tabulated below are initial rate data for the reaction 2 Fe(CN) 3 2 1-1 2 Fe(CN) 1z Initial [I-1] [Fe(CN) '1 Run [Fe(CN)s"3] [V] Rate (Mls) 0.01 0.01 0.02 0.02 0.02 0.01 0.02 0.02 0.02 0.02 0,01 0.01 0.01 0.02 0.02 0.01 0.01 0.01 0.01 0.02 Ix 10-$ 2x 10-$ 8 x 10-$ 8x 10-5 8x 10-$ 3 The value of k is: 107 M-s 8 "1 10* M-3 $-! 10 M-2 s-! 50 M-2 $-1 none of these Pick one



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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 value of the rate constant $k$ of the zero-order reaction?

This is a method of initial rates with two reactant. To determine the order of the reaction with respect to each reactant, we need to treat each reactant separately, namely, if we look at experiments one and two, we see that the concentration of N. 02 is changing, but the concentration of F two is not. So let's just look at what a change to the concentration of N. 02 does to the rate. By looking at experiments one and 2 where F two stays constant. So with these two experiments a doubling of N 02 Only a doubling of eno to keeping F to constant, results in a doubling of the rate going from 20.262 point 51 This is indicative of a first order reaction with respect to N. 02 Now let's look at experiments two and three where the concentration of CO two is stained constant and the concentration of F two is being doubled from 20.12 point two. In this process we see that the rate is doubled from .512.103 Doubling the concentration resulting in doubling the rate is indicative of a first order reaction. So it's also first order with respect to F two. So the rate laws rate is equal to K, multiplied by each concentration raised to the first power. To solve for K will simply plug in the rate for any experiment and the two corresponding concentrations and we'll calculate a K value of 2.61 over Mueller seconds, one over Mueller seconds. Because the overall order is second order. First order with respect to each reactant. Or we could do this for all four experiments, Get four K values and then calculate the average. So the differential rate law is rate is equal to 2.61 over moller seconds, multiplied by each concentration raised to the first power giving us a second order overall

This question should really help you increase your conceptual understanding of the principle of initial rates, which are used in order to determine the order of a reaction for each particular reactant. For this particular reaction, it's just a generic reaction of a plus. To be plus C goes to two d plus e. We can write the rate law or the differential rate law for this particular reaction as rate equals K, the rate constant multiplied by the concentration of a raised to some power concentration of B, raised to some power concentration of C raised to some power with a method of initial rates. We have several experiments where we change the concentration of just one of the reactant and determined the change in the rate. Typically, we double or we cut the concentration in half. If we look closely at experiments one and two, where we are comparing rate one with rate two, we see that the only change is we have decreased the concentration of a by one half. Everything else is the same. So if we decrease the concentration of a by one half and the rate is also decreased by one half, which it is, then when we do one thing to a particular concentration and the same thing happens to the rate namely, if we double the concentration, we double the rate. Or if we cut the concentration in half, we cut the rate in half, then its first order. So depending on how you want to compare these two experiments, compare 1 to 2 or compare 2 to 1. If we compare 2 to 1, then a is doubled and the rate is doubled. If we compare 1 to 2, then a concentration is cut in half and rate is cut in half. Either way, we can deduce that X equals one. The order of the reaction with respect to a is one. If we now look at experiments two and three, we see that our three is 1/4 are too. And all we have done is cut the concentration of be in half. So if we cut the concentration of be in half and the rate ends up being 1/4 then that is indicative of second order or if we compare 3 to 2, we would see that doubling the concentration of be resulted in a quadrupling of the rate. Either way, we would deduce that why is, too, if we double the rate? I'm sorry if we double the concentration and we quadruple the rate, this is indicative of second order. Now, if we look at experiments three and four, we need to find out what is happening to each to each of the concentrations and how this is changing the rate in four. We have double the concentration of a doubling. The concentration of A based on what we determined up here should result in a doubling of the rate comparing experiments 3 to 4. We have also doubled the concentration of B. Doubling the concentration of B should result in a quadrupling of the rate. So these two changes alone should result in the rate being eight times the rate in experiment for being eight times the rate in Experiment three. But results show that the rate is 16 times the rate in Experiment three. So what has happened to the concentration of C C has been cut in half, cutting the concentration of C and half, resulting in a doubling of the rate going from eight times to what is reported of 16 times that is representative of not first order, but actually negative one. And I'll show you why here in just a minute. So we will say that Z is negative one up here. If we were to double the concentration, we double the rate its first order because to raise to the one. If we double the concentration, we raise it to the one we will double the rate. If we double the concentration, we raise it to two. We quadruple the rate. But here we have cut the concentration in half. So if we cut the concentration in half, we have one half the original concentration. What do we need to raise this to in order to get a doubling of the rate we need to raise it to negative 11 half to the negative One is two. That's why the order with respect to see ends up being this strange value. Negative one. Now for Part B, we want to compare the rate of Experiment five to the rate of Experiment one. You'll notice that in all cases the concentrations were cut in half a was cut in half, B was cut in half, see was cut in half well, if a is cut in half, we know the order of the reaction for a is one, so the rate should be cut in half. If B is cut in half, order of the reaction with respect to be as one is second. So a cutting of B and half one half race to the two gives us 1/4 so the new rate will be 1/4. So right now it would be one half multiplied by 1/4 or 18 But if we cut, see in half also then the rate is going to be doubled because it's order is negative one. So the effects of cutting a and half resulting in half the rate and cutting see in half, resulting in double the rate cancel. So the ultimate result is coming from B, where the rate of five ends up being 1/4 the rate of one

For this next question. It asks us to calculate the rate for the second order reaction. In order to calculate the rate we need to revert back to the differential rate law differential rate law, sometimes just called the rate law is simply rate equals the rate constant multiplied by the concentration raised to the order. The order in this case because it's referring to the second order reaction would be to So if you haven't done question 27 yet, you need to do that by plotting concentration. Natural log of concentration, one over concentration as a function of time for all three experiments. Plot these, um, values as a function of time and determine which one gives you the best straight line based upon the R squared value. And you would identify that the second order reaction is the last one. Experiment three then, because it has an R squared value closest to one. If we had also plotted the natural log and went over, the concentration would actually plot all three for all of them. So with this being the second order reaction, this is the equation of the line. Then the K value simply corresponds to the slope for the second order reaction. So rate is going to be equal to K, which is the slope multiplied by the initial concentration, which is one up as shown here squared. So it's going to be 0.1 or more specifically, it would be point 00992 But we'll go with this as I assume the author was trying to make all of them 0.1 the K values for all of them one. So when we do that, we get our initial rate of point Oh one Mueller per second.

Hey, I So in this question, were given a reaction between A B and C forming D and E and were asked to find the orders with respect to our reactant and then find the value of rate five with respect to rate one according to the experience that they gave us in the chart in the textbook. So the first get started, I'm going to find the rate order with respect to A by using experiments one and two. So I'm going to use the rate law equation that I have here to solve for em using you want. And you too. I'm going to put e to on the bottom of each one. Since anyone has larger values, then e to when I do this, I put in my rates. So I have are one for Experiment one, and this is going to be equal to my rate, constant times, a concentration of a that was given to us in Moeller to the Empower Times. A concentration of B times are concentration of C I'm using on exponents of P instead of Oh, since P doesn't look like zero like oh would for e two. My rate is it with respect to or one. So that's our 1/2. And I'm going to write the rest of my equation toe have my very constant multiplied by my concentration for a multiplied by our concentration of B and finally multiplied by our concentration for C. Now, I chose these two equations to solve for the order with respect to A since my concentrations for B and C cancel out and my rates cancel out and I'm left with what I'm looking for, which is the ah, Constance and ah, numbers that are, with respect to my rate constant our with my, uh, order for a so simplifying this down we have, or one who also cancel out because of multiplication. And we're gonna be left with two on the left side as well as to to the M on the right. So when I saw for M, I find that em is equal to one, and this is my order with respect to a showing that this is first order. Now that I have that I can go ahead and do the same things for reactant b and C so for react and be I'm gonna be using experiments two and three for the same reasons that I chose experiments one into last time. And I'm gonna put experiment to over Experiment three plugging in my rate toe have are two for Experiment two and are to over four for Experiment three and then for the rest, my equation. I'm gonna plug in my concentrations for a do and see going to do that for both experiments. So this is for experiment two and right here is going to be for experiment Three of my re constant times My concentration of a times a concentration of B times the concentration of C. When I do simplification, my are twos. Cancel out case, cancel out A's concentration, cancels out and sees concentrations cancel out on the left side. I'm going to be ending up with four. And on the right side, I'll have to to the end power. When I solve for n, I find that an is equal to two making my order to with respect to be. Now that I have that I can go ahead and find my final order for C. So for this one, I'm gonna be using experience of one and four and I'm going to be putting Experiment one over experiment four. When I do that, I'm gonna leave this part, blinked for a second so I can explain how to find the U rates. So I'm gonna go ahead and just plug it in my concentrations first for a being. See, that's experiment one. And then we have experiment to over here. Okay, so when I find my rates, I'm gonna use the, ah table that they provided us. And just look at the relationships between everything. So for the 1st 1 we know that are two is equal to 1/2 or one. So when we just make a little side note over here, we know that are too secret of 1/2 or one. So that makes our one equal to to our two by multiplying this to over. And we also know that are three is equal toe 1/4 or two and that makes our to equal to four or three. Using this relationship, I can go ahead and plug this into this equation, and that's gonna fuck Show that or one is equal to eight are three. Using this relationship I can use plug that into my rates over here. So for the one I wanted with respect to our three, so I can cancel those out. So that's why I found that from earlier. And this 16 r three is what's given to us simplifying everything. I find my case. Cancel out my concentration for A and B cancel out or three cancels out and I'm going to be left with 1/2 is equal to To to the P. Now, this isn't as easy to see as for the answer as Thea. Other previous to so I'm gonna go ahead and show how to do self for this using logs. So the first thing I'm going to do is take the log of both sides of the equation. When I take that my left side is log of 1/2 is going to be equal to p log to. I can take the exponents from the two and place it on the side of the log using log rules, and I can go ahead and sell for P. So P is equal to log 1/2 over, log to and when you plug this into your calculator, you find that P is equal to negative one. So that means reactant c has a order of negative one. And now that I have all of the orders for my rate law, I can go ahead and use that to find K. And when I confined K, I can find the rate with respect to experimental five. So the first thing I'm going to do, I'm going to use Experiment one to find my K value. So are one is equal to K times my concentration of a from experiment one and thats to the one power. So I'm going to leave that blank since we know that one. Ah, power of one is just going to you'd itself. We have a concentration of B which is gonna be squared and then our concentration for seeing that's to the negative one solving for K, we find that K is equal to our one over 2.2 or actually 2.7 for four and dividing this by two point this 2.744 We find that this is equal to 0.3 64 are one just putting it this way so we can easily plug this into our equation and not have to worry about fractions. Now that we have our K value, we confined our our five with respect to are one. So our five is going to be equal to okay that we found, which is 0.364 or one multiplied by our concentrations that are given to us too. 0.7 for a 0.70 squared for B and 0.5 to the negative one for C. And when we plug this all indoor calculator, we find that this is equal to 0.25 are one. And just to make it, ah, fraction eyes, As with the rest of what's in the table, this is equal to 1/4 are one. And this is our answer for what?


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